Minimizing expected makespan in stochastic open shops

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Abstract
Suppose that two machines are available to process n tasks. Each task has to be processed on both machines; the order in which this happens is immaterial. Task j has to be processed on machine 1 (2) for random time Xj (Yj) with distribution Fj (Gj). This kind of model is usually called an open shop. The time that it takes to process all tasks is normally called the makespan. Every time a machine finishes processing a task the decision-maker has to decide which task to process next. Assuming that Xj and Yj have the same exponential distribution we show that the optimal policy instructs the decision-maker, whenever a machine is freed, to start processing the task with longest expected processing time among the tasks still to be processed on both machines. If all tasks have been processed at least once, it does not matter what the decision-maker does, as long as he keeps the machines busy. We then consider the case of n identical tasks and two machines with different speeds. The time it takes machine 1 (2) to process a task has distribution F (G). Both distributions F and G are assumed to be new better than used (NBU) and we show that the decision-maker stochastically minimizes the makespan when he always gives priority to those tasks which have not yet received processing on either machine.