Comparison among three pull control policies: kanban, base stock, and generalized kanban

Annals of Operations Research 93 (2000) 41–69 41

Comparison among three pull control policies: kanban,base stock, and generalized kanban ∗

C. Duri, Y. Frein ∗∗ and M. Di Mascolo

Laboratoire d’Automatique de Grenoble (CNRS-INPG-UJF), ENSIEG/INPG, BP 46,F-38402 Saint Martin d’Heres, France

E-mail: [email protected]: [email protected]

This paper is concerned with make-to-stock pull control policies. A classical policy isthe kanban policy. Another policy, very easy to implement, is the base stock policy. Thesetwo policies contain one design parameter per stage. A general control policy, known asthe generalized kanban policy, can also be used to implement the pull mechanism. Thegeneralized kanban policy includes, as special cases, the kanban and the base stock policies.This policy uses two parameters for each stage of the production system. The aim ofthis paper is to provide qualitative and quantitative comparisons of these three policies.The results of our study will help to choose the policy to implement in order to control aproduction system. We give practical rules. We also show that if there is no delay in fillingorders, all three policies have similar costs. However, for the systems studied, we showthat, if there is a delay in filling orders, generalized kanban systems and base stock systemsyield close to optimal costs, which are lower than costs of kanban systems for the sameservice quality.

Keywords: generalized kanban policy, kanban policy, base stock policy, queueing networks,performance evaluation, design

1. Introduction

The aim of just-in-time policies is to produce exactly what is needed at the righttime. A classical solution is to produce according to real demands using a pull controlpolicy. However, when the delay in filling orders is shorter than the lead time, bufferscontaining semi-finished and/or finished products are required. This paper is concernedwith these make-to-stock pull control policies. An efficient control of a pull systemmust provide a good trade-off between the cost of backordered demands and the costof inventories.

In most cases, to implement such a policy, the initial step is to locate the dif-ferent buffers containing semi-finished parts. This choice implies a decomposition ofthe production system into stages, each stage consisting of an output buffer which

∗Work partially supported by the Region Rhone-Alpes through the ORGALEA project.∗∗ GILCO, ENSGI/INPG, 46 Avenue Felix Viallet, F-38031 Grenoble, France.

J.C. Baltzer AG, Science Publishers

42 C. Duri et al. / Comparison among three pull control policies

contains the finished parts of the stage and a manufacturing process which is usedto supply the output buffer of the stage. In each manufacturing process, we use apush control, but the coordination between the different stages is achieved using a pullmechanism.

A pull mechanism can be implemented in several ways. The best known isthe kanban policy [21], which contains one design parameter per stage and for eachtype of product: the number of kanbans in the stage. This parameter allows us tolimit the maximum level of work-in-process (WIP) and finished parts of the stage.Another policy, very easy to implement, is the base stock policy [8,20]. This policycontains one design parameter per stage and for each type of product: the maximalnumber of finished parts in the stage. This policy is very reactive: when an externaldemand arrives, the information is immediately transferred to all the stages. How-ever, the disadvantage of this method is that the maximum number of WIP is notlimited.

A general control policy, known as the generalized kanban policy, can also beused to implement the pull mechanism [7,28]. The generalized kanban system (GKS)includes, as special cases, the kanban system (KS) and the base stock system (BSS).This policy uses two parameters for each stage of the production system and for eachtype of product. One of the parameters limits the maximum level of WIP in thestage and the other determines the maximum number of finished parts in the stage.The values of these parameters considerably affect the performance measures of theproduction system [19]. Let us note that other policies, not covered in this paper, haverecently been introduced in [10] (extended kanban) and in [5] (hybrid policies). Thesepolicies are combinations of kanban, base stock, or CONWIP policies.

Much work has been carried out on each of these individual operating disci-plines (kanban, generalized kanban and base stock), but relatively few comparisonstudies have been conducted. [6,9,22] introduce and compare various generalizationsor combinations of kanban and base stock. In [1,2], dominance results comparingsome policies are given. [25] shows that the base stock policy is never exactly optimalfor a two-machine line with Poisson demand and exponential processing times.

The aim of this paper is to compare the three policies presented above. We studysingle-product, multi-stage systems with each stage consisting of one or more stationsin series. Processing times are exponentially distributed and demand is Poisson. Forthis class of systems, we can use approximate analytical methods to rapidly evaluateperformance measures, allowing us to find the optimal parameters by search whenthe number of stages is moderate. We also use a new constraint on the service levelthrough a fixed delay. We find parameter values for each of the three policies to mini-mize holding costs under a service quality constraint. Two constraints are considered:a certain proportion of orders must be filled immediately or within an order lead time(OLT). The OLT is approximated by a constraint on the number of backorders. Inmany industrial situations, the supplier has an OLT. For the applications we consider,the OLT is less than the manufacturing lead time. Our approximate analytic methodevaluates these constraints for the three policies. For our examples with an OLT, GKS

C. Duri et al. / Comparison among three pull control policies 43

and BSS give similar costs, which are significantly lower than costs of a KS for thesame service quality.

The paper is organized as follows. Section 2 provides a description of each policywe consider. We present the modeling by queueing networks of a production systemfor the different policies we study. This allows us to obtain a qualitative comparisonof the policies. Section 3 gives a brief presentation of the analytical methods used toevaluate system performance measures. In section 4, we present the design procedureand we explain how to evaluate the performance measures that appear in the designcriteria. Finally, in section 5, we present some results of comparisons between thethree pull control policies, illustrated by examples.

2. Presentation of the three pull control policies

2.1. General remarks

To implement a make-to-stock pull control policy, we must first choose the variousbuffers containing semi-finished parts. This choice implies a decomposition of thesystem into stages, each stage consisting of an output buffer which contains the finishedparts of the stage, and a manufacturing process. The manufacturing process containsproducts which are in service or waiting for a service. These parts represent the WIPof the stage and are used to supply the output buffer (see figure 1).

The production of a new part is ordered by the arrival of a demand. This in-formation is transmitted from the downstream stages to the upstream stages. Thecoordination between the different stages must allow the transfer of products and in-formation throughout the production system. There is a flow of parts moving fromupstream stages to downstream stages, and a flow of demands moving from down-stream stages to upstream stages.

We now provide a brief description of the coordination mechanism for the threepull control policies that we consider.

In this paper we consider production systems made up of N stages in seriesand producing only one type of product. For each Stage i, the manufacturing processcontains mi machines, which can be in series, in parallel, or in any other configuration.We assume that there are always raw parts in front of the first stage of the system and

Figure 1. An example of production systems made up of three stages in series.


Figure 2. Queueing network corresponding to a KS made up of three stages in series.

that the external demands are backordered when no finished part is present in the laststage.

2.2. Kanban policy

Since the kanban policy is a well-known pull control policy, we shall not describethe behavior of a KS in detail. Figure 2 shows the queueing network correspondingto the production system of figure 1 for a kanban policy.

