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Discrete and dynamic versus continuous and static loading policy for a multi-compartment vehicle
European Journal of Operational Research 174 (2006) 1329–1337 www.elsevier.com/locate/ejor
Short Communication
Discrete and dynamic versus continuous and static loading policy for a multi-compartment vehicle Yossi Bukchin b
a,*
, Subhash C. Sarin
b,1
a Department of Industrial Engineering, Faculty of Engineering, Tel-Aviv University, Tel-Aviv 69978, Israel Grado Department of Industrial and Systems Engineering, Virginia Tech, 250 Durham Hall, Blacksburg, VA 24061, USA
Received 15 January 2004; accepted 8 March 2005 Available online 22 June 2005
Abstract In this paper, we address the problem of loading non-intermixable products in a vehicle consisting of compartments of different sizes. The demands of the products are different but uniform over time. The objective is to meet product demands and minimize setup rate (that is, the number of deliveries per unit time). Two approaches, namely, dynamic and static, are investigated and their performances are compared with each other. In the dynamic approach, deliveries are made in several discrete periods and, then, repeated in a cyclic fashion. In each of these deliveries, the allocation of products to compartments can be different. The static approach, on the other hand, assumes a continuous time scale and determines a single assignment of products to compartments that maximizes the time in which the product demands are fully satisfied by this single delivery. The comparison between the two approaches shows that the dynamic approach is superior to the static approach when a discrete time scale is considered. However, even when the discrete time scale constraint is relaxed, the dynamic approach still provides better results for relatively long cycle times. Ó 2005 Elsevier B.V. All rights reserved. Keywords: Assignment problem; Loading problem; Shipping plan
1. Problem definition and literature review We consider a situation in which a vehicle transports several non-intermixable products from * Corresponding author. Tel.: +972 3 640 7941; fax: +972 3 640 7669. E-mail address: [email protected] (Y. Bukchin). 1 Tel.: +1 540 231 7140; fax: +1 540 231 3322.
a supplier to a customer on a vehicle consisting of compartments of different sizes. During the shipment, the products have to be kept separately in different compartments. A known and constant demand for each of the products is assumed, and shortages are not permitted. The objective is to determine a loading policy of the vehicle which minimizes the number of shipments required per unit of time, assuming that the shipment (setup)
0377-2217/$ - see front matter Ó 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2005.03.035
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cost is significantly higher than the inventory holding cost. We also assume that the supplier holds large enough quantities of the products so as to satisfy the allocation of any combination of products to compartments, limited only by the vehicle capacity, and that, the customer has enough storage space to accommodate the quantity of the products that are delivered. Although the vehicle routing and scheduling problems have been covered extensively in the literature, there is little reported on the problem on hand. Fagerholt and Christiansen (2000) have addressed a similar situation in the scheduling and allocation of cargo to a ship. They assume that each ship has a cargo hold which is flexible and can be adjusted to give different size sub-compartments. Only a single product is assumed for loading in each of these compartments during a delivery. Time windows are associated with the loading and unloading times of products, and there is a penalty cost for violating the time windows. They propose a set partitioning based approach for the solution of this problem. A modification of this model, which allows soft time windows due to the introduction of a penalty cost that is varied as a function of the deviation from the targeted time window, is presented in Fagerholt (2001). Another related problem, that has been widely addressed in the literature, is the segregated storage problem (SSP) presented in Shilfer and Naor (1961). In this problem, different types of grains are to be stored in the compartments of a silo such that there is at most one product in each compartment. It is assumed that product demands, storage capacities i.e., the sizes of the compartments in the silo, and the variable storage costs are known. Also, an external storage facility of infinite capacity is assumed to be available to accommodate overflow. The objective is to find an assignment of products to compartments such that the total cost of holding in the external storage, is minimized. Improvements in the solution procedures for SSP have been proposed by White and Francis (1971), Dannenbring and Khumawala (1973), Evans and Cullen (1977), Neebe (1987) and Evans and Tsubakitani (1993). The SSP is a special case of our problem as, in SSP, an assignment is sought for a single period only. However, a solution for the SSP problem
which does not require any external storage is equivalent to a feasible solution to our problem. Yuceer (1997) addresses the problem of allocating products to a vehicle having compartments of different sizes such that at most one product is assigned to a compartment. The objective is to maximize the minimum number of periods over which the demands of all the products are satisfied by a single delivery. No shortages are allowed. Consequently, the solution of this problem gives the time interval between consecutive (and identical) deliveries. Yuceer (1997), assumes a continuous time scale and that a shipment can be performed at any time. We call this policy a continuous and static loading policy, and discuss it in Section 2. We also present a discrete and dynamic loading policy in Section 3. According to this policy, a series of setups are performed and they are repeated cyclically. The length of a cycle consists of a finite number of time periods, while at most one setup can be performed in a period of this cycle. We show that given a large enough number of periods in a cycle, this policy outperforms the continuous and static loading policy of Yuceer (1997). This is demonstrated through a detailed experimentation presented in Section 4.
2. The continuous and static loading policy The continuous and static loading policy provides a loading plan, which maximizes the common replenishment time; namely, the time interval over which the collective demand for all the products is satisfied. This loading plan is then repeated subsequently. Let qi and dj represent the size of compartment i and the demand per period of product j, respectively. Then, the ILP model, for the continuous static policy, is of a max–min type, and is as follows: " # m X qi max min xij ; ð1Þ j dj i¼1 subject to : n X xij ¼ 1;
i ¼ 1; . . . ; m;
ð2Þ
j¼1
xij ¼ 0; 1;
i ¼ 1; . . . ; m; j ¼ 1; . . . ; n;
ð3Þ
Y. Bukchin, S.C. Sarin / European Journal of Operational Research 174 (2006) 1329–1337
where xij is a binary variable and is equal to 1, if product type j, j = 1, . . . , n, is assigned to compartment i, i = 1, . . . , m, and 0, otherwise. We can see that this formulation focuses on the product whose demand is satisfied over the shortest interval of time and the objective function (1), aims at maximizing this interval. The solution value represents the time interval in between two consecutive identical setups. The constraint set (2) assures that at most one product is assigned to a single compartment, and the constraint set (3) captures the binary nature of xij.
3. The discrete and dynamic loading policy The discrete and dynamic loading policy, presented first in Bukchin and Sarin (2004), is different from the continuous and static loading policy due to the following two aspects:
However, since the cycle length is a decision variable, a search over different values of T is required to find a suitable value. For practical reasons, the desired cycle should be relatively short in length. Indeed, later on, we show that satisfactory (close to optimal) solutions can be obtained for cycles of relatively short length. Consequently, the necessary search over T is not too intensive, although the problem is not convex in T, as shown in Bukchin and Sarin (2004). The problem for determining the discrete and dynamic policy can be formulated as follows: min
T X
In order to compare these two approaches, we use the criterion pertaining to setup rate, that is, the number of setups required in a unit of time. Since the problem is solved for the infinite planning horizon, the setup rate associated with the continuous and static loading policy is the inverse of the objective function value presented in (1). However, the situation is a little bit more complicated for the discrete and dynamic loading policy, since it involves different setups. In order to provide a solution for the infinite planning horizon that is comparable to the static policy, we employ a cyclic dynamic loading policy, a policy in which a number of setups over a cycle length is repeated. Hence, the setup rate in this case is the ratio of the number of setups performed in a cycle and the cycle length. The following model characterizes the discrete and dynamic policy, given the cycle length T.
