Research Article Performance Analysis of a Kitting Process...

Hindawi Publishing CorporationMathematical Problems in EngineeringVolume 2013 Article ID 843184 10 pageshttpdxdoiorg1011552013843184

Research ArticlePerformance Analysis of a Kitting Process as a Paired Queue

E De Cuypere K De Turck and D Fiems

Department of Telecommunications and Information Processing Ghent University St Pietersnieuwstraat 41 9000 Gent Belgium

Correspondence should be addressed to E De Cuypere elinedecuyperetelinugentbe

Received 19 February 2013 Revised 14 May 2013 Accepted 14 May 2013

Academic Editor Zhengguang Wu

Copyright copy 2013 E De Cuypere et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

Nowadays customers request more variation in a companyrsquos product assortment leading to an increased amount of parts movingaround on the shop floor To cope with this tendency a kitting process can be implemented Kitting is the operation of collectingthe necessary parts for a given end product in a specific container called a kit prior to arriving at an assembly unit As kittingperformance is critical to the overall cost and performance of the manufacturing system this paper analyses a two-part kittingprocess as a Markovian model In particular kitting is studied as a paired queue thereby accounting for stochastic part arrivalsand kit assembly times Using sparse matrix techniques we assess the impact of kitting interruptions bursty part arrivals andphase-type distributed kit assembly times on the behaviour of the part buffers Finally a cost-profit analysis of kitting processes isconducted and an approximation for a two-part kitting process is established

1 Introduction

Nowadays manufacturing systems are often composed ofmultiple in-house fabrication units [1] The semifinishedproducts stemming from these units are the input materialsfor other fabrication units or for assembly lines Henceefficient transport of materials between the different stagesof the production process is key for overall production costminimisation Kitting is a particular strategy for supplyingmaterials to an assembly line Instead of delivering in contain-ers each holding a single type part and all holding the samenumber of parts kitting collects the necessary set of parts foran individual end product in a specific container referred toas kit prior to arriving at an assembly unit [1ndash6]

Kittingmitigates storage space requirements at the assem-bly station since no part inventories need to be kept thereMoreover parts are placed in proper positions in the con-tainer such that assembly time reductions can be realisedAdditional benefits include reduced learning time of theworkers at the assembly stations and increased quality of theproduct Although kitting is a nonvalue adding activity itsapplication can reduce the overall materials handling time[6] Indeed activities such as selecting and gripping parts areperformed more efficiently Furthermore the whole operatorwalking time is drastically reduced or even eliminated since

kits each containing a complete set of components arebrought to the assembly station [7] The advantages men-tioned above do not come for free since the kitting operationitself incurs additional costs such as the time and effort forplanning the allocation of the parts into kits and the kitpreparation itself Moreover the introduction of a kittingoperation in a production process involves a major invest-ment and the effect on efficiency is uncertain Thereforeit is important to analyse the performance of kitting in aproduction environment prior to its actual introductionThisis the subject of the present paper

In the literature most authors consider a kitting processas a queueing system with stochastic part arrivals and kitassembly times Hopp and Simon [8] developed a model fora kitting process with exponentially distributed processingtimes for kits and Poisson arrivals They found accuratebounds for the required buffer capacity of kitting processeswith two parts Explicitly accounting for finite buffer capac-ities Som et al [3] further refined the results of Hopp andSimon

Of course real buffers always have a finite capacity thecapacity being constrained by the storage room Howeverif the capacity is large enough we can have a good approx-imation of a process with a finite capacity on the basis ofa model with unlimited capacity This means that there is

2 Mathematical Problems in Engineering

always enough space for upcoming parts which simplifies theanalysis Unfortunately the assumption of an infinite bufferis not valid for kitting processes If the capacity is assumedto be infinite then the model will degrade to an unstablestochastic system Harrison [9] showed for a multiple inputgeneralisation of the GIG1 queue that it is necessary toimpose a restriction on the size of the buffer to ensure stabilityin the operations of a kitting process Under this assumptionthe probability to have a certain long-term stock position isequal and independent of the current stock positionThis wasalso demonstrated by Latouche [10] who studiedwaiting lineswith paired customers We can consider this analysis as anabstraction of a kitting process with two types of parts

In this work we focus on a kitting process modulated bya Markovian environment The introduction of this environ-ment allows us to study kitting under more realistic stochas-tic assumptions kitting interruptions bursty part arrivalsphase-type distributed kit assembly times and so forthOur paper extends the results on kitting in a Markovianenvironment [11]

The remainder of this paper is organised as followsSection 2 describes the kitting process at hand In Section 3Chapman-Kolmogorov equations are derived and theirnumerical solution is discussed To illustrate our approachSection 4 considers a number of numerical examples Inparticular we assess the impact of kitting interruptionsbursty part arrivals and phase-type distributed kit assemblytimes on the behaviour of the part buffers Then a cost-profit analysis of kitting processes is conducted and anapproximation for a two-part kitting process is establishedFinally conclusions are drawn in Section 5

2 Model Description

In this paper we study a two-queue kitting process asdepicted in Figure 1 Each queue has a finite capacitymdashlet 119862

ℓ

denote the capacity of buffer ℓ ℓ = 1 2mdashand models theinventory of parts of a single type New parts arrive at thebuffers and if both buffers are nonempty a kit is assembledby collecting a part from each buffer Hence departuresfrom the buffers are synchronised and the buffers are pairedOperation of part buffers therefore considerably differs fromother queueing systems

Arrivals at both buffers are modelled by a Markovianarrival process and kit assembly is not instantaneous Forease of modelling it is assumed that there is a modulatingMarkov process arrival and service rates depending on thestate of this process To be more precise the kitting processis modelled as a continuous-time Markov process with statespace C

1times C2times K whereby C

ℓ= 0 119862

ℓ for ℓ =

1 2 and with K = 1 2 119870 being the state space ofthe modulating process At any time the state of the kittingprocess is described by the triplet [119898 119899 119894] with119898 and 119899 beingthe number of parts in the first and secondbuffer respectivelyand 119894 being the state of the modulating process We nowdescribe the state transitions

(i) The state of the modulating process can changewhen there are neither arrivals nor departures Let

L1

L2

C1

C2

M

Figure 1 Kitting process the buffers are on the left and thetriangularly shaped kitting process is on the right

120572119894119895denote the transition rate from state 119894 to state

119895 (119894 119895 isin K 119894 = 119895) and let A denote the correspondinggenerator matrix

(ii) The state of the modulating process may remain thesame or may change when there is an arrival Let 120582(ℓ)

119894119895

denote the (marked) transition rate from state 119894 tostate 119895 when there is an arrival at buffer ℓ ℓ = 1 2Moreover let L

ℓdenote the corresponding generator

matrix Note that such marked transitions from state119894 to state 119894 are allowed

(iii) Analogously the state of the modulating process mayremain the same or may change when there is adeparture (in each buffer) Let 120583

119894119895and M denote the

corresponding transition rate and generator matrixrespectively

Summarising arrivals at and departures from the buffers aredescribed by the generator matrices A L

1 L2 andM So far

no diagonal elements of A have been defined To simplifynotation it will be further assumed that the diagonal elementsare chosen such that the row sums ofA+L

1+L2+M are zero

The computational method employed here does notrequire any homogeneity of the generator matrices Whenrequired by the applications at hand intensities may dependon the buffer content In this case we introduce super-scripts to make this dependence explicit For exampleM(119898119899)denotes the generator matrix of state transitions with depar-ture when there are119898 parts in buffer 1 and 119899 parts in buffer 2In addition we use arguments for rates as we already usedsuperscripts and subscripts to distinguish the arrival ratesat the different queues For example 120582(ℓ)

119894119895(119898 119899) denotes the

arrival rate at buffer ℓ = 1 2 from state 119894 to state 119895 when thereare119898 parts in buffer 1 and 119899 parts in buffer 2

Example 1 In the most basic setting parts arrive at thebuffers in accordance to an independent Poisson processeswith rate 120582

