Queuing Theory 1

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    QUEUING THEORY

    Introduction

    Queuing theory deals with problems that involve waiting (or queuing). It isquite common that instances of queue occurs everyday in our daily life.

    Examples of queues or long waiting lines might be

    Waiting for service in bank and at reservation counter. Waiting for a train or bus. Waiting at barber saloon. Waiting at doctors clinic.

    Whenever a customer arrives at a service facility, some of them usually have

    to wait before they receive the desired service. This form a queue or waitingline and customer feel discomfort either mentally or physically because of

    long waiting queue.

    We infer that queues from because the service facilities are

    inadequate. If service facilities are increased, then the question arise how

    much to increase? For example, how many buses would be needed to avoid

    queues? How many reservation counters would be needed to reduce the

    queue? Increase in number of buses and reservation counters requires

    additional resources. At the same time, cost due to customer dissatisfactionmust also be considered.

    Symbols and notations:

    n = total number of customers in the system, both waiting and in service

    = average number of customers being serviced per unit of time.

    = average number of customers arriving per unit of time.

    C = number of parallel service channels

    Ls or E(n) = average number of customers in the system, both waiting in the

    service.

    Lq or E(m) = average number of customers waiting in the queue

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    Ws or E(w) = average wating time of a customer in the system both waiting and

    in service

    Wq or E(w) = average waiting time of a customer in the queue

    Pn (t = probability that there are n customer in the queue

    total cost of the system

    cost

    cost of

    service

    cost of waiting

    optical service level level of service

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    Queuing system

    The customers arrive at service counter (single or in a group) and attended by

    one or more servers. A customer served leaves the system after getting the

    service. In general, a queuing system comprise with two components, the

    queue and the service facility. The queue is where the customers are waiting

    to be served. The service facility is customers being served and the individual

    service stations.

    SERVICE SYSTEM

    The service is provided by a service facility (or facilities). This may be a

    person (a bank teller, a barber, a machine (elevator, gasoline pump), or a

    space (airport runway, parking lot, hospital bed), to mention just a few. A

    service facility may include one person or several people operating as a team.

    There are two aspects of a service system(a) the configuration of the servicesystem and (b) the speed of the service.

    Configuration of the service system

    The customers entry into the service system depends upon the queue

    conditions. If at the time of customers arrival, the server is idle, then thecustomer is served immediately. Otherwise the customer is asked to join the

    queue, which can have several configurations. By configuration of the service

    system we mean how the service facilities exist. Service systems are usually

    classified in terms of their number of channels, or numbers of servers.

    Single Server Single Queue

    The models that involve one queue one service station facility are called

    single server models where customer waits till the service point is ready to

    take him for servicing. Students arriving at a library counter is anexample of a single server facility.

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    Several (Parallel) Servers Single Queue

    In this type of model there is more than one server and each server

    provides the same typeof facility. The customers wait in a single queue until one of the service

    channels is ready to take them in for servicing

    Several Servers Several Queues

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    This type of model consists of several servers where each of the servers

    has a different queue. Different cash counters in an electricity office

    where the customers can make payment in respect of their electricity bills

    provide an example of this type of model.

    Service facilities in a series

    In this, a customer enters the first station and gets a portion of service

    and then moves on to the next station, gets some service and then again

    moves on to the next station. . and so on, and finally leaves the system,

    having received the complete service. For example, machining of a

    certain steel item may consist of cutting, turning, knurling, drilling,

    grinding, and packaging operations, each of which is performed by a

    single server in a series. Service Facility.

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    Characteristics of Queuing System

    In designing a good queuing system, it is necessary to have a good

    Information about the model. The characteristic listed below would

    Provide sufficient information.

    1. The Arrival pattern.2. The service mechanism.3.

    The queue discipline.

    4. The number of service channels.5.Number of Service Stages

    1. The Arrival pattern.

    Arrivals can be measured as the arrival rate or the interarrival time

    (time between arrivals).

