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Erratum to: “Simultaneously scheduling n jobs and the preventive maintenance on the two-machine flow shop to minimize the makespan” [Int. J. Prod. Econ. 112 (2008) 161–167]
Int. J. Production Economics 153 (2014) 361–363
Contents lists available at ScienceDirect
Int. J. Production Economics journal homepage: www.elsevier.com/locate/ijpe
Erratum
Erratum to: “Simultaneously scheduling n jobs and the preventive maintenance on the two-machine flow shop to minimize the makespan” [Int. J. Prod. Econ. 112 (2008) 161–167] Ahmed Gara-Ali n, Marie-Laure Espinouse Laboratoire G-SCOP, Institut National Polytechnique de Grenoble (INPG), 46 avenue Félix Viallet, 38031 Grenoble Cedex 1, France
art ic l e i nf o
a b s t r a c t
Article history: Received 4 April 2013 Accepted 23 February 2014 Available online 15 March 2014
In Allaoui et al. (2008), ‘Simultaneously scheduling n jobs and the preventive maintenance on the twomachine flow shop to minimize the makespan’ published in the International Journal of Production Economics, the authors show that the problem of two machine flow shop scheduling with a single maintenance occurring on the second machine is NP-hard (Theorem 7). In this Erratum we establish that this theorem is false and provide a polynomial time algorithm to solve this problem. An extension is proposed for further research in this area. We also highlight that the proof of Proposition 6 is incorrect and we propose a new proof for this proposition. & 2014 Elsevier B.V. All rights reserved.
Keywords: Scheduling Preventive maintenance Flow shop
1. The studied model in Allaoui et al. (2008) In Allaoui et al. (2008), ‘Simultaneously scheduling n jobs and the preventive maintenance on the two-machine flow shop to minimize the makespan’, the model studied is presented in a confusing way. At the end of page 162: “The maximum allowed continuously working time of the first machine or the second machine is T and the maintaining time is g.”, which is clear up to this point. But after this assertion the authors add that “A machine must be maintained exactly once with a constant maintenance length g during the given time interval [0, T]”. This sentence leaves us with two different problems: A. Find a schedule such that there is at most T time units between two consecutive maintenance periods, B. Find a schedule with exactly one maintenance period, and this maintenance starting in time interval [0, T]. Furthermore, the last sentence of the page 162: “To ensure that each machine will be maintained only one time during the scheduling horizon we assume that 2T is an upper bound of the makespan”, is incorrect. In fact, it is easy to see that this upper
bound does not guarantee that only one maintenance takes place (in model (A)). Consider an instance constituted of 3 jobs, each one of duration 2=3T 1, and with a maintenance duration of 1. Clearly 2 maintenances are requested in model (A), and yet the resulting makespan is 2T 1. Based on the results presented in Allaoui et al. (2008), we assume in the remainder of this paper that the original article (Allaoui et al., 2008) deals with problem (B). In particular, the maintenance period may occur at the beginning of the time horizon. This may be an undesirable situation, since the machine then operates all the jobs without maintenance. For this reason, we also consider in Section 4 the case where the maintenance is not allowed at the beginning of the schedule. Whether the maintenance period is allowed at the beginning or not, we prove that the problem can be optimally solved in polynomial time. The remainder of this paper is organized as follows. In Section 2 we propose a new proof for Proposition 6. In Section 3, some important properties are provided. We show that Theorem 7 is incorrect and we provide a polynomial solution algorithm for this problem. We also study an extension of this problem in Section 4.
2. The maintenance period on the first machine ðF 2 nrðM 1 Þ C max Þ DOI of original article: http://dx.doi.org/10.1016/j.ijpe.2006.08.017 Corresponding author. Tel.: þ 33 4 76575055. E-mail addresses: [email protected] (A. Gara-Ali), [email protected] (M.-L. Espinouse). n
http://dx.doi.org/10.1016/j.ijpe.2014.02.018 0925-5273 & 2014 Elsevier B.V. All rights reserved.
In Section 4 of Allaoui et al. (2008), the authors show that the two-machine flow shop problem with a maintenance period on the first machine is NP-hard and they propose an algorithm (H1).
