How to linearize min function as a constraint?

Question

I'm trying to solve an optimization problem including following constraint, and I need to linearize it in a maximization nonlinear programming model. Please help me to reformulate it with mixed integer programming. A part of my model is here: $$\min\ Z=-q_1p_1 \\ q_1=\min\{b,ap_1\}$$
Note that $q_1$ and $p_1$ are non-negative variables and the others are parameters. I know that $$q_1\le b,\, q_1\le ap_1 $$ are helpful. But I'm trying to find another way to linearize it.

Answer 1

Here is an answer that uses the same approach as in this answer, but converted from $\max\{\cdot,\cdot\}$ to $\min\{\cdot,\cdot\}$. I'll write the constraint in a more general form:

$$X = \min\{x_1,x_2\}$$

This method works if $x_1$ and $x_2$ are constants or decision variables (or one of each). (In your question, $X = q_1$, $x_1 = b$ and $x_2 = ap_1$.)

We want a set of constraints that enforces $X = \min\{x_1,x_2\}$. Define a new binary decision variable $y$, which will equal 1 if $x_1 < x_2$, will equal 0 if $x_1 > x_2$, and could equal either if $x_1 = x_2$. Let $M$ be a constant such that $x_1,x_2 \le M$ in any "reasonable" solution to the problem.

The following constraints enforce the definition of $y$: $$\begin{align} x_2 - x_1 & \le My \\ x_1 - x_2 & \le M(1-y) \end{align}$$ Then, the following constraints enforce $X = \min\{x_1,x_2\}$: $$\begin{align} X & \le x_1 \\ X & \le x_2 \\ X & \ge x_1 - M(1-y) \\ X & \ge x_2 - My. \end{align}$$ The first two constraints say $X \le \min\{x_1,x_2\}$. Combined with these constraints, the last two constraints say that $X = x_1$ if $x_1 < x_2$ (so $y=1$) and $X = x_2$ if $x_2 < x_1$ (so $y=0$).