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I am trying to write an ILP for a problem but I have this logical constraint and I'm stuck.

In my model I have:

two binary variables: $x$ and $y$

One Integer variable: $z$

The logical constraint I am trying to write is:

if $x = 1$ and $y = 1$ then $z \le Mx$, else $z = 0$

Thank you for your help.

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  • $\begingroup$ Oh, I think I was assuming $z \ge 0$. Is that correct? $\endgroup$ – RobPratt Aug 14 at 22:10
  • $\begingroup$ yes that is correct but also $z$ has an upper bound Big-M , so $0 \leq z \leq M$ . $\endgroup$ – che Aug 14 at 22:33
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You always want $$z \ge 0 \tag1$$ You want $x = 0 \implies z \le 0$: $$z \le M x \tag2$$ Similarly, you want $y = 0 \implies z \le 0$: $$z \le M y \tag3$$

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  • $\begingroup$ Thank you so much for your answer. Do yo think expressing it like this would work : $ z \leq M (x + y - 1) $ ? $\endgroup$ – che Aug 14 at 19:08
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    $\begingroup$ No. Consider the $(x,y)=(0,0)$ case. $\endgroup$ – RobPratt Aug 14 at 20:06
  • $\begingroup$ Hello again , regarding the constraint that I have asked about in my previous question, I have a slight modification. I want to write the following constraint : let z be an integer variable, and t a binary variable, and a big-M. z is $0\leq z \leq M$. the logical constraint is as following : if $ z \leq M$ and $z > 0 \Rightarrow t = 1 $ ; and if $ z = 0 \Rightarrow t = 0$. Is this $ z \leq M t $ sufficient? the t and z variables are not in my objective function but , variable t is connected to another variable in the objective function? Thank you very much, I appreciate your help. $\endgroup$ – che Aug 15 at 17:14
  • $\begingroup$ Please start a new post for this question. $\endgroup$ – RobPratt Aug 15 at 17:55

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