ð4Þ
st ;
t¼1
subject to: n X xijt 6 1;
i ¼ 1; . . . ; m; j ¼ 1; . . . ; n;
ð5Þ
t ¼ 1; . . . ; T ; i ¼ 1; . . . ; m;
ð6Þ
j¼1
st P 1. The setups that are performed involve different assignments of products to compartments instead of a single assignment as in the continuous static policy. 2. Time is treated as a discrete parameter rather than a continuous one (e.g., weeks, months, quarters and so on).
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n X
xijt ;
j¼1 t X m X k¼1
xijt ;
xijk qi P td j ;
t ¼ 1; . . . ; T ; j ¼ 1; . . . ; n;
i¼1
st ¼ ð0; 1Þ;
ð7Þ ð8Þ
where xijt is a binary variable and is equal to 1 if product type j is assigned to compartment type i in period t, and 0, otherwise, and st is a binary variable which is equal to 1 if a setup is incurred in period t, and 0, otherwise. Constraint set (5) assures that not more than one product is assigned to a single compartment during a setup. The setup variable, st, is set to 1 in period t, if at least one product type is delivered in that period by any of the compartments. This is captured in constraint (6). The capacity constraints (7) guarantee that no shortage occurs in any period for any product. The binary nature of the variables xijk and st is dictated by constraint (8), while the objective of minimizing the number of setups, for a given number of periods, is represented by (4). In the sequel, we show that there always exists a discrete and dynamic solution, which outperforms the continuous and static solution. Subsequently,
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based on a series of experiments, we show that, in most cases, such a solution can be found rather easily under practical operating conditions. Let DPdenote the total demand per period, n namely, j¼1 d j , and Q the total vehicle capacity, Pm q . Clearly, D/Q is a lower bound of the setup i¼1 i rate, if the problem is relaxed by permitting every compartment to accommodate any mixture of products.
The discrete and dynamic loading policy provides a solution in which S setups are performed in a cycle time of T periods. Since we assume that not more than one shipment can occur in a period, therefore S 6 T. Bukchin and Sarin (2004) show that, theoretically, a discrete and dynamic solution can be obtained that satisfies the condition of Proposition 1. This follows from the following result.
Proposition 1. The continuous and static solution is equal to the lower bound, D/Q, if and only if the compartments are fully utilized and the leftover inventory at the arrival of the next shipment is zero.
Proposition 2. The setup rate values of the discrete and dynamic solutions converge to the lower bound, D/Q, as the cycle length approaches infinity. Proof. See Bukchin and Sarin (2004).
Proof. The proof is by contradiction. Assume that the compartments are not fully utilized and contain, in all (Q D1) items, where D1 > 0. As the total demand, D, is satisfied by this shipment, the setup rate is equal to D/(Q D1) > D/Q, and hence, not equal to D/Q. On the other hand, assume that a leftover of D2 items exists when the next shipment arrives. Since the shipments are identical, this leftover inventory does not satisfy any demand and hence can be eliminated from the shipment in the first place. In that case, the shipped quantity would be (Q D2) < Q, thereby leading to a setup rate larger than the lower bound. h We can thus conclude that the continuous and static loading policy can yield a solution equal to the lower bound only when the products are assigned to the compartments in proportion to their demands, and the demands of all the products are exactly satisfied until the arrival of the next identical shipment. However, due to the discrete nature of the problem, and the fact that not more than one product can be assigned to a compartment, this kind of solution is highly unlikely to obtain, if at all. As a result of Proposition 1, one can say that the ratio between the setup rate of a feasible solution and the lower bound gives an indication of underutilization of the vehicle. Consequently, when this ratio is equal to one, the vehicle is fully utilized, and as this ratio increases, the underutilization of the vehicle increases as well due to the non-intermixable nature of the product allocation.
h
In other words, S setups can be performed with completely filled compartments each time and zero leftover inventory at the end of period T. The setup rate of this solution is thus equal to the lower bound D/Q. Proposition 2 indicates that, theoretically, the continuous and static solution is dominated by the discrete and dynamic solution. However, we need to address the following two issues: 1. Length of cycle time (T). Although, from above, the discrete and dynamic solution is superior to the continuous and static solution for a relatively large cycle length (T), yet, because of the dynamic nature of the market in reality, we desire a relatively small value of T. If a large cycle length is determined, only a part of that cycle length may actually get to be implemented before the initiation of a new cycle due to a change in product demand or the addition of a new product. This fact will impair the quality of the solution. Hence, an examination of the shortest cycle length for which a discrete and dynamic solution is superior to the continuous static solution, is needed. 2. Length of a period. In practice, the length of a period (the time unit) takes on discrete values (e.g., day, week, month, quarter, etc.). The discrete and dynamic policy relies on this fact. Note that, the length of a period that is used, in relation to the cycle length, can impact the quality of the solution obtained. We expect that, on the average, the smallest cycle length,
Y. Bukchin, S.C. Sarin / European Journal of Operational Research 174 (2006) 1329–1337
T, for which the discrete and dynamic solution is superior to the continuous and static solution, is smaller for a shorter length of a period. However, we do not allow more than one shipment per period in the discrete and dynamic policy, and the shipment made in a period should satisfy the demand of that period. Consequently, the length of a period should be small enough so that the demand of that period, D, can be satisfied by the vehicle capacity, Q. In the next section, the practicality of the discrete and dynamic solutions is examined, while comparing the corresponding solutions with the continuous and static solutions under various lengths of a period and cycle lengths.
