Is it possible to convert a fractional problem (maximization) with objective function equal to the ratio of a concave function and convex function ? This question sound impossible but I have read this claim in wikipedia without any reference !


<https://en.wikipedia.org/wiki/Nonlinear_programming#Methods_for_solving_the_problem>


"If the objective function is a ratio of a concave and a convex function (in the maximization case) and the constraints are convex, then the problem can be transformed to a convex optimization problem using fractional programming techniques"



convert a special case of nonlinear fractional programming into a convex problem
Here is some insight into Jean's answer, showing what that final problem means. Suppose you transform the problem to finding a vector $(y,t) \in \mathbb{R}^{n+1}$ to solve: 
\begin{align\*}
&\mbox{Max: } & tf(y/t) \\
&\mbox{Subject to: } & tg(y/t) \leq 1 \\
& & y/t \in S \\
& & t>0 \\
& & y \in \mathbb{R}^n 
\end{align\*}
where $S$ is a convex subset of $\mathbb{R}^n$, the functions $f(\cdot)$ and $g(\cdot)$ are defined over $S$, and $f$ is concave while $g$ is convex. To show this is a convex optimization problem, it suffices to show: 


i) The constraints $y/t \in S$, $y \in \mathbb{R}^n$, $t >0$ together specify a convex set. 


ii) The function $tf(y/t)$ is concave in the joint variables $(y,t)$. 


iii) The function $tg(y/t)$ is convex in $(y,t)$. 


To this end, define $A$ as the set of all $(y,t)$ such that $t>0, y \in \mathbb{R}^n, y/t \in S$. 




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Claim 1: The set $A$ is a convex subset of $\mathbb{R}^{n+1}$. 


Proof: Let $(y\_1,t\_1)$ and $(y\_2,t\_2)$ be elements if $A$. Let $\theta \in [0,1]$ and define $\overline{\theta} = 1-\theta$. We want to show that
$$ (\theta y\_1 + \overline{\theta}y\_2, \theta t\_1 + \overline{\theta}t\_2) \in A$$
Since $t\_1>0$ and $t\_2>0$, it follows that $\theta t\_1 + \overline{\theta}t\_2>0$. It remains to check the ratio condition: 
$$ \frac{\theta y\_1 + \overline{\theta} y\_2}{\theta t\_1 + \overline{\theta}t\_2} = \left(\frac{\theta t\_1}{\theta t\_1 + \overline{\theta}t\_2}\right)(y\_1/t\_1) + \left(\frac{\overline{\theta} t\_2}{\theta t\_1 + \overline{\theta}t\_2}\right)(y\_2/t\_2) \in S$$
since $S$ is convex and $y\_1/t\_1 \in S$, $y\_2/t\_2 \in S$, and this is just a convex combination of $y\_1/t\_1$ and $y\_2/t\_2$. $\Box$.




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Claim 2: The function $tf(y/t)$ is concave over $(y,t) \in A$. 


Proof: Fix $(y\_1, t\_1)$ and $(y\_2, t\_2)$ in $A$. Fix $\theta \in [0,1]$ and define $\overline{\theta} = 1-\theta$. Then: 
\begin{align} 
\theta t\_1f(y\_1/t\_1) + \overline{\theta}t\_2f(y\_2/t\_2) &= \left(\theta t\_1 + \overline{\theta}t\_2\right)\left(\frac{\theta t\_1f(y\_1/t\_1) + \overline{\theta}t\_2f(y\_2/t\_2)}{\theta t\_1 + \overline{\theta}t\_2}\right) \\
&\leq \left(\theta t\_1 + \overline{\theta}t\_2\right)f\left( \frac{\theta t\_1 (y\_1/t\_1) + \overline{\theta}t\_2 (y\_2/t\_2)}{\theta t\_1 + \overline{\theta}t\_2}\right)\\
&= \left(\theta t\_1 + \overline{\theta}t\_2\right)f\left( \frac{\theta y\_1 + \overline{\theta}y\_2}{\theta t\_1 + \overline{\theta}t\_2} \right)
\end{align} 
where the inequality holds because the function $f()$ is concave. $\Box$




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That the function $tg(y/t)$ is convex over $(y,t) \in A$ follows by a proof similar to that of Claim 2.