362
A. Gara-Ali, M.-L. Espinouse / Int. J. Production Economics 153 (2014) 361–363
The authors state in Proposition 6 that if all processing times on the first machine are equal then H1 gives the optimal solution. Proposition 6 in Allaoui et al. (2008) is true, but the proof is incorrect. Indeed the proof intends to show that a local exchange between a job scheduled before the maintenance and a job scheduled after the maintenance in H1 can only increase the makespan. However this only establishes that H1 produces a local optimum relatively to this 2-OPT neighborhood. A strong property is missing to assert that such local optimum is necessarily a global optimum. We hereby propose a correct proof. Without loss of generality we assume that the jobs are indexed by non-increasing order of their processing time bi on the second machine, and we denote by a the processing time of each job on the first machine. Consider the schedule obtained by inserting the maintenance period at the last possible instant in this sequence, that is after job k such that k ¼ T=a. We assume in the following that k o n, otherwise all the jobs can fit before time T and the problem is trivial. Note that this is one of the schedules constructed by heuristic H1. We prove below that this schedule is optimal. It is easy to see that the complexity of this heuristic is Oðn log ðnÞÞ. Consider the last block of jobs without idle time on the second machine, and let u be the first job of this block. That is, no idle time appears after the processing of job u on the second machine. If C 1u denotes the completion time of u on the first machine, the makespan of the schedule is then given by C max ¼ C 1u þ∑ni¼ u bi . We consider two cases. Assume first that u 4 k, which means that u starts after the maintenance. As a consequence we have C max ¼ au þg þ ∑ni¼ u bi . Now consider any sequence π. The uth job in π; πðuÞ, necessarily starts after the maintenance since at most k jobs can be processed before time T on the first machine. Hence the makespan of π is at least C πmax Zau þg þ ∑ni¼ u bπðiÞ . By our choice of the indexation, ∑ni¼ u bðiÞ corresponds to the sum of the ðn u þ 1Þ smallest processing times on the second machine. It results that C max r C πmax for any sequence π, and thus our schedule is optimal. Conversely consider that u ok. In particular u is not the last job of the schedule. It implies that we must have bu Z a to avoid any idle time before the next job ðu þ 1Þ. However, due to our indexation, bi Z a then holds for all the jobs sequenced before u. It is immediate to see that necessarily u is in fact the first job, u ¼ 1: Hence the makespan of the sequence is C max ¼ a þ ∑ni¼ u bðiÞ , which is a trivial lower bound. The result follows. □
But if the scheduling problem has a yes answer, the partition problem could have answer: no. So the complexity proof proposed by Allaoui et al. is false. The following example proves that this proof is incorrect: For the partition problem, we consider n ¼ 2, a1 ¼ 1, a2 ¼ 3, so A ¼ 2. The corresponding instance in the reduction is constituted of 3 jobs of processing times p1;1 ¼ 1; p2;1 ¼ 3; p1;2 ¼ 3; p2;2 ¼ 9; p1;3 ¼ 0 and p2;3 ¼ amax ¼ 3: For the maintenance period T ¼ 9 and g ¼ A2 ¼ 4. Consider the sequence fT 3 ; T 1 ; T 2 g presented in the following (Fig. 1). With the solution presented in Fig. 1, the makespan is 3A2 þ 2A þ amax ¼ 19 and S1 a A, S2 a A. Therefore the proof is false. □
3. The maintenance period on the second machine ðF 2 nrðM 2 Þ C max Þ
The question now is to determine which job will be scheduled before the maintenance. The remainder of jobs will be scheduled after the maintenance period according to Johnson's rule (Proposition 8. Page 166 (Allaoui et al., 2008)). The following example proves that the first job in an optimal schedule is not necessarily the first job of Johnson's sequence. Consider the instance with T ¼ 7 and g ¼ 2, we have two jobs to schedule, whose processing times are given below:
In Allaoui et al. (2008), the authors proved by a reduction from the partition problem that the problem of two machine flow shop scheduling is NP-hard in the ordinary sense. We show that this theorem is false. Theorem 7. (H. Allaoui et al. P.166) The problem F 2 jnrðM 2 ÞjC max is NP-hard. In this theorem, authors consider that the second machine must be maintained exactly once with a constant maintenance length g during the given time interval [0, T]. The authors consider the following partition problem: Given a set of n integers a1 ; a2 ; …; an ; which total to 2A, can this set be partitioned into two disjoint subsets S1 and S2 such that the sum of each is equal to A? From this partition problem the authors define the following instance: p1;i ¼ ai , p2;i ¼ ai ðA þ 1Þ ði ¼ 1; …; nÞ and p1;n þ 1 ¼ 0; p2;n þ 1 ¼ amax ; where amax ¼ max fai =i ¼ 1; …; ng. Also set T ¼ A2 þ A þamax and g ¼ A2 . It is asked whether a schedule S exists for the problem such that C smax ¼ 3A2 þ 2A þ amax ?. As stated in Allaoui et al. (2008), if the partition problem has a positive yes answer then the scheduling problem has a yes answer.
