Complexity and Algorithms for Two-Stage Flexible Flowshop Scheduling with Availability Constraints

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1 Complexity and Algorithms or Two-Stage Flexible Flowshop Scheduling with Availability Constraints Jinxing Xie, Xijun Wang Department o Mathematical Sciences, Tsinghua University, Beijing , China jxie@math.tsinghua.edu.cn Abstract This paper considers the two-stage lexible lowshop scheduling problem with availability constraints. We discuss the complexity and the approximability o the problem, and provide some approximation algorithms with inite and tight worst case perormance bounds or some special cases o the problem. Keywords: Scheduling; Flexible lowshop; Availability constraints; Approximability; Approximation algorithms; Worst-case analysis 1 Introduction A lexible lowshop, which is also known as hybrid lowshop or lowshop with parallel machines, consists o a set o machine centers with parallel machines. Scheduling problems or such kind o lowshops are irstly studied by Salvador in 1973 [1]. From then on, the lexible lowshop scheduling problems with makespan objective have attracted a great deal o attention. For example, the two-stage problem F 2 (P ) C max is already proved to be N P-hard in the strong sense, even in the case where there are only two machines in one stage and there is only a single machine in the other stage [2]. Various heuristics or approximation algorithms or these problems have been designed [3, 4, 5], or some o which the corresponding worst case perormance bounds and the average case perormance bounds have been proved. Hall [6], Schuurman and Woeginger [7] have discussed the existence o polynomial time approximation schemes (PTAS) or lexible lowshop scheduling problems. Recently, the scheduling problems or lexible low shops with sequencedependent setup times are considered, where some heuristics are proposed and their computational eectiveness are examined [8, 9]. Most studies on scheduling problems assume that the machines are available at all times. In real industry settings, however, a machine may not always be available in the scheduling period due to, or example, a breakdown (stochastic) or preventive maintenance (deterministic). This paper only considers the scheduling problem under the deterministic case, i.e., the unavailable time intervals called holes are deterministically known beore the decision making or the schedules starts. Corresponding author. This research has been supported by NSFC Project No

2 2 In some models with availability constraints, a job started its processing but not inished beore the unavailable period can be resumed without restarting when the machine becomes available again, which is usually the case in the ood industry. The job is then called resumable. Otherwise, i a job started its processing but not inished beore the unavailable period must be restarted later, which usually occurs in steel industry, it is called nonresumable. There is also a semiresumable case, in which a job started its processing but not inished beore the unavailable period will have to partially restart when the machine becomes available again [10]. This paper only considers the scheduling problems in the resumable case. Scheduling problems with limited machine availability have been studied to a less extent. Adiri et al. [11] showed that a single machine problem with machine breakdowns is N P-hard i the breakdowns are known in advance. Schmidt [12] studied parallel machine scheduling problems with availability constraints. Hwang and Chang [13] proved that the makespan o the PT (ongest Processing Time) schedule is bounded by twice the optimal makespan i no more than hal o the machines are allowed to be shutdown simultaneously. ee [14] proved that the two-machine lowshop problem with makespan objective and one unavailability period is N P-hard in the ordinary sense, and presented a dynamic programming approach and some approximation algorithms. Kubiak et al. [15] proved that the twomachine problem with arbitrary number o unavailability periods on one machine is NP-hard in the strong sense, and proposed a branch and bound algorithm and some heuristic algorithms [16]. ee [10] given the irst discussion about the semiresumable two-machine low shop model and provided worst case perormance analysis or some algorithms. Espinouse et al. [17] studied both the resumable and the nonresumable two-machine lowshop models under the no-wait environment. Recently, the lexible lowshop scheduling problems problems with limited machine availability are irstly studied by Wang and Xie [18], in which a bround and bound algorithm is presented. This paper discusses the complexity and approximability o two-stage lexible lowshop problems or the nonresumable case with deterministic availability constraints. In the next section, we give the notations o the problem. In Section 3, we point out the strongly N P-hardness irstly; then we prove the APX -hardness o the problems with holes on the whole second stage, or on the whole irst stage but with more than one machine at this stage. In Section 4, we analyze the worst case perormance o two algorithms or the case where there is only one machine at the irst stage and only one hole on that machine. In section 5, we provide an algorithm with inite worst case perormance bound or the special case where the unavailable periods apply only on a part o machines at some machine centers. In the last section, we conclude this paper with a short discussion. 2 Problem ormulation et us suppose that the lexible lowshop consists o a set o m 2 machine centers [Z 1, Z 2,..., Z m ] with center Z j having m j 1 identical parallel machines {M j1, M j2,..., M jmj }. There are n jobs {J i 1 i n} to be processed in the lowshop. Function p(j i ) = [p i1, p i2,..., p im ] will denote the processing times required by J i on

