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www.palgrave-journals.com/jors/

Multiple lot-sizing decisions in a two-stage

production with an interrupted geometric

yield and non-rigid demand

M-C Wu∗, L-C Huang, H-M Hsu and T-S Su

National Chiao Tung University, Hsin-Chu, Taiwan

In a production system with random yield, it may be more cost effective to release lots multiple times towards fulfilling a customer order. Such a decision, called the multiple lot-sizing problem, has been investigated in various contexts. This paper proposes an efficient algorithm for solving a new multiple lot-sizing problem defined in the context of a two-stage production system with non-rigid demand when its process yields are governed by interrupted geometric distributions. We formulate this problem as a dynamic program (DP) and develop lemmas to solve it. However, solving such a DP may be computationally extensive, particularly for large-scale cases with a high yield. Therefore, this study proposes an efficient algorithm for resolving computational issues. This algorithm is designed to reduce the DP network into a much simpler algorithm by combining a group of DP branches into a single one. Extensive experiments were carried out. Results indicate that the proposed reduction algorithm is quite helpful for practitioners dealing with large-scale cases characterized by high-yield.

Journal of the Operational Research Society (2011) 62, 1075 – 1084. doi:10.1057/jors.2010.39

Published online 19 May 2010

Keywords: lot-sizing; interrupted geometric distribution; dynamic programming; two-stage system; production/inventory system

1. Introduction

Multiple lot-sizing problems arise when considering varia-tions in process yield. In a production system with a random yield, it may be more cost effective to release lots multiple times fulfill customer orders. In this case, it is necessary to determine the optimal lot size for each lot release. This problem has been called a Multiple Lot-sizing Production to Order (MLPO) problem (Grosfeld-Nir and Gerchak, 1996).

This paper presents a new MLPO problem with the following features. Firstly, completion of a product requires two stages (or two manufacturing processes). The two stages are interconnected and cannot be performed at the same time. Secondly, the yield of each stage is random and governed by an interrupted geometric (IG) distribution. A manufacturing process with an IG distribution denotes that once a defec-tive item is produced by the process, the items produced afterwards are all defective. Thirdly, the delivery due dates for customer orders are strictly imposed; in other words, delivery after the due date is not acceptable, and supply shortages will be penalised. This feature is also called

non-rigid demand because a customer’s demand may not be

completely fulfilled by the due date. Fourthly, the products are ∗Correspondence: M-C Wu, Department of Industrial Engineering and Management, National Chiao Tung University, Hsin-Chu, Taiwan.

E-mail: [email protected]

customised and the salvage values of over-supplied items are negligible.

One example of a two-stage MLPO problem is a capital-intensive machine tool (eg, a steel rolling mill) equipped with two shape-forming dies. Each die is used to perform, can only perform a particular shaping process. The manufacturing of a product involves two shaping operations through the succes-sive use of both dies. Yet, at any instance, only one die can be equipped on the machine. Therefore, the two shaping opera-tions cannot be simultaneously carried out—fulfilling the first requirement above. Secondly, the process yields of both dies are governed by IG distributions, since once a die becomes worn out, all the parts it produces afterwards are defective. Thirdly, the product is highly customised and has a short life cycle. Therefore, customers are highly concerned with the due date and any delivery after the due date is not acceptable.

At each period, the two-stage MLPO problem involves two decisions: (1) which die or manufacturing process to select, and (2) how large to make the lot-size for the selected die. These two decisions are made at each period until the customer order is completely fulfilled or the decision point reaches the due date. In the event that the customer order is only partially fulfilled by the due date, the manufacturer will incur a penalty cost for shortage.

Numerous studies on the MLPO problem with IG distri-butions have been published. Most of these studies assume

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a rigid demand scenario in which a customer order must be completely fulfilled. A few studies examine non-rigid demand scenarios, but limit their scope to a single-stage production system (Guu and Zhang, 2003). To date, the two-stage inter-locked MLPO problem described above has not been investi-gated in the literature.

We formulated the MLPO problem as a dynamic program (DP) and developed lemmas to solve it. For a small-scale case, solving the DP efficiently obtains an optimum solu-tion. However, for large-scale cases with a high yield, solving this DP becomes computationally prohibitive. To address this problem, this study proposes an efficient method of solving the MLPO problem. Our idea is to reduce the complex DP network into a much simpler algorithm by combining a group of DP branches into a single one. Extensive experiments were carried out. Results indicate that the proposed method is quite helpful for practitioners dealing with large-scale cases char-acterized by a high yield.

