If-Then-Else modeling in MILP using the Big M method

Question

I have trouble finding a solution to the following problem. I have a decision variable $x$. If the value of $x$ is between 0 and a constant $A$, then the binary variable $y_1$ must be equal to 1. If $x$ is greater than $A$ but lower than another constant $B$, then the binary variable $y_2$ must be equal to 1 while the $y_1$ must be equal to 0. If $x$ goes above $B$, the binary variable $y_3$ must be equal to 1 while $y_1$ and $y_2$ must be equal to 0.

I tried using the Big $M$ method as follows:

$$x \le Ay_1 \tag{1}$$

The problem here is that if $x$ goes above $A$ then $x$ is infeasible. Then I created three new decision variables $x_1$, $x_2$, and $x_3$ that could "follow" $x$ for a certain amount:

$$x = (x_1 y_1)+(x_2 y_2)+(x_3 y_3) \tag{2}$$

$$ x_1 \le A y_1 \tag{1}$$

$$ x_2 \ge A y_2 \tag{3}$$

$$ x_2 \le B y_2 \tag{4}$$

$$ x_3 \ge B y_3 \tag{5} $$

This does not work: the solver tells me it is infeasible. I'm using OpenSolver in Excel. How can I resolve this?

Answer 1

There is some ambiguity about the strictness of above/beneath, but does the following do what you want? $$ 0y_1 + Ay_2 + By_3 \le x \le Ay_1 + By_2 + Cy_3 \\ y_1 + y_2 + y_3 = 1 $$ Checking, we have: \begin{align} y_1 = 1 &\implies x \in [0,A] \\ y_2 = 1 &\implies x \in [A,B] \\ y_3 = 1 &\implies x \in [B,C] \\ \end{align}