# Strict inequality logical implication in optimization problems

I have $$x \in \{0,1\}$$ and $$y \geq 0$$ and I want to model that $$x=1$$ iff $$y>0$$, is this possible while keeping the constraint linear? Thanks.

One part of the implication is easy $$y \leq Mx$$. The other part I can approximate as $$\epsilon x \leq y$$. Where $$\epsilon$$ is small and $$M$$ is big. But this doesn't work if $$y$$ is very small.

• Your constraint handles $y>0\implies x=1$. To handle the reverse case, you want $x=1$ to be less optimal than $x=0$ when $y=0$. You can handle this by adjusting the objective function by $-Mx$ (in the case of maximisation), as it will force $x=0$ unless it is required to be positive at the optimal solution. – Daryl Apr 13 at 12:04
• thanks, there's no other way? – lucmobz Apr 13 at 12:08
• As you’ve observed, trying to put an upper bound on $x$ changes the range of values available for $y$. I don’t have all of my literature available to check at the moment, but the method I described is my usual approach when the remainder of the problem is linear. – Daryl Apr 13 at 12:15

In practice, if $$y$$ is computed (as opposed to a parameter that is read in and never changed), the chances are high that you will get a nonzero value (possibly even negative) when it should be zero, due to rounding error. Constraints aside, you probably should set a tolerance value $$\epsilon > 0$$, small but not too small, and round $$y$$ to $$0$$ any time $$|y|<\epsilon$$. That will let you proceed as you proposed in your question.