How to model logic constraint: $y=1$ if $a\le x\le b$ and $y=0$ otherwise?

Question

I am trying to formulate indicator-type of constraints. $y$ is binary $0$ or $1$ and $x$ is a continuous variable. $$ y = \begin{cases} 1, & \text{ if } a \leq x \leq b \\ 0, & \text{ Otherwise } \end{cases} $$

The question is how to model this using linear constraints? I have researched and done my homework below with thoughts:

1. Define binary $$ y_1 = \begin{cases} 1, & \text{ if } a \leq x \\ 0, & \text{ Otherwise } \end{cases} $$

Define binary $$ y_2 = \begin{cases} 1, & \text{ if } x \leq b \\ 0, & \text{ Otherwise } \end{cases} $$

Big-M methods can be used to define linear constraints for $ x, y_1 $ and $ y_2.$

2. Enforce $ y=1 $ if both $ y_1=1 $ and $ y_2=1 $; $ y=0 $ otherwise, as follows: $$ y \leq y_1, y \leq y_2, y \geq y_1+y_2-1 $$

However I ended up with two many ( 4 with Big-M methods and 3 above) constraints. I was wondering if anyone has other more efficient or better approaches?

Answer 1

Suppose $\ell \le a \le b \le u$, where $\ell$, $a$, $b$, and $u$ are constants. Introduce binary variables $y^-$ and $y^+$, and impose the following linear constraints: $$\ell y^- + a y + b y^+ \le x \le a y^- + b y + u y^+ \\ y^- + y + y^+ = 1$$ If you want to avoid the ambiguity at $x=a$ and $x=b$, introduce a small constant $\epsilon > 0$ and instead impose $$\ell y^- + a y + (b + \epsilon) y^+ \le x \le (a - \epsilon) y^- + b y + u y^+ \\ y^- + y + y^+ = 1$$

Answer 2

I can think of:

a <= xy, since a/x will b smaller than 1, so y ==1.

Similarly, x <= by.

If want to avoid bilinear terms then z+y =1, then az <= x

If it's AND situation which means both have to be true:

a <= xy1

x <= by2

both y1 & y2 are binary followed by

2y <= y1+y2