1and 120582

2and kit assembly times are exponentially

distributed with parameter 120583 In this case there is no needto have a modulating Markov process and the state iscompletely described by the number of parts in each buffer(119898 119899) We have

M = [120583] L1= [1205821]

L2= [1205822] A = [minus1205821 minus 120582

2minus 120583]

(1)

Example 2 To account for burstiness in the arrival processof the parts at the different buffers the modulating process

Mathematical Problems in Engineering 3

allows the mitigation of the Poissonian arrival assump-tions We can replace the Poisson processes by a two-classMarkovian arrival processes Multiclass Markovian arrivalprocesses allow for intricate correlation and can be efficientlycharacterised from trace data [12 13] As we have two typesof arrivals the Markovian arrival process is described by thegenerator matrix Λ

1of transitions with arrivals at buffer 1

the generator matrix Λ2with arrivals at buffer 2 and the

generator matrix Λ0without arrivals As usual the diagonal

elements of Λ0are negative and ensure that the row sums of

Λ0+Λ1+Λ2are zero Retaining exponentially distributed kit

assembly times we have

L1= Λ1 L

2= Λ2

A = Λ0minus 120583I M = 120583I

(2)

Here I denotes the identity matrix

Example 3 As for the arrival processes the model at handis sufficiently flexible to include phase-type distributed kitassembly times The phase-type distribution is completelycharacterised by an initial probability vector 120591 and thematrixT which corresponds to non-absorbing transitions [14] Lett1015840 = minusTe1015840 be the column vector with the rates to the absorbingstate and let f be a row vector with zero elements exceptthe first one Assuming Poisson arrivals in both buffers (withrates 120582

1and 120582

2 resp) we get the following matrices

L(119898119899)1

= 1205821I (1 minus 1

119898=0119899gt0) + 1205821e10158401205911119898=0119899gt0

L(119898119899)2

= 1205822I (1 minus 1

119898gt0119899=0) + 1205822e10158401205911119898gt0119899=0

A(119898119899) = T1119898gt0119899gt0

minus 1205821I minus 1205822I

M(119898119899) = t10158401205911119898gt1119899gt1

+ t1015840f (1 minus 1119898gt1119899gt1

)

(3)

Here it is assumed that the background state equals 1 ifone of the buffers is empty When service starts again thebackground state is chosen in accordancewith the probabilityvector 120591

3 Analysis

Having established the modelling assumptions and settledour notation we now focus on the analysis of the kittingprocess

31 Balance Equations We aim to define a set of equationsfor the steady-state probability vector for the Markov process[1198761(119905) 1198762(119905) 119878(119905)] 119876

ℓ(119905) being the number of parts in buffer

ℓ at time 119905 and 119878(119905) being the state of the background processat time 119905

Let 120587119894(119898 119899) = lim

119905rarrinfinPr[1198761(119905) = 119898119876

2(119905) = 119899 119878(119905) = 119894]

be the steady-state probability to be in state [119898 119899 119894] and let120587(119898 119899) be the vector with elements 120587

119894(119898 119899) for 119894 isin K

Figure 2 shows a fragment of the transition rate diagram ofthe kitting model in state [119898 119899 119894] As mentioned above twoindependent input streams arrive at the buffers with intensity120582(ℓ)

119894119895and are processed into kits with intensity 120583

119894119895 Upon

completion of a kit the content of both buffers is decreasedby 1 Note that we only show the transitions whereby themodulating Markov process remains in state 119894 Moreoverpossible dependence of the transition rates on the buffer sizesis not indicated

Based on the transition rate diagram we now derive thebalance equations of the kitting process at hand

(i) First consider the case where both buffers are neitherempty nor full 0 lt 119899 lt 119862

1and 0 lt 119898 lt 119862

2 We have

120587119894 (119898 119899)(

119870

sum

119895=1

120582(1)

119894119895(119898 119899) +

119870

sum

119895=1

120582(2)

119894119895(119898 119899)

+120583119894119895 (119898 119899) +

119870

sum

119895=1119895 = 119894

120572119894119895 (119898 119899))

=

119870

sum

119895=1

120587119895 (119898 minus 1 119899) 120582

(1)

119895119894(119898 minus 1 119899)

+

119870

sum

119895=1

120587119895 (119898 119899 minus 1) 120582

(2)

119895119894(119898 119899 minus 1)

+

119870

sum

119895=1

120587119895 (119898 + 1 119899 + 1) 120583119895119894 (119898 + 1 119899 + 1)

+

119870

sum

119895=1119895 = 119894

120587119895 (mn) 120572119894119895 (119898 119899)

(4)

for 119894 119895 isin K or equivalently

120587 (119898 minus 1 119899) L(119898minus1119899)1

+ 120587 (119898 119899 minus 1) L(119898119899minus1)2

+ 120587 (119898 + 1 119899 + 1)M(119898+1119899+1) + 120587 (119898 119899)A(119898119899) = 0(5)

(ii) If buffer 1 is empty and buffer 2 is neither empty norfull (119898 = 0 and 0 lt 119899 lt 119862

2) we have

120587119894 (0 119899)(

119870

sum

119895=1

120582(1)

119894119895(0 119899) +

119870

sum

119895=1

120582(2)

119894119895(0 119899)

+

119870

sum

119895=1119895 = 119894

120572119894119895 (0 119899))

=

119870

sum

119895=1

120587119895 (0 119899 minus 1) 120582

(2)

119895119894(0 119899 minus 1)

+

119870

sum

119895=1

120587119895 (1 119899 + 1) 120583119895119894 (1 119899 + 1)

+

119870

sum

119895=1119895 = 119894

120587119895 (0 119899) 120572119895119894 (0 119899)

(6)


middot middot middot middot middot middot

m minus 1 n minus 1 i

middot middot middot




middot middot middot middot middot middot middot middot middot



middot middot middotm minus 1 n i m minus 1 n + 1 i

m n minus 1 i m n i m n + 1 i

m + 1 n minus 1 i m + 1 n i m + 1 n + 1 i

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii120582(2)ii

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii120582(2)ii

120582(2)ii

120582(2)ii

120582(1)ii120582(1)ii 120582(1)ii

120582(1)ii 120582(1)ii 120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii

120582(2)ii

120583ii120583ii

120583ii120583ii120583ii

120583ii

Figure 2 Fragment of the transition rate diagram for state (119898 119899 119894)


120587 (0 119899 minus 1) L(0119899minus1)2

+ 120587 (1 119899 + 1)M(1119899+1)

+ 120587 (0 119899) (A(0n) + diag (M(0119899)e1015840)) = 0(7)

(iii) Similarly if buffer 2 is empty and buffer 1 is neitherempty nor full (119899 = 0 and 0 lt 119898 lt 119862

1) we have

120587 (119898 minus 1 0) L(119898minus10)1

+ 120587 (119898 + 1 1)M(119898+11)

+ 120587 (119898 0) (A(m0) + diag (M(1198980)e1015840)) = 0(8)

(iv) If both buffers are empty (119898 = 0 and 119899 = 0) we have

120587 (1 1)M(11)

+ 120587 (0 0) (A(00) + diag (M(00)e1015840)) = 0(9)

(v) If buffer 1 is empty and buffer 2 is full (119898 = 0 and119899 = 119862

2) we get

120587 (0 1198622minus 1) L(01198622minus1)

2

+ 120587 (0 1198622) (A(01198622) + diag (M(01198622)e1015840 + L(01198622)

2e1015840)) = 0

(10)

(vi) Similarly if buffer 1 is full and buffer 2 is empty (119898 =

1198621and 119899 = 0) we have

120587 (1198621minus 1 0) L(1198621minus10)

1

+ 120587 (1198621 0) (A(11986210) + diag (M(1198621 0)e1015840 + L(1198621 0)

1e1015840)) = 0

(11)

(vii) Finally if both buffers are full (119898 = 1198621and 119899 = 119862

2)

we find

120587 (1198621minus 1 119862

2) L(1198621minus11198622)1

+ 120587 (1198621 1198622minus 1) L(1198621 1198622minus1)