The remainder of this paper is organised as follows. Section 2 reviews previous multiple-stage MLPO studies. Section 3 introduces how the MLPO problem is formulated as a DP model and develops lemmas to determine the upper bounds of lot-sizes for the DP. Section 4 describes the DP reduction method. Section 5 presents numerical experiments, followed by concluding remarks in Section 6.

2. Research background

Two comprehensive survey papers (Yano and Lee, 1995; Grosfeld-Nir and Gerchak, 2004) on MLPO problems have been published. Prior studies on MLPO problems can be categorised according to four perspectives.

The first perspective is associated with the number of

production/inspection stages, and includes two options: single-stage and multiple-stage. Single-stage studies assume

there is only one inspection station at the end of a production line, and that inspection results are utilised to make lot-sizing decisions for the following period. Alternatively, multiple-stage studies assume that the completion of a product must involve a sequence of manufacturing stages, and that each stage is equipped with its own inspection station. In this case, the inspection results of a particular stage are utilised to make subsequent lot-sizing decisions for upstream and downstream stages.

Examples of single-stage studies include Beja, 1977; Sepheri et al, 1986; Pentico, 1988; Grosfeld-Nir and Gerchak, 1990; Anily, 1995; Grosfeld-Nir and Gerchak, 1996; Zhang and Guu, 1997, 1998; Guu, 1999; Guu and Liou, 1999; Anily

et al, 2002; Guu and Zhang, 2003. Example of

multiple-stage studies include Lee and Yano, 1988; Wein, 1992; Grosfeld-Nir and Ronen, 1993; Pentico, 1994; Grosfeld-Nir, 1995, 2005; Grosfeld-Nir and Robinson, 1995; Barad and Braha, 1996, Barad, 1999; Grosfeld-Nir and Gerchak, 2002; Grosfeld-Nir et al, 2006; Ben-Zvi and Grosfeld-Nir, 2007. In practice, the steel engraving process is like a single-stage

system, while the steel drawing and coating processes are like a multiple-stage system.

The second perspective is associated with delivery requirement, which includes two options: rigid-demand and

non-rigid demand. In the rigid-demand scenarios, partial

fulfillment of an order is not acceptable (Wein, 1992; Grosfeld-Nir and Ronen, 1993; Pentico, 1994; Grosfeld-Nir, 1995, 2005; Nir and Robinson, 1995; Grosfeld-Nir and Gerchak, 2002; Grosfeld-Grosfeld-Nir et al, 2006; Ben-Zvi and Grosfeld-Nir, 2007). In the non-rigid demand scenarios, though customers will not accept any products after their due dates, partial fulfillment of an order is acceptable (Lee and Yano, 1988; Barad and Braha, 1996; Braha, 1999).

Practical examples of rigid-demand scenarios involve the order of making components for use in an assembly line. Partial fulfillment of such an order would result in partial fulfillment of its final assembly, and might idle some other types of components in the assembly. Alternatively, non-rigid demand scenarios emphasise customers’ eager concern in meeting due dates. For example, at the due date, customers have booked a scarce capacity to undergo a batch operation on the products they received. Therefore, products delivered after the due-date will not be accepted.

The third perspective is associated with the probability distributions of process yield. These probability distributions include the general discrete distribution (Grosfeld-Nir, 1995; Grosfeld-Nir and Robinson, 1995; Grosfeld-Nir and Gerchak, 2002; Grosfeld-Nir et al, 2006), the binomial distribution (Nir and Ronen, 1993; Pentico, 1994; Grosfeld-Nir, 1995, 2005; Grosfeld-Nir and Robinson, 1995; Barad and Braha, 1996; Braha, 1999; Grosfeld-Nir and Gerchak, 2002; Grosfeld-Nir et al, 2006; Ben-Zvi and Grosfeld-Nir, 2007), the IG distribution (Grosfeld-Nir and Robinson, 1995; Grosfeld-Nir and Gerchak, 2002; Grosfeld-Nir et al, 2006; Ben-Zvi and Grosfeld-Nir, 2007), the all-or-nothing distribu-tion (Grosfeld-Nir and Robinson, 1995; Grosfeld-Nir, 1995; Grosfeld-Nir and Gerchak, 2002; Grosfeld-Nir et al, 2006; Ben-Zvi and Grosfeld-Nir, 2007), and the stochastically proportional distribution (Lee and Yano, 1988; Wein, 1992; Grosfeld-Nir, 1995).