4. Comparison of the two approaches According to the static approach, the same allocation of products to the compartments is made in all deliveries. In the dynamic approach, on the other hand, we allow different allocations to be made for different deliveries. Consequently, we can view the dynamic approach as a relaxed version of the static approach and therefore, its solution dominates the solution of the static approach. To illustrate this, consider an example in which two products (A and B) are transported to a customer in a vehicle containing two compartments. The demand for these products is identical and is equal to one unit per period. For the sake of simplicity and without loss of generality, we assume that one unit of either product type takes the same volume in the vehicle. Consequently, each compartment can accommodate the same number of units of either product. Suppose one compartment can contain only one unit of a product while the other can contain three units. One can see that by applying the static loading policy, a shipment has to be made in every period, since the demand of the product which is allocated to the smaller compartment can be satisfied for one period only. Hence, the delivery rate for the static policy is equal to one. According to the dynamic loading policy, however, we can satisfy the demands by making two deliveries in every four periods. In the first delivery, performed in period 1, both com-
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partments are fully filled, compartment 1 with one unit of product A and compartment 2 with three units of product B. In the second delivery, the loading is reversed, that is, compartment 1 is filled with one unit of product B while compartment 2 with three units of product A. The demands for periods 3 and 4 are automatically satisfied by these two deliveries. If this policy is repeated every four periods, the average delivery rate, in accordance with the resultant cyclic policy, will be 0.5, namely, an average of one delivery every two periods. Note that, although the dynamic loading policy can be seen as a relaxed version of the static loading policy, there are some advantages of the static policy, and in some situations this policy may be preferred over the dynamic one. For instance, the static policy is much simpler to apply and is easier to manage. Moreover, in some cases, changing the content of a compartment from one material to another may require a setup (i.e., cleaning the compartment and preparing it for the new product). This type of setup is not considered here; however, in cases where a high cost is associated with such a setup, the static policy may turn out to be preferable to the dynamic one. In the following experimentation, we examine the length of the minimum cycle length for which the dynamic policy obtains a lower setup rate than that obtained by the static policy. We also study the impact that the length of a period makes on the quality of the solution obtained. All problems were solved via the ILOG OPL-Studio using the CPLEX solver. In this experimentation, ten 13-compartment, 7product problems were solved. Each problem was solved for two lengths of a period. Since the demand data is expressed as number of items per period, the data associated with different lengths of a period can be generated by multiplying the demand data by a different constant. For example, multiplying the per period demand by four is associated with changing the period length from weekly to monthly. In generating the set of data associated with a longer length of a period, we kept the capacity over demand ratio around 1.5. (Note that a value of this ratio smaller than one cannot yield a feasible solution since the vehicle
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which express the setup rate (2b). Rows (3) contain the setup rate obtained by the dynamic solution for each of the cycle lengths examined in this experimentation; six values are associated with the months in Table 1 and 24 with the weeks in Table 2. Row (4) displays the best dynamic solution obtained in (3). According to Proposition 2, the dynamic solution approaches the lower bound as the number of periods approaches infinity. Accordingly, its solution should outperform the static solution even when a continuous solution is allowed, that is, for a solution in which the number of periods satisfied by the delivery is fractional. Hence, we compare the two approaches, while relaxing the integrality constraint for the static solution. In this case, the continuous static solutions (in Row 2b) are compared with the dynamic solutions values (in Row 4). We can see that in only three out of the 10 cases of the longer length of a period (a month), presented in Table 1, the dynamic solution is superior to the static one (shown by the highlighted figures in Row (3)). On the average, the static solution value is smaller than the dynamic solution value by 1.56% (average of the values in Row (5c) in Table 1), and is 3.81% above the lower bound (5.63% for the dynamic solution), as can be seen in Rows (5a) and (5b). However, the results for the shorter length of a period (weeks), presented in Table 2,
capacity will be too small to satisfy the demand of one period.) In the set of data associated with a shorter length of a period, we assumed a period length of one fourth the size, and hence, the demand per period was divided by four. Assuming the shorter length of a period to be a week and the longer one to be a month, each problem was solved for different values of cycle length, T, for up to a maximal length of six months, under the dynamic policy. Hence, when the length of a period was considered to be one month, each problem was solved six times in accordance with the cycle lengths of one month, two months, and so on up to six months. Accordingly, the problem was solved 24 times for a length of a period of one week. Since the cycle length is irrelevant for static solution, each problem was optimally solved once for each period length. The results of the 10 problems for the longer (one month) and shorter (one week) lengths of a period are presented in Tables 1 and 2, respectively. The rows of these tables are labeled as shown on the right side of the first column. Row (1) of both tables depicts the lower bound, D/Q, on the setup rate. The continuous optimal solution of the static approach is presented in Rows (2a and 2b), first as the (continuous) number of periods over which their demands are satisfied by a single delivery (2a), and then as their inverse values
Table 1 Dynamic versus static approach—longer length of a period (month) Problem #
Row # 1
Problem LB
(1)
0.6638
S1: # Periods-Cont S2: Setup rate-Cont
(2a) (2b)
1.469 1.441 1.762 1.264 1.339 1.675 1.708 1.846 1.446 1.682 0.68085 0.69388 0.56757 0.79121 0.74666 0.59701 0.58537 0.54167 0.69136 0.59459
Cycle length (month) T=1 T=2 T=3 T=4 T=5 T=6
(3)
D: Setup rate
(4)
0.66667 0.75000 0.60000 0.75000 0.75000 0.60000 0.60000 0.60000 0.66667 0.60000
(S2 LB)/LB (%) (D LB)/LB (%) (S2 D)/D (%)
(5a) (5b) (5c)
2.56 0.43 2.13
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667
2
3 0.6720
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333
3.25 11.60 7.48
4 0.5511 0.7253
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
2.99 8.88 5.41
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333
9.09 3.41 5.49
5
6 0.7136
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333
4.63 5.10 0.44
7 0.5877
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
1.59 2.10 0.50
8 0.5675
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
3.14 5.72 2.44
9 0.5272 0.6503
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
2.75 13.81 9.72