    Interarrival time =1/ arrival rate

    These quantities may be deterministic or stochastic (given by a

    propbability distribution).

    Arrivals may also come in batches of multiple customers, which is

    called batch or bulk arrivals. The batch size may be either deter-

    ministic or stochastic.

    (i) Balking: The customer may decide not to enter the queue upon

    Arrival, perhaps because it is too long.

    (ii) Reneging: The customer may decide to leave the queue after

    Waiting a certain time in it.

    (iii) Jockeying: If there are multiple queues in parallel the customers

    May switch between them.

    (iv) Drop-os: Customers may be dropped from the queue for rea-

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    4.The number of service channels

    5. Number of Service Stages

    Customers are served by multiple servers in series.

    In general, a multistage queue may be a complex network with feed-Back

    Application of queuing theory:

    Queing theory has been applied to a great variety of business situations. Here

    we shall discuss a few problem s where the theory may be applied-

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    1)Waiting line theory can be applied to be determine the number ofcheck out counters needed to secure smooth and economic operations

    of its stored at various time during the day of a super market or a

    departmental store .

    2)W

    aiting line theory can be used to analyze the delays at the toll boothsof bridges and tunnels.

    3)Waiting line theory can be used to improve the customers service atrestaurants,cafeteria ,gasoline service station , airline

    counters,hospitals etc,

    4)Waiting line theory can be used to determine the proper determine theproper number docks to be constructed in the building of terminal

    facilities for trucks &ships.

    5)Several manufacturing firms have attacked the problems of machinebreak down &repairs by utilizing this theory .Waiting line theory canbe used to determine the number of personnal to be employed so that

    thee cost of the production loss from down time & the cost f

    repairman is minimized.

    6) Queuing theory has been extended to study a wage incentive planQueuing theory (Limitations)

    1)Most of the queuing models are quite complex & cannot be easilyunderstood.

    2)Many times form of theoretical distribution applicable to givenqueuing situations is not known.

    3)If the queuing discipline is not in first in, first out, the study ofqueuing problems become more difficult.

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    BASIC POINTS

    Customer:> (Arrival)

    The arrival unit that requires some services to performed.

    Queue:>The number of Customer waiting to be served.

    Arrival Rate ():>The rate which customer arrive to the service station.

    Service rate () :> The rate at which the service unit can provide sevices to

    the customer

    If Utilization Ratio Or Traffic intensity i.e /

    / > 1 Queue is growing without end.

    / < 1 Length of Queue is go on diminishing.

    / = 1 Queue length remain constant.

    When Arrival Rate () is less than Service rate () the system is working .

    i.e < (system work)

    Formulas

    =Service Rate

    = Arrival Rate

    1. Traffic Intensity (P)= /

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    2. Probability Of System Is Ideal (P0) =1-P

    P0 = 1- /

    3. ExpectedWaiting Time In The System (Ws) = 1/ (- )

    4. ExpectedWaiting Time In Quie (Wq) = / (- )

    5. Expected Number Of Customer In The System (Ls)= / (-)

    Ls=Length Of System

    6. Expected Number Of Customers In The Quie (Lq)= 2/ (- )

    7. Expected Length Of Non-Empty Quie (Lneq)= / (- )

    8. What Is The Probability Track That That K Or More Than K

    Customers In The System.

    P >= K (P Is Greater Than Equal To K)

    = ( /)K

    9. What Is The Probability That More Than K Customers Are In The

    System ( P>K)= ( /)K+1

    10. What Is The Probability That Atleast One Customer Is Standing In Quie.

    P=K=( /)2

    11. What Is The Probability That Atleast Two Customer In TheSystem

    P=K=( /)2

    Solved Example.

    Question 1.People arrive at a cinema ticket booth in a poisson distributed

    arrival rate of 25per hour. Service rate is exponentially distributed with an

    average time of 2 per min.