The following section describes the various applications of the aforementioned probability distributions. An IG distribu-tion refers to a process that uses a tool or die to manufacture part one at a time. However, the tool/die gradually wears out over time. As soon as the tool/die becomes severely worn, none of the parts produced afterwards will not the specifica-tions. A general discrete distribution also refers to a process which produces parts one by one, but the quality of each part is

independent due to some uncontrollable factors. This leads to

various discrete probabilistic distributions in yield-modelling, one of which is the Binomial distribution. The all-or-nothing distribution refers to a batch process, in which all parts in a lot are processed together; therefore, all the parts in a batch either pass or fail. A stochastically proportional distribution is utilised to model continuous production process.

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The fourth perspective is associated with the solution approach. Most prior formulations of the MLPO problems include recursive formulas that have been widely interpreted as DP problems. Therefore, DP has been widely used to solve the MLPO problem. However, this approach can be very demanding computationally. Some researchers have proposed the use of lemmas to reduce the solution space (Beja, 1977; Anily, 1995; Zhang and Guu, 1998), while others attempt to develop near optimal heuristic rules (Sepheri et al, 1986; Pentico, 1988). A few authors have roughly modelled the DP problem using a simpler non-DP problem for cases with extremely large or small demand quantities (Anily et al, 2002).

3. DP model

This section introduces a way to model the two-stage MLPO problem as a DP. We first assume that the production lead times of both processes being equal (ie, one period). Then, we describe the MLPO problem, analyse the outcomes of a lot-sizing decision, and model its associated cost function using a recursive formula, which reveals that the MLPO problem is a DP. Finally, we discuss how this DP can be used to model scenarios in which the production lead-times of the two processes are different.

Notation

T: total number of periods in the time horizon.

t: period index for t=T, . . . , 2, 1, 0, where T denotes the starting period and t= 0 denotes the due date.

i: index of production stage Mi, i = 1, 2. (i): setup cost required for releasing a lot to stage M

i, i = 1, 2.

(i): variable production cost per unit at stage M

i, i =1, 2. D: initial demand (number of demand units at t= T).

Dt: remaining demand at period t (number of demand units not fulfilled at t).

h: inventory holding cost of finished goods ($/unit-period).

m: shortage cost per unit.

(i): yield parameter of the IG distribution at stage M i, i= 1, 2.

kt(i): size of the lot released to Mi at period t, i = 1, 2 (note that kt(1)·kt(2)=0 due to the interlock constraints imposed on the two processes.)

Wt(i): binary variable indicating whether or not a setup is required at stage Mi Wt(i)=  0 if k(i)t = 0 1 if k(i)t > 0 . Yk(i)

t : random variable denoting the number of good units

produced from lot kt(i), yk(i)t denotes the possible

outcome of Yk(i) t , ykt(i)= 0, 1, . . . , k (i) t , i = 1, 2. M1 M2 R

Figure 1 The two-stage production system.

p(yk(i)

t ): probability of producing exactly ykt(i)

good units from lot kt(i), where k(i)t  y k(ti)=0 p(yk(i) t ) = 1, i = 1, 2.

Bt: inventory level of work-in-process (WIP) buffer R at period t, where

Bt= Bt+1− kt+1(2) + Yk(1)t+1

st= (Dt, Bt): production system status at period t, also called state st.

Nt(st) = (k(1)t , k(2)t ): decision made at state st; that is, kt(1) units are released to M1and kt(2)are released to M2. Note that k(1)t · k(2)t = 0 due to the interlock constraints imposed on the two processes.

Ct(Nt(st)): total expected cost incurred from period t to period 0, for a particular decision Nt(st).

Ct(st) = Mi n

(k(1) t , k(2)t )

{Ct(Nt(st))} minimum total expected cost incurred from period t to period 0, for all possible decisions at state st.

Nt(st) = (k(1)

t , k(2)

t ), an optimal decision made at state st; that is, Ct(Nt(st)) = Mi n (k(1) t ,kt(2)) {Ct(Nt(st))} = Ct(st) 3.1. MLPO problem

The production system of the MLPO problem involves two stages/processes, as Figure 1 shows: M1and M2. Both of these processes are equipped with a work-in-process (WIP) buffer

R, and inspection is performed at the end of each process.

Units determined to be defective are scrapped. The produc-tion lead-time (including inspecproduc-tion) for each process is one period.

To meet a non-rigid demand quantity D, lots can be peri-odically released to M1and proceed to M2. At the end of any period, when the inspection results are revealed, we must first select the process (either M1or M2) to proceed with, and then determine the lot-size for that selected process. These two decisions can be aggregated into a single decision determining

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be maintained due to the interlock constraints imposed on the two processes.

Notice that the lot size for M2 must be less than the currently available WIP in R. Lot-sizing decisions must be made until the due date arrives or the customer order has been completely fulfilled. Finally, the MLPO problem only considers the inventory holding cost of finished goods.

