The easy case is when $L=0$. There are two possibilities for the domain of $y$: either $y\in [0,0]$ (if $x=0$) or $y\in [-U,U]$ (if $x=1$). One binary variable is enough to capture two choices, so we don't need any new variables. We just need to rewrite the constraint as$$-Ux\le y \le Ux.$$If $L>0$, then there are

*three*possibilities for the domain of $y$: $[-U, -L] \cup [0,0] \cup [L, U]$. That means we'll need at least two binary variables to cover all choices. Rather than change the interpretation of $x$ (which may be used elsewhere in the questioner's model), I'll introduce two new binary variables ($z_1$ for the left interval and $z_2$ for the right interval) and link them to $x$ via $x=z_1+z_2$. That leads to the following constraints:$$-Uz_1 \le y \le -Lz_1 + U(1-z_1)$$and$$Lz_2 - U(1-z_2) \le y \le Uz_2.$$ If $x=0$, both $z_1$ and $z_2$ must be 0, and the new constraints force $y=0$. If $x=1$, then either $z_1=1$ or $z_2=1$ (but not both). Left to the reader as an exercise: verify that $z_1=1\implies -U\le y \le -L$ while $z_2=1 \implies L\le y\le U$.

## No comments:

## Post a Comment

Due to recent spamming, comments are being moderated. If this is your first time commenting on the blog, please read the Ground Rules for Comments. In particular, if you want to ask an operations research-related question not relevant to this post, consider asking it on Mathematics Stack Exchange.