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    Calculate the mean number in the waiting line, the mean waiting time , the

    mean number in the system , the mean time in the system and the utilization

    factor?

    Solution:

    Arrival rate =25/hr

    Service rate = 2/min=30/hr

    Length of Queue (Lq)=

    2

    / (- )

    = 252/(30(30-25))

    =4.17 persone

    ExpectedWaiting Time In Quie (Wq) = / (- )

    =25/(30(30-25))

    =1/6 hr= 10 min

    ExpectedWaiting Time In The System (Ws) = 1/ (- )

    =1/(30-25)

    =1/5hr= 12 min

    Utilization Ratio = /

    =25/30

    =0.8334 = 83.34%

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    Question 2. Assume that at a bank teller window the customer arrives at a

    average rate of 20 per hour according to poission distribution .Assume also

    that the bank teller spends an distributed customers who arrive from an

    infinite population are served on a first come first services basis and there is

    no limit to possible queue length.

    1.what is the value of utilization factor?

    2.What is the expected waiting time in the system per customer?

    3.what is the probability of zero customer in the system?

    Solution:

    Arrival rate =20 customer per hour

    Service rate = 30 customer per hour

    1.Utilization Ratio = /

    = 20/30 = 2/3

    2. ExpectedWaiting Time In The System (Ws ) = 1/ (- )

    =1/(30-20)

    =1/10 hour = 6 min

    3. Probability of zero customers in the system P0 = 1 P

    =1- 2/3 = 1/3

    Question 3 : Abc company has one hob regrinding machine. The hobs

    needing grinding are sent from companys tool crib to this machine which is

    operated one shift per day of 8 hours duration. It takes on the average half an

    hour to regrind a hob. The arrival of hobs is random with an average of 8

    hobs per shift.

    1. Calculate the present utilization of hob regrinding machine.2. What is average time for the hob to be in the regrinding section?

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    3. The management is prepared to recruit another grinding operator whenthe utilization of the machine increases to 80%.What should the arrival

    rate of hobs then be?

    Solution: : Let us calculate arrival rate and service rate per shift of 8 hours.

    Arrival rate =8 shift

    Service rate =8x60/30=16 /shift

    1.Percentage of the time the machine is busyPb =arrival rate/service rate=8/16=0.50=50%

    2.Average time for the hob to be in the grinding section.i.e., average time in the queue system=ws

    ws = 1/( - )=1/16-8=1/8 shift=1/8x8=1 hour

    3. Let =arrival rate for which utilization of the machine will be 80%,Therefore, Pb

    =

    /

    i.e., =

    Pb. =0.80x16=12.8 per shift.

    Question: 4

    (a) calculate expected number of persons in the system if average waiting time

    pf a customer is 45 or more than 45 minutes .

    b)if service rate is same.

    c)if arrival rate is same.

    Solution:-(a)expected no. of persons in a system(Ls)=/-

    =45/65-45

    =9/4

    =3/4=1/65-

    =191/3

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    (b)Ws= 1/ =1/65-45

    =1/20 x60/1=3 mins.

    (c)ws =1/ - =1/6-4

    = 3/4=1/ -45

    =3 -135=4

    =3 =139

    =46.33

    Question: 5 In a factory, the machines break down and require service

    according to a poission Distribuation at the average of per day. What is the

    probability that exactly six Machines.

    Solution : Given = 4, n = 6, t = 2 p

    P(n,t) = (6,4) when = 4

    We know, p (n,t) = (t)n e-t/ n!

    p(6,2) = (42)6 e-42/ 6!

    =86 e-8/720

    =0.1221

    Question 6 On an average , 6 customer arrive in a coffee shop per hour.

    Determine the probability that Exactly 3 customers will reach in a 30 minute

    period, assuming that the arrivals follow poisson Distribution.

