Abstract
We introduce the following elementary scheduling problem. We are given a collection of n jobs, where each job J i has an integer length ℓ i as well as a set T i of time intervals in which it can be feasibly scheduled. Given a parameter B, the processor can schedule up to B jobs at a timeslot t so long as it is “active” at t. The goal is to schedule all the jobs in the fewest number of active timeslots. The machine consumes a fixed amount of energy per active timeslot, regardless of the number of jobs scheduled in that slot (as long as the number of jobs is non-zero). In other words, subject to ℓ i units of each job i being scheduled in its feasible region and at each slot at most B jobs being scheduled, we are interested in minimizing the total time during which the machine is active. We present a linear time algorithm for the case where jobs are unit length and each T i is a single interval, assuming that jobs are given in sorted order. For general T i , we show that the problem is NP-complete even for B=3. However when B=2, we show that it can be efficiently solved. In addition, we consider a version of the problem where jobs have arbitrary lengths and can be preempted at any point in time. For general B, the problem can be solved by linear programming. For B=2, the problem amounts to finding a triangle-free 2-matching on a special graph. We extend the algorithm of Babenko et al. (Proceedings of the 16th Annual International Conference on Computing and Combinatorics, pp. 120–129, 2010) to handle our variant, and also to handle non-unit length jobs. This yields an \(O(\sqrt{L} m)\) time algorithm to solve the preemptive scheduling problem for B=2, where L=∑ i ℓ i . We also show that for B=2 and unit length jobs, the optimal non-preemptive schedule has active time at most 4/3 times that of the optimal preemptive schedule; this bound extends to several versions of the problem when jobs have arbitrary length.
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