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667
6.31 2.51 3.70
10
Average
0.5840
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
1.81 2.74 0.90
3.81 5.63 1.56
Y. Bukchin, S.C. Sarin / European Journal of Operational Research 174 (2006) 1329–1337
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Table 2 Dynamic versus static approach—shorter length of a period (week) Problem #
Row # 1
2
3
4
5
6
7
8
9
Problem LB
(1)
0.1660
0.1680
0.1378 0.1813 0.1784
0.1469 0.1419 0.1318
S1: # Periods-Cont S2: Setup rate-Cont
(2a) (2b)
5.875 0.17021
5.765 0.17347
7.048 5.056 5.357 0.14189 0.19780 0.18667
6.700 6.833 7.385 5.786 0.14925 0.14634 0.13542 0.17284
6.727 0.14865
Cycle time (weeks) T=1 T=2 T=3 T=4 T=5 T=6 T=7 T=8 T=9 T = 10 T = 11 T = 12 T = 13 T = 14 T = 15 T = 16 T = 17 T = 18 T = 19 T = 20 T = 21 T = 22 T = 23 T = 24
(3) 1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.167 0.231 0.214 0.200 0.188 0.176 0.222 0.211 0.200 0.190 0.182 0.174 0.208
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.250 0.231 0.214 0.200 0.188 0.176 0.222 0.211 0.200 0.190 0.182 0.174 0.208
1.000 0.500 0.333 0.250 0.200 0.167 0.143 0.250 0.222 0.200 0.182 0.167 0.154 0.143 0.200 0.188 0.176 0.167 0.158 0.150 0.143 0.182 0.174 0.167
1.000 0.500 0.333 0.250 0.200 0.167 0.286 0.250 0.222 0.200 0.182 0.167 0.154 0.214 0.200 0.188 0.176 0.167 0.158 0.150 0.190 0.182 0.174 0.167
1.000 0.500 0.333 0.250 0.200 0.167 0.286 0.250 0.222 0.200 0.182 0.167 0.154 0.214 0.200 0.188 0.176 0.167 0.158 0.150 0.190 0.182 0.174 0.167
D: Setup rate
(4)
0.16667
0.17391
0.14286 0.18182 0.18182
(S2 LB)/LB (%) (D LB)/LB (%) (S2 D)/D (%)
(5a) (5b) (5c)
2.56 0.43 2.13
3.25 3.51 0.26
2.99 3.69 0.68
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.250 0.231 0.214 0.200 0.188 0.235 0.222 0.211 0.200 0.190 0.227 0.217 0.208
9.09 0.28 8.79
show that the dynamic solution is superior to the static solution in six cases (see the highlighted figures in Row (3)). On the average, the dynamic solution value is smaller than the static solution value by 1.9% (Row (5c)), and is 1.9% above the lower bound (Row (5b)). A visual illustration of the solutions obtained for smaller and larger lengths of a period is given in Fig. 1. A reduction in the length of a period makes it easier for the dynamic approach to yield a close-to-optimum setup rate much faster. As can be seen above, the dynamic solutions are much improved when the length of a period becomes smaller. We expect cycle length to influence
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.250 0.231 0.214 0.200 0.188 0.235 0.222 0.211 0.200 0.190 0.182 0.217 0.208
4.63 1.91 2.67
1.000 0.500 0.333 0.250 0.200 0.167 0.286 0.250 0.222 0.200 0.182 0.167 0.154 0.143 0.200 0.188 0.176 0.167 0.158 0.150 0.190 0.182 0.174 0.208
1.000 0.500 0.333 0.250 0.200 0.167 0.143 0.250 0.222 0.200 0.182 0.167 0.154 0.143 0.133 0.188 0.176 0.167 0.158 0.150 0.143 0.136 0.174 0.167
0.1626
10
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.167 0.231 0.214 0.200 0.188 0.176 0.167 0.211 0.200 0.190 0.182 0.174 0.167
0.15000 0.14286 0.113333 0.16667 1.59 2.10 0.50
3.14 0.69 2.44
2.75 1.17 1.56
6.31 2.51 3.70
Average
0.1460
0.15000 1.81 2.74 0.90
3.81 1.90 1.90
solution quality as well. However, we expect solution quality to improve with increment in cycle length. In order to verify this behavior of cycle length on solution quality, additional problem instances were considered, allowing the cycle length to increase to up to 24 months (assuming the length of a period to be one month). The results are shown in Table 3. Note that, now, the resulting dynamic solutions are superior to the static solutions in all the problems (see the highlighted figures in Row (3)). The average difference is about 3%, while the maximum difference goes up to 8.79% in one case (see Row (5c)). The average percentage difference of the dynamic solution over the
Y. Bukchin, S.C. Sarin / European Journal of Operational Research 174 (2006) 1329–1337 1.200
1.2000
1.000
1.0000
Setup rate
Setup rate
1336
0.800 0.600
0.8000 0.6000
0.400
0.4000
0.200
0.2000 0.0000
0.000 1
5
9
13
17
1
21
5
9
Cycle length
13
17
21
Cycle length
Fig. 1. Dynamic versus static loading—different lengths of a period.
Table 3 Long period problem—cycle length smaller than or equal to 24 months Problem #
Row # 1
2
3
4
5
6
7
8
9
10
Problem LB
(1)
0.6638
0.6720
0.5511
0.7253
0.7136
0.5877
0.5675
0.5272
0.6503
0.5840
S1:# Periods-Cont S2: Setup rate-Cont
(2a) (2b)
1.469 1.441 1.762 1.264 1.339 1.675 1.708 1.846 1.446 1.682 0.68085 0.69388 0.56757 0.79121 0.74666 0.59701 0.58537 0.54167 0.69136 0.59459
Cycle length (months) T=1 T=2 T=3 T=4 T=5 T=6 T=7 T=8 T=9 T = 10 T = 11 T = 12 T = 13 T = 14 T = 15 T = 16 T = 17 T = 18 T = 19 T = 20 T = 21 T = 22 T = 23 T = 24
(3)
D: Setup rate
(4)
0.66667 0.68182 0.55556 0.72727 0.71429 0.59091 0.57143 0.52941 0.66667 0.58824
(S2 LB)/LB (%) (D LB)/LB (%) (S2 D)/D (%)
(5a) (5b) (5c)
2.56 0.43 2.13
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667 0.7143 0.7500 0.7778 0.7000 0.7273 0.6667 0.6923 0.7143 0.6667 0.6875 0.7059 0.6667 0.6842 0.7000 0.6667 0.6818 0.6957 0.6667
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333 0.7143 0.7500 0.7778 0.7000 0.7273 0.7500 0.6923 0.7143 0.7333 0.6875 0.7059 0.7222 0.6842 0.7000 0.7143 0.6818 0.6957 0.7083
3.25 1.46 1.77
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.5714 0.6250 0.5556 0.6000 0.6364 0.5833 0.6154 0.5714 0.6000 0.5625 0.5882 0.5556 0.5789 0.6000 0.5714 0.5909 0.5652 0.5833
2.99 0.81 2.16
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333 0.8571 0.7500 0.7778 0.8000 0.7273 0.7500 0.7692 0.7857 0.7333 0.7500 0.7647 0.7778 0.7368 0.7500 0.7619 0.7727 0.7391 0.7500
9.09 0.28 8.79
lower bound is also only 0.8% while that of the static solution is 3.81%.
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333 0.7143 0.7500 0.7778 0.8000 0.7273 0.7500 0.7692 0.7857 0.7333 0.7500 0.7647 0.7222 0.7368 0.7500 0.7619 0.7273 0.7391 0.7500
4.63 0.09 4.53
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.7143 0.6250 0.6667 0.6000 0.6364 0.6667 0.6154 0.6429 0.6000 0.6250 0.6471 0.6111 0.6316 0.6000 0.6190 0.5909 0.6087 0.6250
1.59 0.55 1.03
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.5714 0.6250 0.6667 0.6000 0.6364 0.5833 0.6154 0.5714 0.6000 0.6250 0.5882 0.6111 0.5789 0.6000 0.5714 0.5909 0.6087 0.5833
3.14 0.69 2.44
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.5714 0.6250 0.5556 0.6000 0.5455 0.5833 0.5385 0.5714 0.5333 0.5625 0.5294 0.5556 0.5789 0.5500 0.5714 0.5455 0.5652 0.5417
2.75 0.42 2.32
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667 0.7143 0.7500 0.6667 0.7000 0.7273 0.6667 0.6923 0.7143 0.6667 0.6875 0.7059 0.6667 0.6842 0.7000 0.6667 0.6818 0.6957 0.6667
6.31 2.51 3.70
Average
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.7143 0.6250 0.6667 0.6000 0.6364 0.6667 0.6154 0.6429 0.6000 0.6250 0.5882 0.6111 0.6316 0.6000 0.6190 0.5909 0.6087 0.6250
1.81 0.72 1.08
3.81 0.80 3.00
Hence, we can conclude that under a relatively small period length and a large cycle length, the
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dynamic approach provides in general better solutions than those provided by the static approach, even when a continuous time scale is considered.
References Bukchin, J., Sarin, S.C., 2004. A cyclic policy for the loading of multiple products on a vehicle with different compartment sizes. IIE Transactions 30 (7), 641–653. Dannenbring, D.G., Khumawala, B.M., 1973. An investigation of branch-and-bound methods for solving the segregated storage problem. AIIE Transactions 5 (3), 265– 274. Evans, J.R., Cullen, F.H., 1977. The segregated storage problem: Some properties and an effective heuristic. AIIE Transactions 9 (4), 409–413.
