Research Article A Non-Markovian Multistage Batch Arrival...
Transcript of Research Article A Non-Markovian Multistage Batch Arrival...
Research ArticleA Non-Markovian Multistage Batch Arrival Queue withBreakdown and Reneging
Sivagnanasundararam Maragathasundari12 and Santhanagopalan Srinivasan3
1 Sathyabama University Chennai 600 119 India2Department of Mathematics Velammal Institute of Technology Chennai 601 204 India3 Department of Mathematics B S Abdur Rahman University Chennai 600 048 India
Correspondence should be addressed to Sivagnanasundararam Maragathasundari maragatham01gmailcom
Received 21 July 2014 Accepted 7 October 2014 Published 13 November 2014
Academic Editor Antonina Pirrotta
Copyright copy 2014 S Maragathasundari and S Srinivasan This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited
The present investigation deals with analysis of non-Markovian queueing model with multistage of services When the server isunavailable during the systembreakdown (or) vacation periods we consider reneging to prevail Supplementary variable techniqueshave been adopted to obtain steady state system length distributions The numerical illustrations are provided to validate thetractability of performance measures as far as computational aspect is concerned Numerical results in the form of graphicalrepresentation are also presented Practical large scale industry applications are described to justify our model
1 Introduction
Vacation queueingmodels play amajor role inmanufacturingand production computer and communication and serviceand distribution systems Many models for customerrsquos impa-tience in queueing systems have been studied in the pastand the source of impatience has always been consideredto be either a long wait already experienced at a queueor a long wait anticipated by a customer upon arrivalQueueing models enable organizations to implement criticalproduction strategies or tactics aimed at reducing costs onincreasing revenues By examining themethods of operationsresearch and especially queueing theory new models can bedeveloped and existing ones can be extended
One of the earliest works on balking and reneging was byHaight [1 2] and Barrer [3] which was the first to introducereneging in which they studied deterministic reneging withsingle server Markovian arrival and service rates Montazer-Haghighi et al [4] studied a Markovian mutiserver queueingsystem with balking and reneging A two-stage batch arrivalqueueing system where customers receive a batch service inthe first and individual service in the second stagewas studiedby Doshi [5] in the past
Furthermore [6] Chodhury examined an M[119909]G1queueing system with a set-up period and a vacation periodThis paper deals with an M119883G1 queueing system witha vacation period which comprises an idle period and arandom set-up period The server is turned off each timewhen the system becomes empty At this point of time the idleperiod starts As soon as a customer or a batch of customersarrive the setup of the service facility begins which is neededbefore starting each busy period In this paper the steady-statebehaviour of the queue size distributions at stationary (ran-dom) point of time and at departure point of time is studiedAlso explicit expressions for the system state probabilitiesand some performance measures of this queueing systemare derived analytically Finally the probability generatingfunction of the additional queue size distribution due to thevacation period as the limiting behaviour of theM119883M1 typequeueing system is derived
Madan [7] investigated a batch arrival queueing systemwhere the server provides two stages of heterogeneous servicewith a modified Bernoulli schedule under N-policy Theserver remains idle till the queue size becomes N (ge1)As soon as the queue size becomes at least N the serverinstantly starts working and provides two stages of service
Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2014 Article ID 519579 16 pageshttpdxdoiorg1011552014519579
2 Mathematical Problems in Engineering
in succession to each customer that is the first stage servicefollowed by the second stage service However after thesecond stage service the server may take a vacation or decideto stay in the system to provide service to the next customerif anyThe queue size distributions at a random epoch as wellas a departure epoch under the steady-state conditions werederived
The MG1 queue with impatient customers was studiedby Bae et al [8] the complete formula of the limitingdistribution of the virtual waiting time was derived explicitlyThe expected busy period of the queue was also obtained byusing a martingale argument
In 2002 Krishna Kumar et al [9] studied anMG1 retrialqueueing system with two-phase service and preemptiveresume In this paper for an arbitrarily distributed retrial timedistribution the necessary and sufficient condition for thesystem stability is obtained assuming that only the customerat the head of the orbit has priority access to the server Thesteady-state distributions of the server state and the numberof customers in the orbit are obtained along with otherperformance measures The effects of various parameterson the system performance are analysed numerically Ageneral decomposition law for this retrial queueing systemis established
Furthermore Madan et al [10] studied the server vaca-tions based on Bernoulli schedules and a single vacationpolicy in 2003 In this model a single server queue withoptional server vacations based on exhaustive service isstudied Unlike other vacation policies the assumption thatonly at the completion of service of the last customer in thesystem the server has the option to take a vacation or toremain idle in the system waiting for the next customer toarrive is considered The service times of the customers havebeen assumed to be deterministic and vacations are phasetype exponential Steady-state results for the probability gen-erating functions of the queue length the expected numberof customers in the queue and the expected waiting time ofthe customer are derived explicitly
In addition Choudhury [11] investigated a batch arrivalqueueing system with an additional service channel Thispaper deals with anM119909G1 queueing systemwith two phasesof heterogeneous service under N-policy where the serverremains idle till the queue size becomes N (⩾1) As soonas the queue size becomes N the server immediately startsfirst ldquoessential servicerdquo for all the units After completion ofthe essential service of a unit it may leave the system withprobability (1 minus 120579) or may immediately go for a second phaseof service in an additional service channel with probability 120579
(0 ⩽ 120579 ⩽ 1) For this model the queue size distribution ata random epoch as well as at a departure epoch is obtainedThis is a generalization of recent papers considered byMedhiand Lee et al (queueing systems)
In 2004 Madan and Abu Al-Rub [12] investigated on asingle server queue with optional phase type server vaca-tions based on exhaustive deterministic service and a singlevacation policy Furthermore Madan and Choudhury [13]proposed a queueing system with restricted admissibility ofarriving batches In this model a batch arrival queue with aBernoulli vacation schedule is considered After completion
of the service the server either goes for a vacation in randomlength with probability 120579 (0 le 120579 le 1) or may continue toserve the next unit if any with probability (1 minus 120579) undera restricted admissibility policy of arriving batches Unlikethe usual batch arrival queueing system the restricted admis-sibility policy differs during a busy period and a vacationperiod and hence all arriving batches are not allowed to jointhe system at all time Steady-state queue size distributionat a random point of time as well as a departure epoch isderivedMoreover this paper attempts to unify several classesof related batch arrival queueing system
Choudhury and Madan [14] studied a two-stage batcharrival queueing system with a modified Bernoulli sched-ule vacation under N-policy In this paper a batch arrivalqueueing system where the server provides two stages ofheterogeneous service with a modified Bernoulli scheduleunder N-policy is considered The server remains idle tillthe queue size becomes N (ge1) As soon as the queue sizebecomes at least N the server instantly starts working andprovides two stages of service in succession to each customerthat is the first stage service followed by the second stageservice However after the second stage service the servermay take a vacation or decide to stay in the system toprovide service to the next customer if any The queuesize distribution at a random epoch as well as a departureepoch under the steady-state conditions is derived Furtherthe existence of the stochastic decomposition property toshow that the departure point queue size distribution ofthis model can be decomposed into the distributions ofthree independent randomvariables is demonstrated Finallya simple procedure to obtain optimal stationary operatingpolicy under a suitable linear cost structure is developed
Furthermore Chang and Takine [15] studied factorizationand stochastic decomposition properties in bulk queues withgeneralized vacation This paper considers a class of sta-tionary batch-arrival bulk-service queues with generalizedvacationsThe system consists of a single server and a waitingroom of infinite capacity Arrivals of customers follow abatchMarkovian arrival processThe server is unavailable foroccasional intervals of time called vacations and when it isavailable customers are served in groups of fixed size B Forthis class of queues the vector probability generating functionof the stationary queue length distribution is factored intotwo terms one of which is the vector probability generatingfunction of the conditional queue length distribution giventhat the server is on vacation is shown The special caseof batch Poisson arrivals is examined and a new stochasticdecomposition formula is derived for the stationary queuelength distribution
Altman and Yechiali [16] considered the systems withservers vacations where customersrsquo impatience is due to anabsentee of servers upon arrival Such a model representingfrequent behavior by waiting customers in service systemshad never been treated before in the literature
Altman and Yechiali [17] studied a system that is operat-ing as an MMinfin queue However when it becomes emptyit is assigned to perform another task the duration U ofwhich is random Customers arriving while the system isunavailable for service (ie occupied with a U-task) become
Mathematical Problems in Engineering 3
impatient each individual activates an ldquoimpatience timerrdquohaving random duration T such that if the system doesnot become available by the time the timer expires thecustomer leaves the system never to returnWhen the systemcompletes a U-task and there are waiting customers each oneis taken immediately into service Explicit expressions for thecorresponding mean queue sizes were obtained
In 2012 Kumar and Sharma [18] analyzed a single serverqueue with general service time distribution random systembreakdowns and Bernoulli schedule server vacations whereafter a service completion the server may decide to leave thesystem with probability 119901 or to continue serving customerswith probability 1 minus 119901 It is assumed that the customersarrive to the system in batches of variable size but are servedone by one If the system breaks down it enters a repairprocess immediately It is assumed that the repair time hasgeneral distribution while the vacation time has exponentialdistribution The purpose is to find the steady-state resultsin explicit and closed form in terms of the probability-generating functions for the number of customers in thequeue the average number of customers and the averagewaiting time in the queue
Also Maraghi et al [19] studied an M119909G1 queue withrandom breakdowns and Bernoulli schedule server vacationswhere after a service completion the server may decide toleave the system with probability p or to continue servingcustomers with probability 1minus119901 It is assumed that customersarrive to the system in batches of variable size but are servedone by one
Kumar and Sharma [20] analysed the concept of customerbalking and reneging has been exploited to a great extent inthe recent past by the queuing modelers According to thismodel a reneged customer can be convinced in many casesby employing certain convincing mechanism to stay in thequeue for completion of his serviceThus a reneged customercan be retained in the queuing system with some probability(say 119902) and it may leave the queue without receiving servicewith probability 119901 (= 1 minus 119902) This process was referred to ascustomer retention The effect of probability of retention onthe average system size had been studied
Recently in 2013 Baruah et al [21] studied a two-stagequeuing model where the server provides two stages ofservice one by one in succession They considered renegingto occur when the server is unavailable during the systembreakdown or vacation periods
Kumar and Sharma [22] in their queueing model withretention of reneged customers and balking studied a finitecapacity and the retention of reneged customers The cus-tomerrsquos impatience may occur in the form of balking alsoIn this paper the work of Kumar and Sharma (2013) isextended by including balking to take into consideration thebroader perspective of customerrsquos impatience The model issolved iteratively to obtain the steady-state probabilities ofsystem size Some useful performance measures are derivedand the effect of the probability of retaining the renegedcustomers on various performance measures is studiednumerically
11 Model Description In todayrsquos competitive world cus-tomerrsquos impatience has become a serious problem Firms areemploying a number of strategies to retain their customersTo the best of our knowledge studies on multistage batcharrival of service in reneging do not exist We extendand develop the above models on two stages of serviceto multistage of services In particular a two-stage modelof Monita Baruah Kailash C Madan and Tillal Eldabiin the year 2013 motivated us for developing this newmodel which we consider here Their model on two stagesof service is extended to multistage of service where thearrivals occur in batches Breakdown and reneging play a vitalrole in our model Two stages of service will be benefitedonly for small scale industries factories of low productionlimited network and so forth whereasmultistages of servicesrender a huge amount of beneficial factors in large scaleindustries and in multitier applications in e-commerce inparticular
This results in an introduction of a new model a non-Markovian multistage batch arrival queue with breakdownand reneging In this model once the customers enter theywill be provided with the service in all the stages of servicedepartment one by one in succession Once the servicegets completed customer departs from the system Oncethe multistage service of a unit is completed the server isassumed to take vacation with probability 120579 or may continueto offer service with probability (1 minus 120579) As soon as thevacation period of the server ends it joins the system tocontinue service of the waiting customers During systembreakdown (or) vacation periods there will be unavailabilityof service in that case reneging occurs Reneging is assumedto follow exponential distribution Arrival follows a Poissonprocess and service time follows general distribution Westudy the changes in the probability of idle time and in thetraffic intensity in some particular cases like (1) if there isno vacation and (2) if the service time and vacation timeare exponentially distributed Using supplementary variabletechnique we derive the steady-state probabilities and perfor-mance measures
This paper is organized as follows Applications of modelis presented in Section 2 the mathematical description of themodel is given in Section 3 the definitions and notationsare described in Section 4 the equations governing thesystem are given in detail using birth and death processin Section 5 the queue size distribution at random epochis highlighted in Section 6 and the average queue sizeand average waiting time are given in Section 7 Sections6 and 7 gives a comprehensive idea about the model inwhich step by step method is followed for better under-standing which helps in any extension of this model Theparticular cases are derived in Section 8 in which themodel is seen in various aspects or situations For justifi-cation of the model the numerical illustrations are givenin Section 9 and at last the conclusion is presented inSection 10 In this section future scope of this model isdiscussed
4 Mathematical Problems in Engineering
Policing Load balancer
despatcher
despatcherDrop sessions
(if needed)
Individual server
Tier 1
Tier 1 Tier 2
Tier 2
Tier 3
(Nonreplicated)
Sentry
Figure 1 A three-tier application in e-commerce
bodyStator
windingRotor part fixing
Cooling fan
fixing
End cover fixing
Painting
A complete electric motor
Stator
Figure 2 Multistage motor production process
2 Applications of Model
With the development of computer and communicationnetworks queueing systems and networks have been iden-tified as a powerful analysis and design tool for variousapplications Our model plays a major role in multitierapplications in e-commerce motor production process carproduction process and textile industry Somemultistage reallife applications in large scale manufacturing industry aredescribed in the following sections which come across thenecessity of our new model
21 Three-Tier Application in e-Commerce Consider a three-tier application as shown in Figure 1 The first two tiersare clustered web server and clustered java applicationrespectively and the third tier is a database used in e-commerce applications Each tier is considered to be a stageEach incoming application is subjected to admission controlat the sentry to ensure that the contracted performanceguarantees are met The excess sessions are turned awayduring overloadsThe load balancer efficiently distributes theincoming requests Each input data is considered to be acustomer The server maintenance period is considered to bea vacation period and this application can be extended tomultistage of services
Steel plate
Stamping Welding Painting
Engine assembly
and mountingAssemblyInspectionCar
Figure 3 Multistage car production process
22 Electric Motor Production Industry
Multistage in Motor Production Process (See Figure 2) In theelectric motor manufacturing industries multistage produc-tion process is implemented (Figure 2) The stator body isconsidered as the arrival customer In the first stage the statorbody is wounded by the coil in the second stage the rotorpart is fixed in the stator in the third stage the cooling fanis fixed in one end of the rotor in the fourth stage the endcover is fixed in the other end of the rotor in the fifth stagethe entire motor is painted and finally in the sixth stage acomplete electric motor is produced We consider the statorbody as a batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools stock verification and so forth Here the machinegets idle owing to breakdown and the machine idle period isconsidered as the breakdown period
23 Automobile Industry
Multistage in Car Production Process (See Figure 3) In auto-mobile industry car production process is considered as themultistage production process (Figure 3) Here we considerthe steel plate as the batch arrival customer In the firststage of production a roll of steel plate is cut according tothe shape of the car parts in the second stage and steel
Mathematical Problems in Engineering 5
Drawing machine
Carding machine
machineSpinning machine
Cone winding machine
Yarn production
Bunch of
cottonBlow room
Simplex
Figure 4 Textile yarn production process
plates are welded in the third stage all the car parts arepainted in the painting section in the fourth stage the engineassembly is mounted in the fifth stage extra fittings areassembled according to the customer requirements in thesixth stage quality control inspection is conducted andfinally a complete car is produced We consider the steelplate as the batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools andmachines stock verification and so forth Heremachine gets idle owing to breakdown and the machine idleperiod is considered as the breakdown period