The manufacturing process of Stage i is represented by an oval denoted by MPi,for i = 1, . . . ,N . In a KS, let us recall that a fixed number of kanbans, Ki, isassociated with Stage i. Each part which is present in MPi is assigned a kanban fromStage i. The link between the Stages i and i+ 1 is represented by a synchronizationstation consisting of two queues:

– queue Pi, which represents the output buffer and contains the finished parts ofStage i with their associated kanban (Stage i kanban);

– queue Di+1, which contains the demands for the processing of products by Stagei+ 1 and their associated kanban (Stage i+ 1 kanban).

For the synchronization station of Stage N , queue PN contains the finished partsof the production system, which will be used to satisfy customer demands, and queueDN+1 contains the external demands.

Note that if we stop the arrival process of external demands and if we wait longenough, queues Pi will return to their maximum level Ki and the other queues willbe empty.

2.3. Base stock policy

The base stock policy is a very easy to implement pull control policy. Figure 3shows the queueing network corresponding to the production system represented infigure 1 for a base stock policy.


Figure 3. Queueing network corresponding to a BSS made up of three stages in series.

Each manufacturing process is again represented by an oval denoted by MPi,and the link between stages is modeled by a synchronization station at the output ofeach stage. The synchronization station is made up of two queues, one containing thefinished parts of the stage (Pi) and the other (Ai+1) containing demands for productsfrom the next stage. Note that when a demand arrives in the system, an entity issimultaneously transferred to MP1, as raw material is always available, as well asto A2, A3 and A4.

The purpose of a BSS is to satisfy demands and to lead Pi to its maximumlevel Si. Thus, if we stop the demand arrival process and if we wait long enough,the final state of the system will be: Si products in queues Pi, and the other queues(and MPi) empty for i = 1 to 3. Moreover, we have proved (see [16]) the followingproperty for the state of the queues at each time:

QAi +QMPi +QPi = QAi+1 + Si for i > 1,

QMP1 +QP1 = QA2 + S1,

where the notation QX stands for the number of customers (or parts or kanbans) inany queue X. This notation will be used in the remainder of the paper.

2.4. Generalized kanban policy

The generalized kanban policy is an extension of the kanban policy. To simplifythe presentation, we shall introduce this policy from the kanban policy.

In a GKS, the link between Stages i and i + 1 is also represented by a syn-chronization mechanism (see figure 4). However, this mechanism is made up of twosynchronization stations, each containing two queues: Pi and Ai+1 on the one hand,Di+1 and Bi on the other. The entities of queue Pi represent the finished parts fromStage i. In queue Ai+1 we place the kanbans associated with Stage i+ 1 which repre-sent the authorizations to transfer finished parts of Stage i to Stage i+ 1. The entitiesof queue Di+1 represent the demands for the processing of new parts in Stage i. Fi-


Figure 4. Queueing network representing a GKS made up of three stages in series.

nally, Bi contains the kanbans of Stage i which authorize information to move to theupstream stage.

The GKS can be seen to be an extension of a KS by noting that

(1) queue Pi of the KS is split into two queues, Pi and Bi, in the GKS: we separatethe product from its associated kanban at the output of the manufacturing processMPi;

(2) queue Di+1 of the KS is split into two queues, Di+1 and Ai+1, in the GKS: weseparate the demand from its associated kanban.

The last station is special because it models the link with the external demands.The two simple synchronization stations are made up of the queues PN and AN+1 onthe one hand, and DN+1 and BN on the other, where

PN contains the finished parts of the production system,

AN+1 contains the backordered demands,

DN+1 contains the demands for production of new parts by Stage N ,

BN contains the authorizations to transfer processing orders to Stage N − 1 (freekanbans of Stage N ).

We shall now describe the behavior of the GKS, which depends on the valuesof the parameters of the different stages. Note that if we stop the arrival process ofexternal demands and if we wait long enough, queues Pi will contain parts, and queuesBi will contain kanbans. All the other queues will be empty. A single-product GKShas two parameters per stage: Si, which is the number of parts in queue Pi (at thelatence state), and corresponds to the maximum number of parts in queue Pi, and Ki,which is the number of kanbans in queue Bi (at the latence state), and represents thenumber of kanbans associated with Stage i. A GKS tries to bring back queues Pi totheir maximum level while satisfying the demands and controlling the WIP.

The behavior of the production system is the following. Let us first consider theith linkage station (i = 1, . . . ,N − 1). As soon as there is at least one product in Pi


and at least one kanban associated with Stage i + 1 in Ai+1, a product/kanban pairis transferred to the manufacturing process MPi+1, where it will be treated. Whenprocessing is complete, the kanban is separated from the product to join queue Bi+1

and the product is transferred to queue Pi+1. Now, let us consider the synchronizationbetween queues Di+1 and Bi. As soon as there is at least one demand in Di+1 andat least one Stage i kanban present in Bi, a demand/kanban pair will be transferredto the preceding stage. The demand will then join queue Di and the kanban willjoin queue Ai. In a GKS, consumption of parts and transfer of information are notsimultaneous.

The last synchronization station, which receives the external demands, is special.An external demand gives two pieces of information when it arrives: a customerdemand must be satisfied, which is indicated by an entity joining queue AN+1, anda new product must be processed, which is represented by an entity joining queueDN+1 before being transferred to the upstream stage (as soon as there is a kanban inqueue BN ). If a finished part is present in PN when a demand arrives, the demand isimmediately satisfied and a finished part leaves the system. Otherwise, the demand isbackordered in AN+1 until a product becomes available.

Let us describe the routing of the entities in the network for a GKS or a KS. Inboth cases, the parts move from the upstream stages to the downstream stages, visitingin turn the manufacturing process MP1, queue P1, and then MP2,P2, . . . , MPN ,PN .The demands move from the downstream part to the upstream part of the systemvisiting in turn queues DN+1,DN , . . . ,D2. However, in a GKS, unlike a KS, theproducts and the demands are not always associated with a kanban.

Let us describe the routing of the Stage i kanbans. In a GKS, the kanbans movein a closed network made up of queues Ai, MPi, and Bi. Note that, in a KS, theclosed network is made up of the queues Di, MPi, and Pi. Thus, in each case, we canassociate a closed queueing network with each type of kanban. For a GKS, we thusobtain the following relation:

QAi +QMPi +QBi = Ki for i = 1, . . . ,N. (1)

Remark. As we consider that there are always raw parts in front of the first stage ofthe system, we have QA1 = 0.

Other closed sub-networks appear in this model (see figure 4). The number ofcustomers in these sub-networks is constant and equal to the number of customerspresent at the initial state. We can then deduce some properties. All the propertiesand their proofs are presented in [19]. Here, we give only a few important propertiesand some comments.

