We introduce the following elementary scheduling problem. We are given a collection of n jobs, where each job has an integer length 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 all units of each job 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 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. 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 is the sum of the job lengths. 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.