In the MLPO problem, the process yield is an IG distribution. Such a probability distribution is defined as follows: P(Yk(i) t = ykt(i)) = (1 −(i)) · ((i))y k(ti) y kt(i)= 0, 1, 2, . . . , k (i) t − 1 ((i))kt(i) y kt(i)= k (i) t

This equation indicates that each process operates in two possible states: in-control and out-of-control. While the process is in-control, all products produced are good; while out-of-control, all products produced are defec-tive. To produce exactly yk(i)

t good units, the system must

be in-control for the first yk(i)

t units, until it reaches the

out-of-control state at the (yk(i) t + 1)

t h unit. A mold-based shaping process is a good example of process yield characterised by the IG distribution. Once the mold is worn out, the output produced afterwards will be defective.

3.2. Outcomes of a lot-sizing decision

As stated above, the condition kt(1)·kt(2)=0 must be maintained due to the interlock constraints imposed on the two processes. Figure 2 illustrates the possible outcomes at t-1 of a lotsizing decision(k(1)t , k(2)t ) made at t. At t, the state of the production system (called state hereafter) is st= (Dt, Bt), which implies that the remaining demand is Dt and the WIP level is Bt. Now either kt(1)or kt(2)units are released to M1and M2; their respective outcomes are yk(1)

t and ykt(2). With this input/output

information, we can derive that st−1= (Dt− yk(2)t , Bt− k

(2) t + yk(1)

t ).

The occurrence probabilities for yk(1)

t and ykt(2) are

p(yk(1)

t ) and p(ykt(2)), respectively. With 0 yk(1)t k

(1) t and 0 yk(2)

t k

(2)

t , we an easily infer that the number of possible outcomes for a lot-sizing decision (kt(1), kt(2)) is (kt(1)+ 1) · (k(2)t + 1) where k(1)t · kt(2)= 0. That is, the occur-rence of each outcome has a probability of p(yk(1)

t ) · p(yk(2)t ), andk (1) t y k(t1)=0 k(2)t y k(t2)=0 p(yk(1) t ) · p(yk(2)t ) = 1.

3.3. Cost function and DP

Suppose a lot-sizing decision Nt(st) = (kt(1), kt(2)) is made at st= (Dt, Bt). In this case, the total expected cost incurred

from period t to period 0 (the due date) can be formulated as follows. Ct(Nt(st)) = H1+ H2+ kt(1)  y k(t1)=0 kt(2)  y k(t2)=0 p(yk(1) t ) · p(ykt(2)) · H3+ k(2)t  y k(t2)=0 p(yk(2) t ) · H4 (1) where kt(1)· kt(2)= 0 H1=(1)Wt(1)+(1)kt(1) H2=(2)Wt(2)+(2)kt(2) H3= Ct−1(st−1), where st−1= (Dt− yk(2) t , Bt− k (2) t + yk(1)t ) H4= h · (t − 1) · ykt(2)

In Equation (1), the terms H1 and H2 respectively denote the production cost incurred at st for M1 and M2. The third termk (1) t y k(t1)=0 k(2)t y k(t2)=0 p(yk(1)

t ) · p(yk(2)t ) · H3denotes the

minimum total expected cost for all possible states of st−1. The fourth termk

(2) t

y

k(t2)=0

p(yk(2)

t ) · H4 denotes the expected

inventory holding cost for all possible outcomes of k(2)t . Equation (1) is a recursive formula due to the inclusion of H3= Ct−1(st−1). To make an optimal lot-sizing decision at state st, we must know the optimal decision at st−1. This recursive feature indicates that the MLPO problem is a DP.

The DP has two boundary conditions (BCs). The first BC is defined at st=(0, Bt). When the order is completely fulfilled, no more lots are must be released and the corresponding cost is zero. That is,

Nt(st= (0, Bt)) = (0, 0) (2) Ct(st= (0, Bt)) = 0 (3) The second BC is defined at s0= (D0, B0). At the due date, any further production is useless, and only the shortage cost is incurred. That is,

C0(s0= (D0, B0)) = m D0 (4) 3.4. Upper bounds for the DP

To solve the DP, we developed three lemmas that give upper bounds for k(1)tand k(2)

t . Lemma 1 states that kt(2) min(Dt, Bt). Lemmas 2 and 3 as a whole state that kt(1) min{(ln(1)−ln m

ln(1)  − 1), ((t − 1)Dt − Bt)}. The Appendix provides the proofs of these three lemmas as well as one supporting proposition.

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− − = − − = − − + = = − = − − + ⋅ =

Figure 2 A decision structure in the DP model.

