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I am trying to create a set of constraints that forces a binary variable $y$ to be set to a certain value when a variable $x$ is greater than or equal to zero. I have,

$-1 \leq x \leq 1, \qquad x \in {\rm I\!R}$

$y = \begin{cases} 1 & \text{if } x \geq 0\\ 0 & \text{if } x < 0 \end{cases}$

I believe the way to practically do this is as follows:

$x \leq My - \epsilon$

$-x \leq M(1-y)$

Note that $M$ is a very large number and $\epsilon$ is a very small number.

I am unsure if there is a better way to do this, let me know if there is. I don't like the fact that I need to use the $\epsilon$ value to separate from the $0$ boundary, but I may have no other choice.

Cashew
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  • It's not really clear what you are trying to do here. What do you mean by "constraints that flag when a variable is greater than or equal to zero"? Which variable are you trying to check? Why is there an $x$ and a $y$? – kccu Sep 23 '19 at 01:26
  • So when variable $x$ is greater than or equal to $0$ I need to force $y$ to become equal to $1$. On the contrary, when $x$ is less than $0$, I need to force $y$ to become equal to $0$. – Cashew Sep 23 '19 at 01:31
  • Added this clarification to the question. – Cashew Sep 23 '19 at 01:33

1 Answers1

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What you have is the standard modeling approach, and the $\epsilon$ is unavoidable unless you want the $x=0$ case to allow either value for $y$. Because $x$ has bounds $[-1,1]$, you can (and should) replace $M$ with 1 in both places.

RobPratt
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