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On the relaxation of multi-level dynamic lot-sizing models
Int. J. Production Economics 77 (2002) 53–61
On the relaxation of multi-level dynamic lot-sizing models$ ! . os* . Jozsef Vor ! Rak ! oczi ! ut 80, 7622 Pecs, ! Hungary Faculty of Business and Economics, University of Pecs, Received 19 July 1999; accepted 17 August 2001
Abstract In this paper the multi level dynamic lot sizing problem is analyzed when setup and holding costs do not depend on time. A zero–one integer programming formulation and its linear relaxation are investigated. The paper shows that for certain classes of problems the relaxed models provide integer solution when the number of periods is below a certain threshold. A counter example shows that at least six periods are required to obtain a unique non-integer solution of the relaxed problem. r 2002 Elsevier Science B.V. All rights reserved. Keywords: Lot sizing; Integer programming; Dynamic programming
1. Introduction Flexibility is one of the oldest themes of production and operations management. Lot sizing, as one of the most important elements of flexibility, has been playing an important role in the literature of production and operations management, especially since reducing setup costs and setup times has become one of the most effective tools. Parallel with this process, the notion of mass production began to decline, and today JIT, lean production, flexible manufacturing are among the main competing manufacturing paradigms. These changes in the manufacturing paradigms have been reflected by the EOQ-connected literature as well. The pioneer works of Porteus [1–3] deal first $
Support from the National Research Fund (OTKA) under Grant no. T 037291 is acknowledged. The author also highly appreciates the valuable notes of referees. *Tel.: +36-72-211-524; fax: +36-72-501-553. E-mail address: [email protected] (J. Vo¨ro¨s).
time with the problem of setup cost reduction in the EOQ model, however less attention has been paid to the same problem in dynamic case, i.e. when demand is not constant. As demand can rarely be fully smoothed, knowing the required level of setup cost reduction in order to decrease the number of setups during certain time periods, is an important issue for managers. Thus studying the robustness of a given optimal solution, especially with respect to setup cost, and knowing the nature of the problem are useful. The setup cost stability region (SR) for an optimal solution is defined as the set of all setup cost values for which a given solution remains optimal. The sensitivity analysis of the single level dynamic lot sizing problem arises first time in the paper of Richter [4] and–following the idea of dynamic programming–requiring cost inputs to have the same regeneration set for every time period, he could reveal a subset of SR. Later . os . [5] have analyzed the multiRichter and Vor
0925-5273/02/$ - see front matter r 2002 Elsevier Science B.V. All rights reserved. PII: S 0 9 2 5 - 5 2 7 3 ( 0 1 ) 0 0 2 0 2 - X
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
stage lot sizing problem but they could show results only based on the same assumption extended to the multi level problem. They also expressed that they could reveal the whole SR by using a full enumeration which is obviously not an effective way. The next step in this field was made by Chand . os . [6] when they applied a new approach and Vor to the problem. For the single level case–with backlogging–they proved that the total cost function is convex in the number of setups. (We mention also that the convex nature of the holding cost function–without backlogging–is also derived in the paper of [7]. Similar results for the multi level dynamic lot-sizing case were published by . os . [8]. Vor The nature of the multi level dynamic lot sizing problem has attracted many researchers (see for example [9–12]) and the linear relaxation of the zero–one integer programming formulation of the dynamic lot sizing problem plays an important role in the analysis of the stability regions. When the zero–one formulation can be substituted by linear programming models the proof of some theorems required to identify stability regions becomes simple. When setup and holding costs can vary period by period, Pochet and Wolsey [13] defined a two level, four period problem for which the relaxed version has a unique non-integer solution. This paper shows that with time independent setup and holding costs the relaxed version of the integer problem with four and five periods has always integer solution. The paper presents a six period problem that has no optimal integer solution despite the fact that the setup and holding costs are constant through periods.
2. The multi stage serial assembly problem According to a general description of the serial assembly system [12] the production and inventory system is assumed to have M facilities in series, and the input to facility (m þ 1) comes from the production at facility m: Raw materials are available in unlimited amounts as input to facility 1. Facility M produces assemblies which are used to supply the customer demand. All facilities are
allowed to carry inventories while the Mth facility carries the finished goods inventory. It is assumed that production and shipments are instantaneous, and that one unit of production on facility (m þ 1) requires one unit of input from facility m: Let Dt denote the demand in period t; it is assumed that demand is known for periods 1 to T: Let Xmt denote the production at facility m in period t; and the cost of this production is Cmt ðXmt Þ: Imt is the inventory at the end of period t at facility m and the corresponding cost is Hmt ðImt Þ for all mA/1; MS and tA/1; TS; where /a; bS ¼ fa; a þ 1; y; bg: Then the multi stage assembly problem can be formulated as: MIN
T X M X
½Cmt ðXmt Þ þ Hmt ðImt Þ;
ð1aÞ
t¼1 m¼1
XMþ1;t ¼ Dt ;
ð1bÞ
tA/1; TS;
Im;t 1 þ Xmt ¼ Xmþ1;t þ Imt for all
mA/1; MS;
Im0 ¼ ImT ¼ 0; Imt ; Xmt X0;
tA/1; TS;
mA/1; MS;
mA/1; MS;
tA/1; TS:
ð1cÞ ð1dÞ ð1eÞ
According to Crowston and Wagner [10], problem (1) can be represented as a network flow problem with one source and T sinks (shown by Fig. 1). For concave cost functions C and H respectively, Zangwill [14] has stated that there exists an optimal solution corresponding to an extreme flow, i.e. a node can have at most one positive input (see also the proof of [9] for more general cost functions). This means that Im;t 1 Xmt ¼ 0 for all mA/1; MS and tA/1; TS in at least one optimal solution. Now, let us consider problem (1) with a cost function used widely in the literature: ( Sm þ cm Xmt for Xmt > 0; Cmt ðXmt Þ ¼ ð2Þ 0 for Xmt ¼ 0; and Hmt ðImt Þ ¼ hm Imt for all m and t with hm phmþ1 ;
mA/1; M 1S;
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
ΣDt (0) X12
X11 I11 (11)
X13 I12
(12)
(13)
X22
X21 I21 (21)
X23 I22
(22)
X31
(23)
X32 I31
X33 I32
(31)
(32)
(33)
D1
D2
D3
Fig. 1. The dynamic serial assembly system (M ¼ 3; T ¼ 3).