3.5. Embellishment of the DP model

The DP model can be embellished to model a more complex MLPO problem, in which the production lead-times of the two stages are different. Denote the production lead-time of the first stage by p1and that of the second stage by p2. In such a scenario, not all periods may be used to release a lot because the machine is not always free at the end of each period. This embellished DP model assumes that a lot-sizing decision can be made at each period but some additional constraints must be imposed on k(1)t and kt(2), as Equations (5)–(8) indi-cate Moreover, H4 in Equation (1) should be modified as in Equation (9). If kt(1)> 0, then k(1)t= 0 and kt(2)= 0 for = 1, . . . , p1− 1 (5) If kt(2)> 0, then k(1)t−= 0 and kt−(2)= 0 for = 1, . . . , p2− 1 (6) k(1)t = 0, for period t = p1− 1, . . . , 1 (7) kt(2)= 0, for period t = p2− 1, . . . , 1 (8) H4= h · (t − p2) · yk(2)t (9) Equations (5) and (6) ensure that whenever a manufacturing stage is triggered, no more lots can be released until the machine is free again. Equations (7) and (8) show that we are not allowed to any lot if the remaining time to the due date is less than the production lead-time. Equation (9) highlights the reduction of inventory-holding cost while the production lead-time of stage M2is increased.

Owing to inclusion of additional constraints on k(1)t and k(2)t , the embellished DP network is a subset of the original DP network. Therefore, the lemmas in the Appendix, which

define the upper bounds for kt(1)and k(2)t , are also valid in the embellished DP model.

4. Reduction of DP network

This study proposes a method for reducing the DP network into a much simpler one by consolidating the outcomes of a lot-sizing decision. The proposed method includes three major steps. First, we cluster all the possible outcomes of a lot-sizing decision into several groups. Second, we define a representative outcome for each group. Third, we simplify the DP network by including only the representative outcomes of each group.

The first step is to group outcomes. For a lot-sizing deci-sion (kt(1), kt(2)), the total number of possible outcomes is (k(1)t + 1) · (kt(2)+ 1). For the outcomes of k(1)t , there are kt(1)+ 1 options (ie, 0, 1, . . . , kt(1)). By consolidating every n consecutive outcome into a group, we obtainkt(1)+1

n  groups. For example, when n = 2 and kt(1)= 6, we would create four groups:{0, 1}, {2, 3}, {4, 5}, {6}. Likewise, for the outcomes of kt(2)we would create

k(2)t +1

n  groups.

The second step is to define each group’s representative

outcome. Define Qj = {y|n( j − 1) y nj − 1} as the jth group that comprises n consecutive outcomes of either kt(1) or kt(2), where y denotes the possible outcome and its prob-ability is denoted by p(y). The representative outcome of group Qjis Round(

n j−1

y=n( j−1)y· p(y)/n jy=n( j−1)−1 p(y)), and its corresponding probability isn j−1y=n( j−1)p(y). For example,

consider group j = 2 Q2 = {2, 3} with p(2) = 0.1 and p(3) = 0.2. Then, the representative outcome of group Q2is Round((2·0.1)+(3·0.2)(0.1+0.2) )= Round(2.7)=3, and its corresponding

probability is 0.1+0.2 = 0.3.

The third step is to simplify the DP network by modelling each group with its representative outcomes. After this simpli-fication method, kt(1)has only

k(1)t +1

n  outcomes and k (2) t has

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Table 1 Cpand Tp of the solutions obtained by the comprehensive algorithm

((1),(2)) (T, D)

(10 100) (10 200) (20 100)

Cp($) Tp(sec) Cp($) Tp (sec) Cp($) Tp(sec)

0.9 0.9 8070.76 220 18 070.80 1019 6228.48 1875 0.9 0.99 5058.79 169 15 026.30 707 2151.69 821 0.9 0.999 3578.58 158 13 304.30 710 1437.32 789 0.99 0.9 5016.30 2394 14 972.70 19 489 2629.97 23 309 0.99 0.99 1018.96 2412 2561.08 22 080 1017.97 23 057 0.99 0.999 689.91 2324 1363.89 20 789 689.90 22 979 0.999 0.9 3984.39 2754 13 742.00 39 414 2343.91 57 688 0.999 0.99 847.77 2738 1846.54 43 191 847.77 57 377 0.999 0.999 541.07 2703 1057.13 41 408 541.07 57 056

