Disjunctive Constraint , Using Binary Variable to Replace a If or condition

Question

I am trying to use a binary variable based on an inequality. The value of binary variable $q $ is 1 or 0 based on the following equation.

[ $q $ = \begin{cases} 0,& \text{if } b \geq \pi ,\\ 1, & \text{otherwise} \end{cases}

Here, b and $\pi$ are real numbers. Sample value b = 20 , $\pi$ = 30.

I have tried to represent this by:

\begin{equation} q \geq \dfrac { (\pi - b )} {M} \end{equation}

\begin{equation} q \leq 1 + \dfrac { (\pi - b )} {M} \end{equation}

By using these two equations I am able to cover the cases for when $b > \pi $ and when $b < \pi $. Unfortunately I am unable to set $q$ as 0 when $b=q$ without violating other conditions.

Answer 1

The usual approach to this requires that $b$ be bounded, say $L \le b \le U$ for some constants $L$ and $U.$ You can come close to what you want with the following: $$b \ge \pi (1-q) + L q$$ $$b \le \pi q + U (1-q).$$ If $b > \pi,$ the second constraint forces $q=0.$ If $b < \pi,$ the first constraint forces $q=1.$ The tricky part comes when $b=\pi,$ in which case $q$ can be either 0 or 1. Because you cannot enforce a strict inequality in a MIP model, if you can't accept the ambiguity when $b=\pi$ then you can change the second constraint to $$b \le (\pi - \epsilon) q + U (1-q),$$ where $\epsilon > 0$ is a small constant. Now $$b\ge \pi \implies q=0,$$ $$b \le \pi-\epsilon \implies q = 1,$$ and $\pi - \epsilon < b < \pi$ is forbidden.