“…In case (2), machine M 2 is idle in the interval [2,3), and job 2 has to be completed at time C 2 2 = 4, because at least 2 units of the resource would be wasted in the interval [2,4) otherwise. Hence, one unit of the resource is wasted in the interval [3,4), and in consequence, job 3 (whose resource requirement is 6) has to be executed on M 1 in the interval [2,3). Then, job 3 is scheduled on M 2 in the interval [4,5), which leads to losing one more unit of the resource.…”