3.1. Complexity of problem F 2 nrðM 2 Þ C max Proposition 1. There exists an optimal sequence for F2 nr ðM 2 Þ C max such that the maintenance period starts at the beginning of the schedule and all the jobs are sequenced in Johnson order (Johnson, 1954). Proof. See proof of Lemma 1 of Lee (1997). □ Based on the above analysis, Algorithm 2 (H2) developed by the authors (Allaoui et al., 2008) can solve the problem optimally. Notice that a maintenance at time 0 can make sense. One may imagine that the previous maintenance occurred a long time before, and that the next preventive maintenance is due to be performed no later than T times unit from the current instant, time 0. The case where the maintenance cannot take place at time 0 is studied in the following section. 4. An extension of F 2 nrðM 2 Þ C max We consider in this section that the maintenance is not allowed at the beginning of the schedule. We show that the problem is still polynomially solvable. Proposition 2. There exists an optimal sequence such that exactly one task is scheduled before the maintenance on the second machine. Proof. This assertion can be proved by swapping the last task scheduled before the maintenance with the maintenance. Such a swap will never increase the makespan. □
Jobs p1;i p2;i
1 3 4
Fig. 1. Example.
2 2 1
A. Gara-Ali, M.-L. Espinouse / Int. J. Production Economics 153 (2014) 361–363
363
The JO sequence is (1, 2). The optimal solution is obtained with this sequence (2, 1), the maintenance period is inserted after job 2 and results in a makespan of 9. Based on the above analysis, we propose the following algorithm to solve the problem.
Proof. Step 2 can be solved in OðnÞ time for each J i . Step 1 is executed only once. Step 3 does not require any computational effort. Consequently, the overall time requirement of Algorithm 1 is Oðn2 Þ. □
Algorithm 1 (A1). Let J ¼ fJ 1 ; J 2 ; …; J n g set of n jobs to be processed. Set C A1 max ¼ þ 1, the makespan given by Algorithm 1 (A1). Step 1. Index the jobs according to Johnson order. Step 2. For each J i A J , create a schedule with the job J i scheduled before the maintenance and the remainder of jobs will be scheduled after the maintenance, set C max ðJ i Þ the makespan given by this sequence. If C max ðJ i Þ r C A1 max , then update C A1 max : ¼ C max ðJ i Þ. Step 3. The optimal schedule is the one with the minimum makespan ðC A1 max ).
Acknowledgments
Theorem 1. Algorithm 1 (A1) will find an optimal solution for the problem F 2 jnr ðM k ÞjC max in Oðn2 Þ time.
The authors wish to thank the anonymous reviewer for his helpful comments and in particular on Section 1 and the proof of Proposition 6. References Allaoui, H., Lamouri, S., Artiba, A., Aghezzaf, E., 2008. Simultaneously scheduling n jobs and the preventive maintenance on the two-machine flow shop to minimize the makespan. Int. J. Prod. Econ. 112 (1), 161–167. Johnson, S.M., 1954. Optimal two- and three-stage production schedules with setup times included. Nav. Res. Logist. Q. 1 (1), 61–68. Lee, C.-Y., 1997. Minimizing the makespan in the two-machine flowshop scheduling problem with an availability constraint. Oper. Res. Lett. 20 (3), 129–139.
