I am aware that when we have a linear problem subject to OR constraints, the LP would be a non-convex optimisation problem. For example,

${x = 0}$ OR ${1<=x<=2}$.

My question is in such a situation, is it possible to check if an LP is feasible/infeasible, bounded/unbounded or if it has an optimal value or not

I could not find much explanation on the internet concerning a detailed explanation of this situation. I'd appreciate it if anyone could explain this in more detail.

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  • $\begingroup$ What exactly is your question? I don't see a question here. We are a question-and-answer site, we require you to articulate a specific question that you want answered. When you talk about "a detailed explanation of this situation", I'm not sure what you are hoping for. $\endgroup$
    – D.W.
    Jun 10, 2021 at 5:03
  • $\begingroup$ Hi @D.W. I have edited the problem description. Apologies for the vagueness $\endgroup$ Jun 10, 2021 at 10:49
  • $\begingroup$ math.stackexchange.com/q/4163035/14578 $\endgroup$
    – D.W.
    Jun 12, 2021 at 4:31

1 Answer 1


Yes, it's possible. Let $S$ be the original linear problem, i.e., a conjunction of inequalities. You can form two LPs. One LP has the form $S \land x=0$, which can be solved with a LP solver. The other LP has the form $S \land 1 \le x \le 2$. If either one has a feasible solution, your problem has a feasible solution. This procedure takes polynomial time.

This works if you have a single disjunction. If you have an unlimited number of disjunctions, it is still possible to find a solution, but such a naive strategy will take exponential time. You can convert the problem to an instance of integer linear programming (see Express boolean logic operations in zero-one integer linear programming (ILP)) and then use an ILP solver. ILP is of course NP-hard, so there is no method that will be efficient in general on all problem instances, so if you have many disjunctions, you might run into instances that are intractable to solve within your lifetime.

  • $\begingroup$ Hi, I understand your point. But would we not have multiple global optima and multiple feasible regions? For example, if I have two constraints (OR constraints), if one of the constraints satisfies my objective function (feasible, bounded, and optimal) and the other one does not (infeasible, unbounded, and no optimal value). Are you saying that the LP statement is solvable? $\endgroup$ Jun 11, 2021 at 2:28
  • 1
    $\begingroup$ @AmalSailendran, Yup! If one of those three derived LPs is feasible (there exists a solution to it), then there exists a solution to the full system with those additional constraints. $\endgroup$
    – D.W.
    Jun 11, 2021 at 6:37
  • $\begingroup$ I don't know what it means for a constraint to satisfy an objective function. LPs can always have multiple global optima. $\endgroup$
    – D.W.
    Jun 11, 2021 at 6:37
  • $\begingroup$ Thanks! Accepted your answer. $\endgroup$ Jun 11, 2021 at 7:39

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