. os . [8] proposed a zero–one integer programVor ming formulation for the analysis of problem (1) with the cost functions specified in (2) and proved that picking up pivot elements only from the coefficient matrix, the transformation result in integer numbers. (However, as it can be concluded from this paper, the inclusion of certain unit vectors into the basis can lead to non integer solutions.) In this formulation let Xmjt be one if on facility m in period j the whole Dt is produced, otherwise zero. Similarly, let Ymt be one if there is any production on stage m at period t; otherwise zero. Then the next formulation also describes problem (1) with the cost functions defined by (2): " # M X T t X X MIN Cmjt Xmjt þ Sm Ymt ; ð3aÞ m¼1 t¼1 t X
ð3bÞ
tA/1; TS;
j¼1 k X j¼1
where Sm and hm denote the setup and periodic unit holding cost on stage m; respectively. For simplicity, we also assume that h1 > 0; h1 oh2 and Dt > 0 for all t: Variable production cost (cm ) is considered time independent, thus it can be omitted. We note that to our belief the cost structure defined by (2) does not reduce significantly the industrial relevance of the model. The time horizon of this model is usually one year and the length of the period can be considered one month. Although a declining tendency can be predicted, once a reduction in setup cost is achieved, the reduced setup cost level prevails over certain time. In case of using equal length of periods, time independent holding costs can be assumed. The increasing holding costs by stages come from the fact that cascading downward on stages, the value of product increases. Based on cost functions in (2), problem (1) has a nested optimal schedule [11]. (A schedule is nested if Xmt > 0 implies Xmþ1;t > 0.) So further on, it is sufficient to deal with nested schedules only.
XMjt ¼ 1;
j¼1
Xmjt p
k X
Xm 1;jt ;
j¼1
mA/2; MS;
tA/1; TS;
kA/1; tS
ð3cÞ
tA/1; TS;
jA/1; tS
ð3dÞ
Xmjt pYmj ; mA/1; MS;
Xmjt pAf0; 1g; Ymt Af0; 1g; tAh1; T i; mAh1; M i; jAh1; ti;
ð3eÞ
where Cmjt ¼ ðhm hm 1 ÞDt ðt jÞ; for all m; t; j; and h0 ¼ 0: The equivalence of (1) and (3) origins from the characteristics of the problem. As mentioned above, Zangwill [14] pointed out that there are optimal solutions of (1) with the feature that every node can have at most one positive input. From this it follows that if once a portion of Dt is produced in a certain period, the full quantity of Dt must be produced in that period. This property is secured by the definition of Xmjt : Assumption (3b) formulates that the required quantity is delivered in time. (3c) provides that production can be scheduled in stage m; if the job is completed at stage m 1: In formulation (1) the same idea is expressed by (1c). (3d) is for counting the setups.
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
programming problems, and the second is that the solution really gives a nice variety of possible solutions which one probably cannot imagine. Next, we show that under time independent cost structure S0 can be substituted by integer solution at no higher cost. We emphasize that in our model smt ¼ sm for all t; and the PW example does not apply this restriction. Additionally, in our model hmt ¼ hm holds for all t: Let us take the S0 solution by periods, i.e. let us follow the way of satisfying the demand of a period. Fig. 2. gives the graphical representation of the S0 solution (omitting the flow for D1 ). In Fig. 2, the numbers along vertical lines indicate the amount of total setup cost counted (the values of ymj variables). According to Fig. 2, only 50% of D2 is produced in the first period, despite the fact that the full setup cost is paid in period 1, and 50% of D2 is transported from node (21) to node (22). Similarly, only 50% of D3 is produced in period one at the first stage, and the other half is produced in period two. Second stage works on D3 in periods two and three in 50–50%, respectively. A solution will be called cross solution if at the periodical production plan there is a node that has two positive inputs and two positive outputs. According to Fig. 2, the periodical production plan for period 3 exhibits a cross solution.
3. Analyzing the nature of relaxation 3.1. The presentation of a non-integer solution in case of dynamic costs Pochet and Wolsey [13] have formulated a linear programming model which in its constraints is the same like we would have if we took the linear programming relaxation of problem (3), i.e. when assumptions in (3e) are substituted by simple nonnegativity assumptions. They considered a two level, four period (M ¼ 2; T ¼ 4) problem with the inputs: Dt ¼ 1 for all tA/1; 4S; h1t ¼ ð2; 1; 1; 0Þ; h2t ¼ ð4; 3; 2; 0Þ and s1t ¼ ð0; 4; 6; 2Þ; s2t ¼ ð0; 4; 4; 2Þ where hmt and smt are for the inventory and setup costs, respectively, on level m in period t: It is emphasized again that here their setup and holding costs are dynamic, while in our model they are constant in time. Let us identify this example as ‘‘PW example’’. They show that the solution (let us denote it by S0):: X211 ¼ X111 ¼ 1;
X212 ¼ X222 ¼ X223 ¼ X233 ¼
X234 ¼ X244 ¼ X112 ¼ X122 ¼ X113 ¼ X123 ¼ X124 ¼ X144 ¼ 0:5 ðand as consequence Y21 ¼ Y11 ¼ 1; Y22 ¼ Y23 ¼ Y24 ¼ Y12 ¼ Y14 ¼ 0:5; and the rest of the variables are 0Þ is the only optimal solution of the problem. This solution has two strong characteristics: the first is that it shows, there are multi-level dynamic lot sizing problems whose integer programming formulation cannot be substituted by linear
1 (11)
0.5 (12)
1
0.5
(21) 0.5 (22) D2
1 (11)
0.5 (12) (13) 0.5
(21)
(22)
(11)
0.5 (12)
(13)
0.5 (23)
0.5 (21)
D3 Fig. 2. The S0 solution.
