Proof of the 4/3 conjecture for preemptive vs. nonpreemptive two-processor scheduling

We consider the classical scheduling problem in which a given collection of tasks with lengths tl. t2,..., t , are to be run on two processors, subject to specified precedence constraints among the tasks, so as to minimize the completion time of the last-finishing task, the so-called makespan of the schedule. A schedule is said to be nonpreemptive if each task, once started, is run continuously until its completion t,time units later, whereas a preemptive schedule allows the running of a task to be temporarily suspended and resumed at a later time, that is, run in noncontiguous pieces whose lengths merely sum to the task length t,. A long-standing conjecture is that, for any set of tasks and precedence constraints among them, the least makcspan achievable by a nonpreemptive schedule is no more than 4/3 the least makespan achievable when preemptions are allowed. In this paper, we prove this conjecture.