I am trying to write a linear constraint that computes the absolute value of a difference, only if both the variables $x$ and $y$ are different from zero.
$x,y$ are binary variables while $s$ is a positive integer variable. $a$ and $b$ are positive integer coefficients.
$|xa-yb| \leq s$ only if $x \neq 0$ and $y \neq 0$
for the first part I did:
$(xa-yb) \leq s$ and $-(xa-yb) \leq s$
but I don't know how to proceed with the second part, if that is ever possible.
Thank you in advance
Best Answer
You can enforce $$(x = 1 \land y = 1) \implies |a-b| \le s$$ with linear constraint $$|a-b|(x + y - 1) \le s.$$
Do you instead (or also) want to enforce $$|a-b| \le s \implies (x = 1 \land y = 1),$$ which is the converse?