European Journal of Operational Research 197 (2009) 825–827
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Short Communication
A note on ‘‘scheduling of nonresumable jobs and flexible maintenance activities on a single machine to minimize makespan” Dehua Xu a, Yunqiang Yin a, Hongxing Li b,* a b
School of Mathematical Sciences, Beijing Normal University, Beijing 100875, China School of Electronic and Information Engineering, Dalian University of Technology, Dalian, Liaoning 116024, China
a r t i c l e
i n f o
Article history: Received 21 September 2007 Accepted 8 July 2008 Available online 29 July 2008 Keywords: Scheduling Single machine Maintenance Heuristic algorithm Worst-case analysis
a b s t r a c t In a recent paper, Chen [J.S. Chen, Scheduling of nonresumable jobs and flexible maintenance activities on a single machine to minimize makespan, European Journal of Operational Research 190 (2008) 90–102] proposes a heuristic algorithm to deal with the problem Scheduling of Nonresumable Jobs and Flexible Maintenance Activities on A Single Machine to Minimize Makespan. Chen also provides computational results to demonstrate its effectiveness. In this note, we first show that the worst-case performance bound of this heuristic algorithm is 2. Then we show that there is no polynomial time approximation algorithm with a worst-case performance bound less than 2 unless P ¼ NP, which implies that Chen’s heuristic algorithm is the best possible polynomial time approximation algorithm for the considered scheduling problem. Ó 2008 Elsevier B.V. All rights reserved.
1. Introduction Scheduling problems with machine availability constraints have attracted many researchers’ attention in the last two decades [1,3,4,7,9–20]. However, most of the works are focused on problems with at most one unavailable period on each machine [1,3,4,9– 11,13,15–17]. Note that in practice machines may not be available now and then due to preventive maintenances, periodical repairs or tool changes. So there is a need to develop approaches to handle scheduling problems with more than one unavailable period. In a recent paper, Chen [2] studies the problem Scheduling of Nonresumable Jobs and Flexible Maintenance Activities on A Single Machine to Minimize Makespan, which can be described as follows. We have to schedule n independent nonresumable jobs J 1 ; . . . ; J n on a single machine, where the machine is assumed to be stopped almost periodically for maintenance for a constant time w in the planning horizon. Time uðvÞ is the earliest (latest) time at which the machine starts (stops) its maintenance activity, as measured in relation to the completion of the previous maintenance activity. Assume that the machine has just finished a dummy maintenance activity at time zero, u 6 v w and max16i6n pi 6 u, where pi denotes the processing time of job J i . View each interval between two consecutive maintenance activities as a batch. The objective is to determine the capacity (i.e., length) of each used batch and schedule all the jobs on the machine such that the makespan, * Corresponding author. Tel.: +86 411 84706402; fax: +86 411 84706405. E-mail addresses:
[email protected] (D. Xu),
[email protected]. edu.cn (Y. Yin),
[email protected] (H. Li). 0377-2217/$ - see front matter Ó 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2008.07.021
i.e., the completion time of the last finished job, is minimum. This problem is denoted by 1jnr fpajC max in Chen [2]. If u ¼ v w, we denote this problem as 1jnr pajC max . Without loss of generality, we assume that u, v and w are nonnegative integers and that pi ð1 6 i 6 nÞ are positive integers. Moreover, we assume that v w > 0 since otherwise there is trivially no feasible schedule for the considered scheduling problem. Apparently, in order to minimize makespan, unnecessary machine idle time should be avoided, i.e., the capacity of each used batch should be the maximum of u and the total processing time of the jobs in it, and the jobs in the last batch should be processed consecutively as early as possible. We thus always make such an assumption when considering schedules for 1jnr fpajC max in this note and only need to consider batches with integer capacities. Furthermore, for the convenience of description, we assume that the jobs in each used batch are sequenced in nonincreasing order of their processing times and the jobs with the same processing time are not distinguished for that these assumptions do not increase the makespan. Chen [2] proposes two mixed binary integer programming models and a heuristic algorithm for 1jnr fpajC max and also provides computational results to demonstrate the efficiency of the models and the effectiveness of the heuristic algorithm. Chen’s heuristic algorithm first reorders all the jobs in a sequence in nonincreasing order of their processing times. Then, the batches are filled one by one choosing a subset of jobs each time using the following largest fit (LF) method: repeatedly choose the largest (i.e., the lowest indexed) job from the sequence that fits the remaining capacity of the current batch and then remove it from the job sequence until
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either there is no more job in the sequence that fits the remaining capacity of the current batch or the sequence becomes empty. In this note, we shall show that the worst-case performance bound of Chen’ heuristic algorithm is 2 and there is no polynomial time approximation algorithm with a worst-case performance bound less than 2 for the considered problem unless P ¼ NP, which imply that Chen’ heuristic algorithm is the best possible polynomial time approximation algorithm for 1jnr fpajC max . One may note that if we take the batches and jobs as bins and items, respectively, Chen’s heuristic algorithm is in effect equivalent to the First Fit Decreasing (FFD) algorithm, which is a classical heuristic algorithm for solving the Bin-Packing problem (BPP). BPP (see e.g. [6, p. 46]). Given n items a1 ; a2 ; ; an . The size of ai is sðai Þ 2 ð0; c, 1 6 i 6 n. We are asked to pack them into a minimum number of c-capacity bins (i.e., partition the n items into a minimum number m of subsets B1 ; B2 ; ; Bm such that P ai 2Bj sðai Þ 6 c for 1 6 j 6 m). Algorithm FFD (see e.g. [6, p. 59]). Reorder all the items in nonincreasing order of their sizes, i.e., sða1 Þ P sða2 Þ P P sðan Þ; for i ¼ 1; 2; ; n, item ai is packed into the first partially filled bin Bj whose level levelðBj Þ 6 c sðai Þ, where levelðBj Þ is the sum of the sizes of the items in bin Bj . In fact, let c ¼ v w. View job J i as item ai with sðai Þ ¼ pi , 1 6 i 6 n. Pack these n items into the bins using the FFD algorithm. Take the used bins as batches. We can easily see that under the assumptions we have just made in paragraph 4, the schedule produced by the FFD algorithm coincides with the schedule produced by Chen’s heuristic algorithm. Now, let us turn to another algorithm named largest object first with least absolute waste (LFLAW) [5], which is an approximation algorithm for the well-known variable-sized bin-packing problem (VBPP). The VBPP differs from the BPP in that the capacities of the bins in the VBPP may be different and the objective of the VBPP is to minimize the sum of the capacities of the used bins subject to the total size of the items packed into each bin being not greater than the capacity of the bin. Again, take the batches and jobs as bins and items, respectively. The LFLAW works almost the same as Chen’s heuristic algorithm except that when packing a new bin, the former first packs each capacity of bin with a subset of the current remaining items by the LF method, then it chooses the bin with the least absolute remaining capacity (ties break arbitrarily) and the corresponding items. It is easy to see that under the same assumptions as in the previous paragraph, the LFLAW algorithm and Chen’s heuristic algorithm produce the same schedule if ties break in favor of large capacity bin in the former algorithm. The relationships among Chen’ heuristic algorithm, the FFD algorithm and the LFLAW algorithm are crucial to our worst-case analysis of the former for that they enable us to use some important results of the BPP and the VBPP, and thus make it is possible to present a concise analysis.
