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For a scheduling problem to minimize the makespan on parallel machines, we consider schedules with at most one preemption. We show that in the case of two machines the problem is solvable in polynomial time. For m ≥ 3 uniform parallel machines, we measure the quality of a single preemption as the worst-case ratio of the makespan of an optimal schedule with at most one preemption over the makespan of an optimal preemptive schedule. We show that the global bound on such a ratio is 2 − 2/m.

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We study the two-machine flow shop problem with an uncapacitated interstage transporter. The jobs have to be split into batches, and upon completion on the first machine, each batch has to be shipped to the second machine by a transporter. The best known heuristic for the problem is a [Formula: see text]–approximation algorithm that outputs a two-shipment schedule. We design a [Formula: see text]–approximation algorithm that finds schedules with at most three shipments, and this ratio cannot be improved, unless schedules with more shipments are created. This improvement is achieved due to a thorough analysis of schedules with two and three shipments by means of linear programming. We formulate problems of finding an optimal schedule with two or three shipments as integer linear programs and develop strongly polynomial algorithms that find solutions to their continuous relaxations with a small number of fractional variables.