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I'm wondering, is the following a linear constraint

$$x + ry \geq 12 , \quad r \in [2, 3]$$

$$x,y \in {\rm I\!R}$$

I don't think it is, because it's not defined everywhere where x and y are. If that condition for r was not there, then it would be linear.

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  • $\begingroup$ What are the variables? $\endgroup$ – Rodrigo de Azevedo Oct 28 '19 at 22:38
  • $\begingroup$ @RodrigodeAzevedo real numbers. $\endgroup$ – AlfroJang80 Oct 28 '19 at 22:58
  • $\begingroup$ A linear constraint cannot contain a product of variables. $\endgroup$ – Rob Pratt Oct 28 '19 at 23:10
  • $\begingroup$ @RobPratt My apologies. The variables here x and y. r is just a constant that could be anything from the set [2, 3] $\endgroup$ – AlfroJang80 Oct 28 '19 at 23:35
  • $\begingroup$ Then the constraint is linear for each fixed $r$. $\endgroup$ – Rob Pratt Oct 28 '19 at 23:45
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Yes, it is a linear constraint. $x+ry\geq 12$ is linear for any constant coefficient $r$. The fact that you know additional information about $r$ (namely, that it happens to lie in the set $[2,3]$) does not change the linearity of the constraint.

For example, $px+y\geq 2$ is also linear for a coefficient $p$, even when you know that in particular $p=4$. The key thing is that the coefficient is a constant and not a variable.

You mentioned that the constraint is "not defined everywhere where x and y are". Perhaps a misconception. The constraint is defined using the constant $r$, (which we know lies in $[2,3]$), but that doesn't mean the constraint only holds for $x,y\in[2,3]$ or anything like that. Instead, the constraint is defined for all real $x$ and $y$.

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