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?


1 Answer 1


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.

  • $\begingroup$ 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
    Commented 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
    Commented 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
    Commented 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
    Commented 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
    Commented Jan 14, 2020 at 18:26

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