0

I have a modelling question. I want to model that if my varibale $x$ is in the interval $[a,b]$, where $a,b >0$, my indicator variable $X \in \{0,1\}$ should be 1 (if $x$ is not in $[a,b]$, the indicator variable can attain any value). I'm sure that this kind of problem is quite common in practice, but I couldn't really find anything.

What would be the best way to model this? I guess there are several formulations which model this constraint. Which would be the best regarding computation times? Do you know some papers/publications which cover this topic?

Best regards, ops

ops
  • 1
  • 1

1 Answers1

1

The implication $$ x\in[a,b] \Rightarrow y=1 $$ can be stated as $$ y=0 \Rightarrow x<a \text{ or } x >b$$

Traditionally we would formulate this as: $$\begin{align} &x \le a - 0.001 + M\delta + My\\ &x \ge b + 0.001 - M(1-\delta) -My\\ &\delta \in \{0,1\} \end{align}$$

where $M$ is a large enough constant. This was already answered (but not accepted) here conditional constraint: if $x \in [a,b] => z=1$.

Advanced solvers may support indicator constraints, SOS1 variables or even an or functionality directly.

A version without big-$M$'s but with indicator constraints and SOS1 variables can look like:

$$\begin{align} &y=0 \Rightarrow x \le a - 0.001 + s_1\\ &y=0 \Rightarrow x \ge b + 0.001 - s_2 \\ & s_1, s_2 \ge 0\\ & \text{SOS}1(s_1,s_2) \end{align}$$

The SOS1 set will make sure that only one of the slack variables $s_1, s_2$ will be non-zero. In other words,

$$ s_1 \cdot s_2 = 0$$

Note that I typically drop these $0.001$ terms.

Erwin Kalvelagen
  • 4,267
  • Thanks for your answer. The first formualtion I do understand, but I have problems with the second formulation. I looked up SOS1 and as far as I know it means that either s1 or s2 can be 1 and the other must be 0. But I still have some difficulties to see how those constraints work. – ops Dec 18 '17 at 19:12
  • That looks like a documentation bug. The correct definition of a sos1 set is that only one member can be non-zero. – Erwin Kalvelagen Dec 18 '17 at 19:30