I found this proof for showing that Scheduling with Release Time is NP-Hard by reduction to Subset-sum, but I don't understand it:
Scheduling With Release Time: Given a set of $n$ jobs with processing time $t_i$, release time $r_i$, and eadline $d_i$, is it possible to schedule all jobs on a single machine such that job $i$ is processed with a contiguous slot of $t_i$ time units in the interval [$r_i$, $d_i$]?
Claim: SUBSET-SUM $\leq_p$ SCHEDULE-RELEASE-TIMES. Prof: Given an instance of SUBSET-SUM $w_1, ... , w_n$, and target $W$, Create $n$ jobs with processing time $t_i = w_i$, release time $r_i = 0$, and no deadline ($d_i = 1 + \sum_j w_j$)
Create job 0 with $t_0$ = 1, release time $r_0 = W$, and deadline $d_0 = W + 1$
My understanding: so there is some subset of $t_i$ such that their sum adds up to exact $W$, which is the release time of one special job called job 0 and the rest of the jobs have no release / deadline restrictions. I guess the rest of the jobs that are not part of this set will be scheduled after job 0. But I don't understand how this can ensure they don't overlap and how that notation makes the jobs have no deadline. I dont understand how this proof works in general.