Table 2 RCand RT of the solutions obtained by the reduction algorithm for n= 5

((1),(2)) (T, D) (10 100) (10 200) (20 100) RC(%) RT(%) RC(%) RT(%) RC(%) RT(%) 0.9 0.9 0.28 3.29 0.13 2.05 0.72 2.36 0.9 0.99 0.05 5.19 0.02 4.48 0.33 4.70 0.9 0.999 0.07 4.91 0.00 4.36 0.29 4.45 0.99 0.9 0.58 3.68 0.06 3.34 1.18 3.16 0.99 0.99 0.31 3.75 0.14 3.50 0.17 3.32 0.99 0.999 0.06 3.64 0.03 3.38 0.06 3.15 0.999 0.9 0.33 3.67 0.64 3.24 4.30 2.94 0.999 0.99 0.15 3.75 0.21 3.42 0.15 3.03 0.999 0.999 0.01 3.57 0.03 3.31 0.01 3.03

Table 3 RC and RT of the solution obtained by the reduction algorithm for n= 10

((1),(2)) (T, D) (10 100) (10 200) (20 100) RC(%) RT(%) RC(%) RT(%) RC(%) RT(%) 0.9 0.9 0.27 2.18 0.12 1.25 0.56 1.39 0.9 0.99 0.14 4.45 0.04 2.85 1.70 3.22 0.9 0.999 0.19 3.35 0.03 2.77 2.61 2.92 0.99 0.9 0.92 2.15 0.08 1.92 11.51 1.76 0.99 0.99 1.01 2.23 0.22 2.01 1.11 1.76 0.99 0.999 0.09 2.13 0.07 1.91 0.09 1.73 0.999 0.9 6.10 2.14 0.12 1.77 6.71 1.59 0.999 0.99 0.77 2.23 0.16 1.91 0.77 1.61 0.999 0.999 0.03 2.10 0.17 1.82 0.03 1.60 only kt(2)+1

n  outcomes. The total number of outcomes for (kt(1), kt(2)) is 

k(1)t +1

n  ·  kt(2)+1

n , which greatly reduces the DP network.

Notice that while n = 1, the reduced DP becomes the original DP. To facilitate comparison, the method of solving the original DP is called the comprehensive algorithm,

while that of solving the reduced DP is called the reduction

algorithm.

5. Numerical experiments

This study proposes two methods for solving the MLPO problem—the comprehensive algorithm and the reduction

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Table 4 RC and RT of the solution obtained by the reduction algorithm for n= 15 ((1),(2)) (T, D) (10 100) (10 200) (20 100) RC(%) RT(%) RC(%) RT(%) RC(%) RT(%) 0.9 0.9 0.59 1.49 0.26 0.75 1.06 0.99 0.9 0.99 0.29 2.59 0.07 1.91 3.83 2.23 0.9 0.999 0.54 2.48 0.03 1.89 18.44 2.10 0.99 0.9 0.87 1.65 0.22 1.44 14.67 1.23 0.99 0.99 1.28 1.66 0.58 1.48 1.27 1.27 0.99 0.999 0.85 1.61 0.26 1.41 0.85 1.25 0.999 0.9 0.87 1.61 10.65 1.28 12.36 1.05 0.999 0.99 3.10 1.68 0.92 1.36 3.10 1.12 0.999 0.999 0.05 1.61 0.24 1.32 0.05 1.12

Table 5 RC and RT obtained by the reduction algorithm for n= kt+ 1

((1),(2)) (T, D) (10 100) (10 200) (20 100) RC(%) RT(%) RC(%) RT(%) RC(%) RT(%) 0.9 0.9 1.31 0.18 0.58 0.04 2.96 0.21 0.9 0.99 4.67 0.73 1.73 0.24 30.71 0.26 0.9 0.999 3.88 0.62 1.61 0.42 53.22 0.26 0.99 0.9 3.30 0.38 1.00 0.12 18.22 0.18 0.99 0.99 103.94 0.35 32.27 0.19 104.14 0.20 0.99 0.999 54.80 0.33 90.92 0.17 54.80 0.19 0.999 0.9 2.36 0.56 0.76 0.18 20.47 0.15 0.999 0.99 87.78 0.59 60.76 0.28 87.78 0.18 0.999 0.999 25.02 0.56 91.44 0.26 25.02 0.17

algorithm. Extensive experiments were carried out to justify both algorithms using personal computers equipped with a 3.4 G CPU and 504MB of RAM.

We measured the effectiveness and efficiency of the reduc-tion algorithm using two performance indices: RC=

Cr−Cp

Cp

and RT = TTr

p, where Cr is the total expected cost of the

lot-sizing decision suggested by the reduction algorithm and Tr is its computation time; Cp and Tp are defined accordingly for the comprehensive algorithm.

Experimental results suggest that the reduction algorithm should be used to solve a large-scale problem (ie, large TD) with a high yield in the first stage (eg, (1)0.99). For other scenarios, we suggest the use of the comprehensive algorithm.