2. Worst-case analysis and non-approximability For convenience, we use the following notations in the rest of this note. ðBH1 ; M 1 ; BH2 ; M 2 ; ; M n0 ; BHn0 þ1 Þ or p: the schedule produced by Chen’ heuristic algorithm for 1jnr fpajC max , where BHi , Mi and n0 denote the ith batch, the ith real (i.e., non-dummy) maintenance activity, and the total number of real maintenance activities, respectively.
Before analyzing Chen’s heuristic algorithm, we first present some lemmas. Lemma 1 (see Chu and La [5]). If n0 P 1, then
Pn0 þ1 i¼1
P si < 2 ni¼1 pi .
Lemma 2. n0 6 2n , where n is the minimum number of real maintenance activities required to process all the n jobs J 1 ; ; J n . Proof. Note that there are n0 þ 1 used batches in p while there are at least n þ 1 used batches in any optimal schedule. By the FFD algorithm, we have ðn0 þ 1Þ=ðn þ 1Þ 6 3=2 (see e.g. [6, p. 67]). It follows that n0 6 2n . h Lemma 3 (see Ji et al. [8]). There is no polynomial time approximation algorithm with a worst-case performance bound less than 2 for 1jnr pajC max unless P ¼ NP. Now we are ready to obtain the worst-case performance bound of Chen’s heuristic algorithm. Theorem 1. The worst-case performance bound of Chen’s heuristic algorithm for 1jnr fpajC max is 2 and the bound is tight. Proof. If n0 ¼ 0, it is easy to see that C Hmax ¼ C max . P 0 6 n þ1 s þ n0 w and If n0 P 1, it is easy to see that C H max Pn Pn0 þ1 Pn i¼1 i C max P i¼1 pi þ n w. Note that i¼1 si < 2 i¼1 pi (see Lemma 1) and n0 6 2n (see Lemma 2), so we have C H max < 2C max . Hence, we have completed the proof that the worst-case performance bound of the Chen’s heuristic algorithm is not greater than 2. To show that the bound cannot be smaller than 2, consider the following instance. Let u ¼ 24, v ¼ 25 þ w, p1 ¼ 12, p2 ¼ p3 ¼ p4 ¼ 8, p5 ¼ p6 ¼ 6. Applying Chen’s heuristic algorithm, we obtain a schedule where BH1 contains J 1 and J 2 , BH2 contains J 3 , J 4 , and J 5 , BH3 contains J 6 and the makespan is C H max ¼ 2w þ 54. However, an optimal schedule has two batches, where the first batch contains J 1 , J 5 and J 6 while the second batch contains J 2 , J 3 and J 4 . Hence, C max ¼ w þ 48. It follows that C H max =C max ¼ ð2w þ 54Þ=ðw þ 48Þ ! 2 as w ! 1. This completes the proof. h Note that 1jnr pajC max is a special case of 1jnr fpajC max , thus by Lemma 3, we may conclude that there is no polynomial time approximation algorithm with a worst-case performance bound less than 2 for 1jnr fpajC max unless P ¼ NP either. Combining this with Theorem 1, we establish the following main result. Theorem 2. Unless P ¼ NP, Chen’s heuristic algorithm is the best possible polynomial time approximation algorithm for 1jnr fpajC max . 3. Remarks Chen [2] assumes that max16i6n pi 6 u and u 6 v w in the scheduling problem 1jnr fpajC max . Note that Lemmas 1, 2 and Theorem 1 still hold if we assume that max16i6n pi 6 v w. So, in fact, we have showed that Chen’s heuristic algorithm has a worst-case performance bound of 2 for the more general scheduling problem 1jnr fpa; pi 6 v wjC max and thus, it is also the best possible polynomial time approximation algorithm for 1jnr fpa; pi 6 v wjC max unless P ¼ NP. Acknowledgements
C Hmax : the makespan of p. C max : the optimal makespan. sj : the capacity of batch BHj , i.e., the length of the time interval of batch BHj .
This research was supported in part by the National 973 Fundamental Research Project of China (2002CB312200) and National Natural Science Foundation of China (60774049).
D. Xu et al. / European Journal of Operational Research 197 (2009) 825–827
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