# Another Absolute Value Question

Written by: Paul Rubin

Primary Source: OR in an OB World

Someone asked whether it is possible to eliminate the absolute value from the constraint

$$Lx\le |y| \le Ux$$

(where $$L\ge 0$$ and $$U>0$$ are constants, $$x$$ is a binary variable, and $$y$$ is a continuous variable). The answer is yes, but what it takes depends on whether $$L=0$$ or $$L>0$$.

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 $$z_1=1\Longrightarrow -U\le y \le -L$$ while $$z_2=1 \Longrightarrow L\le y\le U$$.

The following two tabs change content below. #### Paul Rubin

Professor Emeritus at Michigan State University
I'm an apostate mathematician, retired from a business school after 33 years of teaching mostly (but not exclusively) quantitative methods courses. My academic interests lie in operations research. I also study Tae Kwon Do a bit on the side. 