2

+120587 (1198621 1198622) (A(1198621 1198622) + diag (L(1198621 1198622)

1e1015840 + L(11986211198622)

2e1015840))=0

(12)

32 Performance Measures Given the steady-state vectors120587(119898 119899) we can now obtain a number of interesting perfor-mance measures for the kitting system For ease of notationlet 120587(119898 119899) = 120587(119898 119899)e1015840 denote the probability to have119898 partsin buffer 1 and 119899 parts in buffer 2 Moreover let 120587(1)(119898) =

sum119899120587(119898 119899) and 120587(2)(119899) = sum

119898120587(119898 119899) denote the marginal

probability vectors of the buffer 1 and 2 content respectivelyfor the different background states Finally the probabilitymass functions of the buffer contents equal 120587

(1)(119898) =

120587(1)(119898)e1015840 and 120587

(2)(119899) = 120587

(2)(119899)e1015840


The following performance measures are of interest

(i) The mean buffer 1 and 2 content is 1198641198761and 119864119876

2

respectively

1198641198761=

1198621

sum

119898

120587(1)

(119898)119898 1198641198762=

1198622

sum

119899

120587(2)

(119899) 119899 (13)

(ii) The variance of the buffer 1 and 2 is Var1198761and

Var1198762 respectively

Var1198761=

1198621

sum

119898

120587(1)

(119898)1198982minus (119864119876

1)2

Var1198762=

1198622

sum

119899

120587(2)

(119899) 1198992minus (119864119876

2)2

(14)

(iii) The effective load of the system 120588eff is the amount oftime that kitting is ongoing As kitting is only ongoingwhen none of the buffers is empty we have

120588eff = 1 minus 120587(1)

(0) minus 120587(2)

(0) + 120587 (0 0) (15)

(iv) Let the throughput 120578 be defined as the number of kitsdeparting from the system per time unit Taking intoaccount all possible states from which we can have adeparture we find

120578 =

1198621

sum

119898=1

1198622

sum

119899=1

120587 (119898 119899)M(119898119899)e1015840 (16)

(v) The shortage probability119870 is the probability that oneof the buffers is empty as

119870 = 120587(1)

(0) + 120587(2)

(0) minus 120587 (0 0) (17)

(vi) The loss probability 119887119894in buffer 119894 is the probability that

an arriving part cannot be stored in buffer 119894 119894 = 1 2By noting that the accepted arrival load equals thedeparture rate we find

119887119894=120588119894minus 120578

120588119894

(18)

where 120588119894is the arrival load of part 119894 = 1 2 If the

arrival load in both buffers is the same then the lossprobability is also the same 119887

1= 1198872 If the arrival

load is not the same then the excess load in the mostloaded buffer is lost as well

33 Methodology Sparse Matrix Techniques Queueingmod-els for kitting processes are rather complicated Indeedthe modelled kitting process has a multidimensional statespace Even for relative moderate buffer capacity the mul-tidimensionality leads to huge state spaces this is the so-called state-space explosion problem Formany queueing sys-tems infinite-buffer assumptions may mitigate this problem

Given some buffer system with finite capacity more efficientnumerical routines can be constructed for the correspondingqueueing system with infinite capacity Unfortunately asmentioned above the infinite-buffer capacity assumption isnot applicable for kitting processes and therefore cannotsimplify the analysis Recall that the infinite-capacity modelis always unstable For all input parameters except trivial ones(no arrivals) some or all of the queues grow unbounded withpositive probability

Consequently the multidimensionality of the state spaceand the inapplicability of the infinite-buffer assumptionyield Markov processes with a finite but very large statespace However the number of possible state transitionsfrom any specific state is limited This means that most ofthe entries in the generator matrix are zero the matrix issparse As illustrated by the numerical examples using sparsematrices and their associated specialised algorithms resultin manageable memory consumption and processing timescompared to standard algorithms

Themethod used here to solve the sparse matrix equationis the projection method GMRES (generalized minimumresidual) Details can be found in Stewart [15] (p 194ndash197)Philippe et al [16] Buchholz [17] and Saad and Schultz [18]To solve the linear equation 119860119909 = 119887 the GMRES algorithmapproximates 119909 by the vector 119909

119899isin 119870119899in a Krylov subspace

119870119899= span119887 119860119887 119860119899minus1119887 which minimises the norm of

the residual 119860119909119899minus 119887 Since the residual norm is minimised

at every step of the method it is clear that it is nonincreasingHowever the work and storage requirement per iterationincreases linearly with the iteration count Hence the costof 119899 iterations grows by 119874(119899

2) which is a major drawback

of GMRES This limitation is usually overcome by restartingthe algorithm After a chosen number of iterations 119898 theaccumulated data is cleared and the intermediate results areused as the initial data for the next119898 iterations [18]

To ensure fast convergence it is also key to properlychoose the initial vector that is passed on to the algorithmWerely on MATLABrsquos build in GMRES algorithm and assume auniform initial vector if no additional information about thesolution is knownHowever this is often the case Calculationspeed can be further improved as in practice performancemeasures are not calculated for an isolated set of parametersFor example when a plot is created a parameter is varied overa range of values In this case a previously calculated steady-state vector for some set of parameters can be used as a firstestimate of the steady-state vector for a new ldquoperturbedrdquo set ofparameters Using previously calculated steady-state vectorsis trivial if the state spaces corresponding to the parametersets are equal In this case the previously calculated steady-state vector can be passed on unmodified If the state spacechanges the steady-state vector must be rescaled to thenew state space In general adding zero-probability statesif the state space increases or removing states if the statespace decreases turns out to be ineffective This is easilyexplained by a simple example Assume that we increase thequeue capacity of one of the part buffers Typically even formoderate load a considerable amount of probability masscan be found for queue size equal to capacity Increasing the


queue size and assigning zero probability to the new statesis not a good estimate for the new steady-state vector Alsofor the system with higher capacity a considerable amount ofprobability mass can be found when the queue size equals thecapacity (while zero probabilities were assigned)

4 Numerical Results

With the balance equations at hand we now illustrate ournumerical approach by means of some examples

41 Bursty Part Arrivals As a first example we quantify theimpact of production inefficiency on the performance of akitting process To this end we compare part buffers withPoisson arrivals to corresponding kitting systems with inter-rupted Poisson arrivals The arrival interruptions account forinefficiency in the production process Kit assembly timesare assumed to be exponentially distributed with service rateequal to one this value being independent of the number ofparts in the different buffers This is a kitting process withMarkovian arrivals as described in Example 2 in Section 2

The interrupted Poisson process considered here is atwo-state Markovian process In the active state new partsarrive in accordance with a Poisson process with rate 120582

whereas no new parts arrive in the inactive state Let 120572 and120573 denote the rate from the active to the inactive state and viceversa respectively We then use the following parameters tocharacterise the interrupted Poisson process

120590 =120573

120572 + 120573 120581 =

1

120572+1

120573 120588 = 120582120590 (19)

Note that120590 is the fraction of time that the interrupted Poissonprocess is active the absolute time parameter 120581 is the averageduration of an active and an inactive period and 120588 is thearrival load of the parts

Figure 3 shows the mean number of parts in buffer 1versus the arrival load for various values of the buffercapacities 119862

1and 119862

2for Poisson arrivals (for both buffers)

as well as for interrupted Poisson arrivals (again for bothbuffers) The arrival load is set to 120588 = 08 for all curves Inaddition we set 120590 = 04 and 120581 = 10 for the interruptedPoisson processes Clearly the mean buffer content increasesas the arrival load increases as expected Moreover if morebuffer capacity is available it will also be used the meanbuffer content increases for increasing values of 119862

1=

1198622 Comparing interrupted Poisson and Poisson processes

burstiness in the production process has a negative impact onperformancemdashmore buffering is requiredmdashif the queues arenot fully loaded (120588 lt 1) As for ordinary queues the oppositecan be observed for overloaded buffers In this case the effectof a large burst ismainly reflected in loss and not in additionalqueue content Burstiness also yields larger periods withoutarrivals during which the buffer size decreases