3 3 centers [Z 1, Z 2,..., Z m ] respectively. For convenience unavailable periods will be called holes. We assume that the holes do not overlap, i.e., on each machine the holes do not cross over with each other, since otherwise they should be considered as one hole only. Thereore we can denote by s jk,g and l jk,g the starting time and the length, respectively, o the hole g on the machine M jk, numbered according to their starting times. Each machine can process at most one job at a time and each job can be processed by at most one machine at a time. The objective considered is to minimize the makespan o the schedule (the maximum completion time o all jobs). We use in this paper the notation similar to the one described in [15]. Speciically, F 2 (P ), h jk,g C max represents the two-stage lexible lowshop problem with an arbitrary number o holes on each machine. F 2 (P ), h 1k,g C max represents the problem with an arbitrary number o holes on each machine at the irst stage but no holes on the second stage. F 2 (P ), h 11,1 C max represents the problem with one hole on machine M 11 only. In this notation h jk,g speciies the number o the hole(s) and the machine(s) on which they appear. I j or k is replaced by a positive integer, it means that holes are only on those machines. Otherwise, holes will be allowed to be on all machines. I g is replaced by a positive integer, it denotes the number o holes on the corresponding machine. Otherwise this number is arbitrary. 3 N P-hardness and APX -hardness Due to the strongly N P-hardness o both F 2 (P ) m 1 2, m 2 = 1 C max and F 2 (P ) m 1 = 1, m 2 2 C max [2], Theorem 1 stands obviously. Theorem 1 F 2 (P ), h 11,1 m 1 = 1, m 2 2 C max, F 2 (P ), h 21,1 m 1 2, m 2 = 1 C max, F 2 (P ), h 1k,1 m 1 2, m 2 = 1 C max and F 2 (P ), h 2k,1 m 1 = 1, m 2 2 C max are N P-hard in the strong sense. During the approximability discussions in the ollowing, several speciic notations will be used. C max(i) and C H max(i) denote, respectively, the optimal makespan and the makespan given by algorithm H or any instance I. They are oten abbreviated to C max and C max respectively. An algorithm or problems minimizing makespan is called having worst case perormance bound ρ i C max ρc max or all instances. From this deinition it is easily to see that ρ 1 always holds. Furthermore, the bound is called tight i, or all > 0, there exists instance I satisying C max > (ρ )C max. Theorem 2 A polynomial time algorithm or the problem F 2 (P ), h 21,1 m 1 2, m 2 = 1 C max with a inite worst case perormance bound cannot be ound unless P = N P. Proo The theorem will be proved by a contradiction. et us suppose now that there exists an algorithm H which gives a solution in polynomial time or this problem with worst case bound R, i.e., C H max(i) RC max(i) or all instances o F 2 (P ), h 21,1 m 1 2, m 2 = 1 C max. For any instance I o problem F 2 (P ) m 1 2, m 2 = 1 C max and or any integer y > 0, we construct instance I o F 2 (P ), h 21,1 m 1 2, m 2 = 1 C max as ollows: All are the same

4 4 to instance I except that there is one hole with s 21,1 = y, l 21,1 = Ry on machine M 21. We can see that there exists a schedule satisying C max y or instance I i and only i the schedule generated by algorithm H satisies C max y or instance I. This is inconsistent with the N P-hardness o F 2 (P ) m 1 2, m 2 = 1 C max unless P = N P. emma 3 A polynomial time algorithm or the problem P, h k,1 C max with a inite worst case bound cannot be ound unless P = N P. Proo Similarly to the proo o Theorem 2, the existence o a polynomial time algorithm or P, h k,1 C max with a inite worst case bound will be inconsistent with the N P-hardness o P C max unless P = N P [19]. From emma 3, Theorem 4 holds obviously. Theorem 4 A polynomial time algorithm or the problem F 2 (P ), h 1k,1 m 1 2, m 2 = 1 C max and F 2 (P ), h 2k,1 m 1 = 1, m 2 2 C max with a inite worst case bound cannot be ound unless P = N P. Corollary 5 A polynomial time algorithm or the problem F 2 (P ), h 1k,1, h 21,1 m 1 2, m 2 = 1 C max and F 2 (P ), h 11,1, h 2k,1 m 1 = 1, m 2 2 C max with a inite worst case bound cannot be ound unless P = N P. However, unlike F 2 (P ), h 21,1 m 1 2, m 2 = 1 C max, there exist polynomial time algorithms or F 2 (P ), h 11,1 m 1 = 1, m 2 2 C max with inite worst case bounds. This result can be seen rom the discussions in the next section, and it diers very much rom the symmetry between F 2 (P ) m 1 2, m 2 = 1 C max and F 2 (P ) m 1 = 1, m 2 2 C max. 4 Worst case perormance o algorithms or F 2 (P ), h 11,1 m 1 = 1, m 2 2 C max 4.1 ist scheduling algorithm In this algorithm, a list (or permutation) o the job indices 1, 2,..., n is provided. Jobs are ed to the irst machine center Z 1 in the order they appear on the list. Since we have only one machine at Z 1, the jobs processed at Z 1 orm a queue at the buer between the center Z 1 and Z 2 in the same order. Then the jobs are processed in the order, whenever the machine at Z 2 becomes available. Theorem 6 For any instance o F 2 (P ), h 11,1 m 1 = 1, m 2 = µ 2 C max, let and be the inish time o an optimal schedule and the schedule obtained by the list scheduling algorithm respectively. Then bound is tight. 4 1 µ, and this Proo In this problem, there is only one machine M 11 in the irst machine center, and there are µ identical machines in the second machine center. Besides, there is only one hole and it is on M 11, where the starting time and the length o the hole are denoted by s 11,1 and l 11,1 respectively. I the tasks assigned to M 11 by the list scheduling algorithm are all beore s 11,1, and will be the same respectively to those in the case the availability constraint is discarded. From [4], we have 3 1 µ.