In a GKS, we can give limits for the various queues of the network. Let usmention some of these limits:

QMPi 6 Ki for i = 1, . . . ,N , (2)

QPi 6 Si for i = 1, . . . ,N , (3)


QBi > Ki − Si −Ki+1 for i = 1, . . . ,N − 1. (4)

Relation (2) implies that the WIP of Stage i is limited by Ki, which shows theexistence of a WIP control. Relation (3) states that the finished parts of queue Pi arelimited by Si.

Moreover, relation (4) implies that if Ki > Si + Ki+1 then Ki − Si − Ki+1

kanbans will always stay in Bi and will not affect the behavior of the productionsystem. Thus, as already pointed out in [7], these kanbans are useless and we willtherefore only consider the following configurations:

Ki 6 Ki+1 + Si for i = 1, . . . ,N − 1. (5)

Finally, as shown in [18], a GKS with Ki = Si (>0) for i = 1, . . . ,N isequivalent to the KS with the same Ki, and a GKS with Ki = +∞ for i = 1, . . . ,Nis equivalent to the BSS with the same Si.

As KS and BSS are special cases of GKS, we can conclude that GKS can alwaysbe at least as good as KS and BSS. We shall now conduct a quantitative comparisonof these policies in order to see, on examples, whether or not the difference betweenthe policies is significant. We shall use analytical methods to estimate the performancemeasures of GKS, KS and BSS.

3. Analytical methods for performance evaluation

We now summarize analytical methods which allow us to analyze GKS, KS andBSS corresponding to the class of systems we study in this paper: stages in series,Poisson demand arrival process and stations with exponentially distributed processingtimes.

For all the policies, performance measures depend on manufacturing processes,arrival process of external demands, and parameters of the different stages.

3.1. Kanban and generalized kanban systems

To obtain stationary performance measures, we use an approximation analyticalmethod, which has been developed for KS [13,15] and extended to GKS [14,19].

These methods allow us to study systems under the following hypothesis: in agiven stage, all the queues have a capacity higher than the number of kanbans of thestage, thus avoiding blocking; the routing throughout each stage is probabilistic; theservice time distribution of each machine and the arrival process of external demandsare general and are represented by phase-type distributions. The phase-type distribu-tions, introduced by Neuts [23, pp. 41–80], allow us to characterize the behavior of thesystem by a Markov chain. In all the examples considered in this paper, the servicetime distribution of each machine and the interarrival process of external demandsare represented by exponential distributions, which are a special case of phase-typedistributions. We can thus apply the above methods to the analysis of the systemsstudied in this paper.


The queueing networks of a KS and a GKS can be seen as closed multi-classqueueing networks by considering each type of kanban as one class of customers.To analyze these queueing networks, we use an approximation method based onproduct-form approximation. The idea of the method is to approximate the perfor-mance measures of the multi-class queueing network using a set of closed single-classproduct-form queueing networks.

These methods [14,15] enable us to obtain several stationary performance mea-sures, such as average WIP and mean number of finished parts at each stage, meannumber of waiting demands or proportion of backordered demands. These methodshave been tested on many numerical examples. The results obtained were comparedwith simulation ones. The methods appear to be very accurate (a few percents ofrelative error on the performance measures, in general), and very rapid (around onesecond on a SUN/SPARC station). For numerical results see [15] for KS and [19] forGKS. In the present paper, we need a new performance measure (probability of havingn demands waiting for finished parts). In appendix A, we describe how this measurecan be obtained.

3.2. Base stock systems

We have examined two analytical methods encountered in the literature for per-formance evaluation of BSS made up of stages in series. One of them is Buzacottet al.’s method [8] and the other is Svoronos and Zipkin’s method [24]. These twomethods have been applied to stages made up of one station with an exponential ser-vice time distribution and yield the same results [17]. Thus, for the design, we useSvoronos and Zipkin’s method, which is the easiest to extend to more complex stages.At the end of this section, we give the extensions we performed.

We now summarize the method initially proposed by Svoronos and Zipkin.Let us consider the ith manufacturing process with its upstream and downstream

synchronization stations (figure 5).The notations we use are listed below. We have already defined

QPi : number of finished parts in synchronization station i,

QAi+1 : number of waiting demands in synchronization station i,

Si: maximum number of finished parts in synchronization station i.

Let us now define

∆i: waiting time of demands in synchronization station i,

Ti: flow time of the ith manufacturing process

and some combinations of these quantities:

Li: Li = ∆i−1 + Ti,

Ei: Ei = Si −QPi +QAi+1.


Figure 5. ith manufacturing process with its upstream and downstream synchronization stations.

This method is based on the properties of phase-type distributions. We can thusonly use it if the flow times of the different stages have continuous phase-type distri-butions (called CPH). In this case, using the representation introduced by Neuts [23,p. 45], the flow time distribution of Stage i, FTi , is denoted by

FTi ∼ CPH(αi,Hi),

where αi: vector of initial probabilities and Hi: transition matrix.We shall now briefly explain the method. The reader can find details in [24]

which uses properties of phase-type distributions demonstrated in [23]. First of all,the authors consider that they know the distribution of Li denoted by

FLi ∼ CPH(γi,Ci).

The authors use the following key result: The discrete variable Ei has the samedistribution as the number of demands in a random time Li with distribution FLi .

From this key result and from the properties of phase-type distributions [23,p. 59], they obtain the discrete phase-type distribution (called DPH) of Ei:

FEi ∼ DPH(πi,Ri) with Ri = λ(λI − Ci)−1 and πi = γi.Ri.

Furthermore, QAi+1 = max(0,Ei − Si) and phase-type properties allow us todefine

FQAi+1∼ DPH

(πi.R

Sii ,Ri

).

The authors use another key result: The variable QAi+1 has the same distributionas the number of demands in a random time ∆i with a distribution F∆i . Therefore,using phase-type distribution properties [23, p. 59], it can be deduced that

F∆i ∼ CPH(γi.R

Sii ,Ci

).

Moreover, Li+1 is the sum of ∆i (waiting time of demands in synchronizationstation i) and Ti+1 (flow time of the (i + 1)th manufacturing process). Therefore,


assuming the independence of the random variables, they obtain FLi+1 = F∆i ∗ FTi+1

(where ∗ represents the convolution operator). The phase-type distribution convolutionformula is then written as

FLi+1 ∼ CPH(γi+1,Ci+1)

with

γi+1 =[γi.R

Sii ,(1− γi.RSii .e

).αi+1

], Ci+1 =

[Ci −Ci.e.αi+1

0 Hi+1

].

In conclusion, knowing FLi (CPH), FEi , FQAi+1(DPH), F∆i , and FLi+1 (CPH)

are obtained. Moreover, L1 is equal to T1 (because ∆0 = 0) and FT1 is known, soFL1 is known. This allows, by induction, FLi , FEi , FQAi+1

, and F∆i to be evaluatedfor i = 2, . . . ,N . Then, from these distributions, the performance measures of theproduction system can be obtained.

Let us recall that FTi must be known in order to apply the method, and one ofthe main difficulties of the method is to obtain FTi .