24 Textile Yarn Production Industry
Multistage in Production Process (See Figure 4) In the textileindustry yarn production is considered as the multistageproduction process (Figure 4) Here a bunch of cotton isconsidered as the arrival customer In the first stage a bunchof cotton is inserted into the blow room machine in thesecond stage the refined cotton is inserted into the cardingmachine in the third stage the produced cotton sliver isinserted into the drawing machine and is again processedin the fourth stage the processed cotton sliver is insertedinto the simplex machine in the fifth stage the refinedcotton sliver is inserted into the spinningmachine and in thesixth stage the cotton yarn output from spinning machine isinserted into the conewindingmachine and the entire cottonyarn is wounded as a cone These six stages of productionare required to convert a bunch of cotton into cotton yarnIn the production sector vacation denotes obtaining a newraw material maintenance of the tool and machine stockverification and so forth Here machine gets idle owing tobreakdown and the machine idle period is considered as thebreakdown period
3 The Mathematical Description of the Model
31 The Model Is Based on the Following Assumptions (a)Customers arrive in batches following a compound Poissonprocess with a rate of arrival 120582 Let 120582119888119894119889119905 (119894 = 1 2 3 119899) bethe first-order probability that customers arrive at the systemin batches of size 119894 at the system in a short interval of time(119909 119909 + 119889119905) where 0 le 119888119894 le 1 and sum
infin
119894=1 119888119894 = 1(b) The server provides multistage of heterogeneous
services one after the other in succession An arrival batchreceives the service offered at multistage in successiondefined as the first stage second stage and so forth respec-tively The service discipline is assumed to be on a first come
first served basis (FCFS) Let us assume that the service time119878119895 (119895 = 1 to 119899) of the 119895th stage service follows a generalprobability distribution with a distribution function 119870119895(119878119895)where 119896119895(119904119895) is the probability density function and 119864(119878
119899119895 ) is
the 119899th moment of the service time with 119895 = 1 2 119873Let
120583119895 (119909) =
119896119895 (119909)
1 minus 119870119895 (119909)119895 = 1 2 119873
119896119895 (119878119895) = 120583119895 (119909) 119890[minus int119904
0
120583119895(119909)119889119909]
where 119895 = 1 2 119873
(1)
(c) Once the multistage service of a unit is completed theserver is assumed to take vacation with probability 120579 or maycontinue to offer service with probability (1 minus 120579) As soon asthe vacation period of the server ends it joins the system tocontinue the service of the waiting customers
Let us assume the vacation time to be a random variablefollowing general probability law with distribution119872(V) anddensity function by 119898(V) 119864(119881119899) denotes the 119899th moment(119899 = 1 2 ) of the vacation time Here let us consider that120585(119909) is the conditional probability of a vacation period duringthe interval (119909 119909 + 119889119909) given that elapsed time is 119909 whichcan be given as
120585 (119909) =119872 (119909)
1 minus119872 (119909) (2)
Thus
119898(V) = 120585 (V) 119890[minus intV0
120585(119909)119889119909] (3)
(d) Customers arriving for service may become impatientand renege after joining during vacations and breakdownperiods Reneging is assumed to follow exponential distribu-tion with parameter 120595 Thus 119891 (119905) = 120595119890
minus120595119905119889119905 120595 gt 0 Thus
120595119889119905 is the probability that a customer can renege during ashort interval of time (119905 119905 + 119889119905)
The system may fail or be subjected to breakdown atrandom The customer receiving service during breakdownreturns back to the head of the queue Let us assume that theinterval between breakdowns occurs according to a Poissonprocess with a mean rate of breakdown 120573 gt 0
Furthermore the repair times follow a general (arbitrary)distribution with the distribution function 119866(119909) and densityfunction 119892 (119909) Let the conditional probability of comple-tion of the repair process be Υ (119909) 119889119909 such that Υ (119909) =
119866(119909) (1 minus 119866(119909)) and thus 119866 (119903) = Υ (119903) 119890(minus int119903
0
Υ(119909)119889119909)
4 Definitions and Notations
We assume that steady state exists and define 119875119899119895 (119909) as theprobability that there are n (ge1) customers in the systemincluding one customer in type 119895 service 119895 = 1 2 119873and elapsed service time is 119909 119875119899119895 = int
infin
0119875119899119895 (119909) 119889119909 is
the corresponding steady-state probability irrespective ofelapsed time 119909 119881119899 (119909) is the probability that there are n (ge0)customers in the queue and server is on vacation and elapsedvacation time is 119909
6 Mathematical Problems in Engineering
119881119899 = intinfin
0119881119899 (119909) 119889119909 is the corresponding steady-state
probability irrespective of elapsed vacation time 119909Q is the steady-state probability of the server being idle as
the server takes vacationThe probability generating functions are defined as
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 (119909)
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 |119911| le 1 119895 = 1 2 119873
119877 (119909 119911) =
infin
sum
119899=1
119911119899119877119899 (119909)
119877 (119911) =
infin
sum
119899=1
119911119899119877119899 |119911| le 1
119881 (119909 119911) =
infin
sum
119899=1
119911119899119881 (119909)
119881 (119911) =
infin
sum
119899=1
119911119899119881119899 |119911| le 1
119862 (119911) =
infin
sum
119894=1
119911119894119862119894
(4)
5 Equations Governing the System
We frame the difference equations with Poisson input relatedto birth-death process
Let 119899 be the size of customers at time 119909 Let 119875119899(119909) be theprobability of ldquonrdquo customers in the system
Let
(1) 120582119899 be the average arrival rate when ldquo119899rdquo customers arein the system
(2) 120583119899 the average service rate when ldquo119899rdquo customers are inthe system
(3) 119875119899(119909+Δ119909) the probability of ldquo119899rdquo customers at (119909+Δ119909)(4) 120573 the probability of breakdown arrival(5) 120582119862119894 the probability of batch arrival
Consider the following
1198751198991 (119909 Δ119909) = (1 minus 120582 + 1205831 (119909) + 120573 Δ119909)
times 1198751198991 (119909) + 119875119899minus11 (119909) 1205821198621Δ119909
+ 119875119899minus21 (119909) 1205821198622Δ119909
+ sdot sdot sdot + 11987511 (119909) 120582119862119899minus1Δ119909
1198751198991 (119909 + Δ119909) minus 1198751198991 (119909)
Δ119909+ 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
=
119899
sum
119894=1
120582119862119894119875119899minus1198941 (119909)
(5)
From above we have
119889
1198891199091198751198991 (119909 119911) + 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198941 (119909) 119899 ge 1
(6)
Similarly we have
119889
1198891199091198751198992 (119909 119911) + 120582 + 1205832 (119909) + 120573 1198751198992 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198942 (119909) 119899 ge 1
(7)
For Nth stage119873 ge 3
119889
119889119909119875119899119873 (119909 119911) + 120582 + 120583119873 (119909) + 120573 119875119899119873 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus119894119873 (119909) 119899 ge 1
(8)
119889
119889119909119877119899 (119909) + 120582 + Υ (119909) + 120595 119877119899 (119909)
= 120582
119899
sum
119894=1
119888119894119877119899minus119894 (119909) + 120595119877119899 (119909) 119899 ge 1
(9)
119889
1198891199091198770 (119909) + (120582 + Υ (119909)) 1198770 (119909) = 1205951198771 (119909)
(10)
119889
119889119909119881119899 (119909) + 120582 + 120585 (119909) + 120595119881119899 (119909)
= 120582
119899
sum
119894=1
119888119894119881119899minus119894 (119909) + 120595119881119899+1 (119909) 119899 ge 1
(11)
119889
1198891199091198810 (119909) + 120582 + 120585 (119909) + 1205951198810 (119909) = 1205951198810 (119909)
(12)
120582119876 = (1 minus 120579) int
infin
0
1198750119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
1198770 (119909) Υ (119909) 119889119909
(13)
The abovementioned differential equations should be solvedbased on the following boundary conditions
1198751198991 (0) = 120582119888119899119876 + (1 minus 120579) int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
119877119899+1 (119909) Υ (119909) 119889119909
+ int
infin
0
119881119899 (119909) 120595 (119909) 119889119909 119899 ge 1
(14)
1198751198992 (0) = int
infin
0
1198751198991 (119909) 1205831 (119909) 119889119909 119899 ge 1 (15)
Mathematical Problems in Engineering 7
For Nth stage
119875119899119873 (0) = int
infin
0
119875119899119873minus1 (119909) 120583119873minus1 (119909) 119889119909 119899 ge 1 (16)
119881119899 (0) = 120579int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909 119899 ge 0 (17)
119877119899+1 (0) =
119873
sum
119895=1
int
infin
0
119875119899119895 (119909) 119889119909 (18)
1198770 (0) = 0 (19)
Note LHS of (14) indicates that there are n customers in thesystem and the service is about to start in stage 1
RHS of (14) indicates the various situations in which thesystem has n customers and the service is about to start instage 1
The first term of RHS of (14) indicates that the system is inidle (119876) an arrival of batch consisting of 119899 customers occursand 1st stage service is about to start
The second term indicates that there are 119899 + 1 customersin the system including one customer whose service iscompleted in Nth stage (119873 ge 2) with probability 120583119873(119909)
remaining 119899 customers are there in the system 1st stage ofservice is about to start Moreover the server is not going forvacation instead stays in the system with probability 1 minus 120579
The third term indicates the repair process gets completedwith probabilityΥ (119909) So one customer just enters into the 1ststage service Now there are 119899 customers in the system
The fourth term indicates that vacation gets completedwith probability 120595 (119909) There are 119899 customers in the system1st stage of service is about to start
The combination of all the four terms gives the LHS of(14)
The integral in the RHS of (15) states that after comple-tion of 1st stage of service there are 119899 customers in the systemto whom 2nd stage of service is about to start which gives theLHS of (15)
Similarly (16) indicates that the Nth stage of service isabout to start
RHS of (17) explains the fact that after completion ofvacation with probability 120579 the server has just returned to thesystem consisting of n customers and the 1st stage of serviceis about to start This leads to the LHS of (17)
6 Queue Size Distribution at Random Epoch
Weuse supplementary variable technique to get the followingcomprehensive idea about the model
Multiplying (6) to (12) by 119911119899 and summing over n from 1
toinfin yields that we can obtain
119889
1198891199091198751 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205831 (119909) + 120573 1198751 (119909 119911) = 0
(20)119889
1198891199091198752 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205832 (119909) + 120573 1198752 (119909 119911) = 0
(21a)
For Nth stage119889
119889119909119875119873 (119909 119911) + (120582 minus 120582119862 (119911)) + 120583119873 (119909) + 120573 119875119873 (119909 119911) = 0
(21b)
119889
119889119909119877 (119909 119911) + 120582 minus 120582119862 (119911) + Υ (119909) + 120595 minus
120595
2119877 (119909 119911) = 0
(22)
119889
119889119909119881 (119909 119911) + 120582 minus 120582119862 (119911) + 120585 (119909) + 120595 minus
120595
119911 = 0 (23)
Further integration of (20)ndash(23) over limits 0 to 119909 in generalwill result in
119875119895 (119909 119911) = 119875119895 (0 119911) 119890[(120582minus120582119862(119911)+120573)119909minusint
infin
0
120583119895(119905)119889119905]
119895 = 1 to 119873
119877 (119909 119911) = 119877 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
Υ(119905)119889119905]
119881 (119909 119911) = 119881 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
120585(119905)119889119905]
(24)
Next by multiplying the boundary conditions by suitablepowers of 119911119899+1 taking summation over all possible values of119899 and using the probability generating functions we get
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579) int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909
+ int
infin
0
119877 (119909 119911) Υ (119909) 119889119909
+ int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909
1198752 (0 119911) = int
infin
0
1198751 (119909 119911) 1205831 (119909) 119889119909
(25)
For Nth stage
119875119873 (0 119911) = int
infin
0
119875119873minus1 (119909 119911) 120583119873minus1 (119909) 119889119909 (26)
119911119881 (0 119911) = 120579int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909 (27)
119877 (0 119911) = 120572119911
119873
sum
119895=1
119875119895 (119911) (28)
Again integrating (24) with respect to 119909 in general will resultin
119875119895 (119911) = 119875119895 (0 119911) [
1 minus 119870119895 (120582 minus 120582119862 (119911) + 120573)
120582 minus 120582119862 (119911) + 120573] 119895 = 1 to 119873
(29)
119881 (119911) = 119881 (0 119911) [1 minus119872(120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (30)
119877 (119911) = 119877 (0 119911) [1 minus 119866 (120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (31)
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Discrete Dynamics in Nature and Society
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Decision SciencesAdvances in
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
2 Mathematical Problems in Engineering
in succession to each customer that is the first stage servicefollowed by the second stage service However after thesecond stage service the server may take a vacation or decideto stay in the system to provide service to the next customerif anyThe queue size distributions at a random epoch as wellas a departure epoch under the steady-state conditions werederived
The MG1 queue with impatient customers was studiedby Bae et al [8] the complete formula of the limitingdistribution of the virtual waiting time was derived explicitlyThe expected busy period of the queue was also obtained byusing a martingale argument
In 2002 Krishna Kumar et al [9] studied anMG1 retrialqueueing system with two-phase service and preemptiveresume In this paper for an arbitrarily distributed retrial timedistribution the necessary and sufficient condition for thesystem stability is obtained assuming that only the customerat the head of the orbit has priority access to the server Thesteady-state distributions of the server state and the numberof customers in the orbit are obtained along with otherperformance measures The effects of various parameterson the system performance are analysed numerically Ageneral decomposition law for this retrial queueing systemis established
Furthermore Madan et al [10] studied the server vaca-tions based on Bernoulli schedules and a single vacationpolicy in 2003 In this model a single server queue withoptional server vacations based on exhaustive service isstudied Unlike other vacation policies the assumption thatonly at the completion of service of the last customer in thesystem the server has the option to take a vacation or toremain idle in the system waiting for the next customer toarrive is considered The service times of the customers havebeen assumed to be deterministic and vacations are phasetype exponential Steady-state results for the probability gen-erating functions of the queue length the expected numberof customers in the queue and the expected waiting time ofthe customer are derived explicitly
In addition Choudhury [11] investigated a batch arrivalqueueing system with an additional service channel Thispaper deals with anM119909G1 queueing systemwith two phasesof heterogeneous service under N-policy where the serverremains idle till the queue size becomes N (⩾1) As soonas the queue size becomes N the server immediately startsfirst ldquoessential servicerdquo for all the units After completion ofthe essential service of a unit it may leave the system withprobability (1 minus 120579) or may immediately go for a second phaseof service in an additional service channel with probability 120579
(0 ⩽ 120579 ⩽ 1) For this model the queue size distribution ata random epoch as well as at a departure epoch is obtainedThis is a generalization of recent papers considered byMedhiand Lee et al (queueing systems)
In 2004 Madan and Abu Al-Rub [12] investigated on asingle server queue with optional phase type server vaca-tions based on exhaustive deterministic service and a singlevacation policy Furthermore Madan and Choudhury [13]proposed a queueing system with restricted admissibility ofarriving batches In this model a batch arrival queue with aBernoulli vacation schedule is considered After completion
of the service the server either goes for a vacation in randomlength with probability 120579 (0 le 120579 le 1) or may continue toserve the next unit if any with probability (1 minus 120579) undera restricted admissibility policy of arriving batches Unlikethe usual batch arrival queueing system the restricted admis-sibility policy differs during a busy period and a vacationperiod and hence all arriving batches are not allowed to jointhe system at all time Steady-state queue size distributionat a random point of time as well as a departure epoch isderivedMoreover this paper attempts to unify several classesof related batch arrival queueing system
Choudhury and Madan [14] studied a two-stage batcharrival queueing system with a modified Bernoulli sched-ule vacation under N-policy In this paper a batch arrivalqueueing system where the server provides two stages ofheterogeneous service with a modified Bernoulli scheduleunder N-policy is considered The server remains idle tillthe queue size becomes N (ge1) As soon as the queue sizebecomes at least N the server instantly starts working andprovides two stages of service in succession to each customerthat is the first stage service followed by the second stageservice However after the second stage service the servermay take a vacation or decide to stay in the system toprovide service to the next customer if any The queuesize distribution at a random epoch as well as a departureepoch under the steady-state conditions is derived Furtherthe existence of the stochastic decomposition property toshow that the departure point queue size distribution ofthis model can be decomposed into the distributions ofthree independent randomvariables is demonstrated Finallya simple procedure to obtain optimal stationary operatingpolicy under a suitable linear cost structure is developed
Furthermore Chang and Takine [15] studied factorizationand stochastic decomposition properties in bulk queues withgeneralized vacation This paper considers a class of sta-tionary batch-arrival bulk-service queues with generalizedvacationsThe system consists of a single server and a waitingroom of infinite capacity Arrivals of customers follow abatchMarkovian arrival processThe server is unavailable foroccasional intervals of time called vacations and when it isavailable customers are served in groups of fixed size B Forthis class of queues the vector probability generating functionof the stationary queue length distribution is factored intotwo terms one of which is the vector probability generatingfunction of the conditional queue length distribution giventhat the server is on vacation is shown The special caseof batch Poisson arrivals is examined and a new stochasticdecomposition formula is derived for the stationary queuelength distribution
