Suppose we have the constraint $$a_1x_1 + \cdots + a_nx_n \gtreqless b,$$ where $a_i$ and $b$ are constants and $x_i$ are decision variables. Suppose also that we want the constraint to hold if $y=1$ (where $y$ is a binary decision variable), and we don’t care whether it holds if $y=0$.
How can we accomplish this?
Let $M$ be a new parameter (constant) that equals a large number.
Greater-than-or-equal-to constraints:
The constraint is $a_1x_1 + \cdots + a_nx_n \ge b$. Rewrite it as $$a_1x_1 + \cdots + a_nx_n \ge b - M(1-y).$$ Then, if $y = 1$, the constraint is active, and if $y=0$, it has no effect since the right-hand side is very negative.
(If all of the $a_i$ are nonnegative, you can instead use $$a_1x_1 + \cdots + a_nx_n \ge b(1-y),$$ which is tighter.)
Less-than-or-equal-to constraints:
The constraint is $a_1x_1 + \cdots + a_nx_n \le b$. Rewrite it as $$a_1x_1 + \cdots + a_nx_n \le b + M(1-y).$$ Then, if $y=1$, the constraint is active, and if $y=0$, it has no effect since the RHS is very large.
Equality constraints:
The constraint is $a_1x_1 + \cdots + a_nx_n = b$.. Rewrite it as $$\begin{align*} a_1x_1 + \cdots + a_nx_n & \le b + M(1-y) \\ a_1x_1 + \cdots + a_nx_n & \ge b - M(1-y) \end{align*}$$ Then, if $y=1$, the equality constraint is active, and if $y=0$, the constraints have no effect.
Note: If your model is relatively large, i.e., it takes a non-negligible amount of time to solve, then you need to be careful with big-$M$-type formulations. In particular, you want $M$ to be as small as possible while still enforcing the logic of the constraints above.
Related: In an integer program, how I can force a binary variable to equal 1 if some condition holds?
These are know as "indicator constraints" or "on/off" constraints. The best formulation is the convex-hull one, it includes the optimal big-M value plus additional non-redundant constraints, here's a note characterizing this formulation. There's also a generalization for convex nonlinear "on/off" constraints here and recent extensions here.