(22)
(23)
0.5 (14) 0.5 (24) D4
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
while the THC of S1 and S2:
3.2. Analyzing the S0 solution under constant setup and holding costs
THCðS1Þ ¼ 0:5h1 D2 þ ð0:5h1 þ h2 ÞD3 þ h2 D4 ;
Property 1. A cross solution in the S0 solution with constant costs can be substituted by a non-cross solution at no higher cost.
and THCðS2Þ ¼ ð0:5h1 þ 0:5h2 ÞD2 þ ð0:5h1 þ 0:5h2 ÞD3 þ 0:5h2 D4 :
Proof. Let us define two new solutions, S1 and S2. Let S1 be: Y12 ¼ Y14 ¼ Y24 ¼ 0:5 and Y11 ¼ Y21 ¼ Y22 ¼ 1 (consequently the minimum holding cost is provided by the production plan: X112 ¼ X122 ¼ X113 ¼ X123 ¼ X124 ¼ X144 ¼ X244 ¼ 0:5), while S2: Y22 ¼ Y13 ¼ Y23 ¼ Y14 ¼ Y24 ¼ 0:5; Y11 ¼ Y21 ¼ 1 (consequently X112 ¼ X212 ¼ X222 ¼ X113 ¼ X223 ¼ X133 ¼ X233 ¼ X134 ¼ X234 ¼ X144 ¼ X244 ¼ 0:5), and the rest of y variables are set to zero. The minimum cost flows of the solutions are represented in Figs. 3 and 4, respectively.
(For explaining the correct forms of the holding costs in each solution, let us take the holding costs associated with D3 : At the S0 solution, see Fig. 2. for period 3, the 50% of D3 is inventoried both in period 1 and 2, thus its cost at the first level is h1 ; but at stage two the 50% of D3 is also inventoried through one period, so the total inventory cost at the periodical production plan for period 3 is ðh2 þ 0:5h3 ÞD3 : In solution S1, see Fig. 3 for period 3, the 50% of D3 is inventoried at period 1 on stage 1, plus the whole demand of period 3 is inventoried at period 2, thus the resulting holding cost is: ð0:5h1 þ h2 ÞD3 : In solution S2 the 50% of D3 is inventoried once both in stage one (in period 1) and stage two (in period 2). Thus, the inventory cost is ð0:5h1 þ 0:5h2 ÞD3 : Similar process should be
The total holding cost (THC) of the S0 solution is THCðS0Þ ¼ 0:5h2 D2 þ ðh1 þ 0:5h2 ÞD3 þ ð0:5h1 þ 0:5h2 ÞD4 ;
1 (11)
0.5 (12)
1 (11)
0.5
0.5 (12) (13)
(12)
(13)
0.5 (14)
0.5 1
(21)
0.5 (11)
(22)
1 (21)
(22)
1 (23)
(21)
0.5
(22)
(23)
(24)
0.5 D2
D3
D4
Fig. 3. The S1 solution.
1 (11)
(12)
1 (11)
1 (21)
0.5 (22)
(21)
(12) 0.5 (22)
0.5 (13)
(11)
(12)
0.5 0.5 (13) (14)
0.5 (23)
(21)
(22)
0.5 (23)
0.5 (24)
0.5 D2
D3
Fig. 4. The S2 solution.
D4
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
followed for calculating inventory costs associated with demand in other periods. It is easy to see that, not considering the first period, the setup cost is s1 þ 1:5s2 in each solution.) Now let us suppose that the S0 solution is the only optimal solution. Then, as the total setup cost in each of the three solutions is the same, the assumptions ð4aÞ
THCðS0ÞoTHCðS1Þ; and
ð4bÞ
THCðS0ÞoTHCðS2Þ
should be valid. However, substituting the holding costs, we have 0:5½ðh2 h1 ÞD2 þ ðh1 h2 Þ þ ðh1 h2 ÞD4 o0
ð5aÞ ð5bÞ
respectively. As ðh2 h1 Þ > 0; from (5a.) we have D2 oD3 þ D4 ;
ð6aÞ
and from (5b), as h1 > 0; we have D2 > D3 þ D4 ; which is contradiction.
ð6bÞ &
The consequence of this result is that either S1 or S2 outperforms (or is not worse than) the S0 solution when costs are constant, moreover, neither of these contains cross solution. The next property shows that non-integer solutions S1 and S2 can be replaced by an integer solution at no higher cost. Property 2. If a non-cross solution contains noninteger variables, it can be replaced by an integer solution without increase in total costs. Proof. Consider solution S1 and generate new solutions from it by increasing the value of Y12 while decreasing the value of Y14 and Y24 by a fraction of q: Let Dþ be the resulting change in the total cost of the new schedule compared to S1. Then: Dþ ¼ q½s1 þ 2h2 D4 s1 s2 h1 ðD2 þ D3 Þ:
D ¼ q½ s1 22h2 D4 þ s1 þ s2 þ h1 ðD2 þ D3 Þ: Similarly, first generating new solutions from S2 by increasing Y22 ; Y14 ; and Y24 while decreasing Y13 and Y23 with the same fraction q in the solution S2, the resulting change in cost compared to S2 is Dþ ¼ q½s2 þ ðh1 h2 ÞD2 þ ðh1 þ h2 Þ D3 s1 s2 h2 D4 þ s1 þ s2 : When we decrease the value of Y22 ; Y14 ; and Y24 while increasing Y13 and Y23 with the same fraction q in the solution S2, the change is D ¼ q½ s2 þ ðh2 h1 ÞD2 ðh1 þ h2 Þ
and 0:5h1 ð D2 þ D3 þ D4 Þo0;
Now, decreasing the value of Y12 while increasing Y14 and Y24 with the same fraction q; the resulting change in costs is:
D3 þ s1 þ s2 þ h2 D4 s1 s2 : We see that Dþ ¼ D in both cases. Thus there is a way of altering the value of non-integer variables where the total cost of the schedule either does not change or decreases, which means that a noninteger solution can be replaced by integer one at no higher cost. & Corollary 1. The linear relaxation of problem (3) has always an integer solution for M ¼ 2; T ¼ 4: Proof. We classify the problems into two groups. First we consider feasible solutions not containing cross solution. Figs. 3 and 4 present this kind of problems. As it can be seen in the proof of Property 2, generating new solutions from S1 and S2, integer solutions can be always derived at no higher cost. Turning to solutions containing cross solution, above we have analyzed the S0 solution for a two level four period problem in which the first stage in period 1 and 2 produces for period 3, resulting cross solution. As we are considering four period problems, another type of problems is worth mentioning only, namely when period 2 and 3 produce for period 4. Let us call it S3 solution, which is represented on Fig. 5. In this solution Y11 ¼ Y21 ¼ 1; Y12 ¼ Y22 ¼ Y13 ¼ Y23 ¼
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
1 (11) 1 (21) D1
(12)
0.5
0.5 (13)
0.5 (22)
0.5 (23)
0.5 (24)
D3
D4
D2
(14)
1 (11)
1 (12)
(13)
(14)
(22)
(23)
1 (24)
D2
D3
D4
1 (21) D1
1
Fig. 5. The S3 solution.