By numerical examples we can quantify expected bufferbehaviourmdashfor example more production yields higherbuffer content higher buffer capacity mitigates the lossprobability and so forth However less trivial behaviour canbe observed as well Figure 4 depicts the probability that

1 20

20

40

60

Poisson for both partsIPP for both parts

Mean number of parts in buffer 1

C1 = C2 = 60

C1 = C2 = 40

C1 = C2 = 20

1205881 = 1205882

Figure 3 When the queues are not fully loaded burstiness in theproduction process has a negative impact on the performance

the buffer is full versus the buffer capacity 1198621

= 1198622 We

compare performance of kitting with Poisson arrivals tokitting with interrupted Poisson arrivals at one buffer andat both buffers As in the preceding figure the interruptedPoisson processes are characterised by 120590 = 04 and 120581 = 10As expected the probability that the buffer is full decreasesfor increasing values of the buffer capacities Moreover toreduce this probability more buffer capacity is required forthe case of two interrupted Poisson processes than for thecase of two Poisson processes For the kitting process withone Poisson and one interrupted Poisson process non-trivialperformance results can be observed Namely interruptionsin the production of a part more negatively affect bufferperformance of the other part Indeed buffer 1 is full withhigher probability if the arrivals at buffer 2 are interruptedthan if the arrivals at buffer 1 are interrupted First notethat the loss probabilities in both buffers are the same Forthe Poisson buffer parts are rejected at the arrival rate 120588

1

whenever the buffer is full For the IPP buffer parts arerejected when the buffer is full and the arrival process is in theon-state at rate 120588

2120590 As the loss probability in both buffers is

the same we have Pr[1198761= 1198621]1205881= Pr[119876

2= 1198622 119878 = 1]120588

2120590

or equivalently Pr[1198761= 1198621] = Pr[119876

2= 1198622| 119878 = 1] As the

second queue is more likely to be full when there are arrivalswe have Pr[119876

2= 1198622| 119878 = 1] ge Pr[119876

2= 1198622] which shows

Pr[1198761= 1198621] ge Pr[119876

2= 1198622]

42 Phase-Type Distributed Kit Assembly Times The secondnumerical example quantifies the impact of the distributionof the kit assembly times on kitting performance In par-ticular we here study Erlang-distributed kit assembly timesLimiting ourselves to Poisson arrivals to both buffers thisnumerical example fits Example 3 of Section 2


10 50 1000

005

01

015

02

025Probability that buffer 1 is full

Poisson for both partsIPP for buffer 1

IPP for buffer 2IPP for both parts

C1 = C2

Figure 4 Irregularity in the production of a part leads to a higherprobability to have a full buffer for the other part

06 09 120

10

20

Kitting model


1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue

Figure 5 Variation of the kitting time distribution does not have asignificant impact on the mean number of parts

Figures 5 and 6 depict themean number of parts in buffer1 and the loss probability in buffer 1 for the kitting processand as a reference point for the ME1N queue as well Inboth figures the arrival load is varied and different values ofthe variance of the kitting time distribution are assumed asindicated The mean kitting time is equal to 1 for all curvesand the capacity of both buffers is equal to 20 In underload(120588 lt 1) kitting performs worse than theME1N queue themean buffer content and the loss probability have a highervalue This follows from the fact that kitting stops when oneof the buffers is empty By increasing the load it is obvious

06 09 12

Kitting modelLoss probability

100

10minus3

10minus7

1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue

Figure 6 Variation of the kitting time distribution does not have asignificant impact on the loss probability

that the buffer content converges to the capacity and theloss probability to one It is most interesting to observe thatthe shape of the service time distribution only has a smalleffect on these performance measures Indeed there is nosignificant performance difference when 120590

2 equals 14 andwhen it equals 18

43 Cost and Profit Analysis We now add a cost structure tothe kitting process under study In particular cost and profitfor the kitting systems of Sections 41 and 42 are analysed

The proposed cost function is

1198881(1198641198761+ 1198641198762) + 1198882119870 + 1198883(1198871+ 1198872) (20)

where 1198881is the holding cost of a part in the buffer 119888

2is the

shortage cost in one or in both of the buffers and 1198883is the

loss cost Note that for all figures the input parameters aresymmetric for both parts such that 119864119876

1= 1198641198762and 1198871= 1198872

431 Poisson Arrivals In Figure 7 the total cost of applyingkitting systems versus the buffer capacity (varying from 1 to30) is depicted Limiting ourselves to Poisson arrivals for bothbuffers 120588

119894= 120582119894= 08 for part 119894 = 1 2 We compare kitting

performance for different costs as depicted As expectedhigher cost values lead to higher total costThe cost structurealso affects the optimal buffer capacity When looking at thedifferent costs separately we observe that a higher holdingcost decreases the optimal buffer capacity (from 15 to 12)and a higher loss cost increases the optimal capacity (from12 to 14) whereas the optimal buffer capacity remains thesame (12) for a higher shortage cost Obviously bufferingis interesting when the storage cost is low and when thecost of rejecting parts (because of a full buffer) is high It ismost interesting to observe that as the capacity increases the


5 301520

50

80Cost of a kitting process with Poisson arrivals

C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40

Figure 7 Each type of cost has a different impact on the value of the optimal buffer capacity

5 15 3020

50

80Cost analysis

Poisson for both partsIPP for one partIPP for both parts

C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)

Figure 8 Production inefficiency results in higher required storage space

cost models converge when the sum of the holding and theshortage cost (119888

1+ 1198882) is equal Indeed as the loss tends to

zero the loss cost tends to zero as well Furthermore whenthe capacity equals one the state space of the kitting modelhas size 4 and hence the cost is easily calculated explicitly as

total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)

432 Bursty Part Arrivals Next we analyse the cost and theprofit of kitting systemswith differentMarkovian arrivalsWe

use the same values to model the arrivals as in Figure 4 InFigures 8(a) and 8(b) the total cost (left) and the profit (right)versus the buffer capacity are depictedWe consider a holdingcost 1198881equal to 2 a shortage cost 119888

2equal to 55 and a loss cost

1198883equal to 40 On the left figure the optimal buffer capacity

for Poisson arrivals is 12 for an interrupted Poisson arrivalat one buffer it equals 22 and for an interrupted Poissonprocess at both buffers the optimal capacity is 28 As expectedthe optimal buffer capacity increases as the burstiness in theproduction process increases Concerning the profit analysiswe assume that the yield equals the throughput multiplied


5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)

Figure 9 Shape of the service time distribution does not impact significantly the value of the optimal buffer capacity

by a sale unit equal to 100 Assuming a maximum storageroom of 30 parts we observe that the optimal buffer capacityis 5 11 and 12 for the depicted arrival processes respectivelyConsequently kitting systems under production inefficiencyrequire much higher storage space especially when profitis applied as the parameter to determine the optimal buffercapacity Moreover the optimal capacity is very sensitive tothe burstiness parameters 120590 and 120581 In the example at handthe plots suggest that the cost function is a convex functionof the buffer capacity However one can also easily choose thecost parameters to obtain nonconvex cost functions

433 Phase-Type Kit Assembly Times Finally Figures 9(a)and 9(b) depict the cost (left) and the profit (right) for akitting system with Erlang-distributed kit assembly timesversus the buffer capacity As in the preceding figure theprofit equals the difference between the yield (equal to thethroughput multiplied by 100) and the cost (defined by theparameters 119888

1= 2 119888

2= 55 and 119888

3= 40) Out of the numer-

ical examples we observe that the shape of the service timedistribution has no significant impact on profit and costThese results confirm those found in Section 42

44 Performance Analysis of Solution Methods We comparethe performance of (sparse) GMRES and solving the Markovchain by standard LU decomposition Strang [19] on a kittingprocess with Poisson arrivals with rate 120582

