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I try to model a decision process in which I first need to decide about a binary variable $x$. In case $x=1$, I also need to decide upon variable $y$ and the value of $y$ is not defined otherwise. Especially, I do not know how to correctly introduce constraints. Assume that I want to model a constraint that states that $y$ needs to be greater than 5. Of course I could then just state: \begin{equation}\label{eq:cap} y > 5 \end{equation} But this would suggest that $y$ has a defined value, which does not need to be the case (if $x = 0$).

Is there a standard way how to introduce such a variable and associated constraints (e.g. in an IP or MDP)?

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    $\begingroup$ The answer will depend on whether $y$ appears in the objective function and how it appears in the constraints. Basically, there are two possibilities for when $x=0$: either you do not care what value $y$ takes, or you need it to take a specific value (most likely 0, possibly 1 if it multiplies other variables) that would be the algebraic equivalent of omitting it from the constraints. So we need more detail about how $y$ is used in the model. $\endgroup$
    – prubin ♦
    Jan 19, 2022 at 17:35
  • $\begingroup$ You could add $y\ge 5 x$. This way, if $x=1$, you have $y\ge 5$, but if $x=0$, $y$ can take any non negative value. I agree with @prubin, more detail is needed to provide a complete answer though. $\endgroup$
    – Kuifje
    Jan 20, 2022 at 9:38
  • $\begingroup$ Thank you for your answers @prubin. I cannot just set the value of y to 0, because it corresponds to a geographical position and appears in the objective function. Any value of y would lead to costs > 0. A solution would be (similar to the suggestion of @Kuifje) to multiply the term in the objective function by x, so that the value of y only matters if x=1. But I was wondering if there is another standard way of introducing such a relationship mathematically. $\endgroup$
    – PeterD
    Jan 20, 2022 at 17:43
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    $\begingroup$ The usual approaches to that would basically amount to doing what you suggest (multiplying the cost term by $x$) and then linearizing it. $\endgroup$
    – prubin ♦
    Jan 20, 2022 at 18:58

1 Answer 1

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Optional decision variables are part of the building block of scheduling within CPLEX CPOptimizer.

For instance

using CP;

dvar interval s size 1;
dvar interval e size 1;
dvar interval itvs optional size 7;

maximize presenceOf(itvs);
subject to
{
  startOf(s)==1;
  startOf(e)==8;
  startBeforeStart(itvs,s);
  startBeforeEnd(e,itvs);  
}

int isPresent=presenceOf(itvs);
execute
{
  writeln("isPresent=",isPresent);
}

gives

isPresent=1

but if I turn

dvar interval itvs optional size 7;

into

dvar interval itvs optional size 5;

then we get

isPresent=0

The constraints on itvs (start and end) are valid as long as the optional interval is present.

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