Y24 ¼ 0:5; and THCðS3Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ ðh1 þ 0:5h2 ÞD4 : Although S3 contains cross solution, an approach presented in the proof of Property 2. can be utilized directly as the problem terminates in period four (period 3 cannot transport for period 5. In the PW solution period 2 transported for period 4). Hence let us increase the production capacities (the values of Y variables) in period 2 and 4, and decrease those in period 3. The marginal change, denoted by D1 ; of the total cost is
(a)
1 (11)
1 (13)
(12)
1 (21) D1
(22)
1 (23)
D2
D3
(14)
(24) D4
(b) Fig. 6. (a) The S4a solution. (b) The S4b solution.
D1 ¼ s2 h2 D2 þ h2 D3 þ 2h1 D4 h2 D4 : Increasing the values of Y’s in period 2 till one, we obtain the solution that is presented in Fig. 6a. Let it be named S4a solution. Now decrease the production capacities in period 2 and 4, and increase those in period 3. The marginal change of total costs, denoted by D2 ; can be written as D2 ¼ s2 þ h2 D2 h2 D3 2h1 D4 þ h2 D4 : Decreasing the capacities till zero in period 2, we obtain a solution, let it be named S4b, which is presented in Fig. 6b. However, D1 ¼ D2 ; from which it follows that either S4a or S4b outperforms S3. Consequently, the non-integer solution can be substituted by an integer one at lower cost &. 3.3. Properties of larger problems Turning to more general problems, we have to see that a large variety of non-integer basic
solutions can be generated which increases the difficulty of indirect proof. Not the number of stages is the source of difficulties but that of the periods. Cases we are not going to analyse in details are simpler versions of the following situations. Considering five period problems first, we deal with solutions when periods 1 and 2 produce for period 3 (the case presented by Fig. 7), when periods 2 and 3 produce for period 4 (as in Fig. 8), and when periods 3 and 4 produce for period 5 (like in Fig. 9). Based on the approaches used above, the reader can easily resolve problems represented in Fig. 7 (an S0-like solution) and Fig. 9 (similar to the problem discussed in the proof of Corollary 1). Now we give more details on resolving the problem defined by Fig. 8. Let S5 denote this solution. The tie of S5 can be broken up by defining the solutions S6 and S7, showed by Figs. 9 and 10, respectively. It can be calculated that:
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
1 (11)
0.5 (12) (13)
1 (21)
0.5 (22)
D1
D2
(23)
0.5 (14) 0.5 0.5 (24)
D3
D4
0.5 (15)
1 (11)
0.5
1 (21)
(25)
0.5 (12)
D2
Fig. 7. A five period problem.
1 (11)
0.5 (12) 1
(21) D1
0.5 (13) (14)
0.5 0.5 0.5 (22) (23) (24) D2
D3
D4
0.5 (12)
1 (21)
0.5 0.5 0.5 (22) (23) (24)
D1
D2
0.5 0.5 (13) (14)
D3
D4
0.5 (25)
D4
D5
Fig. 10. The S7 solution.
0.5 (15)
1 (11)
0.5 (25)
1 (21)
1 (22)
D5
D1
25
Fig. 8. The S5 solution.
1 (11)
D3
0.5 (15)
(14)
0.5 0.5 (22) (23) (24)
D1
D5
0.5 (13)
0.5 (12)
0.5 (13)
(14)
0.5 0.5 (23) (24) 15
10
0.5 (15)
0.5 (16)
0.5
0.5
(25)
(26)
10
20
Fig. 11. The optimal solution of a six period problem.
would follow. However, from (8a,b), as ðh2 h1 Þ > 0 and h1 > 0; we have (15)
2D4 oD5
0.5 (25)
and D5 o2D4 ;
D5
which is contradiction.
&
Fig. 9. The S6 solution.
THCðS5Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ ðh1 þ 0:5h2 ÞD4 þ 0:5ðh1 þ h2 ÞD5 ;
ð7aÞ
THCðS6Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ 0:5h2 D4 þ ðh1 þ 0:5 h2 ÞD5 ;
ð7bÞ
THCðS7Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ 1:5h2 þ hD5 : ð7cÞ If S5 were the only optimal solution, then THCðS5Þ THCðS6Þ ¼h1 D4 0:5h1 D5 o0;
ð8aÞ
and THCðS5Þ THCðS7Þ ¼ ðh1 h2 ÞD4 þ 0:5ðh2 h1 ÞD5 o0
ð8bÞ
Thus S5 is outperformed by either S6 or S7. As S7 does not contain cross solution, it can be easily outperformed by an integer solution. Although S6 has cross solution, but as in case of the S3 solution it can be traced back to an integer solution. This way the next corollary can be stated. Corollary 2. The linear relaxation of problem (3) has always an integer solution for M ¼ 2; T ¼ 5: However, neither of these approaches works for the following six period problem: s1 ¼ 40; s2 ¼ 20; h1 ¼ 1; h2 ¼ 2; and D1 =arbitrary, D2 ¼ 25; D3 ¼ 15; D4 ¼ D5 ¼ 10; D6 ¼ 20: A non-integer optimal solution is presented by Fig. 11, and the minimum cost is 565/2, while the value of the optimal integer solution is 285.
J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
4. Conclusions In this paper the facilities in series inventory model is analyzed. It is shown that with timeindependent setup and holding costs the linear relaxation of the integer programming model has always integer solution till five period problems, and a six period problem is developed that has no integer solution. These findings extend our knowledge on the features of multi level dynamic lot sizing problems and facilitates the use of linear programming techniques for analyzing the robustness of dynamic lot sizing problems up to the size of M ¼ 2; T ¼ 5:
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. os, . On the stability region for multi level [5] K. Richter, J. Vor inventory problems, European Journal of Operational Research 41 (2) (1989) 169–174. . os, . Setup cost stability region for the [6] S. Chand, J. Vor dynamic lot sizing problem with backlogging, European Journal of Operational Research 58 (1992) 68–77. [7] S. Chand, S. Sethi, A dynamic lot sizing model with learning in setups, Operations Research 38 (4) (1990) 644–656. . os, . [8] J. Vor Setup cost stability regions for multilevel dynamic lot sizing problems, European Journal of Operational Research 87 (1995) 132–141. [9] T. Veinott, Minimum concave-cost solution of Leontief substitution models of multiple facility inventory systems, Operations Research 17 (1969) 262–291. [10] W.B. Crowston, M.H. Wagner, Dynamic lot sizing models for multi-stage assembly systems, Management Science 20 (1973) 14–21. [11] S.F. Love, A facilities in series inventory model with nested schedules, Management Science 18 (1972) 327–338. [12] S. Chand, Rolling horizon procedures for the facilities in series inventory model with nested schedules, Management Science 29 (1983). [13] Y. Pochet, L.A. Wolsey, Polyhedra for lot-sizing with Wagner-Whitin costs, CORE Discussion Paper, University of Louvain, 1993. [14] W.I. Zangwill, A backlogging model and a multi-echelon model of a dynamic economic lot size production system – a network approach, Management Science 15 (1969) 505–527.