    Solution:

    Given, = 6 customers / hour

    t = 30 minutes = 0.5 hour

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    n = 2

    we know, p(n,t) = (t)n e-t/n!

    p(6,2) = (60.5)2 e-60.5/2!

    = 0.22404

    Question 7 In a bank with a single sever, there are two chairs for waiting

    customers. On an average one customer arrives 12 minutes and each

    customer takes 6 minutes for getting served. Make suitable assumption, find

    (i) The probability that an arrival will get a chair to sit on,(ii) The probability that an arrival will have to stand, and(iii) Expected waiting time of a customer.

    Solution following assumption are made for solving the given queuing

    problem :

    1. The arrival rate is randomly distributed according to poissiondistribution.

    2. The mean value of the arrival rate is .3. The services time distribution approximated by an exponiential

    distribution and a nmean rate of services is .

    4. The rate of services is greather than the rate of arrival (>)5. The queue discipline id FIFO.

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    Arrival rate = 12min or 5 customer / hr

    Services rate = 6 min or 10 customer/ hr

    / = 5/10 =

    there are two chairs including services one.

    (i) The probality that an arrival get a chair to seat on is given by:Pn (n2)

    1-(/)3

    1-(1/2)3

    = 7/8

    (II) The probability that an arrival will have to stand is given by

    1-(P0+p1+P2)

    = 1-(7/8)= 1/8

    (III)Expected waiting time of a customer in the queue is given by

    Wq =/(-)

    =5/10(10-5) = 1/(2*5) hr = 6 min

    Question 8 A television repairman finds that the time spent on his jobs has an

    expontial distribution with a mean of 30 minutes. If he repairs sets in the

    order in which they came in, and if the arrival of sets follow a passion

    distribution approximately with an average rate of 10 per 8- hour day, what is

    the repairmans expected idle time each day? How many jobs are ahead of the

    average set just brought in?

    Solution from data of problem, we have

    = 10/8=5/4 set per hour; and =(1/30)60= 2set per hour

    (i) Expected idle time of repairmen each dayNumber of hour for a repairman remains busy in 8 hour day( traffic

    intensity) is given by

    (8) (/)=(8) (5/8)= 5 hour

    Hence , the idle time for a repairman in an 8 hour day will be : (8-5)

    =3 hour

    (ii) Expected (or average) number of TV set in the system

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    LS = /- = 5/4/2-(5/4)

    =5/3

    =2 (APPROX) T.V sets

    Unsolved question

    Question 1 Calculate expected number of person in the system. If average

    waiting time of customer is 30 min or more than 30 min , then services

    provider starts another windows .

    Calculate Arrival rate if service rate is same .

    Calculate service rate if arrival rate is same.

    (answer: Ws=1/5 hr,

    =13

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    = 2)

    Question 2 At a certain petrol pump , Customer arrive according to a passion

    process with a average time at 5 min between the arrivals. The service time is

    exponential distribution with mean 2 mins on the basic of this information.

    Find out:-

    a. Traffic intensityb. What would be the average quieting length?c. What is the expected number of customer at petrol pump?d. What is the expected number time one spend at petrol pump?e. What would we expected waiting time?f. What would be the proportion time the petrol pump is idle?Answer

    a. 0.4b. 0.26c. 0.66d. 0.02e. 0.05f. 0.6

    Question3. The machines in production shop breakdown at an average of 2

    per hour. The non productive time of any machine costs rs.30 per hour. If the

    cost of repairman is Rs.50 per hour.

    Calculate:

    a.Number of machines not working at any point of time.b.Average time that a machine is waiting for the repairman.c. Cost of non-productive time of the machine operator.

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    d.Expected cost of system per hour.Answer. a:: 2 machines

    b :- 2/3 hours

    c: Rs. 60

    d: Rs.110

    Question 4.In a bank cheques are cashed at a single teller counter.