Altman and Yechiali [16] considered the systems withservers vacations where customersrsquo impatience is due to anabsentee of servers upon arrival Such a model representingfrequent behavior by waiting customers in service systemshad never been treated before in the literature
Altman and Yechiali [17] studied a system that is operat-ing as an MMinfin queue However when it becomes emptyit is assigned to perform another task the duration U ofwhich is random Customers arriving while the system isunavailable for service (ie occupied with a U-task) become
Mathematical Problems in Engineering 3
impatient each individual activates an ldquoimpatience timerrdquohaving random duration T such that if the system doesnot become available by the time the timer expires thecustomer leaves the system never to returnWhen the systemcompletes a U-task and there are waiting customers each oneis taken immediately into service Explicit expressions for thecorresponding mean queue sizes were obtained
In 2012 Kumar and Sharma [18] analyzed a single serverqueue with general service time distribution random systembreakdowns and Bernoulli schedule server vacations whereafter a service completion the server may decide to leave thesystem with probability 119901 or to continue serving customerswith probability 1 minus 119901 It is assumed that the customersarrive to the system in batches of variable size but are servedone by one If the system breaks down it enters a repairprocess immediately It is assumed that the repair time hasgeneral distribution while the vacation time has exponentialdistribution The purpose is to find the steady-state resultsin explicit and closed form in terms of the probability-generating functions for the number of customers in thequeue the average number of customers and the averagewaiting time in the queue
Also Maraghi et al [19] studied an M119909G1 queue withrandom breakdowns and Bernoulli schedule server vacationswhere after a service completion the server may decide toleave the system with probability p or to continue servingcustomers with probability 1minus119901 It is assumed that customersarrive to the system in batches of variable size but are servedone by one
Kumar and Sharma [20] analysed the concept of customerbalking and reneging has been exploited to a great extent inthe recent past by the queuing modelers According to thismodel a reneged customer can be convinced in many casesby employing certain convincing mechanism to stay in thequeue for completion of his serviceThus a reneged customercan be retained in the queuing system with some probability(say 119902) and it may leave the queue without receiving servicewith probability 119901 (= 1 minus 119902) This process was referred to ascustomer retention The effect of probability of retention onthe average system size had been studied
Recently in 2013 Baruah et al [21] studied a two-stagequeuing model where the server provides two stages ofservice one by one in succession They considered renegingto occur when the server is unavailable during the systembreakdown or vacation periods
Kumar and Sharma [22] in their queueing model withretention of reneged customers and balking studied a finitecapacity and the retention of reneged customers The cus-tomerrsquos impatience may occur in the form of balking alsoIn this paper the work of Kumar and Sharma (2013) isextended by including balking to take into consideration thebroader perspective of customerrsquos impatience The model issolved iteratively to obtain the steady-state probabilities ofsystem size Some useful performance measures are derivedand the effect of the probability of retaining the renegedcustomers on various performance measures is studiednumerically
11 Model Description In todayrsquos competitive world cus-tomerrsquos impatience has become a serious problem Firms areemploying a number of strategies to retain their customersTo the best of our knowledge studies on multistage batcharrival of service in reneging do not exist We extendand develop the above models on two stages of serviceto multistage of services In particular a two-stage modelof Monita Baruah Kailash C Madan and Tillal Eldabiin the year 2013 motivated us for developing this newmodel which we consider here Their model on two stagesof service is extended to multistage of service where thearrivals occur in batches Breakdown and reneging play a vitalrole in our model Two stages of service will be benefitedonly for small scale industries factories of low productionlimited network and so forth whereasmultistages of servicesrender a huge amount of beneficial factors in large scaleindustries and in multitier applications in e-commerce inparticular
This results in an introduction of a new model a non-Markovian multistage batch arrival queue with breakdownand reneging In this model once the customers enter theywill be provided with the service in all the stages of servicedepartment one by one in succession Once the servicegets completed customer departs from the system Oncethe multistage service of a unit is completed the server isassumed to take vacation with probability 120579 or may continueto offer service with probability (1 minus 120579) As soon as thevacation period of the server ends it joins the system tocontinue service of the waiting customers During systembreakdown (or) vacation periods there will be unavailabilityof service in that case reneging occurs Reneging is assumedto follow exponential distribution Arrival follows a Poissonprocess and service time follows general distribution Westudy the changes in the probability of idle time and in thetraffic intensity in some particular cases like (1) if there isno vacation and (2) if the service time and vacation timeare exponentially distributed Using supplementary variabletechnique we derive the steady-state probabilities and perfor-mance measures
This paper is organized as follows Applications of modelis presented in Section 2 the mathematical description of themodel is given in Section 3 the definitions and notationsare described in Section 4 the equations governing thesystem are given in detail using birth and death processin Section 5 the queue size distribution at random epochis highlighted in Section 6 and the average queue sizeand average waiting time are given in Section 7 Sections6 and 7 gives a comprehensive idea about the model inwhich step by step method is followed for better under-standing which helps in any extension of this model Theparticular cases are derived in Section 8 in which themodel is seen in various aspects or situations For justifi-cation of the model the numerical illustrations are givenin Section 9 and at last the conclusion is presented inSection 10 In this section future scope of this model isdiscussed
4 Mathematical Problems in Engineering
Policing Load balancer
despatcher
despatcherDrop sessions
(if needed)
Individual server
Tier 1
Tier 1 Tier 2
Tier 2
Tier 3
(Nonreplicated)
Sentry
Figure 1 A three-tier application in e-commerce
bodyStator
windingRotor part fixing
Cooling fan
fixing
End cover fixing
Painting
A complete electric motor
Stator
Figure 2 Multistage motor production process
2 Applications of Model
With the development of computer and communicationnetworks queueing systems and networks have been iden-tified as a powerful analysis and design tool for variousapplications Our model plays a major role in multitierapplications in e-commerce motor production process carproduction process and textile industry Somemultistage reallife applications in large scale manufacturing industry aredescribed in the following sections which come across thenecessity of our new model
21 Three-Tier Application in e-Commerce Consider a three-tier application as shown in Figure 1 The first two tiersare clustered web server and clustered java applicationrespectively and the third tier is a database used in e-commerce applications Each tier is considered to be a stageEach incoming application is subjected to admission controlat the sentry to ensure that the contracted performanceguarantees are met The excess sessions are turned awayduring overloadsThe load balancer efficiently distributes theincoming requests Each input data is considered to be acustomer The server maintenance period is considered to bea vacation period and this application can be extended tomultistage of services
Steel plate
Stamping Welding Painting
Engine assembly
and mountingAssemblyInspectionCar
Figure 3 Multistage car production process
22 Electric Motor Production Industry
Multistage in Motor Production Process (See Figure 2) In theelectric motor manufacturing industries multistage produc-tion process is implemented (Figure 2) The stator body isconsidered as the arrival customer In the first stage the statorbody is wounded by the coil in the second stage the rotorpart is fixed in the stator in the third stage the cooling fanis fixed in one end of the rotor in the fourth stage the endcover is fixed in the other end of the rotor in the fifth stagethe entire motor is painted and finally in the sixth stage acomplete electric motor is produced We consider the statorbody as a batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools stock verification and so forth Here the machinegets idle owing to breakdown and the machine idle period isconsidered as the breakdown period
23 Automobile Industry
Multistage in Car Production Process (See Figure 3) In auto-mobile industry car production process is considered as themultistage production process (Figure 3) Here we considerthe steel plate as the batch arrival customer In the firststage of production a roll of steel plate is cut according tothe shape of the car parts in the second stage and steel
Mathematical Problems in Engineering 5
Drawing machine
Carding machine
machineSpinning machine
Cone winding machine
Yarn production
Bunch of
cottonBlow room
Simplex
Figure 4 Textile yarn production process
plates are welded in the third stage all the car parts arepainted in the painting section in the fourth stage the engineassembly is mounted in the fifth stage extra fittings areassembled according to the customer requirements in thesixth stage quality control inspection is conducted andfinally a complete car is produced We consider the steelplate as the batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools andmachines stock verification and so forth Heremachine gets idle owing to breakdown and the machine idleperiod is considered as the breakdown period
24 Textile Yarn Production Industry
Multistage in Production Process (See Figure 4) In the textileindustry yarn production is considered as the multistageproduction process (Figure 4) Here a bunch of cotton isconsidered as the arrival customer In the first stage a bunchof cotton is inserted into the blow room machine in thesecond stage the refined cotton is inserted into the cardingmachine in the third stage the produced cotton sliver isinserted into the drawing machine and is again processedin the fourth stage the processed cotton sliver is insertedinto the simplex machine in the fifth stage the refinedcotton sliver is inserted into the spinningmachine and in thesixth stage the cotton yarn output from spinning machine isinserted into the conewindingmachine and the entire cottonyarn is wounded as a cone These six stages of productionare required to convert a bunch of cotton into cotton yarnIn the production sector vacation denotes obtaining a newraw material maintenance of the tool and machine stockverification and so forth Here machine gets idle owing tobreakdown and the machine idle period is considered as thebreakdown period
3 The Mathematical Description of the Model
31 The Model Is Based on the Following Assumptions (a)Customers arrive in batches following a compound Poissonprocess with a rate of arrival 120582 Let 120582119888119894119889119905 (119894 = 1 2 3 119899) bethe first-order probability that customers arrive at the systemin batches of size 119894 at the system in a short interval of time(119909 119909 + 119889119905) where 0 le 119888119894 le 1 and sum
infin
119894=1 119888119894 = 1(b) The server provides multistage of heterogeneous
services one after the other in succession An arrival batchreceives the service offered at multistage in successiondefined as the first stage second stage and so forth respec-tively The service discipline is assumed to be on a first come
first served basis (FCFS) Let us assume that the service time119878119895 (119895 = 1 to 119899) of the 119895th stage service follows a generalprobability distribution with a distribution function 119870119895(119878119895)where 119896119895(119904119895) is the probability density function and 119864(119878
119899119895 ) is
the 119899th moment of the service time with 119895 = 1 2 119873Let
120583119895 (119909) =
119896119895 (119909)
1 minus 119870119895 (119909)119895 = 1 2 119873
119896119895 (119878119895) = 120583119895 (119909) 119890[minus int119904
0
120583119895(119909)119889119909]
where 119895 = 1 2 119873
(1)
(c) Once the multistage service of a unit is completed theserver is assumed to take vacation with probability 120579 or maycontinue to offer service with probability (1 minus 120579) As soon asthe vacation period of the server ends it joins the system tocontinue the service of the waiting customers
Let us assume the vacation time to be a random variablefollowing general probability law with distribution119872(V) anddensity function by 119898(V) 119864(119881119899) denotes the 119899th moment(119899 = 1 2 ) of the vacation time Here let us consider that120585(119909) is the conditional probability of a vacation period duringthe interval (119909 119909 + 119889119909) given that elapsed time is 119909 whichcan be given as
120585 (119909) =119872 (119909)
1 minus119872 (119909) (2)
Thus
119898(V) = 120585 (V) 119890[minus intV0
120585(119909)119889119909] (3)
(d) Customers arriving for service may become impatientand renege after joining during vacations and breakdownperiods Reneging is assumed to follow exponential distribu-tion with parameter 120595 Thus 119891 (119905) = 120595119890
minus120595119905119889119905 120595 gt 0 Thus
120595119889119905 is the probability that a customer can renege during ashort interval of time (119905 119905 + 119889119905)
The system may fail or be subjected to breakdown atrandom The customer receiving service during breakdownreturns back to the head of the queue Let us assume that theinterval between breakdowns occurs according to a Poissonprocess with a mean rate of breakdown 120573 gt 0
Furthermore the repair times follow a general (arbitrary)distribution with the distribution function 119866(119909) and densityfunction 119892 (119909) Let the conditional probability of comple-tion of the repair process be Υ (119909) 119889119909 such that Υ (119909) =
119866(119909) (1 minus 119866(119909)) and thus 119866 (119903) = Υ (119903) 119890(minus int119903
0
Υ(119909)119889119909)
4 Definitions and Notations
We assume that steady state exists and define 119875119899119895 (119909) as theprobability that there are n (ge1) customers in the systemincluding one customer in type 119895 service 119895 = 1 2 119873and elapsed service time is 119909 119875119899119895 = int
infin
0119875119899119895 (119909) 119889119909 is
the corresponding steady-state probability irrespective ofelapsed time 119909 119881119899 (119909) is the probability that there are n (ge0)customers in the queue and server is on vacation and elapsedvacation time is 119909
6 Mathematical Problems in Engineering
119881119899 = intinfin
0119881119899 (119909) 119889119909 is the corresponding steady-state
probability irrespective of elapsed vacation time 119909Q is the steady-state probability of the server being idle as
the server takes vacationThe probability generating functions are defined as
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 (119909)
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 |119911| le 1 119895 = 1 2 119873
119877 (119909 119911) =
infin
sum
119899=1
119911119899119877119899 (119909)
119877 (119911) =
infin
sum
119899=1
119911119899119877119899 |119911| le 1
119881 (119909 119911) =
infin
sum
119899=1
119911119899119881 (119909)
119881 (119911) =
infin
sum
119899=1
119911119899119881119899 |119911| le 1
119862 (119911) =
infin
sum
119894=1
119911119894119862119894
(4)
5 Equations Governing the System
We frame the difference equations with Poisson input relatedto birth-death process
Let 119899 be the size of customers at time 119909 Let 119875119899(119909) be theprobability of ldquonrdquo customers in the system
Let
(1) 120582119899 be the average arrival rate when ldquo119899rdquo customers arein the system
(2) 120583119899 the average service rate when ldquo119899rdquo customers are inthe system
(3) 119875119899(119909+Δ119909) the probability of ldquo119899rdquo customers at (119909+Δ119909)(4) 120573 the probability of breakdown arrival(5) 120582119862119894 the probability of batch arrival
Consider the following
1198751198991 (119909 Δ119909) = (1 minus 120582 + 1205831 (119909) + 120573 Δ119909)
times 1198751198991 (119909) + 119875119899minus11 (119909) 1205821198621Δ119909
+ 119875119899minus21 (119909) 1205821198622Δ119909
+ sdot sdot sdot + 11987511 (119909) 120582119862119899minus1Δ119909
1198751198991 (119909 + Δ119909) minus 1198751198991 (119909)
Δ119909+ 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
=
119899
sum
119894=1
120582119862119894119875119899minus1198941 (119909)
(5)
From above we have
119889
1198891199091198751198991 (119909 119911) + 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198941 (119909) 119899 ge 1
(6)
Similarly we have
119889
1198891199091198751198992 (119909 119911) + 120582 + 1205832 (119909) + 120573 1198751198992 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198942 (119909) 119899 ge 1
(7)
For Nth stage119873 ge 3
119889
119889119909119875119899119873 (119909 119911) + 120582 + 120583119873 (119909) + 120573 119875119899119873 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus119894119873 (119909) 119899 ge 1
(8)
119889
119889119909119877119899 (119909) + 120582 + Υ (119909) + 120595 119877119899 (119909)
= 120582
119899
sum
119894=1
119888119894119877119899minus119894 (119909) + 120595119877119899 (119909) 119899 ge 1
(9)
119889
1198891199091198770 (119909) + (120582 + Υ (119909)) 1198770 (119909) = 1205951198771 (119909)
(10)
119889
119889119909119881119899 (119909) + 120582 + 120585 (119909) + 120595119881119899 (119909)
= 120582
119899
sum
119894=1
119888119894119881119899minus119894 (119909) + 120595119881119899+1 (119909) 119899 ge 1
(11)
119889
1198891199091198810 (119909) + 120582 + 120585 (119909) + 1205951198810 (119909) = 1205951198810 (119909)
(12)
120582119876 = (1 minus 120579) int
infin
0
1198750119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
1198770 (119909) Υ (119909) 119889119909
(13)
The abovementioned differential equations should be solvedbased on the following boundary conditions
1198751198991 (0) = 120582119888119899119876 + (1 minus 120579) int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
119877119899+1 (119909) Υ (119909) 119889119909
+ int
infin
0
119881119899 (119909) 120595 (119909) 119889119909 119899 ge 1
(14)
1198751198992 (0) = int
infin
0
1198751198991 (119909) 1205831 (119909) 119889119909 119899 ge 1 (15)
Mathematical Problems in Engineering 7
For Nth stage
119875119899119873 (0) = int
infin
0
119875119899119873minus1 (119909) 120583119873minus1 (119909) 119889119909 119899 ge 1 (16)
119881119899 (0) = 120579int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909 119899 ge 0 (17)
119877119899+1 (0) =
119873
sum
119895=1
int
infin
0
119875119899119895 (119909) 119889119909 (18)
1198770 (0) = 0 (19)
Note LHS of (14) indicates that there are n customers in thesystem and the service is about to start in stage 1
RHS of (14) indicates the various situations in which thesystem has n customers and the service is about to start instage 1
The first term of RHS of (14) indicates that the system is inidle (119876) an arrival of batch consisting of 119899 customers occursand 1st stage service is about to start
