On Solving the Set Covering Problem with Conflicts on Sets

TL;DR

This work studies SCP-CS, a Set Covering Problem variant with penalties for selecting conflicting subsets. It develops a compact MILP with binary variables for subset selection and for conflicts, solved via the open-source CP-SAT solver, and evaluates it on Beasley-derived benchmark instances. The approach delivers strong, balanced lower and upper bounds and even improves 9 best-known heuristic solutions, though it does not consistently outperform non-compact dynamic MILPs or dedicated metaheuristics. The study demonstrates the practicality of an accessible, open-source formulation for SCP-CS and suggests directions to improve bound gaps and leverage instance structure. The work has implications for coverage planning in networks where interference and over-coverage must be controlled cost-effectively.

Abstract

A variant of the well-known Set Covering Problem is studied in this paper, where subsets of a collection have to be selected, and pairwise conflicts among subsets of items exist. The selection of each subset has a cost, and the inclusion of conflicting subsets is associated with a penalty to be paid. The problem, which can be used to model real applications, looks for a selection of subsets that cover the original collection, while minimizing the sum of covering and penalty costs. In this paper we consider a compact mixed integer linear program and we solve it with an open-source solver. Computational results on the benchmark instances commonly used in the literature of the problem are reported. The results indicate that the new approach we propose is capable of good results, both in terms of lower and upper bounds, although not matching the state-of-the-art on average. The new approach was, however, able to improve 9 best-known heuristic solutions.

Figures (1)

• Figure 1: On the left an example of a simplified SCP-CS instance is presented, where the elements to be covered are numbered from 0 to 5, and the sets are the rectangular shapes, and we assume that each set has a fixed cost equal to $\alpha$. To simplify the example, no cost is shown, and we assume that two sets sharing nodes pay a penalty proportional to the overlap, and anyway much higher than the costs associated with the single sets, when they are both selected. On the right, an optimal solution with cost $3\alpha$, where three non-overlapping sets are selected to cover all the elements, and therefore no penalty is incurred. Notice that this is not always the case, and in larger problems it is common to have a penalty in the optimal cost