119894= 08 for part

119894 = 1 2 and with exponentially distributed kit assemblytimes with rate 120583 Figure 10 depicts both methods in termsof speed versus the state space for a kitting process startingfrom a symmetric buffer capacity 119862

1= 1198622

= 1 to 1198621

=

1198622

= 60 While LU performs better than GMRES whenthe capacity (and state space) is small the figure clearly

0 500 2000 35000

10

20

30

40

State space

CPU-time (s)

LU-decompositionGMRES

Figure 10 The GMRES method performs well in terms of speed

shows that GMRES performs considerably is better than LUdecomposition for capacity

5 Conclusion

In this paper we evaluate the performance of two-partkitting processes in aMarkovian setting Furthermore a cost-profit analysis is conducted and an approximation for ourmodel is given Note that the particularity of the studiedkitting systems is that the part buffers are paired Thismeans that each demand requires both parts such that a


kit can only be assembled if both inventories are nonemptyMethodologically we have applied sparse matrix techniques(eg GMRES) as most of the entries in the generator matrixhave a value equal to zero The solution is not exact butperforms well in terms of solution speed and accuracy

As our numerical results show the interplay between thedifferent queues leads to complex performance behaviourFor example interruptions in the production of a partmore negatively affect buffer performance of the other partOverall we observe extreme sensitivity of kitting perfor-mance on arrival process parameters while performanceis reasonable insensitive to variation of the kitting timedistribution Furthermore we determine the optimal buffercapacity based on a cost-profit analysis and establish a quitegood approximation for our two-part kitting process

References

[1] L Medbo ldquoAssembly work execution and materials kit func-tionality in parallel flow assembly systemsrdquo International Jour-nal of Industrial Ergonomics vol 31 no 4 pp 263ndash281 2003

[2] Y A Bozer and L F McGinnis ldquoKitting versus line stocking aconceptual framework and a descriptive modelrdquo InternationalJournal of Production Economics vol 28 no 1 pp 1ndash19 1992

[3] P Som W Wilhelm and R Disney ldquoKitting process in astochastic assembly systemrdquo Queueing Systems vol 17 no 3-4pp 471ndash490 1994

[4] H Bryzner and M I Johansson ldquoDesign and performanceof kitting and order picking systemsrdquo International Journal ofProduction Economics vol 41 no 1ndash3 pp 115ndash125 1995

[5] S Ramachandran and D Delen ldquoPerformance analysis of aKitting process in stochastic assembly systemsrdquo Computers ampOperations Research vol 32 no 3 pp 449ndash463 2005

[6] R Ramakrishnan and A Krishnamurthy ldquoAnalytical approxi-mations for kitting systems with multiple inputsrdquo Asia-PacificJournal of Operational Research vol 25 no 2 pp 187ndash216 2008

[7] B Johansson and M I Johansson ldquoHigh automated Kittingsystem for small parts a case study from the Volvo uddevallaplantrdquo in Proceedings of the 23rd International Symposium onAutomotive Technology and Automation pp 75ndash82 ViennaAustria 1990

[8] W J Hopp and J T Simon ldquoBounds and heuristics forassembly-like queuesrdquo Queueing Systems vol 4 no 2 pp 137ndash155 1989

[9] J M Harrison ldquoAssembly-like queuesrdquo Journal of AppliedProbability vol 10 pp 354ndash367 1973

[10] G Latouche ldquoQueueswith paired customersrdquo Journal of AppliedProbability vol 18 no 3 pp 684ndash696 1981

[11] E De Cuypere and D Fiems ldquoPerformance evaluation ofa kitting processrdquo in Proceedings of the 17th InternationalConference on analytical and stochastic modelling techniquesand applications vol 6751 Lecture Notes in Computer ScienceVenice Italy 2011

[12] P Buchholz P Kemper and J Kriege ldquoMulti-class Markovianarrival processes and their parameter fittingrdquo PerformanceEvaluation vol 67 no 11 pp 1092ndash1106 2010

[13] D Fiems B Steyaert and H Bruneel A Genetic ApproachTo Markovian Characterisation of H 264 Scalable VideopagesMultimedia Tools and Applications 2011

[14] G Latouche and V Ramaswami Introduction To Matrix Ana-lytic Methods in Stochastic Modeling ASA-SIAM Series onStatistics and Applied Probability Society for Industrial andApplied Mathematics Philadelphia PA 1999

[15] W J Stewart Introduction To the Numerical Solution of MarkovChains Princeton University Press Princeton NJ USA 1994

[16] B Philippe Y Saad and W Stewart ldquoNumerical methods inmarkov chainmodelingrdquoOperations Research vol 40 no 6 pp1156ndash1179 1992

[17] P Buchholz ldquoStructured analysis approaches for large Markovchainsrdquo Applied Numerical Mathematics vol 31 no 4 pp 375ndash404 1999

[18] Y Saad and M H Schultz ldquoGMRES a generalized minimalresidual algorithm for solving nonsymmetric linear systemsrdquoJournal on Scientific and Statistical Computing vol 7 no 3 pp856ndash869 1986

[19] G Strang Linear Algebra and Its Applications Academic PressOrlando FL 2nd edition 1980

Submit your manuscripts athttpwwwhindawicom

Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

MathematicsJournal of


Mathematical Problems in Engineering

Hindawi Publishing Corporationhttpwwwhindawicom

Differential EquationsInternational Journal of

Volume 2014

Applied MathematicsJournal of


Probability and StatisticsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Journal of


Mathematical PhysicsAdvances in

Complex AnalysisJournal of


OptimizationJournal of


CombinatoricsHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of


Operations ResearchAdvances in

Journal of


Function Spaces

Abstract and Applied AnalysisHindawi Publishing Corporationhttpwwwhindawicom Volume 2014

International Journal of Mathematics and Mathematical Sciences


The Scientific World JournalHindawi Publishing Corporation httpwwwhindawicom Volume 2014


Algebra

Discrete Dynamics in Nature and Society



Decision SciencesAdvances in

Discrete MathematicsJournal of


Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


always enough space for upcoming parts which simplifies theanalysis Unfortunately the assumption of an infinite bufferis not valid for kitting processes If the capacity is assumedto be infinite then the model will degrade to an unstablestochastic system Harrison [9] showed for a multiple inputgeneralisation of the GIG1 queue that it is necessary toimpose a restriction on the size of the buffer to ensure stabilityin the operations of a kitting process Under this assumptionthe probability to have a certain long-term stock position isequal and independent of the current stock positionThis wasalso demonstrated by Latouche [10] who studiedwaiting lineswith paired customers We can consider this analysis as anabstraction of a kitting process with two types of parts

In this work we focus on a kitting process modulated bya Markovian environment The introduction of this environ-ment allows us to study kitting under more realistic stochas-tic assumptions kitting interruptions bursty part arrivalsphase-type distributed kit assembly times and so forthOur paper extends the results on kitting in a Markovianenvironment [11]

The remainder of this paper is organised as followsSection 2 describes the kitting process at hand In Section 3Chapman-Kolmogorov equations are derived and theirnumerical solution is discussed To illustrate our approachSection 4 considers a number of numerical examples Inparticular we assess the impact of kitting interruptionsbursty part arrivals and phase-type distributed kit assemblytimes on the behaviour of the part buffers Then a cost-profit analysis of kitting processes is conducted and anapproximation for a two-part kitting process is establishedFinally conclusions are drawn in Section 5

2 Model Description

In this paper we study a two-queue kitting process asdepicted in Figure 1 Each queue has a finite capacitymdashlet 119862

ℓ

denote the capacity of buffer ℓ ℓ = 1 2mdashand models theinventory of parts of a single type New parts arrive at thebuffers and if both buffers are nonempty a kit is assembledby collecting a part from each buffer Hence departuresfrom the buffers are synchronised and the buffers are pairedOperation of part buffers therefore considerably differs fromother queueing systems