The second term indicates that there are 119899 + 1 customersin the system including one customer whose service iscompleted in Nth stage (119873 ge 2) with probability 120583119873(119909)
remaining 119899 customers are there in the system 1st stage ofservice is about to start Moreover the server is not going forvacation instead stays in the system with probability 1 minus 120579
The third term indicates the repair process gets completedwith probabilityΥ (119909) So one customer just enters into the 1ststage service Now there are 119899 customers in the system
The fourth term indicates that vacation gets completedwith probability 120595 (119909) There are 119899 customers in the system1st stage of service is about to start
The combination of all the four terms gives the LHS of(14)
The integral in the RHS of (15) states that after comple-tion of 1st stage of service there are 119899 customers in the systemto whom 2nd stage of service is about to start which gives theLHS of (15)
Similarly (16) indicates that the Nth stage of service isabout to start
RHS of (17) explains the fact that after completion ofvacation with probability 120579 the server has just returned to thesystem consisting of n customers and the 1st stage of serviceis about to start This leads to the LHS of (17)
6 Queue Size Distribution at Random Epoch
Weuse supplementary variable technique to get the followingcomprehensive idea about the model
Multiplying (6) to (12) by 119911119899 and summing over n from 1
toinfin yields that we can obtain
119889
1198891199091198751 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205831 (119909) + 120573 1198751 (119909 119911) = 0
(20)119889
1198891199091198752 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205832 (119909) + 120573 1198752 (119909 119911) = 0
(21a)
For Nth stage119889
119889119909119875119873 (119909 119911) + (120582 minus 120582119862 (119911)) + 120583119873 (119909) + 120573 119875119873 (119909 119911) = 0
(21b)
119889
119889119909119877 (119909 119911) + 120582 minus 120582119862 (119911) + Υ (119909) + 120595 minus
120595
2119877 (119909 119911) = 0
(22)
119889
119889119909119881 (119909 119911) + 120582 minus 120582119862 (119911) + 120585 (119909) + 120595 minus
120595
119911 = 0 (23)
Further integration of (20)ndash(23) over limits 0 to 119909 in generalwill result in
119875119895 (119909 119911) = 119875119895 (0 119911) 119890[(120582minus120582119862(119911)+120573)119909minusint
infin
0
120583119895(119905)119889119905]
119895 = 1 to 119873
119877 (119909 119911) = 119877 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
Υ(119905)119889119905]
119881 (119909 119911) = 119881 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
120585(119905)119889119905]
(24)
Next by multiplying the boundary conditions by suitablepowers of 119911119899+1 taking summation over all possible values of119899 and using the probability generating functions we get
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579) int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909
+ int
infin
0
119877 (119909 119911) Υ (119909) 119889119909
+ int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909
1198752 (0 119911) = int
infin
0
1198751 (119909 119911) 1205831 (119909) 119889119909
(25)
For Nth stage
119875119873 (0 119911) = int
infin
0
119875119873minus1 (119909 119911) 120583119873minus1 (119909) 119889119909 (26)
119911119881 (0 119911) = 120579int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909 (27)
119877 (0 119911) = 120572119911
119873
sum
119895=1
119875119895 (119911) (28)
Again integrating (24) with respect to 119909 in general will resultin
119875119895 (119911) = 119875119895 (0 119911) [
1 minus 119870119895 (120582 minus 120582119862 (119911) + 120573)
120582 minus 120582119862 (119911) + 120573] 119895 = 1 to 119873
(29)
119881 (119911) = 119881 (0 119911) [1 minus119872(120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (30)
119877 (119911) = 119877 (0 119911) [1 minus 119866 (120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (31)
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
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Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 3
impatient each individual activates an ldquoimpatience timerrdquohaving random duration T such that if the system doesnot become available by the time the timer expires thecustomer leaves the system never to returnWhen the systemcompletes a U-task and there are waiting customers each oneis taken immediately into service Explicit expressions for thecorresponding mean queue sizes were obtained
In 2012 Kumar and Sharma [18] analyzed a single serverqueue with general service time distribution random systembreakdowns and Bernoulli schedule server vacations whereafter a service completion the server may decide to leave thesystem with probability 119901 or to continue serving customerswith probability 1 minus 119901 It is assumed that the customersarrive to the system in batches of variable size but are servedone by one If the system breaks down it enters a repairprocess immediately It is assumed that the repair time hasgeneral distribution while the vacation time has exponentialdistribution The purpose is to find the steady-state resultsin explicit and closed form in terms of the probability-generating functions for the number of customers in thequeue the average number of customers and the averagewaiting time in the queue
Also Maraghi et al [19] studied an M119909G1 queue withrandom breakdowns and Bernoulli schedule server vacationswhere after a service completion the server may decide toleave the system with probability p or to continue servingcustomers with probability 1minus119901 It is assumed that customersarrive to the system in batches of variable size but are servedone by one
Kumar and Sharma [20] analysed the concept of customerbalking and reneging has been exploited to a great extent inthe recent past by the queuing modelers According to thismodel a reneged customer can be convinced in many casesby employing certain convincing mechanism to stay in thequeue for completion of his serviceThus a reneged customercan be retained in the queuing system with some probability(say 119902) and it may leave the queue without receiving servicewith probability 119901 (= 1 minus 119902) This process was referred to ascustomer retention The effect of probability of retention onthe average system size had been studied
Recently in 2013 Baruah et al [21] studied a two-stagequeuing model where the server provides two stages ofservice one by one in succession They considered renegingto occur when the server is unavailable during the systembreakdown or vacation periods
Kumar and Sharma [22] in their queueing model withretention of reneged customers and balking studied a finitecapacity and the retention of reneged customers The cus-tomerrsquos impatience may occur in the form of balking alsoIn this paper the work of Kumar and Sharma (2013) isextended by including balking to take into consideration thebroader perspective of customerrsquos impatience The model issolved iteratively to obtain the steady-state probabilities ofsystem size Some useful performance measures are derivedand the effect of the probability of retaining the renegedcustomers on various performance measures is studiednumerically
11 Model Description In todayrsquos competitive world cus-tomerrsquos impatience has become a serious problem Firms areemploying a number of strategies to retain their customersTo the best of our knowledge studies on multistage batcharrival of service in reneging do not exist We extendand develop the above models on two stages of serviceto multistage of services In particular a two-stage modelof Monita Baruah Kailash C Madan and Tillal Eldabiin the year 2013 motivated us for developing this newmodel which we consider here Their model on two stagesof service is extended to multistage of service where thearrivals occur in batches Breakdown and reneging play a vitalrole in our model Two stages of service will be benefitedonly for small scale industries factories of low productionlimited network and so forth whereasmultistages of servicesrender a huge amount of beneficial factors in large scaleindustries and in multitier applications in e-commerce inparticular
This results in an introduction of a new model a non-Markovian multistage batch arrival queue with breakdownand reneging In this model once the customers enter theywill be provided with the service in all the stages of servicedepartment one by one in succession Once the servicegets completed customer departs from the system Oncethe multistage service of a unit is completed the server isassumed to take vacation with probability 120579 or may continueto offer service with probability (1 minus 120579) As soon as thevacation period of the server ends it joins the system tocontinue service of the waiting customers During systembreakdown (or) vacation periods there will be unavailabilityof service in that case reneging occurs Reneging is assumedto follow exponential distribution Arrival follows a Poissonprocess and service time follows general distribution Westudy the changes in the probability of idle time and in thetraffic intensity in some particular cases like (1) if there isno vacation and (2) if the service time and vacation timeare exponentially distributed Using supplementary variabletechnique we derive the steady-state probabilities and perfor-mance measures
This paper is organized as follows Applications of modelis presented in Section 2 the mathematical description of themodel is given in Section 3 the definitions and notationsare described in Section 4 the equations governing thesystem are given in detail using birth and death processin Section 5 the queue size distribution at random epochis highlighted in Section 6 and the average queue sizeand average waiting time are given in Section 7 Sections6 and 7 gives a comprehensive idea about the model inwhich step by step method is followed for better under-standing which helps in any extension of this model Theparticular cases are derived in Section 8 in which themodel is seen in various aspects or situations For justifi-cation of the model the numerical illustrations are givenin Section 9 and at last the conclusion is presented inSection 10 In this section future scope of this model isdiscussed
4 Mathematical Problems in Engineering
Policing Load balancer
despatcher
despatcherDrop sessions
(if needed)
Individual server
Tier 1
Tier 1 Tier 2
Tier 2
Tier 3
(Nonreplicated)
Sentry
Figure 1 A three-tier application in e-commerce
bodyStator
windingRotor part fixing
Cooling fan
fixing
End cover fixing
Painting
A complete electric motor
Stator
Figure 2 Multistage motor production process
2 Applications of Model
With the development of computer and communicationnetworks queueing systems and networks have been iden-tified as a powerful analysis and design tool for variousapplications Our model plays a major role in multitierapplications in e-commerce motor production process carproduction process and textile industry Somemultistage reallife applications in large scale manufacturing industry aredescribed in the following sections which come across thenecessity of our new model
21 Three-Tier Application in e-Commerce Consider a three-tier application as shown in Figure 1 The first two tiersare clustered web server and clustered java applicationrespectively and the third tier is a database used in e-commerce applications Each tier is considered to be a stageEach incoming application is subjected to admission controlat the sentry to ensure that the contracted performanceguarantees are met The excess sessions are turned awayduring overloadsThe load balancer efficiently distributes theincoming requests Each input data is considered to be acustomer The server maintenance period is considered to bea vacation period and this application can be extended tomultistage of services
Steel plate
Stamping Welding Painting
Engine assembly
and mountingAssemblyInspectionCar
Figure 3 Multistage car production process
22 Electric Motor Production Industry
Multistage in Motor Production Process (See Figure 2) In theelectric motor manufacturing industries multistage produc-tion process is implemented (Figure 2) The stator body isconsidered as the arrival customer In the first stage the statorbody is wounded by the coil in the second stage the rotorpart is fixed in the stator in the third stage the cooling fanis fixed in one end of the rotor in the fourth stage the endcover is fixed in the other end of the rotor in the fifth stagethe entire motor is painted and finally in the sixth stage acomplete electric motor is produced We consider the statorbody as a batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools stock verification and so forth Here the machinegets idle owing to breakdown and the machine idle period isconsidered as the breakdown period
23 Automobile Industry
Multistage in Car Production Process (See Figure 3) In auto-mobile industry car production process is considered as themultistage production process (Figure 3) Here we considerthe steel plate as the batch arrival customer In the firststage of production a roll of steel plate is cut according tothe shape of the car parts in the second stage and steel
Mathematical Problems in Engineering 5
Drawing machine
Carding machine
machineSpinning machine
Cone winding machine
Yarn production
Bunch of
cottonBlow room
Simplex
Figure 4 Textile yarn production process
plates are welded in the third stage all the car parts arepainted in the painting section in the fourth stage the engineassembly is mounted in the fifth stage extra fittings areassembled according to the customer requirements in thesixth stage quality control inspection is conducted andfinally a complete car is produced We consider the steelplate as the batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools andmachines stock verification and so forth Heremachine gets idle owing to breakdown and the machine idleperiod is considered as the breakdown period
24 Textile Yarn Production Industry
Multistage in Production Process (See Figure 4) In the textileindustry yarn production is considered as the multistageproduction process (Figure 4) Here a bunch of cotton isconsidered as the arrival customer In the first stage a bunchof cotton is inserted into the blow room machine in thesecond stage the refined cotton is inserted into the cardingmachine in the third stage the produced cotton sliver isinserted into the drawing machine and is again processedin the fourth stage the processed cotton sliver is insertedinto the simplex machine in the fifth stage the refinedcotton sliver is inserted into the spinningmachine and in thesixth stage the cotton yarn output from spinning machine isinserted into the conewindingmachine and the entire cottonyarn is wounded as a cone These six stages of productionare required to convert a bunch of cotton into cotton yarnIn the production sector vacation denotes obtaining a newraw material maintenance of the tool and machine stockverification and so forth Here machine gets idle owing tobreakdown and the machine idle period is considered as thebreakdown period
3 The Mathematical Description of the Model
31 The Model Is Based on the Following Assumptions (a)Customers arrive in batches following a compound Poissonprocess with a rate of arrival 120582 Let 120582119888119894119889119905 (119894 = 1 2 3 119899) bethe first-order probability that customers arrive at the systemin batches of size 119894 at the system in a short interval of time(119909 119909 + 119889119905) where 0 le 119888119894 le 1 and sum
infin
119894=1 119888119894 = 1(b) The server provides multistage of heterogeneous
services one after the other in succession An arrival batchreceives the service offered at multistage in successiondefined as the first stage second stage and so forth respec-tively The service discipline is assumed to be on a first come
first served basis (FCFS) Let us assume that the service time119878119895 (119895 = 1 to 119899) of the 119895th stage service follows a generalprobability distribution with a distribution function 119870119895(119878119895)where 119896119895(119904119895) is the probability density function and 119864(119878
119899119895 ) is
the 119899th moment of the service time with 119895 = 1 2 119873Let
120583119895 (119909) =
119896119895 (119909)
1 minus 119870119895 (119909)119895 = 1 2 119873
119896119895 (119878119895) = 120583119895 (119909) 119890[minus int119904
0
120583119895(119909)119889119909]
where 119895 = 1 2 119873
(1)
(c) Once the multistage service of a unit is completed theserver is assumed to take vacation with probability 120579 or maycontinue to offer service with probability (1 minus 120579) As soon asthe vacation period of the server ends it joins the system tocontinue the service of the waiting customers
Let us assume the vacation time to be a random variablefollowing general probability law with distribution119872(V) anddensity function by 119898(V) 119864(119881119899) denotes the 119899th moment(119899 = 1 2 ) of the vacation time Here let us consider that120585(119909) is the conditional probability of a vacation period duringthe interval (119909 119909 + 119889119909) given that elapsed time is 119909 whichcan be given as
120585 (119909) =119872 (119909)
1 minus119872 (119909) (2)
Thus
119898(V) = 120585 (V) 119890[minus intV0
120585(119909)119889119909] (3)
(d) Customers arriving for service may become impatientand renege after joining during vacations and breakdownperiods Reneging is assumed to follow exponential distribu-tion with parameter 120595 Thus 119891 (119905) = 120595119890
minus120595119905119889119905 120595 gt 0 Thus
120595119889119905 is the probability that a customer can renege during ashort interval of time (119905 119905 + 119889119905)
The system may fail or be subjected to breakdown atrandom The customer receiving service during breakdownreturns back to the head of the queue Let us assume that theinterval between breakdowns occurs according to a Poissonprocess with a mean rate of breakdown 120573 gt 0
Furthermore the repair times follow a general (arbitrary)distribution with the distribution function 119866(119909) and densityfunction 119892 (119909) Let the conditional probability of comple-tion of the repair process be Υ (119909) 119889119909 such that Υ (119909) =
119866(119909) (1 minus 119866(119909)) and thus 119866 (119903) = Υ (119903) 119890(minus int119903
0
Υ(119909)119889119909)
4 Definitions and Notations
We assume that steady state exists and define 119875119899119895 (119909) as theprobability that there are n (ge1) customers in the systemincluding one customer in type 119895 service 119895 = 1 2 119873and elapsed service time is 119909 119875119899119895 = int
infin
0119875119899119895 (119909) 119889119909 is
the corresponding steady-state probability irrespective ofelapsed time 119909 119881119899 (119909) is the probability that there are n (ge0)customers in the queue and server is on vacation and elapsedvacation time is 119909
6 Mathematical Problems in Engineering
119881119899 = intinfin
0119881119899 (119909) 119889119909 is the corresponding steady-state
probability irrespective of elapsed vacation time 119909Q is the steady-state probability of the server being idle as
the server takes vacationThe probability generating functions are defined as
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 (119909)
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 |119911| le 1 119895 = 1 2 119873
119877 (119909 119911) =
infin
sum
119899=1
119911119899119877119899 (119909)
119877 (119911) =
infin
sum
119899=1
119911119899119877119899 |119911| le 1
119881 (119909 119911) =
infin
sum
119899=1
119911119899119881 (119909)
119881 (119911) =
infin
sum
119899=1
119911119899119881119899 |119911| le 1
119862 (119911) =
infin
sum
119894=1
119911119894119862119894
(4)
5 Equations Governing the System
We frame the difference equations with Poisson input relatedto birth-death process
Let 119899 be the size of customers at time 119909 Let 119875119899(119909) be theprobability of ldquonrdquo customers in the system
Let
(1) 120582119899 be the average arrival rate when ldquo119899rdquo customers arein the system
