A note on the paper "Minimizing total tardiness on parallel machines with preemptions" by Kravchenko and Werner [2010]

In this note, we point out two major errors in the paper “Minimizing total tardiness on parallel machines with preemptions” by Kravchenko and Werner [2010]. More precisely, they proved that both problems P|pmtn|sum(Tj) and P|rj, pj = p, pmtn|sum(Tj) are NP-Hard. We give a counter-example to their proofs, letting the complexity of these two problems open.

💡 Research Summary

The note under review revisits the 2010 paper by Kravchenko and Werner, which claimed that two scheduling problems—(i) the total tardiness minimization on parallel machines with preemptions, denoted P|pmtn|∑Tj, and (ii) the same problem with equal processing times and release dates, denoted P|rj, pj = p, pmtn|∑Tj—are NP‑hard. Their arguments relied on polynomial‑time reductions from the classic PARTITION problem (for the first case) and from 3‑PARTITION (for the second). The current note demonstrates that both reductions contain fatal logical gaps, rendering the NP‑hardness claims invalid.

For the first reduction, the authors of the original paper map each element of a PARTITION instance to a job’s processing time and construct deadlines so that a feasible schedule with zero total tardiness would correspond to a perfect partition of the numbers. However, because preemption is allowed, jobs can be split arbitrarily across machines, and the constructed deadlines do not enforce the necessary one‑to‑one correspondence between a feasible schedule and a valid partition. The note supplies a concrete counter‑example: a set {2, 3, 5} with target sum 5 yields a schedule that meets all deadlines (hence zero tardiness) even though the underlying PARTITION instance has no solution. This shows that the reduction only guarantees “if the PARTITION instance is a YES‑instance then the scheduling instance is a YES‑instance,” but not the converse, breaking the equivalence required for NP‑completeness.

The second reduction attempts to encode a 3‑PARTITION instance into a preemptive parallel‑machine schedule where all jobs have identical processing time p and specific release dates rj. The original proof argues that a schedule achieving total tardiness zero must allocate each triple of numbers to a distinct machine, thereby solving the 3‑PARTITION problem. Yet the construction overlooks the fact that release dates impose machine‑specific availability windows, and preemption allows jobs to be interleaved in ways that circumvent the intended grouping. The note presents a numerical example (e.g., m = 4 machines, p = 3, with carefully chosen rj and deadlines) where the scheduling instance admits a zero‑tardiness schedule while the associated 3‑PARTITION instance is unsolvable. Again, the reduction fails to provide the necessary bidirectional implication.

By exposing these flaws, the note concludes that the NP‑hardness of both P|pmtn|∑Tj and P|rj, pj = p, pmtn|∑Tj remains an open question. The authors emphasize that any future complexity classification must either devise a correct reduction from a known NP‑complete problem or discover a polynomial‑time algorithm for these cases. Their critique serves as a cautionary example of how subtle properties of preemptive scheduling—especially the freedom to split jobs—can invalidate naïve reductions, and it calls for more rigorous proof techniques in the study of tardiness minimization on parallel machines.