How can I linearize the IFF-THEN constraint with binary and continuous variable?

Question

I have an optimization problem where

$a_{m,n}\in\{0,1\}$ is a binary variable and $0\le f_{m,n}\le 1$ is a continuous vaiable

I have an Iff-THEN constraint like this

IFF $a_{m,n}=1$, THEN $f_{m,n}>0$

and

IFF $a_{m,n}=0$, THEN $f_{m,n}=0$

How can I linearize them?

I tried as below:

$f>a-1$

$f\le a$

but it is not working

score 1 · Accepted Answer · answered May 28 '19 at 21:20

1

First, note that you can't impose strict $>$ constraints. Instead, you'll have to choose a small $\epsilon > 0$ and then require $f \ge \epsilon$. Then these constraints should do the trick:

$$\begin{align} f & \le Ma \\ f & \ge \epsilon - M(1-a) \end{align}$$

Here, $M$ is a large constant. You should set it equal to the largest value that $f$ can reasonably take.

The logic is: If $a = 0$, then the first constraint says that $f$ must equal $0$ (and the second constraint has no effect). If $a = 1$, then the second constraint says that $f$ must be greater than or equal to $\epsilon$ (and the first constraint has no effect).

answered May 28 '19 at 21:20

LarrySnyder610

1,131

note that $f$ is upper bounded by 1, so putting $M=1$ does not really give what I want. For some values of $f$, I am getting negative values. – dipak narayanan May 29 '19 at 14:57
If $f \le 1$, then putting $M=1$ gives precisely what you want, does it not? – LarrySnyder610 May 29 '19 at 14:58
lets take the case $a=0$. the feasible range of $f$ then becomes $\epsilon-1\le f\le0$. The optimizer is taking a value very close to -1. – dipak narayanan May 29 '19 at 15:04
But you also have a constraint that says $0 \le f \le 1$, so together these two constraints say that $f=0$, which is what you want. – LarrySnyder610 May 29 '19 at 15:12
1

Sorry, my mistake, I just forgot to put the other constraint. Yes, I am getting what I want. Thanks a lot. – dipak narayanan May 29 '19 at 15:16

How can I linearize the IFF-THEN constraint with binary and continuous variable?

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