(2) 120583119899 the average service rate when ldquo119899rdquo customers are inthe system
(3) 119875119899(119909+Δ119909) the probability of ldquo119899rdquo customers at (119909+Δ119909)(4) 120573 the probability of breakdown arrival(5) 120582119862119894 the probability of batch arrival
Consider the following
1198751198991 (119909 Δ119909) = (1 minus 120582 + 1205831 (119909) + 120573 Δ119909)
times 1198751198991 (119909) + 119875119899minus11 (119909) 1205821198621Δ119909
+ 119875119899minus21 (119909) 1205821198622Δ119909
+ sdot sdot sdot + 11987511 (119909) 120582119862119899minus1Δ119909
1198751198991 (119909 + Δ119909) minus 1198751198991 (119909)
Δ119909+ 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
=
119899
sum
119894=1
120582119862119894119875119899minus1198941 (119909)
(5)
From above we have
119889
1198891199091198751198991 (119909 119911) + 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198941 (119909) 119899 ge 1
(6)
Similarly we have
119889
1198891199091198751198992 (119909 119911) + 120582 + 1205832 (119909) + 120573 1198751198992 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198942 (119909) 119899 ge 1
(7)
For Nth stage119873 ge 3
119889
119889119909119875119899119873 (119909 119911) + 120582 + 120583119873 (119909) + 120573 119875119899119873 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus119894119873 (119909) 119899 ge 1
(8)
119889
119889119909119877119899 (119909) + 120582 + Υ (119909) + 120595 119877119899 (119909)
= 120582
119899
sum
119894=1
119888119894119877119899minus119894 (119909) + 120595119877119899 (119909) 119899 ge 1
(9)
119889
1198891199091198770 (119909) + (120582 + Υ (119909)) 1198770 (119909) = 1205951198771 (119909)
(10)
119889
119889119909119881119899 (119909) + 120582 + 120585 (119909) + 120595119881119899 (119909)
= 120582
119899
sum
119894=1
119888119894119881119899minus119894 (119909) + 120595119881119899+1 (119909) 119899 ge 1
(11)
119889
1198891199091198810 (119909) + 120582 + 120585 (119909) + 1205951198810 (119909) = 1205951198810 (119909)
(12)
120582119876 = (1 minus 120579) int
infin
0
1198750119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
1198770 (119909) Υ (119909) 119889119909
(13)
The abovementioned differential equations should be solvedbased on the following boundary conditions
1198751198991 (0) = 120582119888119899119876 + (1 minus 120579) int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
119877119899+1 (119909) Υ (119909) 119889119909
+ int
infin
0
119881119899 (119909) 120595 (119909) 119889119909 119899 ge 1
(14)
1198751198992 (0) = int
infin
0
1198751198991 (119909) 1205831 (119909) 119889119909 119899 ge 1 (15)
Mathematical Problems in Engineering 7
For Nth stage
119875119899119873 (0) = int
infin
0
119875119899119873minus1 (119909) 120583119873minus1 (119909) 119889119909 119899 ge 1 (16)
119881119899 (0) = 120579int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909 119899 ge 0 (17)
119877119899+1 (0) =
119873
sum
119895=1
int
infin
0
119875119899119895 (119909) 119889119909 (18)
1198770 (0) = 0 (19)
Note LHS of (14) indicates that there are n customers in thesystem and the service is about to start in stage 1
RHS of (14) indicates the various situations in which thesystem has n customers and the service is about to start instage 1
The first term of RHS of (14) indicates that the system is inidle (119876) an arrival of batch consisting of 119899 customers occursand 1st stage service is about to start
The second term indicates that there are 119899 + 1 customersin the system including one customer whose service iscompleted in Nth stage (119873 ge 2) with probability 120583119873(119909)
remaining 119899 customers are there in the system 1st stage ofservice is about to start Moreover the server is not going forvacation instead stays in the system with probability 1 minus 120579
The third term indicates the repair process gets completedwith probabilityΥ (119909) So one customer just enters into the 1ststage service Now there are 119899 customers in the system
The fourth term indicates that vacation gets completedwith probability 120595 (119909) There are 119899 customers in the system1st stage of service is about to start
The combination of all the four terms gives the LHS of(14)
The integral in the RHS of (15) states that after comple-tion of 1st stage of service there are 119899 customers in the systemto whom 2nd stage of service is about to start which gives theLHS of (15)
Similarly (16) indicates that the Nth stage of service isabout to start
RHS of (17) explains the fact that after completion ofvacation with probability 120579 the server has just returned to thesystem consisting of n customers and the 1st stage of serviceis about to start This leads to the LHS of (17)
6 Queue Size Distribution at Random Epoch
Weuse supplementary variable technique to get the followingcomprehensive idea about the model
Multiplying (6) to (12) by 119911119899 and summing over n from 1
toinfin yields that we can obtain
119889
1198891199091198751 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205831 (119909) + 120573 1198751 (119909 119911) = 0
(20)119889
1198891199091198752 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205832 (119909) + 120573 1198752 (119909 119911) = 0
(21a)
For Nth stage119889
119889119909119875119873 (119909 119911) + (120582 minus 120582119862 (119911)) + 120583119873 (119909) + 120573 119875119873 (119909 119911) = 0
(21b)
119889
119889119909119877 (119909 119911) + 120582 minus 120582119862 (119911) + Υ (119909) + 120595 minus
120595
2119877 (119909 119911) = 0
(22)
119889
119889119909119881 (119909 119911) + 120582 minus 120582119862 (119911) + 120585 (119909) + 120595 minus
120595
119911 = 0 (23)
Further integration of (20)ndash(23) over limits 0 to 119909 in generalwill result in
119875119895 (119909 119911) = 119875119895 (0 119911) 119890[(120582minus120582119862(119911)+120573)119909minusint
infin
0
120583119895(119905)119889119905]
119895 = 1 to 119873
119877 (119909 119911) = 119877 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
Υ(119905)119889119905]
119881 (119909 119911) = 119881 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
120585(119905)119889119905]
(24)
Next by multiplying the boundary conditions by suitablepowers of 119911119899+1 taking summation over all possible values of119899 and using the probability generating functions we get
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579) int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909
+ int
infin
0
119877 (119909 119911) Υ (119909) 119889119909
+ int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909
1198752 (0 119911) = int
infin
0
1198751 (119909 119911) 1205831 (119909) 119889119909
(25)
For Nth stage
119875119873 (0 119911) = int
infin
0
119875119873minus1 (119909 119911) 120583119873minus1 (119909) 119889119909 (26)
119911119881 (0 119911) = 120579int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909 (27)
119877 (0 119911) = 120572119911
119873
sum
119895=1
119875119895 (119911) (28)
Again integrating (24) with respect to 119909 in general will resultin
119875119895 (119911) = 119875119895 (0 119911) [
1 minus 119870119895 (120582 minus 120582119862 (119911) + 120573)
120582 minus 120582119862 (119911) + 120573] 119895 = 1 to 119873
(29)
119881 (119911) = 119881 (0 119911) [1 minus119872(120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (30)
119877 (119911) = 119877 (0 119911) [1 minus 119866 (120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (31)
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
4 Mathematical Problems in Engineering
Policing Load balancer
despatcher
despatcherDrop sessions
(if needed)
Individual server
Tier 1
Tier 1 Tier 2
Tier 2
Tier 3
(Nonreplicated)
Sentry
Figure 1 A three-tier application in e-commerce
bodyStator
windingRotor part fixing
Cooling fan
fixing
End cover fixing
Painting
A complete electric motor
Stator
Figure 2 Multistage motor production process
2 Applications of Model
With the development of computer and communicationnetworks queueing systems and networks have been iden-tified as a powerful analysis and design tool for variousapplications Our model plays a major role in multitierapplications in e-commerce motor production process carproduction process and textile industry Somemultistage reallife applications in large scale manufacturing industry aredescribed in the following sections which come across thenecessity of our new model
21 Three-Tier Application in e-Commerce Consider a three-tier application as shown in Figure 1 The first two tiersare clustered web server and clustered java applicationrespectively and the third tier is a database used in e-commerce applications Each tier is considered to be a stageEach incoming application is subjected to admission controlat the sentry to ensure that the contracted performanceguarantees are met The excess sessions are turned awayduring overloadsThe load balancer efficiently distributes theincoming requests Each input data is considered to be acustomer The server maintenance period is considered to bea vacation period and this application can be extended tomultistage of services
Steel plate
Stamping Welding Painting
Engine assembly
and mountingAssemblyInspectionCar
Figure 3 Multistage car production process
22 Electric Motor Production Industry
Multistage in Motor Production Process (See Figure 2) In theelectric motor manufacturing industries multistage produc-tion process is implemented (Figure 2) The stator body isconsidered as the arrival customer In the first stage the statorbody is wounded by the coil in the second stage the rotorpart is fixed in the stator in the third stage the cooling fanis fixed in one end of the rotor in the fourth stage the endcover is fixed in the other end of the rotor in the fifth stagethe entire motor is painted and finally in the sixth stage acomplete electric motor is produced We consider the statorbody as a batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools stock verification and so forth Here the machinegets idle owing to breakdown and the machine idle period isconsidered as the breakdown period
23 Automobile Industry
Multistage in Car Production Process (See Figure 3) In auto-mobile industry car production process is considered as themultistage production process (Figure 3) Here we considerthe steel plate as the batch arrival customer In the firststage of production a roll of steel plate is cut according tothe shape of the car parts in the second stage and steel
Mathematical Problems in Engineering 5
Drawing machine
Carding machine
machineSpinning machine
Cone winding machine
Yarn production
Bunch of
cottonBlow room
Simplex
Figure 4 Textile yarn production process
plates are welded in the third stage all the car parts arepainted in the painting section in the fourth stage the engineassembly is mounted in the fifth stage extra fittings areassembled according to the customer requirements in thesixth stage quality control inspection is conducted andfinally a complete car is produced We consider the steelplate as the batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools andmachines stock verification and so forth Heremachine gets idle owing to breakdown and the machine idleperiod is considered as the breakdown period
24 Textile Yarn Production Industry
Multistage in Production Process (See Figure 4) In the textileindustry yarn production is considered as the multistageproduction process (Figure 4) Here a bunch of cotton isconsidered as the arrival customer In the first stage a bunchof cotton is inserted into the blow room machine in thesecond stage the refined cotton is inserted into the cardingmachine in the third stage the produced cotton sliver isinserted into the drawing machine and is again processedin the fourth stage the processed cotton sliver is insertedinto the simplex machine in the fifth stage the refinedcotton sliver is inserted into the spinningmachine and in thesixth stage the cotton yarn output from spinning machine isinserted into the conewindingmachine and the entire cottonyarn is wounded as a cone These six stages of productionare required to convert a bunch of cotton into cotton yarnIn the production sector vacation denotes obtaining a newraw material maintenance of the tool and machine stockverification and so forth Here machine gets idle owing tobreakdown and the machine idle period is considered as thebreakdown period
3 The Mathematical Description of the Model
31 The Model Is Based on the Following Assumptions (a)Customers arrive in batches following a compound Poissonprocess with a rate of arrival 120582 Let 120582119888119894119889119905 (119894 = 1 2 3 119899) bethe first-order probability that customers arrive at the systemin batches of size 119894 at the system in a short interval of time(119909 119909 + 119889119905) where 0 le 119888119894 le 1 and sum
infin
119894=1 119888119894 = 1(b) The server provides multistage of heterogeneous
services one after the other in succession An arrival batchreceives the service offered at multistage in successiondefined as the first stage second stage and so forth respec-tively The service discipline is assumed to be on a first come
first served basis (FCFS) Let us assume that the service time119878119895 (119895 = 1 to 119899) of the 119895th stage service follows a generalprobability distribution with a distribution function 119870119895(119878119895)where 119896119895(119904119895) is the probability density function and 119864(119878
119899119895 ) is
the 119899th moment of the service time with 119895 = 1 2 119873Let
120583119895 (119909) =
119896119895 (119909)
1 minus 119870119895 (119909)119895 = 1 2 119873
119896119895 (119878119895) = 120583119895 (119909) 119890[minus int119904
0
120583119895(119909)119889119909]
where 119895 = 1 2 119873
(1)
(c) Once the multistage service of a unit is completed theserver is assumed to take vacation with probability 120579 or maycontinue to offer service with probability (1 minus 120579) As soon asthe vacation period of the server ends it joins the system tocontinue the service of the waiting customers
Let us assume the vacation time to be a random variablefollowing general probability law with distribution119872(V) anddensity function by 119898(V) 119864(119881119899) denotes the 119899th moment(119899 = 1 2 ) of the vacation time Here let us consider that120585(119909) is the conditional probability of a vacation period duringthe interval (119909 119909 + 119889119909) given that elapsed time is 119909 whichcan be given as
120585 (119909) =119872 (119909)
1 minus119872 (119909) (2)
Thus
119898(V) = 120585 (V) 119890[minus intV0
120585(119909)119889119909] (3)
(d) Customers arriving for service may become impatientand renege after joining during vacations and breakdownperiods Reneging is assumed to follow exponential distribu-tion with parameter 120595 Thus 119891 (119905) = 120595119890
minus120595119905119889119905 120595 gt 0 Thus
120595119889119905 is the probability that a customer can renege during ashort interval of time (119905 119905 + 119889119905)
The system may fail or be subjected to breakdown atrandom The customer receiving service during breakdownreturns back to the head of the queue Let us assume that theinterval between breakdowns occurs according to a Poissonprocess with a mean rate of breakdown 120573 gt 0
Furthermore the repair times follow a general (arbitrary)distribution with the distribution function 119866(119909) and densityfunction 119892 (119909) Let the conditional probability of comple-tion of the repair process be Υ (119909) 119889119909 such that Υ (119909) =
119866(119909) (1 minus 119866(119909)) and thus 119866 (119903) = Υ (119903) 119890(minus int119903
0
Υ(119909)119889119909)
4 Definitions and Notations
We assume that steady state exists and define 119875119899119895 (119909) as theprobability that there are n (ge1) customers in the systemincluding one customer in type 119895 service 119895 = 1 2 119873and elapsed service time is 119909 119875119899119895 = int
infin
0119875119899119895 (119909) 119889119909 is
the corresponding steady-state probability irrespective ofelapsed time 119909 119881119899 (119909) is the probability that there are n (ge0)customers in the queue and server is on vacation and elapsedvacation time is 119909
6 Mathematical Problems in Engineering
119881119899 = intinfin
0119881119899 (119909) 119889119909 is the corresponding steady-state
probability irrespective of elapsed vacation time 119909Q is the steady-state probability of the server being idle as
the server takes vacationThe probability generating functions are defined as
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 (119909)
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 |119911| le 1 119895 = 1 2 119873
119877 (119909 119911) =
infin
sum
119899=1
119911119899119877119899 (119909)
119877 (119911) =
infin
sum
119899=1
119911119899119877119899 |119911| le 1
119881 (119909 119911) =
infin
sum
119899=1
119911119899119881 (119909)
119881 (119911) =
infin
sum
119899=1
119911119899119881119899 |119911| le 1
119862 (119911) =
infin
sum
119894=1
119911119894119862119894
(4)
5 Equations Governing the System
We frame the difference equations with Poisson input relatedto birth-death process
Let 119899 be the size of customers at time 119909 Let 119875119899(119909) be theprobability of ldquonrdquo customers in the system
Let
(1) 120582119899 be the average arrival rate when ldquo119899rdquo customers arein the system
(2) 120583119899 the average service rate when ldquo119899rdquo customers are inthe system
(3) 119875119899(119909+Δ119909) the probability of ldquo119899rdquo customers at (119909+Δ119909)(4) 120573 the probability of breakdown arrival(5) 120582119862119894 the probability of batch arrival
Consider the following
1198751198991 (119909 Δ119909) = (1 minus 120582 + 1205831 (119909) + 120573 Δ119909)
times 1198751198991 (119909) + 119875119899minus11 (119909) 1205821198621Δ119909
+ 119875119899minus21 (119909) 1205821198622Δ119909
+ sdot sdot sdot + 11987511 (119909) 120582119862119899minus1Δ119909
1198751198991 (119909 + Δ119909) minus 1198751198991 (119909)
Δ119909+ 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
=
119899
sum
119894=1
120582119862119894119875119899minus1198941 (119909)
(5)
From above we have
119889
1198891199091198751198991 (119909 119911) + 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198941 (119909) 119899 ge 1
(6)
Similarly we have
119889
1198891199091198751198992 (119909 119911) + 120582 + 1205832 (119909) + 120573 1198751198992 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198942 (119909) 119899 ge 1
(7)
For Nth stage119873 ge 3
119889
119889119909119875119899119873 (119909 119911) + 120582 + 120583119873 (119909) + 120573 119875119899119873 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus119894119873 (119909) 119899 ge 1
(8)
119889
119889119909119877119899 (119909) + 120582 + Υ (119909) + 120595 119877119899 (119909)
= 120582
119899
sum
119894=1
119888119894119877119899minus119894 (119909) + 120595119877119899 (119909) 119899 ge 1
(9)
119889
1198891199091198770 (119909) + (120582 + Υ (119909)) 1198770 (119909) = 1205951198771 (119909)
(10)
119889
119889119909119881119899 (119909) + 120582 + 120585 (119909) + 120595119881119899 (119909)
= 120582
119899
sum
119894=1
119888119894119881119899minus119894 (119909) + 120595119881119899+1 (119909) 119899 ge 1
(11)
119889
1198891199091198810 (119909) + 120582 + 120585 (119909) + 1205951198810 (119909) = 1205951198810 (119909)
(12)
120582119876 = (1 minus 120579) int
infin
0
1198750119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
1198770 (119909) Υ (119909) 119889119909
(13)
The abovementioned differential equations should be solvedbased on the following boundary conditions
1198751198991 (0) = 120582119888119899119876 + (1 minus 120579) int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
119877119899+1 (119909) Υ (119909) 119889119909
+ int
infin
0
119881119899 (119909) 120595 (119909) 119889119909 119899 ge 1
(14)
1198751198992 (0) = int
infin
0
1198751198991 (119909) 1205831 (119909) 119889119909 119899 ge 1 (15)
Mathematical Problems in Engineering 7
For Nth stage