On the relaxation of multi-level dynamic lot-sizing models$ ! . os* . Jozsef Vor ! Rak ! oczi ! ut 80, 7622 Pecs, ! Hungary Faculty of Business and Economics, University of Pecs, Received 19 July 1999; accepted 17 August 2001
Abstract In this paper the multi level dynamic lot sizing problem is analyzed when setup and holding costs do not depend on time. A zero–one integer programming formulation and its linear relaxation are investigated. The paper shows that for certain classes of problems the relaxed models provide integer solution when the number of periods is below a certain threshold. A counter example shows that at least six periods are required to obtain a unique non-integer solution of the relaxed problem. r 2002 Elsevier Science B.V. All rights reserved. Keywords: Lot sizing; Integer programming; Dynamic programming
1. Introduction Flexibility is one of the oldest themes of production and operations management. Lot sizing, as one of the most important elements of flexibility, has been playing an important role in the literature of production and operations management, especially since reducing setup costs and setup times has become one of the most effective tools. Parallel with this process, the notion of mass production began to decline, and today JIT, lean production, flexible manufacturing are among the main competing manufacturing paradigms. These changes in the manufacturing paradigms have been reflected by the EOQ-connected literature as well. The pioneer works of Porteus [1–3] deal first $
Support from the National Research Fund (OTKA) under Grant no. T 037291 is acknowledged. The author also highly appreciates the valuable notes of referees. *Tel.: +36-72-211-524; fax: +36-72-501-553. E-mail address: [email protected] (J. Vo¨ro¨s).
time with the problem of setup cost reduction in the EOQ model, however less attention has been paid to the same problem in dynamic case, i.e. when demand is not constant. As demand can rarely be fully smoothed, knowing the required level of setup cost reduction in order to decrease the number of setups during certain time periods, is an important issue for managers. Thus studying the robustness of a given optimal solution, especially with respect to setup cost, and knowing the nature of the problem are useful. The setup cost stability region (SR) for an optimal solution is defined as the set of all setup cost values for which a given solution remains optimal. The sensitivity analysis of the single level dynamic lot sizing problem arises first time in the paper of Richter [4] and–following the idea of dynamic programming–requiring cost inputs to have the same regeneration set for every time period, he could reveal a subset of SR. Later . os . [5] have analyzed the multiRichter and Vor
0925-5273/02/$ - see front matter r 2002 Elsevier Science B.V. All rights reserved. PII: S 0 9 2 5 - 5 2 7 3 ( 0 1 ) 0 0 2 0 2 - X
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
stage lot sizing problem but they could show results only based on the same assumption extended to the multi level problem. They also expressed that they could reveal the whole SR by using a full enumeration which is obviously not an effective way. The next step in this field was made by Chand . os . [6] when they applied a new approach and Vor to the problem. For the single level case–with backlogging–they proved that the total cost function is convex in the number of setups. (We mention also that the convex nature of the holding cost function–without backlogging–is also derived in the paper of [7]. Similar results for the multi level dynamic lot-sizing case were published by . os . [8]. Vor The nature of the multi level dynamic lot sizing problem has attracted many researchers (see for example [9–12]) and the linear relaxation of the zero–one integer programming formulation of the dynamic lot sizing problem plays an important role in the analysis of the stability regions. When the zero–one formulation can be substituted by linear programming models the proof of some theorems required to identify stability regions becomes simple. When setup and holding costs can vary period by period, Pochet and Wolsey [13] defined a two level, four period problem for which the relaxed version has a unique non-integer solution. This paper shows that with time independent setup and holding costs the relaxed version of the integer problem with four and five periods has always integer solution. The paper presents a six period problem that has no optimal integer solution despite the fact that the setup and holding costs are constant through periods.
2. The multi stage serial assembly problem According to a general description of the serial assembly system [12] the production and inventory system is assumed to have M facilities in series, and the input to facility (m þ 1) comes from the production at facility m: Raw materials are available in unlimited amounts as input to facility 1. Facility M produces assemblies which are used to supply the customer demand. All facilities are
allowed to carry inventories while the Mth facility carries the finished goods inventory. It is assumed that production and shipments are instantaneous, and that one unit of production on facility (m þ 1) requires one unit of input from facility m: Let Dt denote the demand in period t; it is assumed that demand is known for periods 1 to T: Let Xmt denote the production at facility m in period t; and the cost of this production is Cmt ðXmt Þ: Imt is the inventory at the end of period t at facility m and the corresponding cost is Hmt ðImt Þ for all mA/1; MS and tA/1; TS; where /a; bS ¼ fa; a þ 1; y; bg: Then the multi stage assembly problem can be formulated as: MIN
T X M X
½Cmt ðXmt Þ þ Hmt ðImt Þ;
ð1aÞ
t¼1 m¼1
XMþ1;t ¼ Dt ;
ð1bÞ
tA/1; TS;
Im;t 1 þ Xmt ¼ Xmþ1;t þ Imt for all
mA/1; MS;
Im0 ¼ ImT ¼ 0; Imt ; Xmt X0;
tA/1; TS;
mA/1; MS;
mA/1; MS;
tA/1; TS:
ð1cÞ ð1dÞ ð1eÞ
According to Crowston and Wagner [10], problem (1) can be represented as a network flow problem with one source and T sinks (shown by Fig. 1). For concave cost functions C and H respectively, Zangwill [14] has stated that there exists an optimal solution corresponding to an extreme flow, i.e. a node can have at most one positive input (see also the proof of [9] for more general cost functions). This means that Im;t 1 Xmt ¼ 0 for all mA/1; MS and tA/1; TS in at least one optimal solution. Now, let us consider problem (1) with a cost function used widely in the literature: ( Sm þ cm Xmt for Xmt > 0; Cmt ðXmt Þ ¼ ð2Þ 0 for Xmt ¼ 0; and Hmt ðImt Þ ¼ hm Imt for all m and t with hm phmþ1 ;
mA/1; M 1S;
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
ΣDt (0) X12
X11 I11 (11)
X13 I12
(12)
(13)
X22
X21 I21 (21)
X23 I22
(22)
X31
(23)
X32 I31
X33 I32
(31)
(32)
(33)
D1
D2
D3
Fig. 1. The dynamic serial assembly system (M ¼ 3; T ¼ 3).
