Proof of the 4/3 Conjecture for Preemptive vs. Nonpreemptive Two-Processor Scheduling

Abstract

We consider the classical scheduling problem in which a given collection of tasks with lengths t1, t2.….tn, 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. © 1993, ACM. All rights reserved.