119875119899119873 (0) = int
infin
0
119875119899119873minus1 (119909) 120583119873minus1 (119909) 119889119909 119899 ge 1 (16)
119881119899 (0) = 120579int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909 119899 ge 0 (17)
119877119899+1 (0) =
119873
sum
119895=1
int
infin
0
119875119899119895 (119909) 119889119909 (18)
1198770 (0) = 0 (19)
Note LHS of (14) indicates that there are n customers in thesystem and the service is about to start in stage 1
RHS of (14) indicates the various situations in which thesystem has n customers and the service is about to start instage 1
The first term of RHS of (14) indicates that the system is inidle (119876) an arrival of batch consisting of 119899 customers occursand 1st stage service is about to start
The second term indicates that there are 119899 + 1 customersin the system including one customer whose service iscompleted in Nth stage (119873 ge 2) with probability 120583119873(119909)
remaining 119899 customers are there in the system 1st stage ofservice is about to start Moreover the server is not going forvacation instead stays in the system with probability 1 minus 120579
The third term indicates the repair process gets completedwith probabilityΥ (119909) So one customer just enters into the 1ststage service Now there are 119899 customers in the system
The fourth term indicates that vacation gets completedwith probability 120595 (119909) There are 119899 customers in the system1st stage of service is about to start
The combination of all the four terms gives the LHS of(14)
The integral in the RHS of (15) states that after comple-tion of 1st stage of service there are 119899 customers in the systemto whom 2nd stage of service is about to start which gives theLHS of (15)
Similarly (16) indicates that the Nth stage of service isabout to start
RHS of (17) explains the fact that after completion ofvacation with probability 120579 the server has just returned to thesystem consisting of n customers and the 1st stage of serviceis about to start This leads to the LHS of (17)
6 Queue Size Distribution at Random Epoch
Weuse supplementary variable technique to get the followingcomprehensive idea about the model
Multiplying (6) to (12) by 119911119899 and summing over n from 1
toinfin yields that we can obtain
119889
1198891199091198751 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205831 (119909) + 120573 1198751 (119909 119911) = 0
(20)119889
1198891199091198752 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205832 (119909) + 120573 1198752 (119909 119911) = 0
(21a)
For Nth stage119889
119889119909119875119873 (119909 119911) + (120582 minus 120582119862 (119911)) + 120583119873 (119909) + 120573 119875119873 (119909 119911) = 0
(21b)
119889
119889119909119877 (119909 119911) + 120582 minus 120582119862 (119911) + Υ (119909) + 120595 minus
120595
2119877 (119909 119911) = 0
(22)
119889
119889119909119881 (119909 119911) + 120582 minus 120582119862 (119911) + 120585 (119909) + 120595 minus
120595
119911 = 0 (23)
Further integration of (20)ndash(23) over limits 0 to 119909 in generalwill result in
119875119895 (119909 119911) = 119875119895 (0 119911) 119890[(120582minus120582119862(119911)+120573)119909minusint
infin
0
120583119895(119905)119889119905]
119895 = 1 to 119873
119877 (119909 119911) = 119877 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
Υ(119905)119889119905]
119881 (119909 119911) = 119881 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
120585(119905)119889119905]
(24)
Next by multiplying the boundary conditions by suitablepowers of 119911119899+1 taking summation over all possible values of119899 and using the probability generating functions we get
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579) int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909
+ int
infin
0
119877 (119909 119911) Υ (119909) 119889119909
+ int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909
1198752 (0 119911) = int
infin
0
1198751 (119909 119911) 1205831 (119909) 119889119909
(25)
For Nth stage
119875119873 (0 119911) = int
infin
0
119875119873minus1 (119909 119911) 120583119873minus1 (119909) 119889119909 (26)
119911119881 (0 119911) = 120579int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909 (27)
119877 (0 119911) = 120572119911
119873
sum
119895=1
119875119895 (119911) (28)
Again integrating (24) with respect to 119909 in general will resultin
119875119895 (119911) = 119875119895 (0 119911) [
1 minus 119870119895 (120582 minus 120582119862 (119911) + 120573)
120582 minus 120582119862 (119911) + 120573] 119895 = 1 to 119873
(29)
119881 (119911) = 119881 (0 119911) [1 minus119872(120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (30)
119877 (119911) = 119877 (0 119911) [1 minus 119866 (120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (31)
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 5
Drawing machine
Carding machine
machineSpinning machine
Cone winding machine
Yarn production
Bunch of
cottonBlow room
Simplex
Figure 4 Textile yarn production process
plates are welded in the third stage all the car parts arepainted in the painting section in the fourth stage the engineassembly is mounted in the fifth stage extra fittings areassembled according to the customer requirements in thesixth stage quality control inspection is conducted andfinally a complete car is produced We consider the steelplate as the batch arrival customer In the production sectorvacation denotes obtaining new rawmaterial maintenance ofthe tools andmachines stock verification and so forth Heremachine gets idle owing to breakdown and the machine idleperiod is considered as the breakdown period
24 Textile Yarn Production Industry
Multistage in Production Process (See Figure 4) In the textileindustry yarn production is considered as the multistageproduction process (Figure 4) Here a bunch of cotton isconsidered as the arrival customer In the first stage a bunchof cotton is inserted into the blow room machine in thesecond stage the refined cotton is inserted into the cardingmachine in the third stage the produced cotton sliver isinserted into the drawing machine and is again processedin the fourth stage the processed cotton sliver is insertedinto the simplex machine in the fifth stage the refinedcotton sliver is inserted into the spinningmachine and in thesixth stage the cotton yarn output from spinning machine isinserted into the conewindingmachine and the entire cottonyarn is wounded as a cone These six stages of productionare required to convert a bunch of cotton into cotton yarnIn the production sector vacation denotes obtaining a newraw material maintenance of the tool and machine stockverification and so forth Here machine gets idle owing tobreakdown and the machine idle period is considered as thebreakdown period
3 The Mathematical Description of the Model
31 The Model Is Based on the Following Assumptions (a)Customers arrive in batches following a compound Poissonprocess with a rate of arrival 120582 Let 120582119888119894119889119905 (119894 = 1 2 3 119899) bethe first-order probability that customers arrive at the systemin batches of size 119894 at the system in a short interval of time(119909 119909 + 119889119905) where 0 le 119888119894 le 1 and sum
infin
119894=1 119888119894 = 1(b) The server provides multistage of heterogeneous
services one after the other in succession An arrival batchreceives the service offered at multistage in successiondefined as the first stage second stage and so forth respec-tively The service discipline is assumed to be on a first come
first served basis (FCFS) Let us assume that the service time119878119895 (119895 = 1 to 119899) of the 119895th stage service follows a generalprobability distribution with a distribution function 119870119895(119878119895)where 119896119895(119904119895) is the probability density function and 119864(119878
119899119895 ) is
the 119899th moment of the service time with 119895 = 1 2 119873Let
120583119895 (119909) =
119896119895 (119909)
1 minus 119870119895 (119909)119895 = 1 2 119873
119896119895 (119878119895) = 120583119895 (119909) 119890[minus int119904
0
120583119895(119909)119889119909]
where 119895 = 1 2 119873
(1)
(c) Once the multistage service of a unit is completed theserver is assumed to take vacation with probability 120579 or maycontinue to offer service with probability (1 minus 120579) As soon asthe vacation period of the server ends it joins the system tocontinue the service of the waiting customers
Let us assume the vacation time to be a random variablefollowing general probability law with distribution119872(V) anddensity function by 119898(V) 119864(119881119899) denotes the 119899th moment(119899 = 1 2 ) of the vacation time Here let us consider that120585(119909) is the conditional probability of a vacation period duringthe interval (119909 119909 + 119889119909) given that elapsed time is 119909 whichcan be given as
120585 (119909) =119872 (119909)
1 minus119872 (119909) (2)
Thus
119898(V) = 120585 (V) 119890[minus intV0
120585(119909)119889119909] (3)
(d) Customers arriving for service may become impatientand renege after joining during vacations and breakdownperiods Reneging is assumed to follow exponential distribu-tion with parameter 120595 Thus 119891 (119905) = 120595119890
minus120595119905119889119905 120595 gt 0 Thus
120595119889119905 is the probability that a customer can renege during ashort interval of time (119905 119905 + 119889119905)
The system may fail or be subjected to breakdown atrandom The customer receiving service during breakdownreturns back to the head of the queue Let us assume that theinterval between breakdowns occurs according to a Poissonprocess with a mean rate of breakdown 120573 gt 0
Furthermore the repair times follow a general (arbitrary)distribution with the distribution function 119866(119909) and densityfunction 119892 (119909) Let the conditional probability of comple-tion of the repair process be Υ (119909) 119889119909 such that Υ (119909) =
119866(119909) (1 minus 119866(119909)) and thus 119866 (119903) = Υ (119903) 119890(minus int119903
0
Υ(119909)119889119909)
4 Definitions and Notations
We assume that steady state exists and define 119875119899119895 (119909) as theprobability that there are n (ge1) customers in the systemincluding one customer in type 119895 service 119895 = 1 2 119873and elapsed service time is 119909 119875119899119895 = int
infin
0119875119899119895 (119909) 119889119909 is
the corresponding steady-state probability irrespective ofelapsed time 119909 119881119899 (119909) is the probability that there are n (ge0)customers in the queue and server is on vacation and elapsedvacation time is 119909
6 Mathematical Problems in Engineering
119881119899 = intinfin
0119881119899 (119909) 119889119909 is the corresponding steady-state
probability irrespective of elapsed vacation time 119909Q is the steady-state probability of the server being idle as
the server takes vacationThe probability generating functions are defined as
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 (119909)
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 |119911| le 1 119895 = 1 2 119873
119877 (119909 119911) =
infin
sum
119899=1
119911119899119877119899 (119909)
119877 (119911) =
infin
sum
119899=1
119911119899119877119899 |119911| le 1
119881 (119909 119911) =
infin
sum
119899=1
119911119899119881 (119909)
119881 (119911) =
infin
sum
119899=1
119911119899119881119899 |119911| le 1
119862 (119911) =
infin
sum
119894=1
119911119894119862119894
(4)
5 Equations Governing the System
We frame the difference equations with Poisson input relatedto birth-death process
Let 119899 be the size of customers at time 119909 Let 119875119899(119909) be theprobability of ldquonrdquo customers in the system
Let
(1) 120582119899 be the average arrival rate when ldquo119899rdquo customers arein the system
(2) 120583119899 the average service rate when ldquo119899rdquo customers are inthe system
(3) 119875119899(119909+Δ119909) the probability of ldquo119899rdquo customers at (119909+Δ119909)(4) 120573 the probability of breakdown arrival(5) 120582119862119894 the probability of batch arrival
Consider the following
1198751198991 (119909 Δ119909) = (1 minus 120582 + 1205831 (119909) + 120573 Δ119909)
times 1198751198991 (119909) + 119875119899minus11 (119909) 1205821198621Δ119909
+ 119875119899minus21 (119909) 1205821198622Δ119909
+ sdot sdot sdot + 11987511 (119909) 120582119862119899minus1Δ119909
1198751198991 (119909 + Δ119909) minus 1198751198991 (119909)
Δ119909+ 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
=
119899
sum
119894=1
120582119862119894119875119899minus1198941 (119909)
(5)
From above we have
119889
1198891199091198751198991 (119909 119911) + 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198941 (119909) 119899 ge 1
(6)
Similarly we have
119889
1198891199091198751198992 (119909 119911) + 120582 + 1205832 (119909) + 120573 1198751198992 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198942 (119909) 119899 ge 1
(7)
For Nth stage119873 ge 3
119889
119889119909119875119899119873 (119909 119911) + 120582 + 120583119873 (119909) + 120573 119875119899119873 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus119894119873 (119909) 119899 ge 1
(8)
119889
119889119909119877119899 (119909) + 120582 + Υ (119909) + 120595 119877119899 (119909)
= 120582
119899
sum
119894=1
119888119894119877119899minus119894 (119909) + 120595119877119899 (119909) 119899 ge 1
(9)
119889
1198891199091198770 (119909) + (120582 + Υ (119909)) 1198770 (119909) = 1205951198771 (119909)
(10)
119889
119889119909119881119899 (119909) + 120582 + 120585 (119909) + 120595119881119899 (119909)
= 120582
119899
sum
119894=1
119888119894119881119899minus119894 (119909) + 120595119881119899+1 (119909) 119899 ge 1
(11)
119889
1198891199091198810 (119909) + 120582 + 120585 (119909) + 1205951198810 (119909) = 1205951198810 (119909)
(12)
120582119876 = (1 minus 120579) int
infin
0
1198750119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
1198770 (119909) Υ (119909) 119889119909
(13)
The abovementioned differential equations should be solvedbased on the following boundary conditions
1198751198991 (0) = 120582119888119899119876 + (1 minus 120579) int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
119877119899+1 (119909) Υ (119909) 119889119909
+ int
infin
0
119881119899 (119909) 120595 (119909) 119889119909 119899 ge 1
(14)
1198751198992 (0) = int
infin
0
1198751198991 (119909) 1205831 (119909) 119889119909 119899 ge 1 (15)
Mathematical Problems in Engineering 7
For Nth stage
119875119899119873 (0) = int
infin
0
119875119899119873minus1 (119909) 120583119873minus1 (119909) 119889119909 119899 ge 1 (16)
119881119899 (0) = 120579int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909 119899 ge 0 (17)
119877119899+1 (0) =
119873
sum
119895=1
int
infin
0
119875119899119895 (119909) 119889119909 (18)
1198770 (0) = 0 (19)
Note LHS of (14) indicates that there are n customers in thesystem and the service is about to start in stage 1
RHS of (14) indicates the various situations in which thesystem has n customers and the service is about to start instage 1
The first term of RHS of (14) indicates that the system is inidle (119876) an arrival of batch consisting of 119899 customers occursand 1st stage service is about to start
The second term indicates that there are 119899 + 1 customersin the system including one customer whose service iscompleted in Nth stage (119873 ge 2) with probability 120583119873(119909)
remaining 119899 customers are there in the system 1st stage ofservice is about to start Moreover the server is not going forvacation instead stays in the system with probability 1 minus 120579
The third term indicates the repair process gets completedwith probabilityΥ (119909) So one customer just enters into the 1ststage service Now there are 119899 customers in the system
The fourth term indicates that vacation gets completedwith probability 120595 (119909) There are 119899 customers in the system1st stage of service is about to start
The combination of all the four terms gives the LHS of(14)
The integral in the RHS of (15) states that after comple-tion of 1st stage of service there are 119899 customers in the systemto whom 2nd stage of service is about to start which gives theLHS of (15)
Similarly (16) indicates that the Nth stage of service isabout to start
RHS of (17) explains the fact that after completion ofvacation with probability 120579 the server has just returned to thesystem consisting of n customers and the 1st stage of serviceis about to start This leads to the LHS of (17)
6 Queue Size Distribution at Random Epoch
Weuse supplementary variable technique to get the followingcomprehensive idea about the model
Multiplying (6) to (12) by 119911119899 and summing over n from 1
toinfin yields that we can obtain
119889
1198891199091198751 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205831 (119909) + 120573 1198751 (119909 119911) = 0
(20)119889
1198891199091198752 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205832 (119909) + 120573 1198752 (119909 119911) = 0
(21a)
For Nth stage119889
119889119909119875119873 (119909 119911) + (120582 minus 120582119862 (119911)) + 120583119873 (119909) + 120573 119875119873 (119909 119911) = 0
(21b)
119889
119889119909119877 (119909 119911) + 120582 minus 120582119862 (119911) + Υ (119909) + 120595 minus
120595
2119877 (119909 119911) = 0
(22)
119889
119889119909119881 (119909 119911) + 120582 minus 120582119862 (119911) + 120585 (119909) + 120595 minus
120595
119911 = 0 (23)
Further integration of (20)ndash(23) over limits 0 to 119909 in generalwill result in
119875119895 (119909 119911) = 119875119895 (0 119911) 119890[(120582minus120582119862(119911)+120573)119909minusint
infin
0
120583119895(119905)119889119905]
119895 = 1 to 119873
119877 (119909 119911) = 119877 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
Υ(119905)119889119905]
119881 (119909 119911) = 119881 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
120585(119905)119889119905]
(24)
Next by multiplying the boundary conditions by suitablepowers of 119911119899+1 taking summation over all possible values of119899 and using the probability generating functions we get
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579) int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909
+ int
infin
0
119877 (119909 119911) Υ (119909) 119889119909
+ int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909
1198752 (0 119911) = int
infin
0
1198751 (119909 119911) 1205831 (119909) 119889119909
(25)
For Nth stage
119875119873 (0 119911) = int
infin
0
119875119873minus1 (119909 119911) 120583119873minus1 (119909) 119889119909 (26)
119911119881 (0 119911) = 120579int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909 (27)
119877 (0 119911) = 120572119911
119873
sum
119895=1
119875119895 (119911) (28)
Again integrating (24) with respect to 119909 in general will resultin
119875119895 (119911) = 119875119895 (0 119911) [
1 minus 119870119895 (120582 minus 120582119862 (119911) + 120573)
120582 minus 120582119862 (119911) + 120573] 119895 = 1 to 119873
(29)
119881 (119911) = 119881 (0 119911) [1 minus119872(120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (30)
119877 (119911) = 119877 (0 119911) [1 minus 119866 (120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (31)
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
6 Mathematical Problems in Engineering
119881119899 = intinfin
0119881119899 (119909) 119889119909 is the corresponding steady-state
probability irrespective of elapsed vacation time 119909Q is the steady-state probability of the server being idle as
the server takes vacationThe probability generating functions are defined as
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 (119909)
119875119895 (119909 119911) =
infin
sum
119899=1
119911119899119875119899119895 |119911| le 1 119895 = 1 2 119873
119877 (119909 119911) =
infin
sum
119899=1
119911119899119877119899 (119909)
119877 (119911) =
infin
sum
119899=1
119911119899119877119899 |119911| le 1
119881 (119909 119911) =
infin
sum
119899=1
119911119899119881 (119909)
119881 (119911) =
infin
sum
119899=1
119911119899119881119899 |119911| le 1
119862 (119911) =