. os . [8] proposed a zero–one integer programVor ming formulation for the analysis of problem (1) with the cost functions specified in (2) and proved that picking up pivot elements only from the coefficient matrix, the transformation result in integer numbers. (However, as it can be concluded from this paper, the inclusion of certain unit vectors into the basis can lead to non integer solutions.) In this formulation let Xmjt be one if on facility m in period j the whole Dt is produced, otherwise zero. Similarly, let Ymt be one if there is any production on stage m at period t; otherwise zero. Then the next formulation also describes problem (1) with the cost functions defined by (2): " # M X T t X X MIN Cmjt Xmjt þ Sm Ymt ; ð3aÞ m¼1 t¼1 t X
ð3bÞ
tA/1; TS;
j¼1 k X j¼1
where Sm and hm denote the setup and periodic unit holding cost on stage m; respectively. For simplicity, we also assume that h1 > 0; h1 oh2 and Dt > 0 for all t: Variable production cost (cm ) is considered time independent, thus it can be omitted. We note that to our belief the cost structure defined by (2) does not reduce significantly the industrial relevance of the model. The time horizon of this model is usually one year and the length of the period can be considered one month. Although a declining tendency can be predicted, once a reduction in setup cost is achieved, the reduced setup cost level prevails over certain time. In case of using equal length of periods, time independent holding costs can be assumed. The increasing holding costs by stages come from the fact that cascading downward on stages, the value of product increases. Based on cost functions in (2), problem (1) has a nested optimal schedule [11]. (A schedule is nested if Xmt > 0 implies Xmþ1;t > 0.) So further on, it is sufficient to deal with nested schedules only.
XMjt ¼ 1;
j¼1
Xmjt p
k X
Xm 1;jt ;
j¼1
mA/2; MS;
tA/1; TS;
kA/1; tS
ð3cÞ
tA/1; TS;
jA/1; tS
ð3dÞ
Xmjt pYmj ; mA/1; MS;
Xmjt pAf0; 1g; Ymt Af0; 1g; tAh1; T i; mAh1; M i; jAh1; ti;
ð3eÞ
where Cmjt ¼ ðhm hm 1 ÞDt ðt jÞ; for all m; t; j; and h0 ¼ 0: The equivalence of (1) and (3) origins from the characteristics of the problem. As mentioned above, Zangwill [14] pointed out that there are optimal solutions of (1) with the feature that every node can have at most one positive input. From this it follows that if once a portion of Dt is produced in a certain period, the full quantity of Dt must be produced in that period. This property is secured by the definition of Xmjt : Assumption (3b) formulates that the required quantity is delivered in time. (3c) provides that production can be scheduled in stage m; if the job is completed at stage m 1: In formulation (1) the same idea is expressed by (1c). (3d) is for counting the setups.
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
programming problems, and the second is that the solution really gives a nice variety of possible solutions which one probably cannot imagine. Next, we show that under time independent cost structure S0 can be substituted by integer solution at no higher cost. We emphasize that in our model smt ¼ sm for all t; and the PW example does not apply this restriction. Additionally, in our model hmt ¼ hm holds for all t: Let us take the S0 solution by periods, i.e. let us follow the way of satisfying the demand of a period. Fig. 2. gives the graphical representation of the S0 solution (omitting the flow for D1 ). In Fig. 2, the numbers along vertical lines indicate the amount of total setup cost counted (the values of ymj variables). According to Fig. 2, only 50% of D2 is produced in the first period, despite the fact that the full setup cost is paid in period 1, and 50% of D2 is transported from node (21) to node (22). Similarly, only 50% of D3 is produced in period one at the first stage, and the other half is produced in period two. Second stage works on D3 in periods two and three in 50–50%, respectively. A solution will be called cross solution if at the periodical production plan there is a node that has two positive inputs and two positive outputs. According to Fig. 2, the periodical production plan for period 3 exhibits a cross solution.
3. Analyzing the nature of relaxation 3.1. The presentation of a non-integer solution in case of dynamic costs Pochet and Wolsey [13] have formulated a linear programming model which in its constraints is the same like we would have if we took the linear programming relaxation of problem (3), i.e. when assumptions in (3e) are substituted by simple nonnegativity assumptions. They considered a two level, four period (M ¼ 2; T ¼ 4) problem with the inputs: Dt ¼ 1 for all tA/1; 4S; h1t ¼ ð2; 1; 1; 0Þ; h2t ¼ ð4; 3; 2; 0Þ and s1t ¼ ð0; 4; 6; 2Þ; s2t ¼ ð0; 4; 4; 2Þ where hmt and smt are for the inventory and setup costs, respectively, on level m in period t: It is emphasized again that here their setup and holding costs are dynamic, while in our model they are constant in time. Let us identify this example as ‘‘PW example’’. They show that the solution (let us denote it by S0):: X211 ¼ X111 ¼ 1;
X212 ¼ X222 ¼ X223 ¼ X233 ¼
X234 ¼ X244 ¼ X112 ¼ X122 ¼ X113 ¼ X123 ¼ X124 ¼ X144 ¼ 0:5 ðand as consequence Y21 ¼ Y11 ¼ 1; Y22 ¼ Y23 ¼ Y24 ¼ Y12 ¼ Y14 ¼ 0:5; and the rest of the variables are 0Þ is the only optimal solution of the problem. This solution has two strong characteristics: the first is that it shows, there are multi-level dynamic lot sizing problems whose integer programming formulation cannot be substituted by linear
1 (11)
0.5 (12)
1
0.5
(21) 0.5 (22) D2
1 (11)
0.5 (12) (13) 0.5
(21)
(22)
(11)
0.5 (12)
(13)
0.5 (23)
0.5 (21)
D3 Fig. 2. The S0 solution.
