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I have $3$ parameters $a_1,a_2,a_3$ and a variable $d$ and $3$ binary variables $b_1,b_2,b_3$ and a "result" variable $s$. How do I model constraints so that:

  • If $d$ is between $0$ and $a_1$, then $s=1$
  • If $d$ is between $a_1$ and $a_2$, then $s =$ ... complex calculation ...
  • If $d$ is between $a_2$ and $a_3$, then $s = 0$

I have found a formulation that is as follows: \begin{align} a_1b_1 + a_2b_2 - d \le 0 \\ -a_1b_1 - a_2b_2 - a_3b_3 + d \le 0 \\ b_1 + b_2 + b_3 = 1 \end{align}

But if use this formulation in my model, the b-variables get calculated incorrectly. Does anyone see the reason why and/or has a better idea?

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  • $\begingroup$ If $d$ and $a_i$ are parameters, then you have no modeling to do, the values are known a priori so you can deduce $s$ while pre processing. $\endgroup$
    – Kuifje
    Jul 21 at 13:39
  • $\begingroup$ d is a variable. I made a mistake in the Post. I will edit it sorry. $\endgroup$
    – Koli
    Jul 21 at 13:58
  • $\begingroup$ Does the "complex calculation" for $s$ involve other decision variables, or is it constant? $\endgroup$
    – RobPratt
    Jul 22 at 13:33
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It looks like your first constraint should instead be $$0b_1 + a_1 b_2 + a_2 b_3 - d \le 0$$ With this change, the logical implications are \begin{align} b_1 = 1 &\implies 0 \le d \le a_1 \\ b_2 = 1 &\implies a_1 \le d \le a_2 \\ b_3 = 1 &\implies a_2 \le d \le a_3 \end{align}

To avoid ambiguous borders, introduce a small tolerance $\epsilon>0$ and impose instead $$0 b_1 + (a_1+\epsilon) b_2 + (a_2+\epsilon) b_3 - d \le 0$$

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  • $\begingroup$ I tried hat and it still didn't calculate the b values correctly. Just for clarification do you mean my constraint should be: 0b1+a1b2+a2b3 - d <= 0 or 0b1+a1b2+a2b3 <= 0? And also the interval borders would be problematic. I would need: b1 -> 0 <= d <= a1; b2 -> a1 < d <= a2; b3 -> a2 < d <= a3; $\endgroup$
    – Koli
    Jul 21 at 14:22
  • $\begingroup$ I updated my answer. You cannot avoid the ambiguous borders without introducing a small tolerance $\epsilon>0$. $\endgroup$
    – RobPratt
    Jul 21 at 16:11
  • $\begingroup$ Ok I tried changing the first constraint again and now GUROBI concluded that my program is infeasible $\endgroup$
    – Koli
    Jul 22 at 7:53
  • $\begingroup$ Ok, how would the constraints change if I wanted to simplify my situation: Let's say I want s = ... if d <= a_1 and s=0 if d > a1 I'm sorry if my questions are dumb but I am really new to this topic $\endgroup$
    – Koli
    Jul 22 at 7:59
  • $\begingroup$ I suspect that the infeasibility arises from your constraints that link $s$ to $b$. Try solving without those first to see if the problem is then feasible. $\endgroup$
    – RobPratt
    Jul 22 at 13:01

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