According to Table 1, when (1)0.99 and (T, D) ∈ {(10, 200), (20, 100)}, solving the MLPO problem using the comprehensive algorithm is computationally extensive taking about 5.4–16.0 hours. This is essentially due to that

k(1)t min{(

ln(1)−ln m

ln(1)  − 1), ((t − 1)Dt − Bt)}, which indicates that higher values of (1), D, and T increase the upper bound of k(1)t ∗, which in turn increases computational complexity.

Table 6 Average computation time required by the comprehensive algorithm for solving some low-yield MLPO

problems (unit: sec)

T D

10 50 100 200

10 0.13 4.17 13.13 21.20

15 0.34 12.74 43.03 167.95

20 0.33 26.32 123.21 496.78

In contrast, such complex MLPO problems can be effi-ciently and satisfactorily solved using the reduction algorithm. As shown in Table 2, n = 5 RTranges from 2.94% to 3.50%, while Rcranges only from 0.06% to 4.30%. For the scenario with (T D) = (20,100) and ((1),(2)) = (0.999, 0.999), the computation time is greatly reduced from 16 hours to 30 min. Moreover, as Tables 3–5 indicate, increasing the value of n tends to decrease RT at the cost of increasing Rc.

Alternatively, the comprehensive algorithm is better-suited to scenarios with a relatively low yield in the first process (ie,(1)0.9). Table 6 shows the average computation times for solving some MLPO problems using the comprehensive

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algorithm. Each (T, D) scenario in this table includes 16 instances, which are generated by setting (1),(2){0.6, 0.8}, (1) = (2) ∈ {1, 2}, m ∈ {100, 200}, and

(1)=(2)= 50.

In practice, a manufacturing processes with a yield higher than 0.99 (ie,(1)0.99) is not unusual. In shaping processes, the yield of a die is generally higher than 0.99. Otherwise the die would be frequently changed, increasing processing/production costs. Therefore, the proposed reduc-tion algorithm can help practireduc-tioners quickly and effectively solve the MLPO problem.

6. Conclusions

This paper examines a novel two-stage MLPO problem with an IG yield distribution and non-rigid demand. The processing times for the two stages may be different. Once a stage completes the manufacturing of a lot, we need to decide which of the two stages to proceed further and make the lot-size decision.

We first formulate this MLPO problem as a DP. Then by developing lemmas forr the upper bounds of lot-sizes, we propose an exact approach (ie, the comprehensive algorithm) to solve the DP. However, solving the DP using an exact approach may be computationally extensive, and particularly for large-scale, high-yield scenarios. To resolve this computa-tional difficulty, we further develop an efficient algorithm—an

approximation approach—that reduces the DP network into

a much simpler algorithm by consolidating a group of DP branches into a single one.

Numerical experiments indicate that the approximation

approach (ie, the reduction algorithm) can effectively and

efficiently solve large-scale, high-yield problems. For a prac-tical MLPO problem requiring 16 hours of computation to solve, the reduction algorithm can obtain a satisfac-tory solution in only 30 min. Yet, for small-scale cases, we suggest solving the DP using the comprehensive

algo-rithm because it requires only a few minutes or less of

computation.

A possible extension of this research is to apply the reduc-tion method to solve other MLPO problems. For example, many other MLPO problems in the literature are modelled as a DP, and various heuristic methods have been developed for dealing with large-scale cases. Comparing the performance of these heuristic methods and the reduction method would be an interesting topic for further research.

References

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Appendix

Proposition 1 Ct−1(st−1 = (Dt, Bt − n))Ct−1(st−1 = (Dt, Bt)) for Btn 1 and n is an integer.

Proof Consider a case in which we intentionally hold n WIP units in station R until the due date. In our cost model, the n WIP units will be scrapped after the due date, and therefore add no more cost. By intentionally scrapping the n WIP units, the lot-sizing decision for st−1= (Dt, Bt) in this case becomes the lot-sizing decision for st−1= (Dt, Bt− n). That is, Ct−1(st−1= (Dt, Bt − n)) can always be obtained from the lot-sizing decision at st−1= (Dt, Bt). This implies that Ct−1(st−1= (Dt, Bt− n))Ct∗−1(st−1= (Dt, Bt)) 

Lemma 1 Given t1, Dt1, st = (Dt, Bt), we have k(2)t Dt and k(2)

t  Bt.

Proof kt(2) Bt is trivial because Bt is the available WIP that can be released at stage 2.

