I would like to seek some advice on modeling the following:

I have two integer decisions variables, $x, x'$, that are either equal or greater than zero and either of them is greater than or equal to a third integer decision variable, $z$, which is also equal or greater than zero in accordance to the value of a binary indicator variable, $\beta$.

$\beta=1$ $\implies$ $x\ge z$

$\beta=0$ $\implies$ $x'\ge z$

Also, I would also like to about the case for the opposite too:

$\beta=1$ $\implies$ $x\le z$

$\beta=0$ $\implies$ $x'\le z$

Appreciate your kind guidance.

Thank you!


A non-linear formulation is given for $ x \geq 0$, $ x’ \geq 0$, $z \geq 0$ and $ \beta $ binary.

$\left\{ \begin{array}{l} x \geq \beta \cdot z \\ x’ \geq (1- \beta) \cdot z \\ \end{array} \right. $

If $ \beta = 1 $ we get

$\left\{ \begin{array}{l} x \geq z \\ x’ \geq 0 \\ \end{array} \right. $

If $ \beta = 0 $ we get

$\left\{ \begin{array}{l} x \geq 0 \\ x’ \geq z \\ \end{array} \right. $

Oppositive case

$\left\{ \begin{array}{l} z \geq \beta \cdot x \\ z \geq (1- \beta) \cdot x’ \\ \end{array} \right. $

The following formulation is linear, but it is valid only for $z < 1$.

$\left\{ \begin{array}{l} x \geq (\beta -1) + z \\ x’ \geq - \beta + z \\ \end{array} \right. $

As known, $ (0;1) \cong R$ meaning that a bijective function exists between the open interval and the set of real numbers. Also the generic interval $ (a;b) $ is isomorph to $ (0;1) $, $ (a;b) \cong (0;1) $ by means of the following bijection:

$ f(x):= (x-a)/(b-a) $

Thanks to this observation, the linear formulation of the logical constraints can be given changing the interval where variables are defined. From $ x \geq 0$, $ x’ \geq 0$, $z \geq 0$ and $ \beta $ binary, let introduce $ y := (x-a)/(b-a) $, $ y’ := (x’-a)/(b-a) $$ w := (z-a)/(b-a) $ where $a$ and $b$ are suggested from the context of the examinated problem.

‘> $\left\{ \begin{array}{l} y \geq ( \beta - 1) + w \\ y’ \geq - \beta + w \\ y \in (0;1) \\ y’ \in (0;1) \\ w \in (0;1) \\ \beta binary \\ \end{array} \right. $

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  • $\begingroup$ Dear Marco, Thank you for your timely reply! May I ask If you are any linear formulations as I am dealing with a mixed-integer linear programming formulation. Thanks. $\endgroup$ – Mike Sep 19 at 10:44
  • $\begingroup$ I wrote a linear formulation that is valid only for $z \leq 1$.$\left\{ \begin{array}{l} x \geq (\beta - 1) + z \\ x’ \geq - \beta + z \\ \end{array} \right. $ $\endgroup$ – marco tognoli Sep 19 at 10:59
  • $\begingroup$ Dear Marco, thanks! $\endgroup$ – Mike Sep 20 at 9:23

Let $y$ be a binary variable and let $f$ be a linear function bounded above by some constant $M$. The standard approach to enforce $y=1 \implies f\le 0$ is to impose linear big-M constraint $$f\le M(1-y)\tag1.$$ All four of your implications are of this form. For the first one, take $y=\beta$ and $f=z-x$ in $(1)$, yielding $z-x\le M(1-\beta)$. For the second one, take $y=1-\beta$ and $f=z-x’$ in $(1)$, yielding $z-x’\le M\beta$. The other two are similar.

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  • $\begingroup$ Dear Dr Rob, Thank you! $\endgroup$ – Mike Sep 19 at 16:23

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