1990
DOI: 10.1111/j.1467-9574.1990.tb01276.x
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Scheduling identical jobs on uniform parallel machines

Abstract: We address the problem of scheduling n identical jobs on m uniform parallel machines to optimize scheduling criteria that are nondecreasing in the job completion times. It is well known that this can be formulated as a linear assignment problem, and subsequently solved in O(n3) time. We give a more concise formulation for minsum criteria, and show that general minmax criteria can be minimized in O(n2) time. We present faster algorithms, requiring only O(n +mlog m) time for minimizing makespan and total complet… Show more

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Cited by 68 publications
(55 citation statements)
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“…As shown in [3], [4], [9], [10], [12], one of the reasons why it is easy to schedule equal length jobs is that there are few possible starting times. Indeed, active schedules are dominant and thus starting times are equal to a release date modulo p. This idea can be extended to batching problems.…”
Section: Starting Timesmentioning
confidence: 99%
“…As shown in [3], [4], [9], [10], [12], one of the reasons why it is easy to schedule equal length jobs is that there are few possible starting times. Indeed, active schedules are dominant and thus starting times are equal to a release date modulo p. This idea can be extended to batching problems.…”
Section: Starting Timesmentioning
confidence: 99%
“…We summarize this result in the following proposition. When all jobs have identical ready times or identical due dates, we note that the above procedures are identical to the ones presented by Dessouky et al [3]. Hence, for these cases optimal solutions may be found in polynomial-time.…”
Section: Procedures Lstfmentioning
confidence: 61%
“…Simons [11] presents a recursive polynomial-time algorithm that guarantees optimality even when the ready times and due dates are not integer multiples of the processing time. With uniform parallel machines, Dessouky et al [3] propose an O(n log m) algorithm to minimize the maximum lateness when the jobs have equal ready times. The algorithm uses the earliest completion time (ECT) rule to determine a set of minimum completion times.…”
Section: Introductionmentioning
confidence: 99%
“…Since there is no more than one positive arc in a right-chain cycle, the origin t is followed by the nodes of one zero-component only, say (14) of w * i . We illustrate how the formulated optimality condition can be used for the optimality check in Appendix 4; the resulting algorithm has time complexity O(n), an improvement in comparison with the O(n log n)-time algorithm for finding an optimal schedule for problem P| p j = 1| w j U j (Brucker and Kravchenko 2006;Dessouky et al 1990). Example 8 Consider an instance of the problem P|r j , p j = 1| w j T j with one machine and n = 4 jobs given by the table…”
Section: Theorem 6 Network N (C) Contains a Negative Cycle If And Onlmentioning
confidence: 99%