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Evans, J.R., Tsubakitani, S., 1993. Solving the segregated storage problem with BendersÕ partitioning. Journal of Operational Research Society 44 (2), 175–184. Fagerholt, K., 2001. Ship scheduling with soft time windows: An optimization based approach. European Journal of Operational Research 131, 559–571. Fagerholt, K., Christiansen, M., 2000. A combined ship scheduling and allocation problem. Journal of the Operational Research Society 51 (7), 834–842. Neebe, A.W., 1987. An improved, multiplier adjustment procedure for the segregated storage problem. Journal of Operational Research Society 38 (9), 815–825. Shilfer, E., Naor, P., 1961. Elementary theory of the optimal silo storage problem. Operations Research 12, 54–65. White, J.A., Francis, R.L., 1971. Solving a segregated storage problem using branch-and-bound and extreme point ranking methods. AIIE Transactions 3 (1), 37–44. Yuceer, U., 1997. A multi-product loading problem: A model and solution method. European Journal of Operational Research 101, 519–531.
Short Communication
Discrete and dynamic versus continuous and static loading policy for a multi-compartment vehicle Yossi Bukchin b
a,*
, Subhash C. Sarin
b,1
a Department of Industrial Engineering, Faculty of Engineering, Tel-Aviv University, Tel-Aviv 69978, Israel Grado Department of Industrial and Systems Engineering, Virginia Tech, 250 Durham Hall, Blacksburg, VA 24061, USA
Received 15 January 2004; accepted 8 March 2005 Available online 22 June 2005
Abstract In this paper, we address the problem of loading non-intermixable products in a vehicle consisting of compartments of different sizes. The demands of the products are different but uniform over time. The objective is to meet product demands and minimize setup rate (that is, the number of deliveries per unit time). Two approaches, namely, dynamic and static, are investigated and their performances are compared with each other. In the dynamic approach, deliveries are made in several discrete periods and, then, repeated in a cyclic fashion. In each of these deliveries, the allocation of products to compartments can be different. The static approach, on the other hand, assumes a continuous time scale and determines a single assignment of products to compartments that maximizes the time in which the product demands are fully satisfied by this single delivery. The comparison between the two approaches shows that the dynamic approach is superior to the static approach when a discrete time scale is considered. However, even when the discrete time scale constraint is relaxed, the dynamic approach still provides better results for relatively long cycle times. Ó 2005 Elsevier B.V. All rights reserved. Keywords: Assignment problem; Loading problem; Shipping plan
1. Problem definition and literature review We consider a situation in which a vehicle transports several non-intermixable products from * Corresponding author. Tel.: +972 3 640 7941; fax: +972 3 640 7669. E-mail address: [email protected] (Y. Bukchin). 1 Tel.: +1 540 231 7140; fax: +1 540 231 3322.
a supplier to a customer on a vehicle consisting of compartments of different sizes. During the shipment, the products have to be kept separately in different compartments. A known and constant demand for each of the products is assumed, and shortages are not permitted. The objective is to determine a loading policy of the vehicle which minimizes the number of shipments required per unit of time, assuming that the shipment (setup)
0377-2217/$ - see front matter Ó 2005 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2005.03.035
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cost is significantly higher than the inventory holding cost. We also assume that the supplier holds large enough quantities of the products so as to satisfy the allocation of any combination of products to compartments, limited only by the vehicle capacity, and that, the customer has enough storage space to accommodate the quantity of the products that are delivered. Although the vehicle routing and scheduling problems have been covered extensively in the literature, there is little reported on the problem on hand. Fagerholt and Christiansen (2000) have addressed a similar situation in the scheduling and allocation of cargo to a ship. They assume that each ship has a cargo hold which is flexible and can be adjusted to give different size sub-compartments. Only a single product is assumed for loading in each of these compartments during a delivery. Time windows are associated with the loading and unloading times of products, and there is a penalty cost for violating the time windows. They propose a set partitioning based approach for the solution of this problem. A modification of this model, which allows soft time windows due to the introduction of a penalty cost that is varied as a function of the deviation from the targeted time window, is presented in Fagerholt (2001). Another related problem, that has been widely addressed in the literature, is the segregated storage problem (SSP) presented in Shilfer and Naor (1961). In this problem, different types of grains are to be stored in the compartments of a silo such that there is at most one product in each compartment. It is assumed that product demands, storage capacities i.e., the sizes of the compartments in the silo, and the variable storage costs are known. Also, an external storage facility of infinite capacity is assumed to be available to accommodate overflow. The objective is to find an assignment of products to compartments such that the total cost of holding in the external storage, is minimized. Improvements in the solution procedures for SSP have been proposed by White and Francis (1971), Dannenbring and Khumawala (1973), Evans and Cullen (1977), Neebe (1987) and Evans and Tsubakitani (1993). The SSP is a special case of our problem as, in SSP, an assignment is sought for a single period only. However, a solution for the SSP problem
which does not require any external storage is equivalent to a feasible solution to our problem. Yuceer (1997) addresses the problem of allocating products to a vehicle having compartments of different sizes such that at most one product is assigned to a compartment. The objective is to maximize the minimum number of periods over which the demands of all the products are satisfied by a single delivery. No shortages are allowed. Consequently, the solution of this problem gives the time interval between consecutive (and identical) deliveries. Yuceer (1997), assumes a continuous time scale and that a shipment can be performed at any time. We call this policy a continuous and static loading policy, and discuss it in Section 2. We also present a discrete and dynamic loading policy in Section 3. According to this policy, a series of setups are performed and they are repeated cyclically. The length of a cycle consists of a finite number of time periods, while at most one setup can be performed in a period of this cycle. We show that given a large enough number of periods in a cycle, this policy outperforms the continuous and static loading policy of Yuceer (1997). This is demonstrated through a detailed experimentation presented in Section 4.
2. The continuous and static loading policy The continuous and static loading policy provides a loading plan, which maximizes the common replenishment time; namely, the time interval over which the collective demand for all the products is satisfied. This loading plan is then repeated subsequently. Let qi and dj represent the size of compartment i and the demand per period of product j, respectively. Then, the ILP model, for the continuous static policy, is of a max–min type, and is as follows: " # m X qi max min xij ; ð1Þ j dj i¼1 subject to : n X xij ¼ 1;
i ¼ 1; . . . ; m;
ð2Þ
j¼1
xij ¼ 0; 1;
i ¼ 1; . . . ; m; j ¼ 1; . . . ; n;
ð3Þ
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where xij is a binary variable and is equal to 1, if product type j, j = 1, . . . , n, is assigned to compartment i, i = 1, . . . , m, and 0, otherwise. We can see that this formulation focuses on the product whose demand is satisfied over the shortest interval of time and the objective function (1), aims at maximizing this interval. The solution value represents the time interval in between two consecutive identical setups. The constraint set (2) assures that at most one product is assigned to a single compartment, and the constraint set (3) captures the binary nature of xij.
