Linearizing constraint with continuous and boolean variables

Question

I have two continuous variables $A$, $B$ and two binary variables $x$, $y$.

Condition: if $A = B \wedge x = 1 \wedge y=1$ then $z = 1$ else $z = 0$ from

• In an integer program, how I can force a binary variable to equal 1 if some condition holds?

• In an integer program, how can I “activate” a constraint only if a decision variable has a certain value?

My current attempt is:

\begin{align} A-B + \delta+x+y &\leq 2 + M \cdot k_1\\ B - A - \delta-x-y &\leq - 2 + M \cdot(1 - k_1)\\ B-A + \delta+x+y &\leq 2 + M \cdot k_2\\ A - B - \delta-x-y &\leq -2 + M \cdot(1 - k_2)\\ k_1 + k_2 - 1 &\leq z \\k_1 &\geq z \\k_2 &\geq z \end{align}

where $k_1, k_2$ are boolean variables, but I am not getting the expected result.

Answer 1

You want to model $$z=1 \iff (A=B\land x=1 \land y=1).$$

To enforce $z=1 \implies (A=B\land x=1 \land y=1)$: \begin{align} -M(1-z) \le A - B &\le M(1-z)\\ z &\le x\\ z &\le y \end{align}

To enforce the converse $(A=B\land x=1 \land y=1) \implies z=1$, equivalently, $A>B\lor A<B\lor x=0 \lor y=0 \lor z=1$: \begin{align} B-A+\delta &\le M_1 (1-w_1)\\ A-B+\delta &\le M_2 (1-w_2)\\ w_1+w_2+(1-x)+(1-y) + z&\ge 1\\ w_1,w_2&\in\{0,1\} \end{align}