Lee and Zipkin [20] have applied the method to a system with one exponentialstation in each stage. To obtain the distribution FTi , the authors have assumed that thearrival process at each stage is a Poisson arrival process. Thus, if Stage i is made upof an exponential machine (with a service rate of µi) and if the demand arrival processis a Poisson arrival process (with an arrival rate of λ), the manufacturing process MPican be studied in the same way as an M/M/1 queue with an arrival rate of λ and aservice rate of µi. Therefore, Ti has an exponential distribution with a rate of µi − λ.As we know the rate µi (for i = 1 to N ) and λ, the distribution FTi is known fori = 1 to N .

We then extended the analysis to several exponential machines and a probabilisticrouting in a stage. We also proposed another extension which allows us to take intoaccount more complex service time distributions. The machines can have a Coxian-2service time distribution. We can then identify the first two moments of a generaldistribution (if the squared coefficient of variation is greater than or equal to 0.5). Thisextended method yields good results (for numerical results see [17]).

4. Optimization

To compare the policies, we must first design the system for each one of thepolicies. Many papers in the literature deal with KS design, and more particularlyoptimization of the number of kanbans in each stage. See for example [3,4,11,26].Details on these papers are given in [12]. These optimizations aim to provide a goodtrade-off between the cost of producing ahead of what is needed and therefore buildingup inventory (which implies a holding cost) and the cost of not being able to satisfythe demands on time (i.e., service quality). For example in [12], the author designs KSby minimizing a cost function equal to the weighted sum of the number of kanbans


in each stage (as a measure of the holding cost) and the mean number of waitingdemands (as a measure of service quality). He also looks for the configuration withthe proportion of backordered demands lower than a given level with the minimal cost(the cost function is equal to the weighted sum of the number of kanbans in each stage).

To the best of our knowledge, such optimization methods do not exist for gen-eralized kanban systems. Therefore we must first develop a new design method. Thismethod will be adapted for kanban and base stock systems, since they are particularcases of generalized kanban systems. Let us begin by presenting the design criteriathat we have chosen.

4.1. Criteria

We produce a design under a service quality constraint with minimization of acost function, i.e., we search for the configuration satisfying a given constraint andwhich has the minimum cost. We use the following two service quality performancemeasures:

(1) the probability that an arriving demand is backordered (denoted by PArupt),

(2) the probability that an arriving demand sees more than n waiting demands, ex-cluding itself (denoted by PA(Q > n)).

Performance measure 1 is used when we are concerned with demands that mustbe immediately satisfied and performance measure 2 allows us to introduce a delayin filling orders. Giving a delay in filling orders is equivalent to authorizing somedemands to wait. We have chosen to introduce performance measure 2 because, inmany industries, when a customer sends an order, the industry has a delay (whichdepends on the type of industries) in which it must be satisfied. However, at theend of the delay, virtually all the demands must be satisfied if possible. Likewise, insubcontracting relations, the subcontractor has a delay in filling orders.

As our aim in this paper is to compare the three policies in steady-state behavior,we shall try to include the estimation of these performance measures in the two analyt-ical methods described in subsections 3.1 and 3.2. We limited our study to exponentialdemand arrival distributions. To obtain the performance measures, we must study thelast station of the production system for the three policies.

PArupt is the probability that an arriving demand is backordered, which is equal, in

our case of a Poisson demand arrival process, to the stationary probability of havingno finished parts in the station (Arrival theorem [27]), denoted by Prupt.

Similarly, PA(Q > n) is equal to the stationary probability of having more thann customers which are waiting (Arrival theorem), denoted by P (Q > n).

4.1.1. Calculation of Prupt and P (Q > n) in KS and GKSPrupt is directly obtained using the method described in subsection 3.1 and

P (Q > n) is equal to


P (Q > n) =∞∑

x=n+1

P (QDN+1 = x) for a KS,

P (Q > n) =∞∑

x=n+1

P (QAN+1 = x) for a GKS.

We describe in appendix A how to obtain P (QAN+1 = n) for a GKS, andP (QDN+1 = n) for a KS.

4.1.2. Calculation of Prupt and P (Q > n) in BSSIn subsection 3.2 we saw that:

FQAi+1 ∼ DPH(πi.R

Sii ,Ri

)for i = 1, . . . ,N ,

F∆i ∼ CPH(γi.R

Sii ,Ci

)for i = 1, . . . ,N.

Using these two definitions and some properties of phase-type distributions [23],we can evaluate the two performance measures.

We denote by e the column vector where all the components are equal to 1.First, using a result given in [23, p. 46] on discrete phase-type distributions we

can directly obtain P (Q > n):

P (Q > n) = P (QAN+1 > n) =(πN .R

SNN

).RnNe = πNR

n+SNN .e.

To obtain Prupt, we use F∆i because Prupt is equal to the probability of no delay.Moreover, using a result given in [23, p. 45] on continuous phase-type distributions,we know that

P (∆i 6 t) = 1− γi.RSii . exp(Cit).e.

We can thus deduce Prupt:

Prupt = P (∆N = 0) = P (∆N 6 0) = 1− γN .RSNN .e.

Then, if we denote by “Lim” the maximum accepted value for the performancemeasures Prupt or P (Q > n), the constraints are: Prupt 6 Lim or P (Q > n) 6 Lim.The cost function depends on the number of entities in the various buffers. We havechosen the following cost function:

N∑i=1

ci.QMPi + hi.QPi ,

where ci represents the holding cost associated with the WIP of Stage i, hi representsthe holding cost associated with the finished parts of Stage i.

With this function, we can associate a cost with each configuration and thencompare the different configurations that satisfy the constraint. We will use the word“solution” for a configuration which satisfies the chosen constraint (for example P (Q >10) 6 0.01), and “optimal solution” (or “optimal configuration”) for the solution with


the minimum cost. Let us specify that since the performance evaluation methods(described in section 3) are approximate ones, the optimal solution could be slightlydifferent if we used an exact performance evaluation method. In the following we willnevertheless call “optimal configuration” the solution obtained using our procedure.We shall now see how it can be found without testing all the configurations.

4.2. Design procedure for generalized kanban systems

First we must set the initial configuration and limits of the parameters values.We have chosen 100 as the maximum value of each parameter. We initialize theparameters Si to 0 (for i = 1 to N ). To obtain the initial values of the parameters Ki,we can, for example, study each stage separately. For each stage, we calculate theproduction capacity of the stage for different values of the number of kanbans in thestage. Indeed, if we study a single stage the production capacity depends only on thisparameter [19]. The initial value of Ki is set to the minimum number of kanbans forwhich the production capacity of Stage i (operated in isolation) is strictly higher thanthe arrival rate of demands.

To find the optimal configuration, we use an enumerative method: we test all thepossible configurations (Ki varying from its initial value to 100 and Si varying fromits initial value to 100 for i = 1, . . . ,N ) in the order given by algorithm 1. We shalllater show that many configurations can be excluded due to special properties of theproblem.