infin
sum
119894=1
119911119894119862119894
(4)
5 Equations Governing the System
We frame the difference equations with Poisson input relatedto birth-death process
Let 119899 be the size of customers at time 119909 Let 119875119899(119909) be theprobability of ldquonrdquo customers in the system
Let
(1) 120582119899 be the average arrival rate when ldquo119899rdquo customers arein the system
(2) 120583119899 the average service rate when ldquo119899rdquo customers are inthe system
(3) 119875119899(119909+Δ119909) the probability of ldquo119899rdquo customers at (119909+Δ119909)(4) 120573 the probability of breakdown arrival(5) 120582119862119894 the probability of batch arrival
Consider the following
1198751198991 (119909 Δ119909) = (1 minus 120582 + 1205831 (119909) + 120573 Δ119909)
times 1198751198991 (119909) + 119875119899minus11 (119909) 1205821198621Δ119909
+ 119875119899minus21 (119909) 1205821198622Δ119909
+ sdot sdot sdot + 11987511 (119909) 120582119862119899minus1Δ119909
1198751198991 (119909 + Δ119909) minus 1198751198991 (119909)
Δ119909+ 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
=
119899
sum
119894=1
120582119862119894119875119899minus1198941 (119909)
(5)
From above we have
119889
1198891199091198751198991 (119909 119911) + 120582 + 1205831 (119909) + 120573 1198751198991 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198941 (119909) 119899 ge 1
(6)
Similarly we have
119889
1198891199091198751198992 (119909 119911) + 120582 + 1205832 (119909) + 120573 1198751198992 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus1198942 (119909) 119899 ge 1
(7)
For Nth stage119873 ge 3
119889
119889119909119875119899119873 (119909 119911) + 120582 + 120583119873 (119909) + 120573 119875119899119873 (119909)
= 120582
119899
sum
119894=1
119888119894119875119899minus119894119873 (119909) 119899 ge 1
(8)
119889
119889119909119877119899 (119909) + 120582 + Υ (119909) + 120595 119877119899 (119909)
= 120582
119899
sum
119894=1
119888119894119877119899minus119894 (119909) + 120595119877119899 (119909) 119899 ge 1
(9)
119889
1198891199091198770 (119909) + (120582 + Υ (119909)) 1198770 (119909) = 1205951198771 (119909)
(10)
119889
119889119909119881119899 (119909) + 120582 + 120585 (119909) + 120595119881119899 (119909)
= 120582
119899
sum
119894=1
119888119894119881119899minus119894 (119909) + 120595119881119899+1 (119909) 119899 ge 1
(11)
119889
1198891199091198810 (119909) + 120582 + 120585 (119909) + 1205951198810 (119909) = 1205951198810 (119909)
(12)
120582119876 = (1 minus 120579) int
infin
0
1198750119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
1198770 (119909) Υ (119909) 119889119909
(13)
The abovementioned differential equations should be solvedbased on the following boundary conditions
1198751198991 (0) = 120582119888119899119876 + (1 minus 120579) int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909
+ int
infin
0
119877119899+1 (119909) Υ (119909) 119889119909
+ int
infin
0
119881119899 (119909) 120595 (119909) 119889119909 119899 ge 1
(14)
1198751198992 (0) = int
infin
0
1198751198991 (119909) 1205831 (119909) 119889119909 119899 ge 1 (15)
Mathematical Problems in Engineering 7
For Nth stage
119875119899119873 (0) = int
infin
0
119875119899119873minus1 (119909) 120583119873minus1 (119909) 119889119909 119899 ge 1 (16)
119881119899 (0) = 120579int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909 119899 ge 0 (17)
119877119899+1 (0) =
119873
sum
119895=1
int
infin
0
119875119899119895 (119909) 119889119909 (18)
1198770 (0) = 0 (19)
Note LHS of (14) indicates that there are n customers in thesystem and the service is about to start in stage 1
RHS of (14) indicates the various situations in which thesystem has n customers and the service is about to start instage 1
The first term of RHS of (14) indicates that the system is inidle (119876) an arrival of batch consisting of 119899 customers occursand 1st stage service is about to start
The second term indicates that there are 119899 + 1 customersin the system including one customer whose service iscompleted in Nth stage (119873 ge 2) with probability 120583119873(119909)
remaining 119899 customers are there in the system 1st stage ofservice is about to start Moreover the server is not going forvacation instead stays in the system with probability 1 minus 120579
The third term indicates the repair process gets completedwith probabilityΥ (119909) So one customer just enters into the 1ststage service Now there are 119899 customers in the system
The fourth term indicates that vacation gets completedwith probability 120595 (119909) There are 119899 customers in the system1st stage of service is about to start
The combination of all the four terms gives the LHS of(14)
The integral in the RHS of (15) states that after comple-tion of 1st stage of service there are 119899 customers in the systemto whom 2nd stage of service is about to start which gives theLHS of (15)
Similarly (16) indicates that the Nth stage of service isabout to start
RHS of (17) explains the fact that after completion ofvacation with probability 120579 the server has just returned to thesystem consisting of n customers and the 1st stage of serviceis about to start This leads to the LHS of (17)
6 Queue Size Distribution at Random Epoch
Weuse supplementary variable technique to get the followingcomprehensive idea about the model
Multiplying (6) to (12) by 119911119899 and summing over n from 1
toinfin yields that we can obtain
119889
1198891199091198751 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205831 (119909) + 120573 1198751 (119909 119911) = 0
(20)119889
1198891199091198752 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205832 (119909) + 120573 1198752 (119909 119911) = 0
(21a)
For Nth stage119889
119889119909119875119873 (119909 119911) + (120582 minus 120582119862 (119911)) + 120583119873 (119909) + 120573 119875119873 (119909 119911) = 0
(21b)
119889
119889119909119877 (119909 119911) + 120582 minus 120582119862 (119911) + Υ (119909) + 120595 minus
120595
2119877 (119909 119911) = 0
(22)
119889
119889119909119881 (119909 119911) + 120582 minus 120582119862 (119911) + 120585 (119909) + 120595 minus
120595
119911 = 0 (23)
Further integration of (20)ndash(23) over limits 0 to 119909 in generalwill result in
119875119895 (119909 119911) = 119875119895 (0 119911) 119890[(120582minus120582119862(119911)+120573)119909minusint
infin
0
120583119895(119905)119889119905]
119895 = 1 to 119873
119877 (119909 119911) = 119877 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
Υ(119905)119889119905]
119881 (119909 119911) = 119881 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
120585(119905)119889119905]
(24)
Next by multiplying the boundary conditions by suitablepowers of 119911119899+1 taking summation over all possible values of119899 and using the probability generating functions we get
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579) int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909
+ int
infin
0
119877 (119909 119911) Υ (119909) 119889119909
+ int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909
1198752 (0 119911) = int
infin
0
1198751 (119909 119911) 1205831 (119909) 119889119909
(25)
For Nth stage
119875119873 (0 119911) = int
infin
0
119875119873minus1 (119909 119911) 120583119873minus1 (119909) 119889119909 (26)
119911119881 (0 119911) = 120579int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909 (27)
119877 (0 119911) = 120572119911
119873
sum
119895=1
119875119895 (119911) (28)
Again integrating (24) with respect to 119909 in general will resultin
119875119895 (119911) = 119875119895 (0 119911) [
1 minus 119870119895 (120582 minus 120582119862 (119911) + 120573)
120582 minus 120582119862 (119911) + 120573] 119895 = 1 to 119873
(29)
119881 (119911) = 119881 (0 119911) [1 minus119872(120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (30)
119877 (119911) = 119877 (0 119911) [1 minus 119866 (120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (31)
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 7
For Nth stage
119875119899119873 (0) = int
infin
0
119875119899119873minus1 (119909) 120583119873minus1 (119909) 119889119909 119899 ge 1 (16)
119881119899 (0) = 120579int
infin
0
119875119899+1119873 (119909) 120583119873 (119909) 119889119909 119899 ge 0 (17)
119877119899+1 (0) =
119873
sum
119895=1
int
infin
0
119875119899119895 (119909) 119889119909 (18)
1198770 (0) = 0 (19)
Note LHS of (14) indicates that there are n customers in thesystem and the service is about to start in stage 1
RHS of (14) indicates the various situations in which thesystem has n customers and the service is about to start instage 1
The first term of RHS of (14) indicates that the system is inidle (119876) an arrival of batch consisting of 119899 customers occursand 1st stage service is about to start
The second term indicates that there are 119899 + 1 customersin the system including one customer whose service iscompleted in Nth stage (119873 ge 2) with probability 120583119873(119909)
remaining 119899 customers are there in the system 1st stage ofservice is about to start Moreover the server is not going forvacation instead stays in the system with probability 1 minus 120579
The third term indicates the repair process gets completedwith probabilityΥ (119909) So one customer just enters into the 1ststage service Now there are 119899 customers in the system
The fourth term indicates that vacation gets completedwith probability 120595 (119909) There are 119899 customers in the system1st stage of service is about to start
The combination of all the four terms gives the LHS of(14)
The integral in the RHS of (15) states that after comple-tion of 1st stage of service there are 119899 customers in the systemto whom 2nd stage of service is about to start which gives theLHS of (15)
Similarly (16) indicates that the Nth stage of service isabout to start
RHS of (17) explains the fact that after completion ofvacation with probability 120579 the server has just returned to thesystem consisting of n customers and the 1st stage of serviceis about to start This leads to the LHS of (17)
6 Queue Size Distribution at Random Epoch
Weuse supplementary variable technique to get the followingcomprehensive idea about the model
Multiplying (6) to (12) by 119911119899 and summing over n from 1
toinfin yields that we can obtain
119889
1198891199091198751 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205831 (119909) + 120573 1198751 (119909 119911) = 0
(20)119889
1198891199091198752 (119909 119911) + (120582 minus 120582119862 (119911)) + 1205832 (119909) + 120573 1198752 (119909 119911) = 0
(21a)
For Nth stage119889
119889119909119875119873 (119909 119911) + (120582 minus 120582119862 (119911)) + 120583119873 (119909) + 120573 119875119873 (119909 119911) = 0
(21b)
119889
119889119909119877 (119909 119911) + 120582 minus 120582119862 (119911) + Υ (119909) + 120595 minus
120595
2119877 (119909 119911) = 0
(22)
119889
119889119909119881 (119909 119911) + 120582 minus 120582119862 (119911) + 120585 (119909) + 120595 minus
120595
119911 = 0 (23)
Further integration of (20)ndash(23) over limits 0 to 119909 in generalwill result in
119875119895 (119909 119911) = 119875119895 (0 119911) 119890[(120582minus120582119862(119911)+120573)119909minusint
infin
0
120583119895(119905)119889119905]
119895 = 1 to 119873
119877 (119909 119911) = 119877 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
Υ(119905)119889119905]
119881 (119909 119911) = 119881 (0 119911) 119890[(120582minus120582119862(119911)+120595minus(120595119911))119909minusint
infin
0
120585(119905)119889119905]
(24)
Next by multiplying the boundary conditions by suitablepowers of 119911119899+1 taking summation over all possible values of119899 and using the probability generating functions we get
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579) int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909
+ int
infin
0
119877 (119909 119911) Υ (119909) 119889119909
+ int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909
1198752 (0 119911) = int
infin
0
1198751 (119909 119911) 1205831 (119909) 119889119909
(25)
For Nth stage
119875119873 (0 119911) = int
infin
0
119875119873minus1 (119909 119911) 120583119873minus1 (119909) 119889119909 (26)
119911119881 (0 119911) = 120579int
infin
0
119875119873 (119909 119911) 120583119873 (119909) 119889119909 (27)
119877 (0 119911) = 120572119911
119873
sum
119895=1
119875119895 (119911) (28)
Again integrating (24) with respect to 119909 in general will resultin
119875119895 (119911) = 119875119895 (0 119911) [
1 minus 119870119895 (120582 minus 120582119862 (119911) + 120573)
120582 minus 120582119862 (119911) + 120573] 119895 = 1 to 119873
(29)
119881 (119911) = 119881 (0 119911) [1 minus119872(120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (30)
119877 (119911) = 119877 (0 119911) [1 minus 119866 (120582 minus 120582119862 (119911) + 120595 minus (120595119911))
120582 minus 120582119862 (119911) + 120595 minus (120595119911)] (31)
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
8 Mathematical Problems in Engineering
where
119870119895 (120582 minus 120582119862 (119911) + 120573) = int
infin
0
119890minus(120582minus120582119862(119911)+120573)119909
119889119870119895 (119909)
119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119866 (119909)
119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911) = int
infin
0
119890minus(120582minus120582119862(119911)+120595minus(120595119911))119909
119889119872(119909)
(32)
are the Laplace-Stieltjes transform of the jth stage repair timeand vacation time respectively
By multiplying the RHS of (24) by 120583119895 (119909) Υ (119909) and 120585(119909)respectively and integrating with respect to 119909 we can obtain
int
infin
0
119875119895 (119909 119911) 120583119895 (119909) 119889119909 = 119875119895 (0 119911)119870119895 (120582 minus 120582119862 (119911) + 120573)
119895 = 1 to 119873
int
infin
0
119877 (119909 119911) Υ (119909) 119889119909 = 119877 (0 119911) 119866 (120582 minus 120582119862 (119911) + 120595 minus120595
119911)
int
infin
0
119881 (119909 119911) 120585 (119909) 119889119909 = 119881 (0 119911)119872(120582 minus 120582119862 (119911) + 120595 minus120595
119911)
(33)
Let us take
120582 minus 120582119862 (119911) + 120573 = ℎ1 120582 minus 120582119862 (119911) + 120595 minus120595
119911= ℎ2 (34)
By using (33) in (25)ndash(27) we can obtain
1199111198751 (0 119911) = (120582119862 (119911) minus 120582)119876 + (1 minus 120579)119870119873 (ℎ1) 119875119873 (0 119911)
+ 119877 (0 119911) 119866 (ℎ2) + 119911119881 (0 119911)119872 (ℎ2)
(35)
1198752 (0 119911) = 1198751 (0 119911)1198701 (ℎ1) (36a)
For Nth stage
119875119873minus1 (0 119911) = 119875119873minus1 (0 119911)119870119873minus1 (ℎ1) (36b)
119881 (0 119911) = 120579119875119873 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (37)
By employing (36b) in (37) we can get
119911119881 (0 119911) = 1205791198751 (0 119911)
119873
prod
119895=1
119870119895 (ℎ1) (38)
Again by using (28) in (29) we get
119877 (0 119911) =120573119911
ℎ1
[
[
119873
sum
119895=1
119875119895 (0 119911) (1 minus 119870119895 (119898))]
]
(39)
Now by employing (36b) (38) and (39) in (35) we can solvefor 1198751(0 119911)
1198751 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119876
119863 (119911) (40)
119863 (119911)
= ℎ1[
[
119911 minus (1 minus 120579)
119873
prod
119895=1
119870119895 (ℎ1) minus 120579
119873
prod
119895=1
119870119895 (ℎ1)119872 (119896)]
]
minus 120573119911119866 (119896)
1 minus
119873
prod
119895=1
119870119895 (ℎ1)
(41)
1198752 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)1198701 (ℎ1) 119876
119863 (119911) (42)
119875119873 (0 119911) =ℎ1 (120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) 119876
119863 (119911) (43)
119881 (0 119911) =
120579ℎ1 (120582119862 (119911) minus 120582)prod119873
119895=1119870119895 (ℎ1) 119876
119863 (119911) (44)
By substituting (40) (42) (43) and (44) in (29)ndash(31) we canobtain
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(45)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911) (46)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
(47)
119881 (119911) =
120579ℎ1 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)
times [1 minus119872(ℎ2)
ℎ2
]
(48)
Let 119875119876 (119911) denote the probability generating function of thequeue size irrespective of the state of the system
119875119876 (119911) = 1198751 (119911) + 1198752 (119911) + 1198753 (119911) + 119877 (119911) + 119881 (119911) =119873 (119911)
119863 (119911)
(49)
To determine the probability of idle time 119876 we use thenormalizing condition
119875119876 (1) + 119876 = 1 (50)
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 9
Again as (49) is indeterminate of the form 00 at 119911 = 1LrsquoHopitalrsquos rule is employed in (49) to obtain
1198751 (1) =
120582119864 (119868) (1 minus 1198701 (120573))119876
119863119877
(51)
1198752 (1) =
120582119864 (119868)1198701 (120573) (1 minus 1198702 (120573))119876
119863119877
(52)
119875119873 (1) =
120582119864 (119868)119876prod119873
119895=1119870119895minus1 (120573) (1 minus 119870119895 (120573))
119863119877
(53)
Equations (51) (52) and (53) indicate the steady-state proba-bility of the server in stages 1 2 and119873 respectively
It must be noted that
119877 (1) =
120573120582119864 (119868) 119864 (119877) [1 minus prod119873
119895=1119870119895 (120573)]119876
119863119877
119881 (1) =
120579120573120582119864 (119868) 119864 (119881)prod119873
119895=1119870119895 (120573)119876
119863119877
(54)
where 119863119877 = minus(120582119864 (119868) + 120573(120582119864 (119868) minus 120595)119864 (119877)) + [120573 + 120582119864 (119868) +
120573 (120582119864 (119868) minus 120595) 119864 (119877) minus 120579120573(120582119864 (119868) minus 120595)119864(V)]prod119873119895=1119870119895 (120573)Thus the normalization condition yields
119876 = 1
minus ((120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
)
(55)
and the utilization factor 120588 = 1 minus 119876 is
120588 = (120582119864 (119868) [
[
1 + 120573119864 (119877)
minus 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
)
times (120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120579120573120595
119873
prod
119895=1
119870119895 (120573))
minus1
lt 1
(56)
7 Average Queue Size and AverageWaiting Time
As 119871119902 = (119889119889119911) 119875119902 (119911) |119911=1with the mean queue size is of the00 form the LrsquoHopitalrsquos rule is applied twice to obtain
119871119902 = lim119911rarr1
1198631015840(119911)11987310158401015840(119911) minus 119873
1015840(119911)11986310158401015840(119911)
2 (1198631015840(119911))2
(57)
where the primes and double primes denote the first andsecond derivatives respectively
1198731015840(1) = 119876[
[
120582119864 (119868) 1 + 120573119864 (119877)
minus 120582119864 (119868) 1 + 120573119864 (119877) minus 120579120573119864 (119881)
times
119873
prod
119895=1
119870119895 (120573)]
]
11987310158401015840(1) = 119876[
[
120582119864(119868
119868 minus 1)
times
[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ 120573[
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
119864 (119877)
+ 120579120573
119873
prod
119895=1
119870119895 (120573) 119864 (119881)
+ 120573 (120582119864 (119868) minus 120595) 119864 (1198772)
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
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OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
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Operations ResearchAdvances in
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Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
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Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
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Decision SciencesAdvances in
Discrete MathematicsJournal of
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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
10 Mathematical Problems in Engineering
times [
[
1 minus
119873
prod
119895=1
119870119895 (120573)]
]
+ (120582119864 (119868) minus 120595) 119864 (1198772)
times
119873
prod
119895=1
