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In MIP, I have a constraint that is validated only if the demand $d_i\leq Q$: C1 : $d_i \leq x_i \leq Q$ where $x_i \geq 0$. I tried to introduce big M with binary variables as follows: $My \leq x_i \leq Q - d_i$ Where $M=|Q-\min{(d_i)}|$ The problem where $d_i > Q$ then $x_i$ is equal to zero but if $d_i \leq Q$, $x_i$ should be between $d_i$ and $Q$. Any help please to how formulate this constraint. Thank you

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I'm going to drop the subscript $i$ to save a little typing, and I will assume that both $d$ and $Q$ are nonnegative (which was not actually stated). I will also assume that $Q$ is a constant. $d$ can be either a constant or a variable. Given a valid upper bound $M$ (presumably $\ge Q)$ for $d,$ we introduce a binary variable $y$ together with the constraints $$d \le Qy + M(1-y)\quad (1)$$ $$d \ge Q(1-y)\quad (2)$$ $$0 \le x \le Qy \quad (3)$$ and $$x \ge d-M(1-y). \quad (4)$$ If $y=1,$ then (1) and (2) imply $0 \le d \le Q$ and (3) and (4) imply $d \le x \le Q.$ If $y=0,$ then (1) and (2) imply $Q \le d \le M$ and (3) and (4) imply $x = 0.$

Note that $d = Q$ can occur with either value of $y,$ meaning $d=Q$ and $x=0$ is feasible (with $y=0$). Depending on the objective function and the rest of the model, this many not be a problem. If it is, your best recourse is to change (2) to $$d \ge (Q+\epsilon)(1-y)\quad (2')$$for some small positive value of $\epsilon.$ This will ensure that $y=0$ (and thus $x=0)$ occurs only when $d$ is strictly greater than $Q.$ Unfortunately, it will also make any solution with $Q < d < Q+\epsilon$ infeasible.

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  • $\begingroup$ Thank you so much for theses explanations. I finnaly understand how it works. $\endgroup$
    – MAYA
    Dec 29, 2023 at 15:00

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