Contents lists available at ScienceDirect
Int. J. Production Economics journal homepage: www.elsevier.com/locate/ijpe
Erratum
Erratum to: “Simultaneously scheduling n jobs and the preventive maintenance on the two-machine flow shop to minimize the makespan” [Int. J. Prod. Econ. 112 (2008) 161–167] Ahmed Gara-Ali n, Marie-Laure Espinouse Laboratoire G-SCOP, Institut National Polytechnique de Grenoble (INPG), 46 avenue Félix Viallet, 38031 Grenoble Cedex 1, France
art ic l e i nf o
a b s t r a c t
Article history: Received 4 April 2013 Accepted 23 February 2014 Available online 15 March 2014
In Allaoui et al. (2008), ‘Simultaneously scheduling n jobs and the preventive maintenance on the twomachine flow shop to minimize the makespan’ published in the International Journal of Production Economics, the authors show that the problem of two machine flow shop scheduling with a single maintenance occurring on the second machine is NP-hard (Theorem 7). In this Erratum we establish that this theorem is false and provide a polynomial time algorithm to solve this problem. An extension is proposed for further research in this area. We also highlight that the proof of Proposition 6 is incorrect and we propose a new proof for this proposition. & 2014 Elsevier B.V. All rights reserved.
Keywords: Scheduling Preventive maintenance Flow shop
1. The studied model in Allaoui et al. (2008) In Allaoui et al. (2008), ‘Simultaneously scheduling n jobs and the preventive maintenance on the two-machine flow shop to minimize the makespan’, the model studied is presented in a confusing way. At the end of page 162: “The maximum allowed continuously working time of the first machine or the second machine is T and the maintaining time is g.”, which is clear up to this point. But after this assertion the authors add that “A machine must be maintained exactly once with a constant maintenance length g during the given time interval [0, T]”. This sentence leaves us with two different problems: A. Find a schedule such that there is at most T time units between two consecutive maintenance periods, B. Find a schedule with exactly one maintenance period, and this maintenance starting in time interval [0, T]. Furthermore, the last sentence of the page 162: “To ensure that each machine will be maintained only one time during the scheduling horizon we assume that 2T is an upper bound of the makespan”, is incorrect. In fact, it is easy to see that this upper
bound does not guarantee that only one maintenance takes place (in model (A)). Consider an instance constituted of 3 jobs, each one of duration 2=3T 1, and with a maintenance duration of 1. Clearly 2 maintenances are requested in model (A), and yet the resulting makespan is 2T 1. Based on the results presented in Allaoui et al. (2008), we assume in the remainder of this paper that the original article (Allaoui et al., 2008) deals with problem (B). In particular, the maintenance period may occur at the beginning of the time horizon. This may be an undesirable situation, since the machine then operates all the jobs without maintenance. For this reason, we also consider in Section 4 the case where the maintenance is not allowed at the beginning of the schedule. Whether the maintenance period is allowed at the beginning or not, we prove that the problem can be optimally solved in polynomial time. The remainder of this paper is organized as follows. In Section 2 we propose a new proof for Proposition 6. In Section 3, some important properties are provided. We show that Theorem 7 is incorrect and we provide a polynomial solution algorithm for this problem. We also study an extension of this problem in Section 4.
2. The maintenance period on the first machine ðF 2 nrðM 1 Þ C max Þ DOI of original article: http://dx.doi.org/10.1016/j.ijpe.2006.08.017 Corresponding author. Tel.: þ 33 4 76575055. E-mail addresses: [email protected] (A. Gara-Ali), [email protected] (M.-L. Espinouse). n
http://dx.doi.org/10.1016/j.ijpe.2014.02.018 0925-5273 & 2014 Elsevier B.V. All rights reserved.
In Section 4 of Allaoui et al. (2008), the authors show that the two-machine flow shop problem with a maintenance period on the first machine is NP-hard and they propose an algorithm (H1).