3. The discrete and dynamic loading policy The discrete and dynamic loading policy, presented first in Bukchin and Sarin (2004), is different from the continuous and static loading policy due to the following two aspects:
However, since the cycle length is a decision variable, a search over different values of T is required to find a suitable value. For practical reasons, the desired cycle should be relatively short in length. Indeed, later on, we show that satisfactory (close to optimal) solutions can be obtained for cycles of relatively short length. Consequently, the necessary search over T is not too intensive, although the problem is not convex in T, as shown in Bukchin and Sarin (2004). The problem for determining the discrete and dynamic policy can be formulated as follows: min
T X
In order to compare these two approaches, we use the criterion pertaining to setup rate, that is, the number of setups required in a unit of time. Since the problem is solved for the infinite planning horizon, the setup rate associated with the continuous and static loading policy is the inverse of the objective function value presented in (1). However, the situation is a little bit more complicated for the discrete and dynamic loading policy, since it involves different setups. In order to provide a solution for the infinite planning horizon that is comparable to the static policy, we employ a cyclic dynamic loading policy, a policy in which a number of setups over a cycle length is repeated. Hence, the setup rate in this case is the ratio of the number of setups performed in a cycle and the cycle length. The following model characterizes the discrete and dynamic policy, given the cycle length T.
ð4Þ
st ;
t¼1
subject to: n X xijt 6 1;
i ¼ 1; . . . ; m; j ¼ 1; . . . ; n;
ð5Þ
t ¼ 1; . . . ; T ; i ¼ 1; . . . ; m;
ð6Þ
j¼1
st P 1. The setups that are performed involve different assignments of products to compartments instead of a single assignment as in the continuous static policy. 2. Time is treated as a discrete parameter rather than a continuous one (e.g., weeks, months, quarters and so on).
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n X
xijt ;
j¼1 t X m X k¼1
xijt ;
xijk qi P td j ;
t ¼ 1; . . . ; T ; j ¼ 1; . . . ; n;
i¼1
st ¼ ð0; 1Þ;
ð7Þ ð8Þ
where xijt is a binary variable and is equal to 1 if product type j is assigned to compartment type i in period t, and 0, otherwise, and st is a binary variable which is equal to 1 if a setup is incurred in period t, and 0, otherwise. Constraint set (5) assures that not more than one product is assigned to a single compartment during a setup. The setup variable, st, is set to 1 in period t, if at least one product type is delivered in that period by any of the compartments. This is captured in constraint (6). The capacity constraints (7) guarantee that no shortage occurs in any period for any product. The binary nature of the variables xijk and st is dictated by constraint (8), while the objective of minimizing the number of setups, for a given number of periods, is represented by (4). In the sequel, we show that there always exists a discrete and dynamic solution, which outperforms the continuous and static solution. Subsequently,
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based on a series of experiments, we show that, in most cases, such a solution can be found rather easily under practical operating conditions. Let DPdenote the total demand per period, n namely, j¼1 d j , and Q the total vehicle capacity, Pm q . Clearly, D/Q is a lower bound of the setup i¼1 i rate, if the problem is relaxed by permitting every compartment to accommodate any mixture of products.
The discrete and dynamic loading policy provides a solution in which S setups are performed in a cycle time of T periods. Since we assume that not more than one shipment can occur in a period, therefore S 6 T. Bukchin and Sarin (2004) show that, theoretically, a discrete and dynamic solution can be obtained that satisfies the condition of Proposition 1. This follows from the following result.
Proposition 1. The continuous and static solution is equal to the lower bound, D/Q, if and only if the compartments are fully utilized and the leftover inventory at the arrival of the next shipment is zero.
Proposition 2. The setup rate values of the discrete and dynamic solutions converge to the lower bound, D/Q, as the cycle length approaches infinity. Proof. See Bukchin and Sarin (2004).
Proof. The proof is by contradiction. Assume that the compartments are not fully utilized and contain, in all (Q D1) items, where D1 > 0. As the total demand, D, is satisfied by this shipment, the setup rate is equal to D/(Q D1) > D/Q, and hence, not equal to D/Q. On the other hand, assume that a leftover of D2 items exists when the next shipment arrives. Since the shipments are identical, this leftover inventory does not satisfy any demand and hence can be eliminated from the shipment in the first place. In that case, the shipped quantity would be (Q D2) < Q, thereby leading to a setup rate larger than the lower bound. h We can thus conclude that the continuous and static loading policy can yield a solution equal to the lower bound only when the products are assigned to the compartments in proportion to their demands, and the demands of all the products are exactly satisfied until the arrival of the next identical shipment. However, due to the discrete nature of the problem, and the fact that not more than one product can be assigned to a compartment, this kind of solution is highly unlikely to obtain, if at all. As a result of Proposition 1, one can say that the ratio between the setup rate of a feasible solution and the lower bound gives an indication of underutilization of the vehicle. Consequently, when this ratio is equal to one, the vehicle is fully utilized, and as this ratio increases, the underutilization of the vehicle increases as well due to the non-intermixable nature of the product allocation.
h
In other words, S setups can be performed with completely filled compartments each time and zero leftover inventory at the end of period T. The setup rate of this solution is thus equal to the lower bound D/Q. Proposition 2 indicates that, theoretically, the continuous and static solution is dominated by the discrete and dynamic solution. However, we need to address the following two issues: 1. Length of cycle time (T). Although, from above, the discrete and dynamic solution is superior to the continuous and static solution for a relatively large cycle length (T), yet, because of the dynamic nature of the market in reality, we desire a relatively small value of T. If a large cycle length is determined, only a part of that cycle length may actually get to be implemented before the initiation of a new cycle due to a change in product demand or the addition of a new product. This fact will impair the quality of the solution. Hence, an examination of the shortest cycle length for which a discrete and dynamic solution is superior to the continuous static solution, is needed. 2. Length of a period. In practice, the length of a period (the time unit) takes on discrete values (e.g., day, week, month, quarter, etc.). The discrete and dynamic policy relies on this fact. Note that, the length of a period that is used, in relation to the cycle length, can impact the quality of the solution obtained. We expect that, on the average, the smallest cycle length,
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T, for which the discrete and dynamic solution is superior to the continuous and static solution, is smaller for a shorter length of a period. However, we do not allow more than one shipment per period in the discrete and dynamic policy, and the shipment made in a period should satisfy the demand of that period. Consequently, the length of a period should be small enough so that the demand of that period, D, can be satisfied by the vehicle capacity, Q. In the next section, the practicality of the discrete and dynamic solutions is examined, while comparing the corresponding solutions with the continuous and static solutions under various lengths of a period and cycle lengths.
