4
$\begingroup$

I have the constraint \begin{align}\max&\quad\gamma\\\text{s.t.}&\quad a\ge\gamma b\\&\quad\gamma\le 1\end{align} where $\gamma$ is an optimization variable and $a$ is a function of some other variable.

So, can I just write \begin{align}\max&\quad\gamma\\\text{s.t.}&\quad a\le b\end{align} instead?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

No. The constraint $a\ge\gamma b$ sets no upper bound on $a$ so it cannot be bounded above by $b$ as your second formulation suggests.

There are a few posts here on the linearisation of the product of two variables; e.g.

Improve this answer
$\endgroup$
5
  • $\begingroup$ but what about $\gamma\le 1$? So we have $a\ge\gamma b$ and $\gamma\le 1$. If we combine these two, what do we have? $\endgroup$
    – KGM
    Jan 14, 2020 at 17:40
  • $\begingroup$ Combining your constraints you get $a \geq \gamma b $ and $ \gamma b \leq b$, but $a$ is still not bounded above by $b$. $\endgroup$
    – Libra
    Jan 14, 2020 at 17:51
  • $\begingroup$ @dipaknarayanan I assumed in my comment that $b$ is a parameter, however, since you explicited that only $a$ and $\gamma$ are variables. $\endgroup$
    – Libra
    Jan 14, 2020 at 18:02
  • $\begingroup$ @Libra, you are right. $b$ is a known parameter. But I still do not get how do you get $\gamma b\le b$. Combination of $\ge$ and $\le$ sometimes confuse me a lot! $\endgroup$
    – KGM
    Jan 14, 2020 at 18:05
  • $\begingroup$ @dipaknarayanan if $\gamma = 1$ then $1\cdot b = b$, whereas if $0 \leq \gamma < 1$ then $\gamma b < b$. Combining the two you get $\gamma b \leq b$. $\endgroup$
    – Libra
    Jan 14, 2020 at 18:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.