362
A. Gara-Ali, M.-L. Espinouse / Int. J. Production Economics 153 (2014) 361–363
The authors state in Proposition 6 that if all processing times on the first machine are equal then H1 gives the optimal solution. Proposition 6 in Allaoui et al. (2008) is true, but the proof is incorrect. Indeed the proof intends to show that a local exchange between a job scheduled before the maintenance and a job scheduled after the maintenance in H1 can only increase the makespan. However this only establishes that H1 produces a local optimum relatively to this 2-OPT neighborhood. A strong property is missing to assert that such local optimum is necessarily a global optimum. We hereby propose a correct proof. Without loss of generality we assume that the jobs are indexed by non-increasing order of their processing time bi on the second machine, and we denote by a the processing time of each job on the first machine. Consider the schedule obtained by inserting the maintenance period at the last possible instant in this sequence, that is after job k such that k ¼ T=a. We assume in the following that k o n, otherwise all the jobs can fit before time T and the problem is trivial. Note that this is one of the schedules constructed by heuristic H1. We prove below that this schedule is optimal. It is easy to see that the complexity of this heuristic is Oðn log ðnÞÞ. Consider the last block of jobs without idle time on the second machine, and let u be the first job of this block. That is, no idle time appears after the processing of job u on the second machine. If C 1u denotes the completion time of u on the first machine, the makespan of the schedule is then given by C max ¼ C 1u þ∑ni¼ u bi . We consider two cases. Assume first that u 4 k, which means that u starts after the maintenance. As a consequence we have C max ¼ au þg þ ∑ni¼ u bi . Now consider any sequence π. The uth job in π; πðuÞ, necessarily starts after the maintenance since at most k jobs can be processed before time T on the first machine. Hence the makespan of π is at least C πmax Zau þg þ ∑ni¼ u bπðiÞ . By our choice of the indexation, ∑ni¼ u bðiÞ corresponds to the sum of the ðn u þ 1Þ smallest processing times on the second machine. It results that C max r C πmax for any sequence π, and thus our schedule is optimal. Conversely consider that u ok. In particular u is not the last job of the schedule. It implies that we must have bu Z a to avoid any idle time before the next job ðu þ 1Þ. However, due to our indexation, bi Z a then holds for all the jobs sequenced before u. It is immediate to see that necessarily u is in fact the first job, u ¼ 1: Hence the makespan of the sequence is C max ¼ a þ ∑ni¼ u bðiÞ , which is a trivial lower bound. The result follows. □
But if the scheduling problem has a yes answer, the partition problem could have answer: no. So the complexity proof proposed by Allaoui et al. is false. The following example proves that this proof is incorrect: For the partition problem, we consider n ¼ 2, a1 ¼ 1, a2 ¼ 3, so A ¼ 2. The corresponding instance in the reduction is constituted of 3 jobs of processing times p1;1 ¼ 1; p2;1 ¼ 3; p1;2 ¼ 3; p2;2 ¼ 9; p1;3 ¼ 0 and p2;3 ¼ amax ¼ 3: For the maintenance period T ¼ 9 and g ¼ A2 ¼ 4. Consider the sequence fT 3 ; T 1 ; T 2 g presented in the following (Fig. 1). With the solution presented in Fig. 1, the makespan is 3A2 þ 2A þ amax ¼ 19 and S1 a A, S2 a A. Therefore the proof is false. □
3. The maintenance period on the second machine ðF 2 nrðM 2 Þ C max Þ
The question now is to determine which job will be scheduled before the maintenance. The remainder of jobs will be scheduled after the maintenance period according to Johnson's rule (Proposition 8. Page 166 (Allaoui et al., 2008)). The following example proves that the first job in an optimal schedule is not necessarily the first job of Johnson's sequence. Consider the instance with T ¼ 7 and g ¼ 2, we have two jobs to schedule, whose processing times are given below:
In Allaoui et al. (2008), the authors proved by a reduction from the partition problem that the problem of two machine flow shop scheduling is NP-hard in the ordinary sense. We show that this theorem is false. Theorem 7. (H. Allaoui et al. P.166) The problem F 2 jnrðM 2 ÞjC max is NP-hard. In this theorem, authors consider that the second machine must be maintained exactly once with a constant maintenance length g during the given time interval [0, T]. The authors consider the following partition problem: Given a set of n integers a1 ; a2 ; …; an ; which total to 2A, can this set be partitioned into two disjoint subsets S1 and S2 such that the sum of each is equal to A? From this partition problem the authors define the following instance: p1;i ¼ ai , p2;i ¼ ai ðA þ 1Þ ði ¼ 1; …; nÞ and p1;n þ 1 ¼ 0; p2;n þ 1 ¼ amax ; where amax ¼ max fai =i ¼ 1; …; ng. Also set T ¼ A2 þ A þamax and g ¼ A2 . It is asked whether a schedule S exists for the problem such that C smax ¼ 3A2 þ 2A þ amax ?. As stated in Allaoui et al. (2008), if the partition problem has a positive yes answer then the scheduling problem has a yes answer.
