A model contains constraints of the following form

$R(k) \leq X(k) G(k)$

where $X(k)$ binary and $G(k)$, $R(k)$ non-negative variables.

The index $k$ runs from $1$ to $50$.

I linearise the equations as they are products of binary and continuous variables.

The question is, is there anything I can exploit to tighten the region these constraints define?


1 Answer 1


You want to enforce $X(k) = 0 \implies R(k) = 0$ and $X(k) = 1 \implies R(k) \le G(k)$. You can use indicator constraints for that.

Alternatively, a straightforward big-M formulation yields \begin{align} R(k) &\le M_1(k) X(k) \tag1 \\ R(k) - G(k) &\le M_2(k) (1 - X(k)) \tag2 \\ \end{align}

A natural choice for $M_1(k)$ is a small constant upper bound $\bar{G}(k)$ on $G(k)$, and a natural choice for $M_2(k)$ is $\bar{G}(k) - 0$. But notice that you can do better because $M_2(k)$ needs to be an upper bound on $R(k)-G(k)$ only when $X(k)=0$; in that case, $(1)$ forces $R(k)=0$, so you can take $M_2(k)=0$:

\begin{align} R(k) &\le \bar{G}(k) X(k) \tag1 \\ R(k) &\le G(k) \tag2 \\ \end{align}

Depending on other constraints in your model, you might be able to tighten further.

  • $\begingroup$ Hi Rob. Thank you very much again! Your idea was very useful. Through further tightening I was able to solve my problem, which otherwise "refused" to converge to the global optimum. $\endgroup$
    – Clement
    Commented Apr 23, 2021 at 16:21

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