Algorithm 1.

For SN = initial value to 100 do. . .

For S1 = initial value to 100 doFor KN = initial value to 100 do

. . .For K1 = initial value to 100 do

Performance evaluation of the current configurationEnd for

. . .End for

End for. . .

End for

For each configuration, we estimate the performance measures using the analyticalmethod described in subsection 3.1 to determine the configurations which are solutions(i.e., satisfy the constraint). Then, among the configurations that are solutions, wechoose the configuration with the lowest cost.


In reality, we do not test all the configurations. We now describe two conditions,which allow us to eliminate configurations without testing them because they cannotbe an optimal configuration:

– For all the parameters, before the first incrementation we go to the maximal valueof the parameter. If we cannot obtain a solution for this value, it means that wecan move to the following parameter (for example, we move to Ki+1 if the currentparameter is Ki) because the incrementation of the current parameter will not leadto a solution. On the other hand, if we obtain a solution, we return to the initialconfiguration and continue the execution of the algorithm.

– The second point concerns the value of the incrementation. For the parameters Si(for i = 1 to N in a GKS), the value is 1, but for the parameters Ki the value of theincrementation can be greater than 1, which allows us to skip some configurationsand thus to save time. If we are interested in parameter Ki, when we have twovalues of the design performance measures Pb1 and Pb2, which correspond to twosuccessive values of Ki, we can deduce that an incrementation of 1 of Ki (withthe current value of Ki) allows us to obtain a variation ∆Pb = Pb1 − Pb2. Wehave observed that, when Ki increases, the value of the variation ∆Pb decreases(concavity of the performance measures). Thus, the value of the variation is lowerthan or equal to ∆Pb for a variation of Ki of 1. Thus, we can carry out a skipequal to x without dropping below Lim, with the following definition of x:

x = ENT

(Pb2 − Lim

∆Pb

),

where ENT(y) is the lowest integer closest to y.

Finally, let us note that the developed design procedure, which is based on anenumeration, is well adapted to the systems studied in section 5. However, for systemscontaining a large number of parameters, there will be too many configurations to test.

4.3. Design procedure for kanban and base stock systems

To design a KS or a BSS, the principle is the same as for a GKS (see sub-section 4.2). Thus, the algorithm that summarizes the different steps of the de-sign is the same with some simplifications. The main simplification stems from thefact that a KS contains only parameters Ki and a BSS uses only parameters Si fori = 1, . . . ,N .

To evaluate the performance measures of the system, we use the method describedin subsection 3.1 for a KS and in subsection 3.2 for a BSS.

We have also made the modification proposed for GKS, which consists of testingthe maximum value of a parameter before beginning its incrementation. If no solutionexists with this maximum value, the incrementation of this parameter in the currentcontext is useless and we must increment the following parameter.


5. Comparisons

We compare the optimal costs of the three policies, CGKS, CKS, CBSS for thesame service quality constraint. As GKS includes KS and BSS as special cases, thefollowing inequalities hold: CGKS 6 CKS and CGKS 6 CBSS. We consider successivelyexamples made up of one stage to four stages. Let us note that for the cases 1, 2 and 4stages the production system is the same (four stations in series). This will help us tosee the influence of the number of stages for a given production system. For all ourdesigns, the service level constraint is Prupt or P (Q > n) less than or equal to 0.02(i.e., Lim = 0.02), and n varying from 1 to 10.

5.1. Production systems made up of one stage, containing four machines

Let us first study a stage made up of four stations in series; the service time ofeach station has an exponential distribution with a mean service time equal to 1. Thearrival process of demands is Poisson, and we are concerned with two values of thearrival rate of demands denoted by λ: 0.5 and 0.8.

With a make-to-order policy, we obtain the following probabilities:

For λ = 0.5, P (Q > 1) = 0.812, P (Q > 5) = 0.254, P (Q > 10) = 0.029.

For λ = 0.8, P (Q > 1) = 0.993, P (Q > 5) = 0.914, P (Q > 10) = 0.698.

Therefore, make-to-order policies do not satisfy the service quality constraint.

5.1.1. Optimal configurationsFor each value of λ, we design the production system for a kanban policy (KS),

a generalized kanban policy (GKS) and a base stock policy (BSS) with, in turn, theperformance measures Prupt and P (Q > n) for n = 0, 2, 5, and 10. Note that the costfunction used for the design is defined as follows for a single-stage production system:c.QMP1 + h.QP1 . For our study, we have set c to 1 and we have studied the designfor several values of h greater than or equal to 1 (cost of finished parts greater thanor equal to the cost of WIP). For h equal to 1, the optimal configurations are given intable 1.

Table 1Optimal configurations for a single-stage production system composed of four exponential stations in

series.

Design λ = 0.5 λ = 0.8

criterion KS BSS GKS KS BSS GKS

Prupt 6 0.02 K = 12 S = 12 K = 11, S = 12 K = 40 S = 40 K = 37, S = 40P (Q > 0) 6 0.02 K = 11 S = 11 K = 11, S = 11 K = 39 S = 39 K = 37, S = 39P (Q > 2) 6 0.02 K = 10 S = 9 K = 11, S = 9 K = 37 S = 37 K = 37, S = 37P (Q > 5) 6 0.02 K = 8 S = 6 K = 11, S = 6 K = 35 S = 34 K = 37, S = 34P (Q > 10) 6 0.02 K = 6 S = 1 K = 11, S = 1 K = 31 S = 29 K = 37, S = 29


First, for the chosen criteria, we have always observed that S in the GKS is equalto S in the BSS for the optimal configurations.

We can also see that, for a KS, the higher the number of demands that canwait (n high), the lower the optimal number of kanbans. Similarly, for the GKS andthe BSS, the higher n, the lower the optimal value of S. When we increase n, wecan decrease the maximum number of finished parts which are necessary to satisfythe design criteria as we will need parts later. Moreover, we can see that, when λincreases, the values of the optimal configuration parameters increase.

We have experimentally observed that, for each of the three policies, the optimalconfigurations are the same whatever h > 1. Let us try to intuitively explain thisresult.

For a single-stage KS, we have only one parameter (K) to design: we thereforesearch for the minimum value of K satisfying the design criteria. For a greater valueof K, the WIP and the finished parts will increase and the cost of the solution willbe greater. Thus, the minimum value of K, which is a solution, is independent of h.h will only change the value of the cost.

For a BSS, we also have only one parameter (S) to design, thus we look for theminimum value of S which is a solution. For a higher value of S, the WIP is the samebut the number of finished parts increases and implies that the cost of the solution isgreater. The design is thus independent of the value of h.

