How To Linearize $X = \max\{x_1,x_2\}$

Question

I am new to thinking about math programming and I have a particular constraint I am hoping to reformulate, I just don't know the proper mathematical translation for what I am hoping to do. Enforcing the non-negativity constraint on $X$ in Constraint (1) seems to make my model infeasible, and so I am hoping that with a reformulation of my problem the solver I am using (knitro) will be able to report feasibility.

My original constraint looks like this:

$$ X_{t} = X_{t-1} + aY_{t-1} - Z_{t-1}, \quad t\in[1,T] \tag1 \\X_{t} \ge 0 \\ \\ X_{0}, Y_{0}, Z_{0} \ge0 $$

The way that I am thinking about the new formulation is shown in Constraint (2), but I feel that this is where I will need some correction on the formulation:

$$ X_t = \begin{cases} 0, & \text{if} \quad X_{t-1} + aY_{t-1} - Z_{t-1} \le 0 \\ X_{t-1} + aY_{t-1} - Z_{t-1}, & \text{otherwise} \end{cases} \tag2 \\t\in[1,T] \\ X_{0}, Y_{0}, Z_{0} \ge0 $$

I apologize if this is straightforward, I just don't know the precise wording to Google my problem because this is not my field of study.

If someone would be able to (1) help me classify what type of constraint this would be called and possibly a way to linearize it or a useful resource I can follow up on, I would be appreciative of that.

Answer 1

Solution provided by LarrySnyder610, full original post can be found here:

We want a set of constraints that enforces $X = \max\{x_1,x_2\}$. 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 $y$: $$\begin{align} x_1 - x_2 & \le My \\ x_2 - x_1 & \le M(1-y) \end{align}$$ Then, the following constraints enforce $X = \max\{x_1,x_2\}$: $$\begin{align} X & \ge x_1 \\ X & \ge x_2 \\ X & \le x_1 + M(1-y) \\ X & \le x_2 + My. \end{align}$$ The first two constraints say $X \ge \max\{x_1,x_2\}$, as you suggested in the question. 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 $y=0$).