5 5 Now suppose that the tasks assigned to M 11 by the list scheduling algorithm are not all beore s 11,1. et 1 be the inish time on M 11 o the schedule obtained by the list scheduling algorithm. First we process all n tasks {p i1 } on Z 1. And we start the processing o any tasks o {p i2 } only ater the completions o all o {p i1 } on Z 1. Then the processing o {p i2 } on Z 2 can be treated as a parallel-machine shop. The tasks p i1 and p i2, 1 i n, have to be carried out in the order speciied in the list. et 2 be the inish time o all o {p i2 } when they are processed in the parallel-machine shop Z 2 in the order o the list. It should be clear that n 1 s 11,1 + l 11,1 + p i1 ; (1) (2) According to the assumptions o the problem, we have s 11,1 + l 11,1 ; (3) n h 11,1 + p i1. (4) Please note that there are µ identical machines in the second machine center. From [20], we have µ, (5) where 2 is the optimal inish time o the operations set {p i2 } in the parallel-machine shop Z 2. Besides, it is obvious that 2. (6) Using inequalities (1) (6), we obtain s 11,1 + l 11,1 + n p i µ. A worst case instance is constructed using the ollowing job set J: p[j a ] = [2µ, ]; p[j b ] = [µ (2µ 1), ]; p[j c (i)] = [, (µ 1)], 1 i µ 1; p[j d (i)] = [, ], 1 i µ 1; p[j e ] = [, µ];

6 6 where and the only hole on M 11 is s 11,1 = µ, l 11,1 =. Using the processing sequence [J a, J b, J c, J d, J e ], the schedule obtained by the list scheduling algorithm is shown in Figure 1(a). The optimal schedule is shown in Figure 1(b) in the processing sequence [J d, J e, J c, J b, J a ]. Thereore, we have = = (4µ 1) (2µ 3) µ + (2µ + 2) (4µ 1) (2µ 3)/. µ + (2µ + 2)/ Thus, lim / 0 = 4 1 µ. This completes the proo. M 11 M 21 M 22 M 23 M 2,µ 1 M 2µ 2µ µ µ+ µ (2µ 1) µ (a) Schedule by list scheduling algorithm: = (4µ 1) (2µ 3) M 11 M 21 M 22 M 23 M 2,µ 1 M 2µ µ (2µ 1) µ µ µ+ 2µ (b) Optimal schedule: = µ + (2µ + 2) Figure 1: A worst case instance or list scheduling algorithm 4.2 PT algorithm In the PT (ongest Processing Time) algorithm, we sort the job set according to the processing time on M 11 in non-increasing order, then we use the list scheduling algorithm. Theorem 7 For any instance o F 2 (P ), h 11,1 m 1 = 1, m 2 = µ 2 C max, let and be the inish time o an optimal schedule and the schedule obtained by the PT algorithm respectively. Then tight µ, and this bound is Proo The problem is the same as that o Theorem 6: There is only one machine M 11 in the irst machine center, and there are µ identical machines in the second machine center. Besides, there is only one hole and it is on M 11, where the starting time and the length o the hole are denoted by s 11,1 and l 11,1 respectively. The only dierence is that now we use PT algorithm instead o a general list scheduling algorithm. Similarly to the proo o Theorem 6, i the tasks assigned to M 11 by PT algorithm are all beore s 11,1, we have 3 1 µ rom [4].