For a GKS, we have observed that, for h equal to 1, the first solution obtainedis the optimal solution. If we look at the classification of the parameters (see subsec-tion 5.2), we can conclude that the optimal configuration is the configuration that is asolution and has the smallest value of S. This implies that, if for h equal to 1 the firstsolution is (K1, S1) with an associated cost C1, then for any other solution (Kk, Sk)with an associated cost Ck, we have Ck > C1 and Sk > S1. When we increase h,we add a cost to finished parts. We shall thus try to decrease the number of finishedparts. Moreover, experimentally, we have observed that, in order to decrease the meannumber of finished parts, we must decrease S. Thus, the optimal configuration willhave a value S 6 S1. Furthermore we have seen above that the optimal configurationfor h equal to 1 is the configuration that is a solution and has the minimum value of S.Therefore, the optimal configuration is the same whatever h > 1.

5.1.2. Optimal costsLet us now compare the optimal cost of the three policies. Let us consider, for

example, the cost obtained for n = 5 and λ = 0.5. For the optimal configuration, thevalues of the WIP and mean number of finished parts are given in table 2.

We can then write the following relations:

CGKS = 3.97 + h× 2.47, CKS = 3.87 + h× 4.13, CBSS = 4 + h× 2.48,

CKS − CGKS = −0.10 + h× 1.66, CBSS − CGKS = 0.03 + h× 0.01,

CKS − CBSS = −0.126923 + h× 1.65.


Table 2WIP and mean number of finished parts corresponding to the

optimal configuration obtained for λ = 0.5 and n = 5.

WIP Finished parts

KS 3.87 4.13BSS 4.00 2.48GKS 3.97 2.47

Table 3Costs associated with the optimal configurations given in table 1, for h = 10.

Design λ = 0.5 λ = 0.8

criterion KS BSS GKS KS BSS GKS

Prupt 6 0.02 84.2 84.3 84.1 256.7 257.0 256.4P (Q > 0) 6 0.02 74.3 74.4 74.3 246.8 247.2 246.6P (Q > 2) 6 0.02 64.4 55.2 55.1 227.1 227.6 227.1P (Q > 5) 6 0.02 45.2 28.8 28.7 207.5 198.6 198.1P (Q > 10) 6 0.02 26.7 4.62 4.59 168.8 151.7 151.3

The latter relation allows us to conclude that, for this example and for h > 1, thebase stock policy is better than the kanban policy since CKS−CBSS is always positive.These relations also show that the generalized kanban policy is better than the kanbanpolicy and the base stock policy. We can also deduce that the higher h is, the moreimportant the difference between the policies becomes. However, we can note that thecosts of GKS and BSS are very close.

More generally, table 3 lists the costs associated with the optimal configurationsgiven in table 1 for h equal to 10.

For the chosen criteria, the optimal costs are very close for the base stock policyand the generalized kanban policy. However, with a cost depending on the maximumvalue of WIP, the difference between BSS and GKS would be greater. Indeed, in aBSS, there is no limit for the WIP.

We can also note that, when parameter S of the GKS is equal to parameter K ofthe KS, the costs of the optimal configurations for GKS and for KS are very close; thegeneralized kanban policy allows us only to remove the kanbans that may be useless.We can also see that the difference between GKS and KS is significant when, in theoptimal configuration, parameter S of the GKS is strictly less than parameter K ofthe KS. In this case, the generalized kanban policy allows us to reduce the numberof finished parts by counterbalancing with kanbans, which makes the system morereactive. For example, when we take a high value of n, for instance 10, the generalizedkanban policy allows us to decrease the cost by 83% for λ = 0.5.

Finally, we can see that, when λ increases, the difference between the generalizedkanban policy and the kanban policy decreases.

The one-stage system studied here contains four exponential stations, howeverthe same type of observation can be made for other systems. We carried out exten-


Table 4Optimal configurations and associated costs for h = 1 and λ = 0.5.

Design criterion KS BSS GKS

Prupt 6 0.02 K = (3, 11) S = (0, 12) K = (6, 13), S = (0, 12)C = 13.3 C = 12.0 C = 11.9

P (Q > 1) 6 0.02 K = (3, 9) S = (0, 10) K = (6, 13), S = (0, 10)C = 11.4 C = 10.1 C = 9.98

P (Q > 5) 6 0.02 K = (3, 6) S = (0, 6) K = (6, 13), S = (0, 6)C = 8.46 C = 6.48 C = 6.40

P (Q > 10) 6 0.02 K = (3, 4) S = (0, 1) K = (6, 13), S = (0, 1)C = 6.59 C = 4.06 C = 3.99

sive numerical tests with two, four and six stations in the stage, which confirmedthis.

5.2. Production systems made up of two stages, each containing two machines

We shall now consider the same production system (made up of four stations inseries) but with the additional possibility of stocking semi-finished parts at the outputof the second machine. The service rate of each machine is 1. Many cost functionscan be used for the design. We have chosen to present results obtained with the costfunction QMP1 + QMP2 + QP1 + hQP2 for h > 1. We put a higher cost only on thefinished parts of the production system.

5.2.1. Design for h = 1Let us begin by the design of the system for each of the three policies and for

h = 1. First, let us set the arrival rate of demands λ to 0.5. Table 4 contains theoptimal configurations and the associated costs. We denote by K the vector withthe two components K1 and K2, and by S the vector with the two components S1

and S2.We carry out the same study for λ =0.8 and we report the results in table 5.We can see that, when we authorize a large number of waiting demands (n high),

we need fewer finished parts in the output buffer of the second stage. This impliesthat, when n increases, in a KS the optimal value of K2 decreases and, in a GKS orin a BSS the optimal value of S2 decreases.

For both designs, h is equal to 1: WIP, semi-finished parts and finished parts havethe same stocking cost. Thus, for GKS and BSS, the optimal configuration contains nosemi-finished parts (S1 = 0); indeed, it is better to stock products close to the externaldemands to ensure service quality. We can thus note that the optimal BSS is the sameas the optimal BSS obtained with only one stage. For KS, K1 cannot be equal to 0because it must provide the production capacity necessary, in the first stage, to satisfydemands.




Prupt 6 0.02 K = (12, 32) S = (0, 40) K = (20, 41), S = (0, 40)C = 42.4 C = 40.1 C = 39.7

P (Q > 1) 6 0.02 K = (12, 30) S = (0, 38) K = (20, 41), S = (0, 38)C = 40.4 C = 38.1 C = 37.8

P (Q > 5) 6 0.02 K = (12, 26) S = (0, 34) K = (20, 41), S = (0, 34)C = 36.5 C = 34.3 C = 33.9

P (Q > 10) 6 0.02 K = (11, 23) S = (0, 29) K = (20, 41), S = (0, 29)C = 32.2 C = 29.6 C = 29.2



Prupt 6 0.02 K = (5, 9) S = (10, 8) K = (7, 11), S = (10, 8)C = 75.46 C = 72.18 C = 72.11

P (Q > 1) 6 0.02 K = (5, 7) S = (10, 6) K = (7, 11), S = (10, 6)C = 55.81 C = 52.73 C = 52.66

P (Q > 5) 6 0.02 K = (5, 4) S = (4, 3) K = (6, 8), S = (4, 3)C = 27.75 C = 19.47 C = 19.32

P (Q > 10) 6 0.02 K = (4, 3) S = (1, 0) K = (6, 12), S = (1, 0)C = 17.53 C = 4.25 C = 4.18

Moreover, we can see in tables 4 and 5 that the vector of parameters K in theGKS depends on λ and not on n. The kanbans are used to provide the productioncapacity necessary to satisfy demands.

