I would like to model the following:

$B \le \alpha \implies \sum_i W(i) \ge \beta$, where $B$ a continuous variable, $W(i)$ binary variables, $\alpha$ a real constant number, $\beta$ an integer constant number

I could do this by introducing a new binary variable $Z$ and requiring

$B \le \alpha \implies Z = 1$ and $Z =1 \implies \sum_i W(i) \ge \beta$

My question is, is there a possibility of avoiding the introduction of $Z$?

  • $\begingroup$ Related: or.stackexchange.com/questions/10172/… $\endgroup$
    – RobPratt
    Jan 31 at 0:33
  • $\begingroup$ Here the constraint $\sum_i W(i) \ge \beta $ is an all integer constraint. Could this somehow avoid the introduction of $Z$? $\endgroup$
    – Clement
    Jan 31 at 8:44
  • $\begingroup$ Dear @RobPratt, in a case, where one of $w_i$ is equal to one, then $B \leq \alpha \implies w = 1$. That can be linearized without any more binary variables. Would you say, does not any general procedure exist to do that? $\endgroup$
    – A.Omidi
    Jan 31 at 9:39
  • $\begingroup$ I should state that $1 \le \beta \le \lvert I \rvert, i \in I$ $\endgroup$
    – Clement
    Jan 31 at 15:24
  • $\begingroup$ Do you also have bounds on $B$ and $\alpha$? $\endgroup$
    – RobPratt
    Jan 31 at 17:15

1 Answer 1


As in How to enforce logical implication $\sum_j a_j x_j \le b \implies \sum_j c_j x_j \le d$, (temporarily) introduce constant $\epsilon>0$ and binary variable $Z$, and impose linear big-M constraints \begin{align} \alpha + \epsilon - B &\le (\alpha + \epsilon - 0) Z \tag1\label1 \\ \beta - \sum_i W_i &\le (\beta - 0) (1-Z) \tag2\label2 \end{align} Now eliminate $Z$ by multiplying \eqref{1} by $\beta$ and \eqref{2} by $\alpha + \epsilon$ and adding them up to obtain the valid inequality $$\beta(\alpha + \epsilon - B) + (\alpha + \epsilon)\left(\beta - \sum_i W_i\right) \le \beta(\alpha + \epsilon) Z + (\alpha + \epsilon) \beta (1-Z),$$ which simplifies to $$B + \frac{\alpha + \epsilon}{\beta}\sum_i W_i \ge \alpha + \epsilon. \tag3\label3$$

But this is too weak unless $B=0$ or $\sum_i W_i = 0$. The original disjunction $$B \ge \alpha + \epsilon \lor \sum_i W_i \ge \beta$$ is nonconvex, and \eqref{3} is a convex relaxation.

  • $\begingroup$ Thank you very much Rob . I will implement it and see. $\endgroup$
    – Clement
    Jan 31 at 20:08

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