Arrivals at both buffers are modelled by a Markovianarrival process and kit assembly is not instantaneous Forease of modelling it is assumed that there is a modulatingMarkov process arrival and service rates depending on thestate of this process To be more precise the kitting processis modelled as a continuous-time Markov process with statespace C

1times C2times K whereby C

ℓ= 0 119862

ℓ for ℓ =

1 2 and with K = 1 2 119870 being the state space ofthe modulating process At any time the state of the kittingprocess is described by the triplet [119898 119899 119894] with119898 and 119899 beingthe number of parts in the first and secondbuffer respectivelyand 119894 being the state of the modulating process We nowdescribe the state transitions

(i) The state of the modulating process can changewhen there are neither arrivals nor departures Let

L1

L2

C1

C2

M

Figure 1 Kitting process the buffers are on the left and thetriangularly shaped kitting process is on the right

120572119894119895denote the transition rate from state 119894 to state

119895 (119894 119895 isin K 119894 = 119895) and let A denote the correspondinggenerator matrix

(ii) The state of the modulating process may remain thesame or may change when there is an arrival Let 120582(ℓ)

119894119895

denote the (marked) transition rate from state 119894 tostate 119895 when there is an arrival at buffer ℓ ℓ = 1 2Moreover let L

ℓdenote the corresponding generator

matrix Note that such marked transitions from state119894 to state 119894 are allowed

(iii) Analogously the state of the modulating process mayremain the same or may change when there is adeparture (in each buffer) Let 120583

119894119895and M denote the

corresponding transition rate and generator matrixrespectively

Summarising arrivals at and departures from the buffers aredescribed by the generator matrices A L

1 L2 andM So far

no diagonal elements of A have been defined To simplifynotation it will be further assumed that the diagonal elementsare chosen such that the row sums ofA+L

1+L2+M are zero

The computational method employed here does notrequire any homogeneity of the generator matrices Whenrequired by the applications at hand intensities may dependon the buffer content In this case we introduce super-scripts to make this dependence explicit For exampleM(119898119899)denotes the generator matrix of state transitions with depar-ture when there are119898 parts in buffer 1 and 119899 parts in buffer 2In addition we use arguments for rates as we already usedsuperscripts and subscripts to distinguish the arrival ratesat the different queues For example 120582(ℓ)

119894119895(119898 119899) denotes the

arrival rate at buffer ℓ = 1 2 from state 119894 to state 119895 when thereare119898 parts in buffer 1 and 119899 parts in buffer 2

Example 1 In the most basic setting parts arrive at thebuffers in accordance to an independent Poisson processeswith rate 120582

1and 120582

2and kit assembly times are exponentially

distributed with parameter 120583 In this case there is no needto have a modulating Markov process and the state iscompletely described by the number of parts in each buffer(119898 119899) We have

M = [120583] L1= [1205821]

L2= [1205822] A = [minus1205821 minus 120582

2minus 120583]

(1)

Example 2 To account for burstiness in the arrival processof the parts at the different buffers the modulating process









L1= Λ1 L

2= Λ2

A = Λ0minus 120583I M = 120583I

(2)



1and 120582


L(119898119899)1

= 1205821I (1 minus 1

119898=0119899gt0) + 1205821e10158401205911119898=0119899gt0

L(119898119899)2

= 1205822I (1 minus 1

119898gt0119899=0) + 1205822e10158401205911119898gt0119899=0

A(119898119899) = T1119898gt0119899gt0


M(119898119899) = t10158401205911119898gt1119899gt1

+ t1015840f (1 minus 1119898gt1119899gt1

)

(3)


3 Analysis





Let 120587119894(119898 119899) = lim

119905rarrinfinPr[1198761(119905) = 119898119876

2(119905) = 119899 119878(119905) = 119894]


119894(119898 119899) for 119894 isin K



119894119895 Upon




1and 0 lt 119898 lt 119862

2 We have

120587119894 (119898 119899)(

119870

sum

119895=1

120582(1)

119894119895(119898 119899) +

119870

sum

119895=1

120582(2)

119894119895(119898 119899)

+120583119894119895 (119898 119899) +

119870

sum

119895=1119895 = 119894

120572119894119895 (119898 119899))

=

119870

sum

119895=1

120587119895 (119898 minus 1 119899) 120582

(1)

119895119894(119898 minus 1 119899)

+

119870

sum

119895=1

120587119895 (119898 119899 minus 1) 120582

(2)

119895119894(119898 119899 minus 1)

+

119870

sum

119895=1

120587119895 (119898 + 1 119899 + 1) 120583119895119894 (119898 + 1 119899 + 1)

+

119870

sum

119895=1119895 = 119894

120587119895 (mn) 120572119894119895 (119898 119899)

(4)


120587 (119898 minus 1 119899) L(119898minus1119899)1

+ 120587 (119898 119899 minus 1) L(119898119899minus1)2

+ 120587 (119898 + 1 119899 + 1)M(119898+1119899+1) + 120587 (119898 119899)A(119898119899) = 0(5)


2) we have

120587119894 (0 119899)(

119870

sum

119895=1

120582(1)

119894119895(0 119899) +

119870

sum

119895=1

120582(2)

119894119895(0 119899)

+

119870

sum

119895=1119895 = 119894

120572119894119895 (0 119899))

=

119870

sum

119895=1

120587119895 (0 119899 minus 1) 120582

(2)

119895119894(0 119899 minus 1)

+

119870

sum

119895=1

120587119895 (1 119899 + 1) 120583119895119894 (1 119899 + 1)

+

119870

sum

119895=1119895 = 119894

120587119895 (0 119899) 120572119895119894 (0 119899)

(6)













m + 1 n minus 1 i m + 1 n i m + 1 n + 1 i

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii120582(2)ii

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii120582(2)ii

120582(2)ii

120582(2)ii

120582(1)ii120582(1)ii 120582(1)ii

120582(1)ii 120582(1)ii 120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii

120582(2)ii

120583ii120583ii

120583ii120583ii120583ii

120583ii



120587 (0 119899 minus 1) L(0119899minus1)2

+ 120587 (1 119899 + 1)M(1119899+1)

+ 120587 (0 119899) (A(0n) + diag (M(0119899)e1015840)) = 0(7)


1) we have

120587 (119898 minus 1 0) L(119898minus10)1

+ 120587 (119898 + 1 1)M(119898+11)

+ 120587 (119898 0) (A(m0) + diag (M(1198980)e1015840)) = 0(8)


120587 (1 1)M(11)

+ 120587 (0 0) (A(00) + diag (M(00)e1015840)) = 0(9)


2) we get

120587 (0 1198622minus 1) L(01198622minus1)

2

+ 120587 (0 1198622) (A(01198622) + diag (M(01198622)e1015840 + L(01198622)

2e1015840)) = 0

(10)


1198621and 119899 = 0) we have

120587 (1198621minus 1 0) L(1198621minus10)

1

+ 120587 (1198621 0) (A(11986210) + diag (M(1198621 0)e1015840 + L(1198621 0)

1e1015840)) = 0

(11)


2)

we find

120587 (1198621minus 1 119862

2) L(1198621minus11198622)1

+ 120587 (1198621 1198622minus 1) L(1198621 1198622minus1)

2

+120587 (1198621 1198622) (A(1198621 1198622) + diag (L(1198621 1198622)

1e1015840 + L(11986211198622)

2e1015840))=0

(12)


sum119899120587(119898 119899) and 120587(2)(119899) = sum



(1)(119898) =

120587(1)(119898)e1015840 and 120587

(2)(119899) = 120587

(2)(119899)e1015840




2

respectively

1198641198761=

1198621

sum

119898

120587(1)

(119898)119898 1198641198762=

1198622

sum

119899

120587(2)

(119899) 119899 (13)



Var1198761=

1198621

sum

119898

120587(1)

(119898)1198982minus (119864119876

1)2

Var1198762=

1198622

sum

119899

120587(2)

(119899) 1198992minus (119864119876

2)2

(14)


120588eff = 1 minus 120587(1)

(0) minus 120587(2)

(0) + 120587 (0 0) (15)


