I've two variables: C which is binary, and X which is non-negative
If C = 0 then X = 0; If C = 1 then X = X (meaning there's no constraint on X)
How should I format this conditional constraint into a linear constraint for LP?
1 Answer
Note that strictly speaking LP models only contain continuous variables. So we assume this is a MIP model to be solved with a MIP solver.
Here are three ways to attack this, depending on the capabilities of the solver.
(1) If you use a solver that supports indicator constraints, you can simply use:
c=0 ==> x=0
(2) For other solvers you can use:
x <= M*c
where M is a (tight as possible) upper bound on x.
(3) Finally, if your solver supports SOS1 (Special Ordered Sets of type 1) sets, you can use:
d = 1-c
{d,x} ∈ SOS1
d >= 0
(1) and (3) have the advantage that no bound is needed. If you have a good, tight bound on x, (2) is a good option.
