An instance of interval scheduling with eligible machine sets has $n$ jobs and $m$ machines; job $j$ has a processing time $1 \le p_j \le d_j$, a deadline $d_j$, a weight $w_j$ and a set of eligible machines. A schedule runs some of the jobs, each on an eligible machine during exactly $[d_j - p_j, d_j)$, such that jobs on one machine do not overlap; the question is whether the scheduled jobs can have weight at least $W$.

Read the same data as an instance of $P \mid M_j \mid \sum w_j Z_j$: identical machines, job $j$ with processing time $p_j$, due date $d_j$, weight $w_j$ and machine set the eligible machines. Here every job is scheduled, and the cost is the total weight of the jobs that do not complete exactly at their due date.

Theorem. If every job has an eligible machine, the interval scheduling instance has a schedule of weight at least $W$ if and only if the zoo instance has a schedule of cost at most $\sum_j w_j - W$.

The jobs completing at their due dates form the interval schedule; conversely the jobs left out of an interval schedule are run one after another, on an eligible machine each, after all due dates.

Encoding the zoo instance and threshold $\sum_j w_j - W$ by the word of the interval scheduling instance and $W$, the zoo problem on the instances in which every job has an eligible machine is a parameterized problem on words.

Theorem. This problem, parameterized by $m + p_{\max}$, is fixed-parameter tractable.

An interval scheduling instance may have jobs without an eligible machine, which no schedule runs. Padding the instance gives each of them one new machine, eligible for them alone, and weight $0$; the other jobs keep their data. The number of machines grows by one.

Theorem. Padding keeps the answer, and in the padded instance every job has an eligible machine. Hence every interval scheduling instance with threshold $W$ has the same answer as the zoo instance of its padding with threshold $\sum_j w'_j - W$, where $w'$ are the padded weights.

Padding is computed on words by a word RAM program in one pass, linear in the length of the word, and raises the number of machines by one.

Theorem. Interval scheduling with eligible machine sets, parameterized by the number of machines, fpt-reduces to $P \mid M_j \mid \sum w_j Z_j$ parameterized by the number of machines.

Multicoloured Clique fpt-reduces to the former (Theorem 1 of lax-888481), so composing the two reductions makes $P \mid M_j \mid \sum w_j Z_j$ W[1]-hard for the number of machines; the composition of two such reductions is not stated here.