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I have the decision variable $X_{iz}$

And I have two parameters $T_i\in\{0,1\}$ and $IT_z\in\{0,1,2\}$. I can only assign $i$ to $z$ if the following holds:

  • for $T_i=0$, $IT_z$ needs to be $0$ or $2$
  • for $T_i=1$, $IT_z$ needs to be $1$ or $2$
  • So for every value of $T_i$, a value of $2$ for $IT_z$ must satisfy the constraint
  • Or $T_i$ needs to be equal to $IT_z$

I cannot seem to figure out how to make a valid constraint for this problem, any tips?

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  • $\begingroup$ Do you have 2 $T$ and 3 $IT$ or do you have one $T$ and one $IT$ where $T$ can be $\{0, 1\}$ and $IT$ can be $\{0, 1, 2\}$. If the latter, then you don't need the indices. Just say $T \in \{0,1\}$ and $IT \in \{0,1,2\}$ $\endgroup$ – EhsanK Nov 22 at 14:50
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    $\begingroup$ No I have multiple $i$'s and multiple $z$'s, each has their own value of $T\in\{0,1\}$ and $IT\in\{0,1,2\}$. $\endgroup$ – Harry van t Kamp Nov 22 at 14:53
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    $\begingroup$ Check this answer from Decision Variable Value from a Set $\endgroup$ – EhsanK Nov 22 at 15:01
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For the new problem description, it seems like you just want to fix $X_{i,z}=0$ for some disallowed $(i,z)$ pairs. An even better approach is to avoid defining $X_{i,z}$ in those cases.

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Introduce a new binary variable $z_i$ and linear constraints: \begin{align} -z_i \le \text{IT}_i - \text{T}_i &\le 2 z_i \\ \text{IT}_i &\ge 2 z_i \end{align} If $z_i=0$ then $\text{IT}_i = \text{T}_i$. If $z_i=1$ then $\text{IT}_i \ge 2$, hence $=2$.

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  • $\begingroup$ I think this is not correct, because when $T_i=0$ and $IT_i=1$ the constraint also satisfies. And this should not be the case. That is what I am struggling with. $\endgroup$ – Harry van t Kamp Nov 22 at 14:25
  • $\begingroup$ Sorry, I misread $0$ or $2$ as $0$ to $2$. $\endgroup$ – Rob Pratt Nov 22 at 14:54
  • $\begingroup$ Corrected just now. $\endgroup$ – Rob Pratt Nov 22 at 15:04
  • $\begingroup$ Note: all your indices are of type $i$, but actually $IT$ has index $z$. Is the $z$ (you introduced) of type $i$ or of type $z$? I tried some things now, but the model seems infeasible $\endgroup$ – Harry van t Kamp Nov 22 at 16:44
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    $\begingroup$ Hmm, you changed the question after I posted my answer, and I don't quite understand the new question. My binary variable $z_i$ has nothing to do with your new index $z$. Is $X_{i,z}$ a binary decision variable? Are $\text{T}_i$ and $\text{IT}_{z}$ now decision variables or constants? $\endgroup$ – Rob Pratt Nov 22 at 18:18

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