To prove k(2)t Dt, consider two cases. Case 1 k(1)t> 0 We have k(2)t ∗= 0 Dt because k(1)t · k(2)t = 0 Case 2 k(1)t ∗= 0 Let Nt(st) = (0, Dt) and Nt(st) = (0, Dt+ n) Then Ct(Nt(st)) =(2)+(2)· (Dt) + h(t − 1) · E[YDt] + Dt  yDt=0 p(yDt) · Ct−1(st−1=(Dt−yDt,Bt−Dt)) and Ct(Nt(st)) =(2)+(2)· (Dt+ n) + h(t − 1) ·E[YDt+n] + Dt+n yDt +n=0 p(yDt+n) · Ct−1(st−1=(Dt−yDt+n, Bt−(Dt+n))) Ct(Nt(st)) − Ct(Nt(st)) = n ·(2)+ h · (t − 1) · (E[YDt+n] − E[YDt]) + A − B, where A= Dt−1 yDt +n=0 p(yDt+n) · Ct−1(st−1= (Dt− yDt+n, Bt− (Dt+ n))) B= Dt−1 yDt=0 p(yDt) · Ct−1(st−1= (Dt− yDt, Bt− (Dt)))

Based on Proposition 1, A B. Therefore,

Ct(Nt(st)) − Ct(Nt(st))n · (2)+ h

· (t−1)·(E[YDt+n]−E[YDt])>0

This implies that k(2)t Dt.  Lemma 2 k(1)t(t − 1) · Dt− Bt.

For any t, Dt Dt−1. Based on Lemma 1, we can infer t−1

=1k(2)

 (t − 1) · Dt. That is, for stage 2, the sum of lot-sizes from period t− 1 to 1 are at most (t − 1) · Dt. This implies that stage 2 at period t−1 at needs a WIP level of

(t − 1) · Dt at most. At period t, the WIP level at stage 2 is Bt. Therefore, the units to be produced from stage 1 at period t−1 are at most (t − 1) · Dt− Bt. Consider stage 2 as a customer, the customer’s demand at period t−1 is at most

(t − 1) · Dt− Bt. According to a prior study (Guu and Zhang, 2003), in a single-stage production system with an IG distri-bution, the optimal lot-size will not exceed the demand. This implies that the lot-size released from stage 1 at period t is at most(t − 1) · Dt− Bt. That is, the upper bound for k(1)

t is (t − 1) · Dt− Bt.

Lemma 3 If t1, Dt1, and st= (Dt, Bt) where Bt0 then kt(1)∗

ln(1)−ln m ln(1)  − 1. Proof Let Nt(st) = (k, 0) for k1

Ct(Nt(st)) =(1)+(1)· (k) + k  yk=0 p(yk) · Ct−1(st−1= (Dt, Bt+ yyk)) Let Nt(st) = (k + 1, 0) for k1 Ct(Nt(st)) =(1)+(1)· (k + 1) + k+1  yk+1=0 p(yk+1) · Ct−1(st−1= (Dt, Bt+ yk+1)) Ct(Nt(st)) − Ct(Nt(st)) =(1)− ((1))k+1· [Ct−1(st−1 = (Dt, Bt+ k)) − Ct−1(st−1= (Dt, Bt+ k + 1))] Consider a particular state(Dt, Bt+ k), which is changed by adding one more unit of WIP, becoming a new state(Dt, Bt+ k + 1). This newly added WIP unit could become a unit

of finished goods before the due date. In the best case, this decreases the shortage cost by one unit. In other words,

Ct−1(st−1= (Dt, Bt+ k)) − Ct−1(st−1 = (Dt, Bt+ k + 1))m

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Namely, if(1)((1))k+1· m we conclude Ct(Nt(st)) − Ct(Nt(st))0.

This implies that if lot-size k is in the range imposed by

(1)((1))k+1· m then we cannot achieve a better solution by increasing one more unit in lot-size.

Therefore,(1)= ((1))k+1· m defines the upper bound of lot-size k.

That is, k= ln

(1)−ln m

ln(1) − 1 is the upper bound of k (1)

t . Thus, we conclude that k(1)t ∗

ln(1)−ln m

ln(1)  − 1. 

Received December 2008; accepted February 2010 after two revisions

數據

Figure 1 The two-stage production system.
Figure 2 A decision structure in the DP model.
Table 3 R C and R T of the solution obtained by the reduction algorithm for n = 10
Table 4 R C and R T of the solution obtained by the reduction algorithm for n = 15 (  (1) ,  (2) ) (T, D) (10 100) (10 200) (20 100) R C (%) R T (%) R C (%) R T (%) R C (%) R T (%) 0.9 0.9 0.59 1.49 0.26 0.75 1.06 0.99 0.9 0.99 0.29 2.59 0.07 1.91 3.83 2

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