We also note that S1 plus S2 in the GKS is less than K1 plus K2 in the KS.We can verify that, as λ increases, the values of the optimal configuration para-

meters increase.The difference between the cost of the optimal configurations of GKS and KS

increases significantly when n increases. More precisely, in table 4 we see that thisdifference varies from 10.5% (Prupt) to 39.4% (P (Q > 10)) for h = 1 and λ = 0.5.

5.2.2. Design for h = 10We shall now study the influence of h with the design of GKS, KS and BSS.

First, let us give the optimal configurations and the associated costs for the previousproduction system with h = 10 and λ = 0.5 (table 6) then 0.8 (table 7).

When h is equal to 10, the cost of stocking finished parts is ten times higherthan for the other parts. This implies that it is cheaper to stock products in the outputbuffer of the first stage than in the output buffer of the second stage. The policies thus




Prupt 6 0.02 K = (20, 27) S = (25, 26) K = (64, 54), S = (25, 26)C = 215.6449 C = 212.9951 C = 212.9946

P (Q > 1) 6 0.02 K = (20, 25) S = (25, 24) K = (64, 54), S = (25, 24)C = 195.9895 C = 193.4362 C = 193.4358

P (Q > 5) 6 0.02 K = (22, 21) S = (25, 20) K = (64, 54), S = (25, 20)C = 160.0293 C = 154.9999 C = 154.9994

P (Q > 10) 6 0.02 K = (21, 17) S = (25, 15) K = (30, 34), S = (18, 16)C = 120.8944 C = 109.4413 C = 109.3692

try to decrease S2. However, there must be enough finished parts in the second stageto ensure service quality.

Again, we can note that in most cases, parameters S for the GKS are equal toparameters S for the BSS and the costs are very close for these two policies.

Finally, note that for n = 10 and λ = 0.5, we obtain S2 = 0 for BSS and GKS.This means that the second stage works with a make-to-order policy.

5.2.3. Influence of the number of stagesLet us recall that the production system we have just studied (two stages in

series and two machines in tandem in each stage) corresponds physically to the firstproduction system we have considered in this part (one stage made up of four stationsin series). Indeed, the two systems contain four exponential stations in series. We cannote that for h equal to 10 and for the three policies, it is better to divide the systeminto two stages (see table 3 for the one stage case and tables 6 and 7 for the two-stagecase). This is because, with two stages, we can stock semi-finished parts, which areless expensive than finished parts.

For h equal to 1, the previous remark on the decomposition of the system stillapplies for a GKS, but not for a KS, for which it is better to have only one stage. Fora BSS, we obtain the same cost with one or two stages because we can see in table 5that in the case of two stages, S1 is always equal to 0: the system is equivalent to asingle-stage system.

5.3. Production systems made up of three stages, each containing one machine

We now consider a production system made up of three stages, each containingone machine. The service rate of each machine is 1. We consider the cost functioncQMP1 + QMP2 + QMP3 + QP1 + QP2 + QP3 , where c = 0 or c = 1. Let us set thearrival rate of demands λ to 0.5. Table 8 contains the optimal configurations and theassociated costs when c = 1 and table 9 gives the same information for c = 0. Wedenote by K the vector with the three components K1, K2, and K3, and by S thevector with the three components S1, S2, and S3.


Table 8Optimal configurations and associated costs for c = 1 and λ = 0.5.


Prupt 6 0.02 K = (1, 2, 9) S = (0, 0, 10) K = (1, 8, 12), S = (0, 0, 10)C = 11.02 C = 10.0 C = 9.51

P (Q > 1) 6 0.02 K = (1, 2, 7) S = (0, 0, 8) K = (1, 8, 12), S = (0, 0, 8)C = 9.05 C = 8.08 C = 7.56

P (Q > 5) 6 0.02 K = (1, 2, 3) S = (0, 0, 4) K = (1, 8, 12), S = (0, 0, 4)C = 5.27 C = 4.60 C = 4.08

P (Q > 10) 6 0.02 K = (1, 2, 1) S = (0, 0, 0) K = (1, 5, 7), S = (0, 0, 0)C = 3.54 C = 3.00 C = 2.40



Prupt 6 0.02 K = (2, 1, 9) S = (0, 0, 10) K = (1, 8, 12), S = (0, 0, 10)C = 10.4 C = 9.03 C = 9.01

P(Q>1)60.02 K = (2, 1, 7) S = (0, 0, 8) K = (1, 8, 12), S = (0, 0, 8)C = 8.43 C = 7.08 C = 7.06

P (Q > 5) 6 0.02 K = (1, 2, 3) S = (0, 0, 4) K = (1, 8, 12), S = (0, 0, 4)C = 4.77 C = 3.60 C = 3.58

P (Q > 10) 6 0.02 K = (1, 2, 1) S = (0, 0, 0) K = (1, 5, 7), S = (0, 0, 0)C = 3.04 C = 2.00 C = 1.90

Based on the results given in table 8, we can make similar remarks to those of theprevious cases (with one or two stages). In particular, GKS becomes more interestingwhen we authorize a large number of waiting demands (n high). Since the cost ofthe finished parts of each stage (QP1 , QP2 , QP3) is the same in the considered costfunction, we observe that for the BSS and the GKS, S1 = S2 = 0. That means thatno semi-finished parts are stored in the first two stages.

Note that table 9 is the first time we consider that the cost of the parts presentin the first machine is equal to zero. Nevertheless, we can again observe the samekind of behavior as before. In particular, the difference between the costs of GKSand KS is as great as before. However, the BSS costs are very close to GKS costs.Indeed, a drawback of the BSS is the non-limitation of the WIP, particularly on thefirst machine. Also, in the cost function used for table 9, there is no cost on the WIPof the first machine.

5.4. Production systems made up of four stages, each containing one machine

We now consider a production system made up of four stages, each containingone machine. The service rate of each machine is 1. We consider the cost function




Prupt 6 0.02 K = (1, 2, 2, 9) S = (0, 0, 0, 12) K = (1, 6, 10, 14), S = (0, 0, 0, 12)C = 12.7 C = 12.0 C = 11.46

P (Q > 1) 6 0.02 K = (1, 2, 2, 7) S = (0, 0, 0, 10) K = (1, 6, 10, 14), S = (0, 0, 0, 10)C = 10.7 C = 10.1 C = 9.51

P (Q > 5) 6 0.02 K = (1, 1, 4, 2) S = (0, 0, 0, 6) K = (1, 6, 10, 14), S = (0, 0, 0, 6)C = 6.61 C = 6.48 C = 5.93

P (Q > 10) 6 0.02 K = (1, 1, 3, 1) S = (0, 0, 0, 1) K = (1, 6, 10, 14), S = (0, 0, 0, 1)C = 4.84 C = 4.06 C = 3.52

QMP1 +QMP2 +QMP3 +QMP4 +QP1 +QP2 +QP3 +QP4 . The arrival rate of demands λis equal to 0.5. Table 10 contains the optimal configurations and the associated costs.We denote by K the vector with the four components K1, K2, K3 and K4, and by Sthe vector with the four components S1, S2, S3 and S4.

