**Algorithm** *H*<sup>1</sup>

At current time *t*, if some machine is idle, *U*(*t*) = ∅, when *t* ≥ (1 + *βm*)*r*max(*t*) + *βm*; then, start the jobs in *U*(*t*) as a batch on the idle machine that has the minimum machine cost at the moment. Otherwise, do nothing but wait.
