Abstract The task of scheduling jobs to machines while minimizing the total makespan, the sum of weighted completion times, or a norm of the load vector are among the oldest and most fundamental tasks in combinatorial optimization. Since all of these problems are in general -hard, much attention has been given to the regime where there is only a small number k of job types, but possibly the number of jobs n is large; this is the few job types, high-multiplicity regime. Despite many positive results, the hardness boundary of this regime was not understood until now. We show that makespan minimization on uniformly related machines ( $$Q|HM|C_{\max }$$ Q | H M | C max ) is -hard already with 6 job types, and that the related Cutting Stock problem is -hard already with 8 item types. For the more general unrelated machines model ( $$R|HM|C_{\max }$$ R | H M | C max ), we show that if the largest job size $$p_{\max }$$ p max or the number of jobs n is polynomially bounded in the instance size |I|, there are algorithms with complexity $$|I|^{{{\,\mathrm{\textrm{poly}}\,}}(k)}$$ | I | poly ( k ) . Our main result is that this is unlikely to be improved because $$Q||C_{\max }$$ Q | | C max is $$\mathsf {W[1]}$$ W [ 1 ] -hard parameterized by k already when n, $$p_{\max }$$ p max , and the numbers describing the machine speeds are polynomial in |I|; the same holds for $$R||C_{\max }$$ R | | C max (without machine speeds) when the job sizes matrix has rank 2. Our positive and negative results also extend to the objectives $$\ell _2$$ ℓ 2 -norm minimization of the load vector and, partially, sum of weighted completion times $$\sum w_j C_j$$ ∑ w j C j . Along the way, we answer affirmatively the question whether makespan minimization on identical machines ( $$P||C_{\max }$$ P | | C max ) is fixed-parameter tractable parameterized by k, extending our understanding of this fundamental problem. Together with our hardness results for $$Q||C_{\max }$$ Q | | C max , this implies that the complexity of $$P|HM|C_{\max }$$ P | H M | C max is the only remaining open case.
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