Interval scheduling with eligible machine sets is the problem $P \mid M_j \mid \sum w_j Z_j$ read on the words of lax-888481 ($IsemLink$). Its hardness carries to the zoo's problem under the standard reading by a word map computed in polynomial time on the word RAM: drop the machines no job may use and renumber the others, then write the standard word of the instance with the threshold $\sum_j w_j - W$. Dropping the unused machines keeps the answer and does not raise the number of machines or the largest processing time.

Theorem. Under the standard reading, $P \mid M_j \mid \sum w_j Z_j$ is W[1]-hard parameterized by the number of machines.

Theorem. Under the standard reading, $P \mid M_j \mid \sum w_j Z_j$ is NP-hard on the instances whose processing times are at most $25$, so para-NP-hard parameterized by the largest processing time.

Fixed-parameter tractability goes the other way, from the standard reading to the words of interval scheduling, where every job has positive length. Under the standard reading a job may have length zero: it is then a point, which a job running strictly across it blocks, and which no other point blocks. So the points with the same due date and the same machines are one point with their total weight. Scaling all times by $K = 2r + 3$, where $r = \min(n, 2^m)$ bounds the number of such points at one time, turns every job of positive length $[a, d]$ into $[Ka + r + 1, Kd - r - 1)$ and the $s$-th point at time $t$ into $[Kt + s, Kt + s + 1)$: two jobs overlap after scaling exactly when they clash before. Jobs that can never be on time, and points other than the first of their kind, become jobs without a machine, which padding (as in $IsemLink$) gives a new machine of their own. The number of machines grows by one and the largest processing time by the factor $K \le 2 \cdot 2^m + 3$.

Theorem. Under the standard reading, $P \mid M_j \mid \sum w_j Z_j$ is fixed-parameter tractable parameterized by the number of machines plus the largest processing time.