Abstract

The paper considers the open shop scheduling problem to minimize the make-span, provided that one of the machines has to process the jobs according to a given sequence. We show that in the preemptive case the problem is polynomially solvable for an arbitrary number of machines. If preemption is not allowed, the problem is NP-hard in the strong sense if the number of machines is variable, and is NP-hard in the ordinary sense in the case of two machines. For the latter case we give a heuristic algorithm that runs in linear time and produces a schedule with the makespan that is at most 5/4 times the optimal value. We also show that the two-machine problem in the nonpreemptive case is solvable in pseudopolynomial time by a dynamic programming algorithm, and that the algorithm can be converted into a fully polynomial approximation scheme. © 1998 John Wiley & Sons, Inc. Naval Research Logistics 45: 705–731, 1998

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