4. Comparison of the two approaches According to the static approach, the same allocation of products to the compartments is made in all deliveries. In the dynamic approach, on the other hand, we allow different allocations to be made for different deliveries. Consequently, we can view the dynamic approach as a relaxed version of the static approach and therefore, its solution dominates the solution of the static approach. To illustrate this, consider an example in which two products (A and B) are transported to a customer in a vehicle containing two compartments. The demand for these products is identical and is equal to one unit per period. For the sake of simplicity and without loss of generality, we assume that one unit of either product type takes the same volume in the vehicle. Consequently, each compartment can accommodate the same number of units of either product. Suppose one compartment can contain only one unit of a product while the other can contain three units. One can see that by applying the static loading policy, a shipment has to be made in every period, since the demand of the product which is allocated to the smaller compartment can be satisfied for one period only. Hence, the delivery rate for the static policy is equal to one. According to the dynamic loading policy, however, we can satisfy the demands by making two deliveries in every four periods. In the first delivery, performed in period 1, both com-
1333
partments are fully filled, compartment 1 with one unit of product A and compartment 2 with three units of product B. In the second delivery, the loading is reversed, that is, compartment 1 is filled with one unit of product B while compartment 2 with three units of product A. The demands for periods 3 and 4 are automatically satisfied by these two deliveries. If this policy is repeated every four periods, the average delivery rate, in accordance with the resultant cyclic policy, will be 0.5, namely, an average of one delivery every two periods. Note that, although the dynamic loading policy can be seen as a relaxed version of the static loading policy, there are some advantages of the static policy, and in some situations this policy may be preferred over the dynamic one. For instance, the static policy is much simpler to apply and is easier to manage. Moreover, in some cases, changing the content of a compartment from one material to another may require a setup (i.e., cleaning the compartment and preparing it for the new product). This type of setup is not considered here; however, in cases where a high cost is associated with such a setup, the static policy may turn out to be preferable to the dynamic one. In the following experimentation, we examine the length of the minimum cycle length for which the dynamic policy obtains a lower setup rate than that obtained by the static policy. We also study the impact that the length of a period makes on the quality of the solution obtained. All problems were solved via the ILOG OPL-Studio using the CPLEX solver. In this experimentation, ten 13-compartment, 7product problems were solved. Each problem was solved for two lengths of a period. Since the demand data is expressed as number of items per period, the data associated with different lengths of a period can be generated by multiplying the demand data by a different constant. For example, multiplying the per period demand by four is associated with changing the period length from weekly to monthly. In generating the set of data associated with a longer length of a period, we kept the capacity over demand ratio around 1.5. (Note that a value of this ratio smaller than one cannot yield a feasible solution since the vehicle
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which express the setup rate (2b). Rows (3) contain the setup rate obtained by the dynamic solution for each of the cycle lengths examined in this experimentation; six values are associated with the months in Table 1 and 24 with the weeks in Table 2. Row (4) displays the best dynamic solution obtained in (3). According to Proposition 2, the dynamic solution approaches the lower bound as the number of periods approaches infinity. Accordingly, its solution should outperform the static solution even when a continuous solution is allowed, that is, for a solution in which the number of periods satisfied by the delivery is fractional. Hence, we compare the two approaches, while relaxing the integrality constraint for the static solution. In this case, the continuous static solutions (in Row 2b) are compared with the dynamic solutions values (in Row 4). We can see that in only three out of the 10 cases of the longer length of a period (a month), presented in Table 1, the dynamic solution is superior to the static one (shown by the highlighted figures in Row (3)). On the average, the static solution value is smaller than the dynamic solution value by 1.56% (average of the values in Row (5c) in Table 1), and is 3.81% above the lower bound (5.63% for the dynamic solution), as can be seen in Rows (5a) and (5b). However, the results for the shorter length of a period (weeks), presented in Table 2,
capacity will be too small to satisfy the demand of one period.) In the set of data associated with a shorter length of a period, we assumed a period length of one fourth the size, and hence, the demand per period was divided by four. Assuming the shorter length of a period to be a week and the longer one to be a month, each problem was solved for different values of cycle length, T, for up to a maximal length of six months, under the dynamic policy. Hence, when the length of a period was considered to be one month, each problem was solved six times in accordance with the cycle lengths of one month, two months, and so on up to six months. Accordingly, the problem was solved 24 times for a length of a period of one week. Since the cycle length is irrelevant for static solution, each problem was optimally solved once for each period length. The results of the 10 problems for the longer (one month) and shorter (one week) lengths of a period are presented in Tables 1 and 2, respectively. The rows of these tables are labeled as shown on the right side of the first column. Row (1) of both tables depicts the lower bound, D/Q, on the setup rate. The continuous optimal solution of the static approach is presented in Rows (2a and 2b), first as the (continuous) number of periods over which their demands are satisfied by a single delivery (2a), and then as their inverse values
Table 1 Dynamic versus static approach—longer length of a period (month) Problem #
Row # 1
Problem LB
(1)
0.6638
S1: # Periods-Cont S2: Setup rate-Cont
(2a) (2b)
1.469 1.441 1.762 1.264 1.339 1.675 1.708 1.846 1.446 1.682 0.68085 0.69388 0.56757 0.79121 0.74666 0.59701 0.58537 0.54167 0.69136 0.59459
Cycle length (month) T=1 T=2 T=3 T=4 T=5 T=6
(3)
D: Setup rate
(4)
0.66667 0.75000 0.60000 0.75000 0.75000 0.60000 0.60000 0.60000 0.66667 0.60000
(S2 LB)/LB (%) (D LB)/LB (%) (S2 D)/D (%)
(5a) (5b) (5c)
2.56 0.43 2.13
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667
2
3 0.6720
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333
3.25 11.60 7.48
4 0.5511 0.7253
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
2.99 8.88 5.41
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333
9.09 3.41 5.49
5
6 0.7136
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333
4.63 5.10 0.44
7 0.5877
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
1.59 2.10 0.50
8 0.5675