7 7 Now suppose that [J 1, J 2,, J b ] and [J b+1,, J n ] are assigned respectively to M 11 beore and ater the hole in that order, where 0 b < n. Similarly to Theorem 6, we deine 1, 2 and 2. Then inequalities (1) (6) still hold. Furthermore, we have Using equation (7), we obtain n 1 = s 11,1 + l 11,1 + p i1. (7) i=b = s 11,1 b p i1 + l n 11,1 + p i1 + 2 s 11,1 b p i µ. (8) I b = 0, we have s 11,1 p 11, (9) s 11,1 + l 11,1 + p 11. (10) Then s 11,1 b p i1 = s 11,1 s 11,1 s 11,1 + l 11,1 + p (11) Then I b 1, rom our algorithm we have b b+1 p i1 s 11,1 < p i1. (12) s 11,1 b p i1 p b+1,1 l 11,1 + n p i1 p b+1,1 p b,1 + p b+1, (13) So we can see that µ holds or all instances. The ollowing instance shows that this bound is tight. The job set is given as: p[j a ] = [ µ 2 +, ]; p[j b ] = [ µ 2 (2µ 1), ]; p[j c (i)] = [, (µ 1)], 1 i µ 1; p[j d (i)] = [, ], 1 i µ 1; p[j e ] = [, µ];

8 8 M 11 M 21 M 22 M 23 M 2,µ 1 M 2µ µ 2 µ 2 + µ 2 + µ 2 (2µ 1) µ (a) Schedule by PT algorithm: = ( 7 µ 1) (2µ 4) 2 M 11 M 21 M 22 M 23 M 2,µ 1 M 2µ µ 2 (2µ 1) µ 2 + µ 2 µ 2 + µ (b) Optimal schedule: = µ + µ Figure 2: A worst case instance or PT algorithm where and the only hole is on M 11 with s 11,1 = µ 2, l 11,1 =. The schedule obtained by PT algorithm is shown in Figure 2(a) in the processing sequence [J a, J b, J c, J d, J e ]. The optimal schedule is shown in Figure 2(b) in the processing sequence [J d, J e, J c, J b, J a ]. Thereore, we have lim / 0 = µ. This completes the proo. 5 Algorithm or a special case o F 2 (P ), h 1k,1 m 1 2, m 2 = 1 C max From the previous discussions, we know that the lexible lowshop sheduling problems with availability constraints are diicult to be solved because o their APX -hardness properties. According to Theorem 4, both the problems F 2 (P ), h 1k,1 m 1 2, m 2 = 1 C max and F 2 (P ), h 2k,1 m 1 = 1, m 2 2 C max are APX -hard. However, in the industrial reality, it s unusual that all the machines at one machine center are breakdown in the same time. By making use o some special limitations on the availability constraints on machines, the corresponding problem might be much easier to be solved. For example, Hwang and Chang [13], Cheng and Wang [21] make some eorts toward this direction or parallel machine scheduling and lowshop scheduling, respectively. According to their work, it will be possible to approximate the problems with unavailable periods only on a part o machines at some machine centers. However, they do not consider the lexible lowshop problems which are the ocus o this paper. In act, as we can see rom the ollowing discussion, i we relax the constraints that the holes can be presented on all the machines at all machine centers, then the lexible lowshop problems might not be APX -hard anymore. For example, under the situation where at any time the number o unavailable machines does not exceed one hal o the number o all the machines in each machine center, there exist polynomial time algorithms with inite worst case bounds or the

9 9 problem F 2 (P ), h 1k,1 m 1 2, m 2 = 1 C max. One algorithm or this problem can be described as ollows: Step 0: Sort all the jobs in the non-increasing order in terms o p i1 and denote the sequence o jobs as J1. That s to say, J1 is the PT sequence o jobs in terms o p i1. Step 1: Assign the jobs to the irst machine center Z 1 or processing as early as possible in according to the sequence J1. et the completion time or job i at Z 1 be C i1. Sort all the jobs in the non-decreasing order in terms o C i1 and denote the new sequence o jobs as J2. Step 2: Assign the jobs to the machine in machine center Z 2 or processing as early as possible in according to the sequence J2. Step 3: Calculate the makespan and stop. Theorem 8 For any instance o F 2 (P ), h 1k,1 m 1 2, m 2 = 1 C max where at any time the number o unavailable machines does not exceed one hal o the number o the machines (at the irst stage), let and be the makespan o an optimal schedule and the schedule obtained by the above algorithm respectively. Then 3. Proo Consider the sub-problem with only the irst machine center as a classical parallel scheduling problem P C max. et 1 and 1 be the makespan o the optimal schedule and the schedule obtained by the PT algorithm [13] respectively. Then we have 1, (14) n p i2, (15) 1 + According to [13], n p i2. (16) (17) From (14) (17), n 1 + p i2 n 1 + p i2 3. (18) This completes the proo. 6 Discussion In this paper, we present irstly the ormulation o two-stage lexible lowshop problems with available constraints. Then we discuss the complexity and approximability o these models. We show that they are much more diicult to approximate than the case without availability constraints, except F 2 (P ), h 11,1 m 1 = 1, m 2 2 C max. However,

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