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

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.

  • $\begingroup$ I haven't checked your formulation, but what value of M did you use, and how did you choose that value? $\endgroup$ Jan 31 '20 at 19:37
  • $\begingroup$ I am choosing $ M = 1000$ which is higher than the upper bound of the problem $\endgroup$
    – ooo
    Jan 31 '20 at 19:42
  • $\begingroup$ Perhaps you should spell out exactly in what way(s) the result is different than you expected. $\endgroup$ Jan 31 '20 at 20:19
  • $\begingroup$ Is $\delta$ a constant or variable? Is it nonnegative, strictly positive or what? $\endgroup$
    – prubin
    Jan 31 '20 at 21:29
  • 1
    $\begingroup$ The last two constraints are intended to enforce $z=1\implies (k_1=1 \land k_2=1)$, but it is better to instead have $z\le k_1$ and $z\le k_2$, with no big-M. $\endgroup$
    – RobPratt
    Jan 31 '20 at 21:36

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}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.