(22)
(23)
0.5 (14) 0.5 (24) D4
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
while the THC of S1 and S2:
3.2. Analyzing the S0 solution under constant setup and holding costs
THCðS1Þ ¼ 0:5h1 D2 þ ð0:5h1 þ h2 ÞD3 þ h2 D4 ;
Property 1. A cross solution in the S0 solution with constant costs can be substituted by a non-cross solution at no higher cost.
and THCðS2Þ ¼ ð0:5h1 þ 0:5h2 ÞD2 þ ð0:5h1 þ 0:5h2 ÞD3 þ 0:5h2 D4 :
Proof. Let us define two new solutions, S1 and S2. Let S1 be: Y12 ¼ Y14 ¼ Y24 ¼ 0:5 and Y11 ¼ Y21 ¼ Y22 ¼ 1 (consequently the minimum holding cost is provided by the production plan: X112 ¼ X122 ¼ X113 ¼ X123 ¼ X124 ¼ X144 ¼ X244 ¼ 0:5), while S2: Y22 ¼ Y13 ¼ Y23 ¼ Y14 ¼ Y24 ¼ 0:5; Y11 ¼ Y21 ¼ 1 (consequently X112 ¼ X212 ¼ X222 ¼ X113 ¼ X223 ¼ X133 ¼ X233 ¼ X134 ¼ X234 ¼ X144 ¼ X244 ¼ 0:5), and the rest of y variables are set to zero. The minimum cost flows of the solutions are represented in Figs. 3 and 4, respectively.
(For explaining the correct forms of the holding costs in each solution, let us take the holding costs associated with D3 : At the S0 solution, see Fig. 2. for period 3, the 50% of D3 is inventoried both in period 1 and 2, thus its cost at the first level is h1 ; but at stage two the 50% of D3 is also inventoried through one period, so the total inventory cost at the periodical production plan for period 3 is ðh2 þ 0:5h3 ÞD3 : In solution S1, see Fig. 3 for period 3, the 50% of D3 is inventoried at period 1 on stage 1, plus the whole demand of period 3 is inventoried at period 2, thus the resulting holding cost is: ð0:5h1 þ h2 ÞD3 : In solution S2 the 50% of D3 is inventoried once both in stage one (in period 1) and stage two (in period 2). Thus, the inventory cost is ð0:5h1 þ 0:5h2 ÞD3 : Similar process should be
The total holding cost (THC) of the S0 solution is THCðS0Þ ¼ 0:5h2 D2 þ ðh1 þ 0:5h2 ÞD3 þ ð0:5h1 þ 0:5h2 ÞD4 ;
1 (11)
0.5 (12)
1 (11)
0.5
0.5 (12) (13)
(12)
(13)
0.5 (14)
0.5 1
(21)
0.5 (11)
(22)
1 (21)
(22)
1 (23)
(21)
0.5
(22)
(23)
(24)
0.5 D2
D3
D4
Fig. 3. The S1 solution.
1 (11)
(12)
1 (11)
1 (21)
0.5 (22)
(21)
(12) 0.5 (22)
0.5 (13)
(11)
(12)
0.5 0.5 (13) (14)
0.5 (23)
(21)
(22)
0.5 (23)
0.5 (24)
0.5 D2
D3
Fig. 4. The S2 solution.
D4
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
followed for calculating inventory costs associated with demand in other periods. It is easy to see that, not considering the first period, the setup cost is s1 þ 1:5s2 in each solution.) Now let us suppose that the S0 solution is the only optimal solution. Then, as the total setup cost in each of the three solutions is the same, the assumptions ð4aÞ
THCðS0ÞoTHCðS1Þ; and
ð4bÞ
THCðS0ÞoTHCðS2Þ
should be valid. However, substituting the holding costs, we have 0:5½ðh2 h1 ÞD2 þ ðh1 h2 Þ þ ðh1 h2 ÞD4 o0
ð5aÞ ð5bÞ
respectively. As ðh2 h1 Þ > 0; from (5a.) we have D2 oD3 þ D4 ;
ð6aÞ
and from (5b), as h1 > 0; we have D2 > D3 þ D4 ; which is contradiction.
ð6bÞ &
The consequence of this result is that either S1 or S2 outperforms (or is not worse than) the S0 solution when costs are constant, moreover, neither of these contains cross solution. The next property shows that non-integer solutions S1 and S2 can be replaced by an integer solution at no higher cost. Property 2. If a non-cross solution contains noninteger variables, it can be replaced by an integer solution without increase in total costs. Proof. Consider solution S1 and generate new solutions from it by increasing the value of Y12 while decreasing the value of Y14 and Y24 by a fraction of q: Let Dþ be the resulting change in the total cost of the new schedule compared to S1. Then: Dþ ¼ q½s1 þ 2h2 D4 s1 s2 h1 ðD2 þ D3 Þ:
D ¼ q½ s1 22h2 D4 þ s1 þ s2 þ h1 ðD2 þ D3 Þ: Similarly, first generating new solutions from S2 by increasing Y22 ; Y14 ; and Y24 while decreasing Y13 and Y23 with the same fraction q in the solution S2, the resulting change in cost compared to S2 is Dþ ¼ q½s2 þ ðh1 h2 ÞD2 þ ðh1 þ h2 Þ D3 s1 s2 h2 D4 þ s1 þ s2 : When we decrease the value of Y22 ; Y14 ; and Y24 while increasing Y13 and Y23 with the same fraction q in the solution S2, the change is D ¼ q½ s2 þ ðh2 h1 ÞD2 ðh1 þ h2 Þ
and 0:5h1 ð D2 þ D3 þ D4 Þo0;
Now, decreasing the value of Y12 while increasing Y14 and Y24 with the same fraction q; the resulting change in costs is:
D3 þ s1 þ s2 þ h2 D4 s1 s2 : We see that Dþ ¼ D in both cases. Thus there is a way of altering the value of non-integer variables where the total cost of the schedule either does not change or decreases, which means that a noninteger solution can be replaced by integer one at no higher cost. & Corollary 1. The linear relaxation of problem (3) has always an integer solution for M ¼ 2; T ¼ 4: Proof. We classify the problems into two groups. First we consider feasible solutions not containing cross solution. Figs. 3 and 4 present this kind of problems. As it can be seen in the proof of Property 2, generating new solutions from S1 and S2, integer solutions can be always derived at no higher cost. Turning to solutions containing cross solution, above we have analyzed the S0 solution for a two level four period problem in which the first stage in period 1 and 2 produces for period 3, resulting cross solution. As we are considering four period problems, another type of problems is worth mentioning only, namely when period 2 and 3 produce for period 4. Let us call it S3 solution, which is represented on Fig. 5. In this solution Y11 ¼ Y21 ¼ 1; Y12 ¼ Y22 ¼ Y13 ¼ Y23 ¼
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
1 (11) 1 (21) D1
(12)
0.5
0.5 (13)
0.5 (22)
0.5 (23)
0.5 (24)
D3
D4
D2
(14)
1 (11)
1 (12)
(13)
(14)
(22)
(23)
1 (24)
D2
D3
D4
1 (21) D1
1
Fig. 5. The S3 solution.