119870119895 (120573)
]
]
1198631015840(1) = minus 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
+ 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595) 119864 (119877)
minus 120579120573 (120582119864 (119868) minus 120595) 119864 (119881) + 120573
times
119873
prod
119895=1
119870119895 (120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times [
[
(1 + 120573119864 (119877)) minus
119873
prod
119895=1
119870119895 (120573)
times 1 minus 120573119864 (119877) + 120579120573119864 (119881) ]
]
minus 2120573120595119864 (119877)
1 minus
119873
prod
119895=1
119870119895 (120573)
+ 120573 (120582119864 (119868) minus 120595)2119864 (1198772)
times (1 minus
119873
prod
119895=1
119870119895 (120573))
minus 120579120573 2120595119864 (119881) minus (120582119864 (119868) minus 120595)2119864 (1198812)
times
119873
prod
119895=1
119870119895 (120573)
(58)
By utilizing (58) in (57) we can obtain the mean length of thequeue size 119871119902 while the mean waiting time of the queue 119882119902
can be derived by using119882119902 = 119871119902120582 Furthermore we can alsodetermine 119871 = 119871119902 +120588 the mean queue size of the system and119882 = 119871120582 the mean waiting time of the system
8 Particular Cases
Case 1 (no server vacations) In this case the server has novacation Hence 119881 (119911) = 0 Thus 120579 = 0 in (45)ndash(47) Our
model reduces to multistage batch arrivals with renegingduring breakdowns and we have
1198751 (119911) =
(120582119862 (119911) minus 120582) [1 minus 1198701 (ℎ1)]119876
119863 (119911)
1198752 (119911) =
(120582119862 (119911) minus 120582)1198701 (ℎ1) [1 minus 1198702 (ℎ1)]119876
119863 (119911)
(59)
For Nth stage
119875119873 (119911) =
(120582119862 (119911) minus 120582)119870119873minus1 (ℎ1) [1 minus 119870119873 (ℎ1)]119876
119863 (119911)
119877 (119911) =
120573119911 (120582119862 (119911) minus 120582) [1 minus prod119873
119895=1119870119895 (ℎ1)]119876
119863 (119911)[1 minus 119866 (ℎ2)
ℎ2
]
119863 (119911) = ℎ1[
[
119911 minus
119873
prod
119895=1
119870119895 (ℎ1)]
]
minus 120573119911119866 (ℎ2)
119873
prod
119895=1
119870119895 (ℎ1)
(60)
The probability of idle time is
119876 = 1 minus
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(61)
and the traffic intensity 120588 is
120588 =
120582119864 (119868) [1 + 120573119864 (119877) minus 1 + 120573119864 (119877)prod119873
119895=1119870119895 (120573)]
120573120595119864 (119877) 1 minus prod119873
119895=1119870119895 (120573)
(62)
Furthermore by considering 120579 = 0 in (58) we can obtain themean queue size 119871119902 and mean waiting time119882119902
Case 2 (exponential service time and exponential vacationtime) In this case we assume that the service time for themultistage of service with service rate 120583119895 gt 0 119895 = 1 to 119873 isexponentially distributed Furthermore the repair time andvacation time are exponentially distributed with a repair rateΥ gt 0 and vacation time 120585 gt 0
Thus
1198701 (ℎ1) =
120583119895
120583119895 + ℎ1
119895 = 1 to 119873
119866 (ℎ2) =Υ
Υ + ℎ2
119872 (ℎ2) =120585
120585 + ℎ2
119864 (119877) =1
Υ 119864 (119877
2) =
2
Υ2
119864 (119881) =1
120585 119864 (119881
2) =
2
1205852
(63)
where ℎ1 = 120582 minus 120582119862 (119911) + 120573 ℎ2 = 120582 minus 120582119862 (119911) + 120595 minus (120595119911)
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
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Mathematical Problems in Engineering
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Differential EquationsInternational Journal of
Volume 2014
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Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
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Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
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CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 11
By utilizing the abovementioned relations in (44)ndash(46)we can obtain
1198751 (119911) =(120582119862 (119911) minus 120582) [1 minus (1205831 (1205831 + ℎ1))] 119876
119863 (119911)
1198752 (119911)=(120582119862 (119911) minus 120582) (1205831 (1205831 + ℎ1)) [1 minus (1205832 (1205832 + ℎ1))] 119876
119863 (119911)
(64)
For Nth stage
119875119873 (119911) = ((120582119862 (119911) minus 120582)1205831
120583119873minus1 + ℎ1
[1 minus120583119873
120583119873 + ℎ1
]119876)
times (119863 (119911))minus1
119877 (119911) = ((120582119862 (119911) minus 120582)[
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
times1
(Υ + ℎ2)119876) times (119863 (119911))
minus1
119881 (119911) = (120579ℎ1 (120582119862 (119911) minus 120582)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1) (120595 + ℎ2)
119876)
times (119863 (119911))minus1
(65)
where
119863 (119911) = (120582 minus 120582119862 (119911) + 120573)
times [
[
119911 minus (1 minus 120579) + 120579120585
120585 + ℎ2
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
]
]
minus 120573119911
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + ℎ1)
Υ
(Υ + ℎ2)
(66)
Therefore the probability that the server is providing servicein the first second and Nth stages at a random point of timecan be given as presented in (67) (68) and (69) respectivelyConsider the following
1198751 (1) = (120582119864 (119868) [1 minus1205831
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(67)
1198752 (1) = (120582119864 (119868)1205831
1205831 + 120573[1 minus
1205832
1205831 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120595] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(68)
119875119873 (1) = (120582119864 (119868)120583119873minus1
1205831 + 120573[1 minus
120583119873
120583119873 + 120573]119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(69)
Furthermore the probability that server is under repairs atrandom point of time is
119877 (1) = ([120573119864 (119868)
Υ][
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(70)
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
12 Mathematical Problems in Engineering
The probability that server is on vacation at random point oftime is given by
119881 (1) = (120579120573 [120582119864 (119868)
120585]
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
119876)
times minus[
[
(120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ)
+ [120573 + 120582119864 (119868) + 120573 (120582119864 (119868) minus 120595)
times1
Υminus
120579
120585] times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus1
(71)
The probability that the server is idle but available in thesystem is given by
119876 = 1 minus 120582119864 (119868)
times [
[
1 +120573
Υ minus 1 +
120573
Υminus
120579120573
120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
times (120573120595
Υ
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
+ 120579120573120595
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
minus1
(72)
Similarly the mean queue size and mean waiting time can bederived by determining1198731015840 (1)11987310158401015840 (1)1198631015840 (1) and119863
10158401015840(1) and
utilizing them in (57) Consider the following
1198731015840(1) = 119876[
[
120582119864 (119868) 1 +120573
Υ minus 120582119864 (119868) 1 +
120573
Υminus
120579120573
Υ
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
11987310158401015840(1)
= 119876[
[
120582119864(119868
119868 minus 1)
times
(1 +120573
Υ)
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579120573
Υ
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 2120582119864 (119868)
times
120573 (120582119864 (119868) minus 120595)
Υ2
times (1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
)
+120579 (120582119864 (119868) minus 120595)
1205852
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
1198631015840(1)
= minus [120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ]
+ 120573 + 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
11986310158401015840(1) = minus120582119864(
119868
119868 minus 1)
times
(1 +120573
Υ)
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
(1 minus120573
Υ+
120579120573
120585)
minus 2[120573120595
Υ+
(120582119864 (119868) minus 120595)2
Υ2
]
times [
[
1 minus
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
]
]
minus 2120579120573[120595
120585+
(120582119864 (119868) minus 120595)2
1205852
]
times
prod119873
119895=1120583119895
prod119873
119895=1 (120583119895 + 120573)
minus 120582119864 (119868) +120573 (120582119864 (119868) minus 120595)
Υ
minus120579120573 (120582119864 (119868) minus 120595)
120585+ 120573
times
119873
prod
119895=1
120583119895 times (1
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 13
0
01
02
03
04
05
06
07
08
5 9 10 12
07733
0429603866
03222
02267
0570406134
06888
Reneging 120595
Q
120588
Figure 5 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 1 on the proportion of idle time119876 and utilization factor 120588 Fromthe figure it is clear that as the idle time of the server 119876 increases itleads to a decrease in utilization factor 120588
times [
[
(1205831 + 120573)2119873
prod
119895=2
(120583119895 + 120573)
+ (1205831 + 120573) (1205832 + 120573)2
times
119873
prod
119895=3
(120583119895 + 120573) + sdot sdot sdot
+ [ (1205831 + 120573) (1205832 + 120573) sdot sdot sdot
times (120583119873 + 120573)2] ]
]
minus1
)
(73)
9 Numerical Illustration
To compute the numerical results we consider the number ofstages to be119873 = 3 we assume service time vacation time andrepair time to be exponentially distributed Furthermore letus assume that the arrivals come one after the other that is119864 (119868) = 1 and 119864(119868119868 minus 1) = 0
To monitor the effect of reneging 120595 and breakdown 120573 onthe behavior of the queueing model let us consider 120582 = 2120579 = 05 120582 = 2 1205831 = 3 1205832 = 4 1205833 = 5 Υ = 8 and 120585 = 6 withthe values of 120595 being 5 9 10 and 12 and 120573 varying between 1and 4 (Table 1)
From the values presented in Table 1 it can be concludedthat with the increase in the value of the parameter ofcustomerrsquos impatience reneging 120595 for varying values of thebreakdown parameter 120573 the utilization factor 120588 decreaseswhereas the probability of the server idle time 119876 increases
0
01
02
03
04
05
06
07
08
09 08102
0450104051
03376
01898
0549905949
06634
5 9 10 12
Q
120588
Reneging 120595
Figure 6 Effect of reneging 120595 = 5 9 10 and 12 and breakdown at120573 = 2 on the proportion of idle time 119876 and utilization factor Thegraph indicates that as the idle time of the server 119876 increases theutilization factor 120588 decreases
08155
045304077
03397
01845
054705923
06603
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 7 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 3 on the proportion of idle time 119876 and utilization factor 120588The figure explains that as the idle time of the server119876 increases theutilization factor 120588 decreases
Figures 5 6 7 and 8 clearly show that owing to thebreakdown of the system and reneging the proportion ofthe idle time of the server increases and the utilization factordecreases Furthermore Figures 9 10 11 and 12 demonstratethe effect of reneging and breakdown on the mean queue size119871119902 and the mean waiting time119882119902 and evidently indicate thatas the breakdown occurs as well as customers reneging fromthe queue the average length of the queue decreases and theaverage waiting time decreases
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
14 Mathematical Problems in Engineering
Table 1 Effect of reneging and breakdown
120595 120573 120588 119876 119871119902 119871 119882119902 119882
5
1 07733 02267 43678 5141 2184 25712 08102 01898 36436 4454 1822 22273 08155 01845 30592 3875 1530 19374 08233 01767 21651 2988 1083 1494
9
1 04296 05704 65243 6954 3262 34772 04501 05499 55682 6018 2784 30093 04530 05470 48194 5272 2410 26364 04571 05429 41254 4583 2063 2291
10
1 03866 06134 58164 6203 2908 31022 04051 05949 45921 4997 2296 24993 04077 05923 32457 3653 1623 18274 04080 05920 21352 2543 1068 1272
12
1 03222 06888 45824 4905 2291 24522 03376 06634 40261 4364 2013 21823 03397 06603 35024 3842 1751 19214 03398 06602 29254 3265 1463 1633
08233
045710408
03398
01767
054290592
06602
0
01
02
03
04
05
06
07
08
09
5 9 10 12
Q
120588
Reneging 120595
Figure 8 Effect of reneging 120595 = 5 9 10 and 12 and breakdownat 120573 = 4 on the proportion of idle time 119876 and utilization factor 120588The figure gives the fact that as idle time of the server119876 increases itleads to a decrease in utilization factor 120588
10 Conclusion
The present study clearly analyzed the multistage batcharrival queue with reneging during vacation and breakdownperiods The steady-state solutions are obtained by usingsupplementary variable technique and themean queue lengthand mean waiting time are calculated Furthermore someparticular cases are also discussed and numerical illustrationsare presented It can be concluded that with the increase inthe value of the parameter of customerrsquos impatience reneging120595 for varying values of the breakdown parameter 120573 the
0
05
1
15
2
25
3
35
4
4543678
36436
30592
2165121841822
153
1083
Lq
Wq
Breakdown 120573
1 2 3 4
Figure 9 Effect of 120595 = 5 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue so the average waiting time 119882119902 alsodecreases
utilization factor 120588 decreases whereas the probability of theserver idle time 119876 increases In this model due to renegingand breadown the mean queue size 119871119902 and the mean waitingtime 119882119902 decteases and the average waiting time decreasesFor the future study compulsory vacation can be includedin the present model and the effect of reneging duringbreakdown can be obtained The model presented in thisstudy can be utilized for communication networks and large-scale manufacturing industries
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Mathematical Problems in Engineering 15
0
1
2
3
4
5
6
7 65243
55682
48194
41254
32622784
241
2063
Lq
Wq
1 2 3 4Breakdown 120573
Figure 10 Effect of 120595 = 9 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The graph indicates thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
0
1
2
3
4
5
658164
45921
32457
21352
2908
2296
16231068
Lq
Wq
1 2 3 4Breakdown 120573
Figure 11 Effect of 120595 = 10 and 120573 = 1 2 3 and 4 on the mean queuesize 119871119902 and mean waiting time 119882119902 The figure gives the fact thatthe average length of the queue 119871119902 decreases owing to customersreneging from the queue which leads to decrease in average waitingtime119882119902
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper
0
05
1
15
2
25
3
35
4
45
5 45824
40261
35024
29254
22912013
17511463
Lq
Wq
1 2 3 4Breakdown 120573
Figure 12 Effect of 120595 = 12 and 120573 = 1 2 3 and 4 on the meanqueue size 119871119902 and mean waiting time 119882119902 The figure explains thatas the average length of the queue 119871119902 decreases owing to customersreneging from the queue the averagewaiting time119882119902 also decreases
Acknowledgments
Authors would like to express appreciation to the anonymousreviewers and the editor Antonina Pirrotta for their veryhelpful comments that improved the paper
References
[1] F A Haight ldquoQueueing with balkingrdquo Biometrika vol 44 no3-4 pp 360ndash369 1957
[2] F A Haight ldquoQueueing with renegingrdquo Metrika vol 2 no 1pp 186ndash197 1959
[3] D Y Barrer ldquoQueuing with impatient customers and orderedservicerdquo Operations Research vol 5 pp 650ndash656 1957
[4] A Montazer-Haghighi J Medhi and S G Mohanty ldquoOna multiserver Markovian queueing system with balking andrenegingrdquo Computers amp Operations Research vol 13 no 4 pp421ndash425 1986
[5] B Doshi ldquoAnalysis of a two phase queueing systemwith generalservice timesrdquo Operations Research Letters vol 10 no 5 pp265ndash272 1991
[6] G Choudhury ldquoAn M119909G1 queueing system with a setupperiod and a vacation periodrdquo Queueing Systems vol 36 no1-3 pp 23ndash38 2000
[7] K C Madan ldquoOn a single server queue with two-stage hetero-geneous service and binomial schedule server vacationsrdquo TheEgyptian Statistical Journal vol 40 no 1 pp 39ndash55 2000
[8] J Bae S Kim and E Y Lee ldquoThe virtual waiting time of theMG1 queue with impatient customersrdquoQueueing Systems vol38 no 4 pp 485ndash494 2001
[9] B Krishna Kumar A Vijayakumar and D Arivudainambi ldquoAn1198721198661 retrial queueing system with two-phase service andpreemptive resumerdquo Annals of Operations Research vol 113 pp61ndash79 2002
[10] K C Madan W Abu-Dayyeh and F Taiyyan ldquoA two serverqueue with Bernoulli schedules and a single vacation policyrdquo
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
16 Mathematical Problems in Engineering
Applied Mathematics and Computation vol 145 no 1 pp 59ndash71 2003
[11] G Choudhury ldquoA batch arrival queueing system with anadditional service channelrdquo International Journal of Informationand Management Sciences vol 14 no 2 pp 17ndash30 2003
[12] K C Madan and A Z Abu Al-Rub ldquoOn a single server queuewith optional phase type server vacations based on exhaustivedeterministic service and a single vacation policyrdquo AppliedMathematics and Computation vol 149 no 3 pp 723ndash7342004
[13] K C Madan and G Chodhury ldquoAn M[119909]G1 queue withBernoulli vacation schedule under restricted admissibility pol-icyrdquo Sankhaya vol 66 pp 172ndash193 2004
[14] G Choudhury and K C Madan ldquoA two-stage batch arrivalqueueing system with a modified Bernoulli schedule vacationunder 119873-policyrdquo Mathematical and Computer Modelling vol42 no 1-2 pp 71ndash85 2005
[15] S H Chang and T Takine ldquoFactorization and stochasticdecomposition properties in bulk queues with generalizedvacationsrdquoQueueing Systems vol 50 no 2-3 pp 165ndash183 2005
[16] E Altman and U Yechiali ldquoAnalysis of customersrsquo impatiencein queues with server vacationsrdquoQueueing Systems Theory andApplications vol 52 no 4 pp 261ndash279 2006
[17] E Altman and U Yechiali ldquoInfinite-server queues with systemrsquosadditional tasks and impatient customersrdquo Probability in theEngineering and Informational Sciences vol 22 no 4 pp 477ndash493 2008
[18] R Kumar and S K Sharma ldquoAn MM1N Queueing modelwith retention of reneged customers and balkingrdquoTheAmericanJournal of Operations Research vol 2 no 1 pp 1ndash5 2012
[19] F AMaraghi K CMadan and K Darby-Dowman ldquoBernoullischedule vacation queue with batch arrivals and random systembreakdowns having general repair time distributionrdquo Interna-tional Journal of Operational Research vol 7 no 2 pp 240ndash2562010
[20] R Kumar and S K Sharma ldquoA Markovian feedback queuewith retention of reneged customers and balkingrdquo AdvancedModeling and Optimization vol 14 no 3 pp 681ndash688 2012
[21] M Baruah K C Madan and T Eldabi ldquoA two stage batcharrival queue with reneging during vacation and breakdownperiodsrdquo The American Journal of Operations Research vol 3pp 570ndash580 2013
[22] R Kumar and S K Sharma ldquoAMarkovianmulti-server queuingmodel with retention of reneged customers and balkingrdquoInternational Journal of Operations Research vol 20 no 4 pp427ndash438 2014
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of
Submit your manuscripts athttpwwwhindawicom
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical Problems in Engineering
Hindawi Publishing Corporationhttpwwwhindawicom
Differential EquationsInternational Journal of
Volume 2014
Applied MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Mathematical PhysicsAdvances in
Complex AnalysisJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
OptimizationJournal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Operations ResearchAdvances in
Journal of
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Function Spaces
Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014
International Journal of Mathematics and Mathematical Sciences
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Algebra
Discrete Dynamics in Nature and Society
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Decision SciencesAdvances in
Discrete MathematicsJournal of
Hindawi Publishing Corporationhttpwwwhindawicom
Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014
Stochastic AnalysisInternational Journal of