3.1. Complexity of problem F 2 nrðM 2 Þ C max Proposition 1. There exists an optimal sequence for F2 nr ðM 2 Þ C max such that the maintenance period starts at the beginning of the schedule and all the jobs are sequenced in Johnson order (Johnson, 1954). Proof. See proof of Lemma 1 of Lee (1997). □ Based on the above analysis, Algorithm 2 (H2) developed by the authors (Allaoui et al., 2008) can solve the problem optimally. Notice that a maintenance at time 0 can make sense. One may imagine that the previous maintenance occurred a long time before, and that the next preventive maintenance is due to be performed no later than T times unit from the current instant, time 0. The case where the maintenance cannot take place at time 0 is studied in the following section. 4. An extension of F 2 nrðM 2 Þ C max We consider in this section that the maintenance is not allowed at the beginning of the schedule. We show that the problem is still polynomially solvable. Proposition 2. There exists an optimal sequence such that exactly one task is scheduled before the maintenance on the second machine. Proof. This assertion can be proved by swapping the last task scheduled before the maintenance with the maintenance. Such a swap will never increase the makespan. □
Jobs p1;i p2;i
1 3 4
Fig. 1. Example.
2 2 1
A. Gara-Ali, M.-L. Espinouse / Int. J. Production Economics 153 (2014) 361–363
363
The JO sequence is (1, 2). The optimal solution is obtained with this sequence (2, 1), the maintenance period is inserted after job 2 and results in a makespan of 9. Based on the above analysis, we propose the following algorithm to solve the problem.
Proof. Step 2 can be solved in OðnÞ time for each J i . Step 1 is executed only once. Step 3 does not require any computational effort. Consequently, the overall time requirement of Algorithm 1 is Oðn2 Þ. □
Algorithm 1 (A1). Let J ¼ fJ 1 ; J 2 ; …; J n g set of n jobs to be processed. Set C A1 max ¼ þ 1, the makespan given by Algorithm 1 (A1). Step 1. Index the jobs according to Johnson order. Step 2. For each J i A J , create a schedule with the job J i scheduled before the maintenance and the remainder of jobs will be scheduled after the maintenance, set C max ðJ i Þ the makespan given by this sequence. If C max ðJ i Þ r C A1 max , then update C A1 max : ¼ C max ðJ i Þ. Step 3. The optimal schedule is the one with the minimum makespan ðC A1 max ).
Acknowledgments
Theorem 1. Algorithm 1 (A1) will find an optimal solution for the problem F 2 jnr ðM k ÞjC max in Oðn2 Þ time.
The authors wish to thank the anonymous reviewer for his helpful comments and in particular on Section 1 and the proof of Proposition 6. References Allaoui, H., Lamouri, S., Artiba, A., Aghezzaf, E., 2008. Simultaneously scheduling n jobs and the preventive maintenance on the two-machine flow shop to minimize the makespan. Int. J. Prod. Econ. 112 (1), 161–167. Johnson, S.M., 1954. Optimal two- and three-stage production schedules with setup times included. Nav. Res. Logist. Q. 1 (1), 61–68. Lee, C.-Y., 1997. Minimizing the makespan in the two-machine flowshop scheduling problem with an availability constraint. Oper. Res. Lett. 20 (3), 129–139.