# How to express this constraint?

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?

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.

• 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?
– KGM
Commented Jan 14, 2020 at 17:40
• Combining your constraints you get $a \geq \gamma b$ and $\gamma b \leq b$, but $a$ is still not bounded above by $b$. Commented Jan 14, 2020 at 17:51
• @dipaknarayanan I assumed in my comment that $b$ is a parameter, however, since you explicited that only $a$ and $\gamma$ are variables. Commented Jan 14, 2020 at 18:02
• @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!
– KGM
Commented Jan 14, 2020 at 18:05
• @dipaknarayanan if $\gamma = 1$ then $1\cdot b = b$, whereas if $0 \leq \gamma < 1$ then $\gamma b < b$. Combining the two you get $\gamma b \leq b$. Commented Jan 14, 2020 at 18:26