120578 =

1198621

sum

119898=1

1198622

sum

119899=1

120587 (119898 119899)M(119898119899)e1015840 (16)


119870 = 120587(1)

(0) + 120587(2)

(0) minus 120587 (0 0) (17)



119887119894=120588119894minus 120578

120588119894

(18)


















4 Numerical Results





120590 =120573

120572 + 120573 120581 =

1

120572+1

120573 120588 = 120582120590 (19)



1and 119862



1=




1 20

20

40

60



C1 = C2 = 60

C1 = C2 = 40

C1 = C2 = 20

1205881 = 1205882



= 1198622 We


1




2= 1198622 119878 = 1]120588

2120590


2= 1198622| 119878 = 1] As the


2= 1198622| 119878 = 1] ge Pr[119876


Pr[1198761= 1198621] ge Pr[119876

2= 1198622]



10 50 1000

005

01

015

02




C1 = C2


06 09 120

10

20

Kitting model


1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue



06 09 12


100

10minus3

10minus7

1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue






1198881(1198641198761+ 1198641198762) + 1198882119870 + 1198883(1198871+ 1198872) (20)


2is the



1= 1198641198762and 1198871= 1198872





5 301520

50


C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40


5 15 3020

50

80Cost analysis


C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)





total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)







5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra

















L1= Λ1 L

2= Λ2

A = Λ0minus 120583I M = 120583I

(2)



1and 120582


L(119898119899)1

= 1205821I (1 minus 1

119898=0119899gt0) + 1205821e10158401205911119898=0119899gt0

L(119898119899)2

= 1205822I (1 minus 1

119898gt0119899=0) + 1205822e10158401205911119898gt0119899=0

A(119898119899) = T1119898gt0119899gt0


M(119898119899) = t10158401205911119898gt1119899gt1

+ t1015840f (1 minus 1119898gt1119899gt1

)

(3)


3 Analysis





Let 120587119894(119898 119899) = lim

119905rarrinfinPr[1198761(119905) = 119898119876

2(119905) = 119899 119878(119905) = 119894]


119894(119898 119899) for 119894 isin K



119894119895 Upon




1and 0 lt 119898 lt 119862

2 We have

120587119894 (119898 119899)(

119870

sum

119895=1

120582(1)

119894119895(119898 119899) +

119870

sum

119895=1

120582(2)

119894119895(119898 119899)

+120583119894119895 (119898 119899) +

119870

sum

119895=1119895 = 119894

120572119894119895 (119898 119899))

=

119870

sum

119895=1

120587119895 (119898 minus 1 119899) 120582

(1)

119895119894(119898 minus 1 119899)

+

119870

sum

119895=1

120587119895 (119898 119899 minus 1) 120582

(2)

119895119894(119898 119899 minus 1)

+

119870

sum

119895=1

120587119895 (119898 + 1 119899 + 1) 120583119895119894 (119898 + 1 119899 + 1)

+

119870

sum

119895=1119895 = 119894

120587119895 (mn) 120572119894119895 (119898 119899)

(4)


120587 (119898 minus 1 119899) L(119898minus1119899)1

+ 120587 (119898 119899 minus 1) L(119898119899minus1)2

+ 120587 (119898 + 1 119899 + 1)M(119898+1119899+1) + 120587 (119898 119899)A(119898119899) = 0(5)


2) we have

120587119894 (0 119899)(

119870

sum

119895=1

120582(1)

119894119895(0 119899) +

119870

sum

119895=1

120582(2)

119894119895(0 119899)

+

119870

sum

119895=1119895 = 119894

120572119894119895 (0 119899))

=

119870

sum

119895=1

120587119895 (0 119899 minus 1) 120582

(2)

119895119894(0 119899 minus 1)

+

119870

sum

119895=1

120587119895 (1 119899 + 1) 120583119895119894 (1 119899 + 1)

+

119870

sum

119895=1119895 = 119894

120587119895 (0 119899) 120572119895119894 (0 119899)

(6)













m + 1 n minus 1 i m + 1 n i m + 1 n + 1 i

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii120582(2)ii

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii120582(2)ii

120582(2)ii

120582(2)ii

120582(1)ii120582(1)ii 120582(1)ii

120582(1)ii 120582(1)ii 120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii

120582(2)ii

120583ii120583ii

120583ii120583ii120583ii

120583ii



120587 (0 119899 minus 1) L(0119899minus1)2

+ 120587 (1 119899 + 1)M(1119899+1)

+ 120587 (0 119899) (A(0n) + diag (M(0119899)e1015840)) = 0(7)


1) we have

120587 (119898 minus 1 0) L(119898minus10)1

+ 120587 (119898 + 1 1)M(119898+11)

+ 120587 (119898 0) (A(m0) + diag (M(1198980)e1015840)) = 0(8)


120587 (1 1)M(11)

+ 120587 (0 0) (A(00) + diag (M(00)e1015840)) = 0(9)


2) we get

120587 (0 1198622minus 1) L(01198622minus1)

2

+ 120587 (0 1198622) (A(01198622) + diag (M(01198622)e1015840 + L(01198622)

2e1015840)) = 0

(10)


1198621and 119899 = 0) we have

120587 (1198621minus 1 0) L(1198621minus10)

1

+ 120587 (1198621 0) (A(11986210) + diag (M(1198621 0)e1015840 + L(1198621 0)

1e1015840)) = 0

(11)


2)

we find

120587 (1198621minus 1 119862

2) L(1198621minus11198622)1

+ 120587 (1198621 1198622minus 1) L(1198621 1198622minus1)

2

+120587 (1198621 1198622) (A(1198621 1198622) + diag (L(1198621 1198622)

1e1015840 + L(11986211198622)

2e1015840))=0

(12)


sum119899120587(119898 119899) and 120587(2)(119899) = sum



(1)(119898) =

120587(1)(119898)e1015840 and 120587

(2)(119899) = 120587

(2)(119899)e1015840




2

respectively

1198641198761=

1198621

sum

119898

120587(1)

(119898)119898 1198641198762=

1198622

sum

119899

120587(2)

(119899) 119899 (13)



Var1198761=

1198621

sum

119898

120587(1)

(119898)1198982minus (119864119876

1)2

Var1198762=

1198622

sum

119899

120587(2)

(119899) 1198992minus (119864119876

2)2

(14)


120588eff = 1 minus 120587(1)

(0) minus 120587(2)

(0) + 120587 (0 0) (15)


120578 =

1198621

sum

119898=1

1198622

sum

119899=1

120587 (119898 119899)M(119898119899)e1015840 (16)


119870 = 120587(1)

(0) + 120587(2)

(0) minus 120587 (0 0) (17)



119887119894=120588119894minus 120578

120588119894

(18)


















4 Numerical Results





120590 =120573

120572 + 120573 120581 =

1

120572+1

120573 120588 = 120582120590 (19)



1and 119862



1=




1 20

20

40

60



C1 = C2 = 60

C1 = C2 = 40

C1 = C2 = 20

1205881 = 1205882



= 1198622 We


1




2= 1198622 119878 = 1]120588

2120590


2= 1198622| 119878 = 1] As the


2= 1198622| 119878 = 1] ge Pr[119876


Pr[1198761= 1198621] ge Pr[119876

2= 1198622]



10 50 1000

005

01

015

02




C1 = C2


06 09 120

10

20

Kitting model


1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue



06 09 12


100

10minus3

10minus7

1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue






1198881(1198641198761+ 1198641198762) + 1198882119870 + 1198883(1198871+ 1198872) (20)


2is the



1= 1198641198762and 1198871= 1198872





5 301520

50


C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40


5 15 3020

50

80Cost analysis


C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)





total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)







5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra





















m + 1 n minus 1 i m + 1 n i m + 1 n + 1 i

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii120582(2)ii

120582(1)ii 120582(1)ii120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii120582(2)ii