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
3.14 5.72 2.44
9 0.5272 0.6503
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
2.75 13.81 9.72
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667
6.31 2.51 3.70
10
Average
0.5840
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667
1.81 2.74 0.90
3.81 5.63 1.56
Y. Bukchin, S.C. Sarin / European Journal of Operational Research 174 (2006) 1329–1337
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Table 2 Dynamic versus static approach—shorter length of a period (week) Problem #
Row # 1
2
3
4
5
6
7
8
9
Problem LB
(1)
0.1660
0.1680
0.1378 0.1813 0.1784
0.1469 0.1419 0.1318
S1: # Periods-Cont S2: Setup rate-Cont
(2a) (2b)
5.875 0.17021
5.765 0.17347
7.048 5.056 5.357 0.14189 0.19780 0.18667
6.700 6.833 7.385 5.786 0.14925 0.14634 0.13542 0.17284
6.727 0.14865
Cycle time (weeks) T=1 T=2 T=3 T=4 T=5 T=6 T=7 T=8 T=9 T = 10 T = 11 T = 12 T = 13 T = 14 T = 15 T = 16 T = 17 T = 18 T = 19 T = 20 T = 21 T = 22 T = 23 T = 24
(3) 1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.167 0.231 0.214 0.200 0.188 0.176 0.222 0.211 0.200 0.190 0.182 0.174 0.208
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.250 0.231 0.214 0.200 0.188 0.176 0.222 0.211 0.200 0.190 0.182 0.174 0.208
1.000 0.500 0.333 0.250 0.200 0.167 0.143 0.250 0.222 0.200 0.182 0.167 0.154 0.143 0.200 0.188 0.176 0.167 0.158 0.150 0.143 0.182 0.174 0.167
1.000 0.500 0.333 0.250 0.200 0.167 0.286 0.250 0.222 0.200 0.182 0.167 0.154 0.214 0.200 0.188 0.176 0.167 0.158 0.150 0.190 0.182 0.174 0.167
1.000 0.500 0.333 0.250 0.200 0.167 0.286 0.250 0.222 0.200 0.182 0.167 0.154 0.214 0.200 0.188 0.176 0.167 0.158 0.150 0.190 0.182 0.174 0.167
D: Setup rate
(4)
0.16667
0.17391
0.14286 0.18182 0.18182
(S2 LB)/LB (%) (D LB)/LB (%) (S2 D)/D (%)
(5a) (5b) (5c)
2.56 0.43 2.13
3.25 3.51 0.26
2.99 3.69 0.68
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.250 0.231 0.214 0.200 0.188 0.235 0.222 0.211 0.200 0.190 0.227 0.217 0.208
9.09 0.28 8.79
show that the dynamic solution is superior to the static solution in six cases (see the highlighted figures in Row (3)). On the average, the dynamic solution value is smaller than the static solution value by 1.9% (Row (5c)), and is 1.9% above the lower bound (Row (5b)). A visual illustration of the solutions obtained for smaller and larger lengths of a period is given in Fig. 1. A reduction in the length of a period makes it easier for the dynamic approach to yield a close-to-optimum setup rate much faster. As can be seen above, the dynamic solutions are much improved when the length of a period becomes smaller. We expect cycle length to influence
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.250 0.231 0.214 0.200 0.188 0.235 0.222 0.211 0.200 0.190 0.182 0.217 0.208
4.63 1.91 2.67
1.000 0.500 0.333 0.250 0.200 0.167 0.286 0.250 0.222 0.200 0.182 0.167 0.154 0.143 0.200 0.188 0.176 0.167 0.158 0.150 0.190 0.182 0.174 0.208
1.000 0.500 0.333 0.250 0.200 0.167 0.143 0.250 0.222 0.200 0.182 0.167 0.154 0.143 0.133 0.188 0.176 0.167 0.158 0.150 0.143 0.136 0.174 0.167
0.1626
10
1.000 0.500 0.333 0.250 0.200 0.333 0.286 0.250 0.222 0.200 0.182 0.167 0.231 0.214 0.200 0.188 0.176 0.167 0.211 0.200 0.190 0.182 0.174 0.167
0.15000 0.14286 0.113333 0.16667 1.59 2.10 0.50
3.14 0.69 2.44
2.75 1.17 1.56
6.31 2.51 3.70
Average
0.1460
0.15000 1.81 2.74 0.90
3.81 1.90 1.90
solution quality as well. However, we expect solution quality to improve with increment in cycle length. In order to verify this behavior of cycle length on solution quality, additional problem instances were considered, allowing the cycle length to increase to up to 24 months (assuming the length of a period to be one month). The results are shown in Table 3. Note that, now, the resulting dynamic solutions are superior to the static solutions in all the problems (see the highlighted figures in Row (3)). The average difference is about 3%, while the maximum difference goes up to 8.79% in one case (see Row (5c)). The average percentage difference of the dynamic solution over the
Y. Bukchin, S.C. Sarin / European Journal of Operational Research 174 (2006) 1329–1337 1.200
1.2000
1.000
1.0000
Setup rate
Setup rate
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0.800 0.600
0.8000 0.6000
0.400
0.4000
0.200
0.2000 0.0000
0.000 1
5
9
13
17
1
21
5
9
Cycle length
13
17
21
Cycle length
Fig. 1. Dynamic versus static loading—different lengths of a period.
Table 3 Long period problem—cycle length smaller than or equal to 24 months Problem #
Row # 1
2
3
4
5
6
7
8
9
10
Problem LB
(1)
0.6638
0.6720
0.5511
0.7253
0.7136
0.5877
0.5675
0.5272
0.6503
0.5840
S1:# Periods-Cont S2: Setup rate-Cont
(2a) (2b)
1.469 1.441 1.762 1.264 1.339 1.675 1.708 1.846 1.446 1.682 0.68085 0.69388 0.56757 0.79121 0.74666 0.59701 0.58537 0.54167 0.69136 0.59459
Cycle length (months) T=1 T=2 T=3 T=4 T=5 T=6 T=7 T=8 T=9 T = 10 T = 11 T = 12 T = 13 T = 14 T = 15 T = 16 T = 17 T = 18 T = 19 T = 20 T = 21 T = 22 T = 23 T = 24
(3)
D: Setup rate
(4)
0.66667 0.68182 0.55556 0.72727 0.71429 0.59091 0.57143 0.52941 0.66667 0.58824
(S2 LB)/LB (%) (D LB)/LB (%) (S2 D)/D (%)
(5a) (5b) (5c)
2.56 0.43 2.13
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667 0.7143 0.7500 0.7778 0.7000 0.7273 0.6667 0.6923 0.7143 0.6667 0.6875 0.7059 0.6667 0.6842 0.7000 0.6667 0.6818 0.6957 0.6667
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333 0.7143 0.7500 0.7778 0.7000 0.7273 0.7500 0.6923 0.7143 0.7333 0.6875 0.7059 0.7222 0.6842 0.7000 0.7143 0.6818 0.6957 0.7083
3.25 1.46 1.77
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.5714 0.6250 0.5556 0.6000 0.6364 0.5833 0.6154 0.5714 0.6000 0.5625 0.5882 0.5556 0.5789 0.6000 0.5714 0.5909 0.5652 0.5833
2.99 0.81 2.16
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333 0.8571 0.7500 0.7778 0.8000 0.7273 0.7500 0.7692 0.7857 0.7333 0.7500 0.7647 0.7778 0.7368 0.7500 0.7619 0.7727 0.7391 0.7500
9.09 0.28 8.79
lower bound is also only 0.8% while that of the static solution is 3.81%.
1.0000 1.0000 1.0000 0.7500 0.8000 0.8333 0.7143 0.7500 0.7778 0.8000 0.7273 0.7500 0.7692 0.7857 0.7333 0.7500 0.7647 0.7222 0.7368 0.7500 0.7619 0.7273 0.7391 0.7500
4.63 0.09 4.53
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.7143 0.6250 0.6667 0.6000 0.6364 0.6667 0.6154 0.6429 0.6000 0.6250 0.6471 0.6111 0.6316 0.6000 0.6190 0.5909 0.6087 0.6250
1.59 0.55 1.03
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.5714 0.6250 0.6667 0.6000 0.6364 0.5833 0.6154 0.5714 0.6000 0.6250 0.5882 0.6111 0.5789 0.6000 0.5714 0.5909 0.6087 0.5833
3.14 0.69 2.44
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.5714 0.6250 0.5556 0.6000 0.5455 0.5833 0.5385 0.5714 0.5333 0.5625 0.5294 0.5556 0.5789 0.5500 0.5714 0.5455 0.5652 0.5417
2.75 0.42 2.32
1.0000 1.0000 0.6667 0.7500 0.8000 0.6667 0.7143 0.7500 0.6667 0.7000 0.7273 0.6667 0.6923 0.7143 0.6667 0.6875 0.7059 0.6667 0.6842 0.7000 0.6667 0.6818 0.6957 0.6667
6.31 2.51 3.70
Average
1.0000 1.0000 0.6667 0.7500 0.6000 0.6667 0.7143 0.6250 0.6667 0.6000 0.6364 0.6667 0.6154 0.6429 0.6000 0.6250 0.5882 0.6111 0.6316 0.6000 0.6190 0.5909 0.6087 0.6250
1.81 0.72 1.08
3.81 0.80 3.00
Hence, we can conclude that under a relatively small period length and a large cycle length, the
Y. Bukchin, S.C. Sarin / European Journal of Operational Research 174 (2006) 1329–1337
dynamic approach provides in general better solutions than those provided by the static approach, even when a continuous time scale is considered.
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