# How to model this expression?

Suppose $$0\le x \le 1$$ is a decision variable and $$\gamma(x)$$ is defined as follows: $$\gamma(x)= \begin{cases} \theta & x>0\\ 0 & x=0 \end{cases}$$ where $$0\le \theta\le 1$$.

In my model, I have both $$\gamma(x)$$ and $$x \gamma(x)$$ and I want to convert them to linear programs. I used the following constraints: \begin{align} \Gamma&\ge \theta x - (1- y ) \\ \Gamma&\le y \\ \gamma&=\theta y \\ y &\ge x \end{align}

Here, the problem is that $$x$$ can take value 0 and then, $$\gamma = \theta$$ at the same time. I can add the constraint $$y \le 10000 x$$, but it is will exclude some parts of the solution space.

You need a tolerance $$\epsilon>0$$, and you can strengthen your first two inequality constraints: \begin{align} \Gamma &= \theta x \\ \gamma &= \theta y \\ \epsilon y \le x &\le y \end{align}
• Thanks. I tried it but there is a problem. My model tends to maximize $\gamma$. So, it set $x=\epsilon$ to make $y=1$. However, based on logics, it should not do this. Aug 20 '21 at 1:11
• Maybe your $\epsilon$ is not big enough to be considered "really" positive. Aug 20 '21 at 1:33
There is no perfect fix for this, since strict inequalities are not supported in MILP models. So you will have to either live with the ambiguity when $$x=0$$ or exclude a portion of the solution space.