Again GKS becomes very interesting when we authorize a large number of wait-ing demands (n high). We observe that for the BSS and the GKS, S1 = S2 = S3 = 0.That means that no semi-finished parts are stored in the first two stages. This is be-cause the cost of the finished parts of each stage is the same in the considered costfunction.

The production system we have just studied (four stages in series, each containingone machine) corresponds physically to the production system we have considered insubsections 5.1 and 5.2. For the BSS, we obtain the same cost with one, two or fourstages. Indeed we can see in tables 4 and 10 Si = 0, i = 1,N − 1, and therefore,in the two cases, the system is equivalent to a single-stage system. Let us recall thatis because the cost of the finished parts of each stage is the same in the consideredcost function. We have already observed (see subsection 5.2) that this remark doesnot hold if h is greater than 1 (the cost of the finished parts in the last stage is greaterthan the cost of the finished parts in the previous stages). But for a GKS, even in thecase of identical costs of the finished parts in each stage, we observe that it is betterto divide the system into four stages than into two stages (we have already observedthat the cost was lower in the two-stage case than in the one-stage case).

6. Conclusion

First, as KS and BSS are particular cases of GKS, the GKS can always give costswhich are not greater than the other two. However, the generalized kanban policy ismore difficult to implement than kanban and base stock policies.

On the examples studied, we noted that, if there is no delay in filling orders, thecosts of the three policies are similar. It thus seems more natural to use a KS, whichis easier to implement and limits the maximum amount of WIP.


If there is a delay in filling orders, GKS and BSS give close optimal costs, whichare lower than KS costs for the same service quality. This difference increases whenstorage cost of finished parts increases or when the arrival rate of demands decreases.We shall thus choose the BSS if it is not important to limit the maximum level of WIP,and otherwise choose the GKS.

Finally, we studied the influence of the cost of products in the various bufferson the optimal values of the design parameters. We also observed that for a storagecost of finished parts greater than the storage cost of the other parts, it is better todecompose the system into several stages for the three policies.

In future research, it would be interesting to see if the remarks we have made onour examples continue to apply for more complex production systems (for example,production systems with assembly). Another interesting work would consist in thecomparison with other policies such as, for example, the extended kanban [10] whichis a combination of kanban and base stock policies.

Acknowledgements

We are grateful to the anonymous referees and to the associate editor for theirhelpful and constructive comments.

Appendix A: Calculation of the probability P (Q = n) for a kanban system or ageneralized kanban system [18]

Firstly, we shall see how to evaluate the probability P (Q = n) in a KS and thenwe shall examine the GKS.

Kanban systemsWhen we study a KS or a GKS, we use an approximation analytical method

(see subsection 3.1 for references). For the studied configuration, when the analyticalmethod has converged, we obtain the performance measures of the system. Moreover,on convergence of the method used, we obtain in particular the arrival rates of cus-tomers (kanbans from Stage N ) in the last synchronization station. These arrival rates,denoted by λ(nP ), depend on the number nP of customers in the station. In figure 6,we can see the station to study for the calculation of P (Q = n) in a KS: the N thsynchronization station.

We shall now calculate the probability P (nD = n), where n > 0, which repre-sents the probability of having n waiting demands (i.e., P (Q = n)).

The Markov chain corresponding to the synchronization station of figure 6 is abirth–death process, with state vector (nP ,nD).

Its stationary distribution has the following expression of P (0, 0):

P (0, 0) =1

11− λ

λ(0)

+∑K

x=1

( 1λx∏x−1i=0 λ(i)

)


Figure 6. Last synchronization station of a KS.

Figure 7. Last synchronization station of a GKS.

and the probability of having n waiting demands (n > 0) is given by:

P (Q = n) = P (0,n) =

(λ

λ(0)

)nP (0, 0).

Generalized kanban systemsJust as for the KS, when the analytical method has converged, we know the

load-dependent arrival rates of kanbans in queue BN (see figure 7) denoted by λ(nB).In figure 7, we have represented the last station of a GKS. This station is made up oftwo synchronization stations containing two queues: PN and AN+1 on the one hand,DN+1 and BN on the other. The entities of queue PN represent the finished partsof the system. The customers’ demands are placed in queue AN+1. The entities ofqueue DN+1 represent the processing orders for products of Stage N . Finally, BNcontains the kanbans of Stage N which represent authorizations to transfer demandsin Stage N − 1. To simplify the presentation, K denotes the number of kanbansassociated with the last stage, and S denotes the maximum number of finished partsin the last stage.

To obtain the performance measures needed for the design, we must evaluateP (nA = n), which corresponds to the probability of having n waiting demands.Therefore, let us construct the Markov chain modeling the behavior of the station rep-resented in figure 7. The chosen state vector is (nB,nD,nA,nP ). For the constructionof the Markov chain, two cases must be distinguished according to the values of K


Figure 8. Markov chain associated with the station in figure 7 for K > S.

and S:

First case: K > S. The Markov chain is represented in figure 8.To calculate P (nA = n), we must distinguish two cases according to the value

of n:

– If 0 < n < K − S (quadrant at the top and on the right)

P (nA = n) =

∏K−S−n−1i=0 λ(i)λK−S−n

P (0, 0,K − S, 0);

– If n > K − S (quadrant at the top and on the left)

P (nA = n) =

(λ

λ(0)

)n−K+S

P (0, 0,K − S, 0).

To evaluate these probabilities, we must know P (0, 0,K − S, 0). Writing allthe states of the Markov chain as functions of this probability and using the property


Figure 9. Markov chain associated with the station in figure A2 for K 6 S.

that the sum of the probabilities of all the possible states must be equal to 1, wededuce:

P (0, 0,K − S, 0) =1

11− λ

λ(0)

+∑K

x=1

(1λx∏x−1i=0 λ(i)

) .Second case: K 6 S. We obtain the Markov chain of figure 9.

Using the same steps as for the first case, we obtain:

P (nA = n) =

(λ

λ(0)

)nP (0,S −K, 0, 0) for n > 0,

where

P (0,S −K, 0, 0) =

(λλ(0)

)S−K1

1− λλ(0)

+∑K

x=1

(1λx∏x−1i=0 λ(i)

) .


References

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Comparison among three pull control policies: kanban, base stock, and generalized kanban

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