I got a model of MILP but I want to add extra constraints.

$$ \min\{c^T x + d^T y:Ax+By=a,Cx+Dy \leq b,x,y\geq0,y \in \{0,1\}\} $$

More specificlly, I got extra original constraints with indicator function in the following form

$$ \frac{1}{N}\sum_{i=1}^{N}\mathbb{I}(z\mu_i-1) \geq 1-\epsilon,Ex+Fy+Gz\geq0 $$ where $z$ is extra matrix variable to be optimized.

The indicator function can be approximated by logistic function $g(x)=\frac{e^{-x/{\epsilon}}}{1+e^{-x/{\epsilon}}}$, transforming the above constrints to

$$ \frac{1}{N}\sum_{i=1}^{N}g(z\mu_i-1) \geq 1-\epsilon $$

So how could the resulting problem be solved?

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    $\begingroup$ Please give your definition of $\mathbb{I}(x).$ $\endgroup$
    – prubin
    Commented Apr 8 at 15:25
  • $\begingroup$ @prubin $\mathbb{I}(x)$ is the indicator function which takes 0 when $x$ is greater than 0 and 1 otherwise. $\endgroup$ Commented Apr 17 at 1:23
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    $\begingroup$ What are $z$ and $\mu_i$? On the one hand, you say $z$ a matrix variable. On the other hand, the constraint $Ex + Fy + Gz \geq 0$ indicates it's a vectorial variable, i.e. $z \in \mathbb{R}^n$. $\endgroup$
    – joni
    Commented Apr 17 at 9:30
  • $\begingroup$ @KaimingZhang what are you modelling with the indicator constraint? Should it be interpreted as something along the lines of "$z\mu_i\leq 1$ for $1-\epsilon$ percent of the $N$ constraints? $\endgroup$
    – Sune
    Commented Apr 19 at 9:15
  • $\begingroup$ @Sune Yes, that is the case. $\endgroup$ Commented Apr 25 at 3:22

1 Answer 1


I think you can achieve what you want as follows: Let $u_i$ be upper bounds on $z\mu_i$, for all $i=1,...,N$. Let furthermore $\alpha_i$ be binary variables equalling one, if $z\mu_i>1$ is allowed. Then you may formulate your requirement as \begin{align} &z\mu_i\leq 1+(u_i-1)\alpha_i,&&\text{for } i=1,..,N&(1)\\ &\sum_{i=1}^N\alpha_i\leq \lfloor N\epsilon\rfloor&&&(2) \end{align} This way, you allow violation of only $N\epsilon$ of the constraints ($\alpha_i=1$), stating that at least $N(1-\epsilon)$ of the constraints must be satisfied. You can round down the right hand side of (2) as the left hand side is an integer.

  • $\begingroup$ Thank you. I got what you mean. $\endgroup$ Commented Apr 29 at 2:49
  • $\begingroup$ @KaimingZhang, if it solves your problem, then please accept the answer for future visitors. $\endgroup$
    – Sune
    Commented Apr 29 at 6:23

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