Abstract

We consider the following scheduling problem. We are given a set S of jobs which are to be scheduled sequentially on a single processor. Each job has an associated processing time which is required for its processing. Given a particular permutation of the jobs in S, the jobs are processed in that order with each job started as soon as possible, subject only to the following constraint: For a fixed integer $$B \ge 2$$B?2, no unit time interval $$[x, x+1)$$[x,x+1) is allowed to intersect more than B jobs for any real x. There are several real world situations for which this restriction is natural. For example, suppose in addition to the jobs being executed sequentially on a single main processor, each job also requires the use of one of B identical subprocessors during its execution. Each time a job is completed, the subprocessor it was using requires one unit of time to reset itself. In this way, it is never possible for more than B jobs to be worked on during any unit interval. In Braun et al. (J Sched 17: 399---403, 2014a) it is shown that this problem is NP-hard when the value B is variable and a classical worst-case analysis of List Scheduling for this situation has been carried out. We prove a tighter bound for List Scheduling for $$B\ge 3$$B?3 and we analyze the worst-case behavior of the makespan $$\tau _\mathrm{LPT}(S)$$?LPT(S) of LPT (longest processing time first) schedules (where we rearrange the set S of jobs into non-increasing order) in relation to the makespan $$\tau _o(S)$$?o(S) of optimal schedules. We show that LPT ordered jobs can be processed within a factor of $$2-2/B$$2-2/B of the optimum (plus 1) and that this factor is best possible.

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