120582(2)ii

120582(2)ii

120582(1)ii120582(1)ii 120582(1)ii

120582(1)ii 120582(1)ii 120582(1)ii

120582(2)ii

120582(2)ii 120582(2)ii

120582(2)ii

120583ii120583ii

120583ii120583ii120583ii

120583ii



120587 (0 119899 minus 1) L(0119899minus1)2

+ 120587 (1 119899 + 1)M(1119899+1)

+ 120587 (0 119899) (A(0n) + diag (M(0119899)e1015840)) = 0(7)


1) we have

120587 (119898 minus 1 0) L(119898minus10)1

+ 120587 (119898 + 1 1)M(119898+11)

+ 120587 (119898 0) (A(m0) + diag (M(1198980)e1015840)) = 0(8)


120587 (1 1)M(11)

+ 120587 (0 0) (A(00) + diag (M(00)e1015840)) = 0(9)


2) we get

120587 (0 1198622minus 1) L(01198622minus1)

2

+ 120587 (0 1198622) (A(01198622) + diag (M(01198622)e1015840 + L(01198622)

2e1015840)) = 0

(10)


1198621and 119899 = 0) we have

120587 (1198621minus 1 0) L(1198621minus10)

1

+ 120587 (1198621 0) (A(11986210) + diag (M(1198621 0)e1015840 + L(1198621 0)

1e1015840)) = 0

(11)


2)

we find

120587 (1198621minus 1 119862

2) L(1198621minus11198622)1

+ 120587 (1198621 1198622minus 1) L(1198621 1198622minus1)

2

+120587 (1198621 1198622) (A(1198621 1198622) + diag (L(1198621 1198622)

1e1015840 + L(11986211198622)

2e1015840))=0

(12)


sum119899120587(119898 119899) and 120587(2)(119899) = sum



(1)(119898) =

120587(1)(119898)e1015840 and 120587

(2)(119899) = 120587

(2)(119899)e1015840




2

respectively

1198641198761=

1198621

sum

119898

120587(1)

(119898)119898 1198641198762=

1198622

sum

119899

120587(2)

(119899) 119899 (13)



Var1198761=

1198621

sum

119898

120587(1)

(119898)1198982minus (119864119876

1)2

Var1198762=

1198622

sum

119899

120587(2)

(119899) 1198992minus (119864119876

2)2

(14)


120588eff = 1 minus 120587(1)

(0) minus 120587(2)

(0) + 120587 (0 0) (15)


120578 =

1198621

sum

119898=1

1198622

sum

119899=1

120587 (119898 119899)M(119898119899)e1015840 (16)


119870 = 120587(1)

(0) + 120587(2)

(0) minus 120587 (0 0) (17)



119887119894=120588119894minus 120578

120588119894

(18)


















4 Numerical Results





120590 =120573

120572 + 120573 120581 =

1

120572+1

120573 120588 = 120582120590 (19)



1and 119862



1=




1 20

20

40

60



C1 = C2 = 60

C1 = C2 = 40

C1 = C2 = 20

1205881 = 1205882



= 1198622 We


1




2= 1198622 119878 = 1]120588

2120590


2= 1198622| 119878 = 1] As the


2= 1198622| 119878 = 1] ge Pr[119876


Pr[1198761= 1198621] ge Pr[119876

2= 1198622]



10 50 1000

005

01

015

02




C1 = C2


06 09 120

10

20

Kitting model


1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue



06 09 12


100

10minus3

10minus7

1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue






1198881(1198641198761+ 1198641198762) + 1198882119870 + 1198883(1198871+ 1198872) (20)


2is the



1= 1198641198762and 1198871= 1198872





5 301520

50


C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40


5 15 3020

50

80Cost analysis


C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)





total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)







5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












2

respectively

1198641198761=

1198621

sum

119898

120587(1)

(119898)119898 1198641198762=

1198622

sum

119899

120587(2)

(119899) 119899 (13)



Var1198761=

1198621

sum

119898

120587(1)

(119898)1198982minus (119864119876

1)2

Var1198762=

1198622

sum

119899

120587(2)

(119899) 1198992minus (119864119876

2)2

(14)


120588eff = 1 minus 120587(1)

(0) minus 120587(2)

(0) + 120587 (0 0) (15)


120578 =

1198621

sum

119898=1

1198622

sum

119899=1

120587 (119898 119899)M(119898119899)e1015840 (16)


119870 = 120587(1)

(0) + 120587(2)

(0) minus 120587 (0 0) (17)



119887119894=120588119894minus 120578

120588119894

(18)


















4 Numerical Results





120590 =120573

120572 + 120573 120581 =

1

120572+1

120573 120588 = 120582120590 (19)



1and 119862



1=




1 20

20

40

60



C1 = C2 = 60

C1 = C2 = 40

C1 = C2 = 20

1205881 = 1205882



= 1198622 We


1




2= 1198622 119878 = 1]120588

2120590


2= 1198622| 119878 = 1] As the


2= 1198622| 119878 = 1] ge Pr[119876


Pr[1198761= 1198621] ge Pr[119876

2= 1198622]



10 50 1000

005

01

015

02




C1 = C2


06 09 120

10

20

Kitting model


1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue



06 09 12


100

10minus3

10minus7

1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue






1198881(1198641198761+ 1198641198762) + 1198882119870 + 1198883(1198871+ 1198872) (20)


2is the



1= 1198641198762and 1198871= 1198872





5 301520

50


C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40


5 15 3020

50

80Cost analysis


C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)





total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)







5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











4 Numerical Results





120590 =120573

120572 + 120573 120581 =

1

120572+1

120573 120588 = 120582120590 (19)



1and 119862



1=




1 20

20

40

60



C1 = C2 = 60

C1 = C2 = 40

C1 = C2 = 20

1205881 = 1205882



= 1198622 We


1




2= 1198622 119878 = 1]120588

2120590


2= 1198622| 119878 = 1] As the


2= 1198622| 119878 = 1] ge Pr[119876


Pr[1198761= 1198621] ge Pr[119876

2= 1198622]



10 50 1000

005

01

015

02




C1 = C2


06 09 120

10

20

Kitting model


1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue



06 09 12


100

10minus3

10minus7

1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue






1198881(1198641198761+ 1198641198762) + 1198882119870 + 1198883(1198871+ 1198872) (20)


2is the



1= 1198641198762and 1198871= 1198872





5 301520

50


C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40


5 15 3020

50

80Cost analysis


C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)





total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)







5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










10 50 1000

005

01

015

02




C1 = C2


06 09 120

10

20

Kitting model


1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue



06 09 12


100

10minus3

10minus7

1205881 = 1205882

1205902 = 1

1205902 = 14

1205902 = 18

ME1n queue






1198881(1198641198761+ 1198641198762) + 1198882119870 + 1198883(1198871+ 1198872) (20)


2is the



1= 1198641198762and 1198871= 1198872





5 301520

50


C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40


5 15 3020

50

80Cost analysis


C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)





total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)







5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










5 301520

50


C1 = C2

c1 = 2 c2 = 55 c3 = 40

c1 = 2 c2 = 55 c3 = 60

c1 = 1 c2 = 55 c3 = 40c1 = 2 c2 = 45 c3 = 40


5 15 3020

50

80Cost analysis


C1 = C2

(a)

5 15 30

0

60Profit analysis

minus40

minus80


C1 = C2

(b)





total cost = (2 (1198881+ 1198883) (120583 + 2120582) + 3120583119888

2) (3120583 + 2120582)

minus1

(21)







5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra










5 15 30

55

60

65

70

Cost analysis

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(a)

5 15 30

0

30Profit analysis

minus30

C1 = C2

1205902 = 1

1205902 = 14

1205902 = 18

(b)




1= 2 119888

2= 55 and 119888




119894= 08 for part


1= 1198622

= 1 to 1198621

=

1198622


0 500 2000 35000

10

20

30

40

State space

CPU-time (s)




5 Conclusion





References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra












References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
















Volume 2014




Journal of











Journal of


Function Spaces






Algebra









Research Article Performance Analysis of a Kitting Process...

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