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We introduce the following elementary scheduling problem. We are given a collection of n jobs, where each job Ji has an integer length ℓi as well as a set Ti 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 ℓi 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 Ti is a single interval. For general Ti, we show that the problem is NP-complete even for B=3. However when B=2, we show that it can be 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. [3] 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 ℓi. We also show that for B=2 and unit length jobs, the optimal non-preemptive schedule has at most 4/3 times the active time of the optimal preemptive schedule; this bound extends to several versions of the problem when jobs have arbitrary length.