# Job Shop Scheduling: what objectives are harder?

It is well known that job shop scheduling problems are notoriously hard from a computational point of view. Many papers exist for the makespan objective, and some report on min sum objectives (like minimize weighted completion time, minimize earliness/tardiness penalties etc).

My question: does anyone know whether makespan or min sum objectives are computationally more difficult when using MILP approaches?

• Not directly answering your question, but I believe that such objectives are generally hard for polyhedral approaches such as MILP. My guess is that, since the resource constraints on JSPs are disjunctive in nature, it makes it somewhat unavoidable to use using big M variables in MILP formulations. Ever in literature, we often find CP approaches fare better than MILP approaches for obtaining the optimal solution of JSPs. Jan 23 '20 at 18:06
• Thanks, I usually use "configuration" based formulations (that are solved by branch-and-price), so I can avoid the big $M$. But you are right, of course, MILP is not very suited -- yet, this is exactly what I am interested in (I edit my post, thanks!). Jan 23 '20 at 18:33
• It seems that for the decision problem it can be shown that the total lateness objective is at least as general as the makespan objective: Assume we want to know if there exists a schedule with a makespan of at most M. When setting the due dates of all jobs to M, a schedule with a makespan of at most M exists if there is a schedule with a total lateness of zero. I don't see such a reduction for the other direction. Jan 24 '20 at 15:23
• Another indicator why min sum objectives are computationally more difficult is that the makespan depends only on the completion time of the latest job, whereas for min sum objectives the completion times of all jobs might have an influence. Jan 24 '20 at 15:23
• @SebastianK, would you like to turn your comments into an answer? this is the best we have so far... Jan 29 '20 at 12:05

• For the decision problem, it can be shown that the total lateness objective is at least as general as the makespan objective: Assume we want to know if there exists a schedule with a makespan of at most $$M$$. When setting the due dates of all jobs to $$M$$, a schedule with a makespan of at most $$M$$ exists if there is a schedule with a total lateness of zero. I don't see such a reduction for the other direction.