    Customers arrived at the counter in a Poisson manner at and average

    rate of 30 customers /hour. The teller takes on an average, a minute and a

    half to cash cheque. The service time has been shown to be exponentially

    distributed

    a) Calculate the percentage of time the teller is busy.b) Calculate the average time a person is expected to wait.

    Answer

    a)3/4

    b)6 minutes

    Question 5 Telephone users arrive at a booth following a Poisson distribution

    with an average time of 5 minutes between one arrival and the next. The time

    taken for a telephone call is on a average 3 minutes and it follows an

    exponential distribution. What is the probability that the booth is busy? How

    many more booths should be established to reduce the waiting time less than

    or equal to half of the present waiting time.

    Answer a)0.6

    b)wq=3/40hrs.

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    Question 6 Assume that goods trains are coming in a yard @ 30 trains per day

    and suppose that the inter arrival times follow an exponential distribution .

    the service time for each train is assumed to be exponential with an average of

    36 minutes if the yard can admit 9 trains at a time(there being 10 lines one of

    which is reserved for shunting purpose).calculate the probability that the yard

    is empty and find the average queue length.

    Answer

    =1/48

    =/16

    p=0.75

    Po=o.28

    Lq=1.55

    Question 7 At what average rate must a clerk at a supermarket in order toensure a probability of 0.90 so that the customer will not wait longer than 12

    minutes ? It is assumed that there is only one counter at which customers

    arrive in a Poisson fashion at an average rate of 15/hour. The length of

    service by the clerk has an exponential distribution.

    Answer: 2.48 minutes /service

    Question 8 The beta company s quality control deptt. Is managed by a single

    clerk, who takes an average 5 minutes in checking part of each of the

    machine coming for inspection. The machine arrive once in every 10 min. on

    the average one hour of the machine is valued at Rs 25 and cost for the clerk

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    significantlyand for firemen, perilouslyduring a period of hours every Sunday evening. Astudy revealed an unusual culprit: Milton Berle.

    The comedian hosted an immensely popular weekly television show every Sunday that was

    watched by nearly everyone with a set. When the show went to a commercial break, tens ofthousands of people, having finished dinner, retreated to their bathrooms at the same time.

    With thousands of toilets being flushed within minutes of each other, sewers were inundated.

    More importantly, toilet tanks were refilling, each consuming two or three gallons of fresh

    water. The coordinated demand for water in a brief period of time virtually eliminated water

    pressure. In fact, some toilets took a half hour to refill, and water pressure took hours to

    recover.

    Serious consideration was given to canceling the show. The solution, however, was relativelysimple. The addition of only a few more water towers was sufficient to maintain adequate

    water pressure. In essence, the system was reengineered to handle more demanding peaks.

    This situation may be repeated in a telephone system when everyone is motivated to place a

    call at the same time. During the 1989 San Francisco earthquake, vast numbers of people in

    the metropolitan area attempted to make a call at the same timeimmediately after the quake

    subsidedhoping to learn whether friends and relatives were safe. Although the switching

    systems were automated, they were completely unable to handle the volume of requests for

    dial tone.

    Only a small percentage of calls (enough to meet the capacity of the system) were allowed to

    go through. Indeed, radio and television reporters urged people to stay off the lines so that

    emergency calls could be handled.

    There was no need to reengineer the system because the occurrence of earthquakes, while

    random, are not consistently repeated. It would be uneconomic to engineer the telephonenetwork for peak usages that occur only once every decade or so. As a result, every

    earthquake yields a temporary breakdown in the telephone network.

    Other slightly less offensive instances occur every time a radio host offers a prize to "caller

    number x." Telephone companies and public officials have convinced many radio stations to

    discontinue the practice.

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    GROUPS MEMBER NAME

    1.Shikha2.Dayashankar Yadav3.

    Suriender Singh Prajapati

    4. Gaurav Gupta5.Himanshu Saxena6. Gauri Shankar Mishra7.Pankaj Gangwar8.Sanjeev kumar9.Amit .kr yadav10.Amit sinha

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