Y24 ¼ 0:5; and THCðS3Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ ðh1 þ 0:5h2 ÞD4 : Although S3 contains cross solution, an approach presented in the proof of Property 2. can be utilized directly as the problem terminates in period four (period 3 cannot transport for period 5. In the PW solution period 2 transported for period 4). Hence let us increase the production capacities (the values of Y variables) in period 2 and 4, and decrease those in period 3. The marginal change, denoted by D1 ; of the total cost is
(a)
1 (11)
1 (13)
(12)
1 (21) D1
(22)
1 (23)
D2
D3
(14)
(24) D4
(b) Fig. 6. (a) The S4a solution. (b) The S4b solution.
D1 ¼ s2 h2 D2 þ h2 D3 þ 2h1 D4 h2 D4 : Increasing the values of Y’s in period 2 till one, we obtain the solution that is presented in Fig. 6a. Let it be named S4a solution. Now decrease the production capacities in period 2 and 4, and increase those in period 3. The marginal change of total costs, denoted by D2 ; can be written as D2 ¼ s2 þ h2 D2 h2 D3 2h1 D4 þ h2 D4 : Decreasing the capacities till zero in period 2, we obtain a solution, let it be named S4b, which is presented in Fig. 6b. However, D1 ¼ D2 ; from which it follows that either S4a or S4b outperforms S3. Consequently, the non-integer solution can be substituted by an integer one at lower cost &. 3.3. Properties of larger problems Turning to more general problems, we have to see that a large variety of non-integer basic
solutions can be generated which increases the difficulty of indirect proof. Not the number of stages is the source of difficulties but that of the periods. Cases we are not going to analyse in details are simpler versions of the following situations. Considering five period problems first, we deal with solutions when periods 1 and 2 produce for period 3 (the case presented by Fig. 7), when periods 2 and 3 produce for period 4 (as in Fig. 8), and when periods 3 and 4 produce for period 5 (like in Fig. 9). Based on the approaches used above, the reader can easily resolve problems represented in Fig. 7 (an S0-like solution) and Fig. 9 (similar to the problem discussed in the proof of Corollary 1). Now we give more details on resolving the problem defined by Fig. 8. Let S5 denote this solution. The tie of S5 can be broken up by defining the solutions S6 and S7, showed by Figs. 9 and 10, respectively. It can be calculated that:
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J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
1 (11)
0.5 (12) (13)
1 (21)
0.5 (22)
D1
D2
(23)
0.5 (14) 0.5 0.5 (24)
D3
D4
0.5 (15)
1 (11)
0.5
1 (21)
(25)
0.5 (12)
D2
Fig. 7. A five period problem.
1 (11)
0.5 (12) 1
(21) D1
0.5 (13) (14)
0.5 0.5 0.5 (22) (23) (24) D2
D3
D4
0.5 (12)
1 (21)
0.5 0.5 0.5 (22) (23) (24)
D1
D2
0.5 0.5 (13) (14)
D3
D4
0.5 (25)
D4
D5
Fig. 10. The S7 solution.
0.5 (15)
1 (11)
0.5 (25)
1 (21)
1 (22)
D5
D1
25
Fig. 8. The S5 solution.
1 (11)
D3
0.5 (15)
(14)
0.5 0.5 (22) (23) (24)
D1
D5
0.5 (13)
0.5 (12)
0.5 (13)
(14)
0.5 0.5 (23) (24) 15
10
0.5 (15)
0.5 (16)
0.5
0.5
(25)
(26)
10
20
Fig. 11. The optimal solution of a six period problem.
would follow. However, from (8a,b), as ðh2 h1 Þ > 0 and h1 > 0; we have (15)
2D4 oD5
0.5 (25)
and D5 o2D4 ;
D5
which is contradiction.
&
Fig. 9. The S6 solution.
THCðS5Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ ðh1 þ 0:5h2 ÞD4 þ 0:5ðh1 þ h2 ÞD5 ;
ð7aÞ
THCðS6Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ 0:5h2 D4 þ ðh1 þ 0:5 h2 ÞD5 ;
ð7bÞ
THCðS7Þ ¼ 0:5h2 D2 þ 0:5h2 D3 þ 1:5h2 þ hD5 : ð7cÞ If S5 were the only optimal solution, then THCðS5Þ THCðS6Þ ¼h1 D4 0:5h1 D5 o0;
ð8aÞ
and THCðS5Þ THCðS7Þ ¼ ðh1 h2 ÞD4 þ 0:5ðh2 h1 ÞD5 o0
ð8bÞ
Thus S5 is outperformed by either S6 or S7. As S7 does not contain cross solution, it can be easily outperformed by an integer solution. Although S6 has cross solution, but as in case of the S3 solution it can be traced back to an integer solution. This way the next corollary can be stated. Corollary 2. The linear relaxation of problem (3) has always an integer solution for M ¼ 2; T ¼ 5: However, neither of these approaches works for the following six period problem: s1 ¼ 40; s2 ¼ 20; h1 ¼ 1; h2 ¼ 2; and D1 =arbitrary, D2 ¼ 25; D3 ¼ 15; D4 ¼ D5 ¼ 10; D6 ¼ 20: A non-integer optimal solution is presented by Fig. 11, and the minimum cost is 565/2, while the value of the optimal integer solution is 285.
J. Vo¨ro¨s / Int. J. Production Economics 77 (2002) 53–61
4. Conclusions In this paper the facilities in series inventory model is analyzed. It is shown that with timeindependent setup and holding costs the linear relaxation of the integer programming model has always integer solution till five period problems, and a six period problem is developed that has no integer solution. These findings extend our knowledge on the features of multi level dynamic lot sizing problems and facilitates the use of linear programming techniques for analyzing the robustness of dynamic lot sizing problems up to the size of M ¼ 2; T ¼ 5:
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