my the optimization problem as follows:

$f\left(x \right) = \frac{\log_2\left( 1+ x\right)}{1+\exp\left(x\right) + 0.1 \log_2\left( 1+ x\right)} $

The presented problem is the fractional programming problem so I can use e.g. the Dinkelbach method but in order to do it the numerator has to be nonnegative and concave while the denominator has to be positive and convex or if the numerator is affine, the denominator does not have to be restricted in sign.

In my case, the numerator is concave while the denominator is a nonconvex function. So, can I add the constraint which removes a convex function from the numerator and looks like this:

$f\left(x \right) = \frac{y}{1+\exp\left(x\right) + 0.1 y}$


$\log_2\left( 1+ x\right) \geq y$

After this transformation the objective function is convex and the constraint is convex, thus the optimization problem is convex. Is it correct?


The function

$$ f(x,y) = \frac{y}{1+e^x+0.1y} $$

has not compact support as can be depicted from the region (in light blue) obeying

$$ f(x,y) > 0 $$

enter image description here

  • $\begingroup$ Ok, so let's consider a problem $f\left(x,y\right) = \frac{y}{1+x+y}$ s.t. $\log_2\left(1+x\right) \geq y$ for $x > 0$ and $y > 0$. What do you think? $\endgroup$ – user3648211 May 29 '18 at 16:17
  • $\begingroup$ $\frac{y}{1+x+y}$ also don't have compact support in $R^2$ neither in the feasible region $\log_2 (x+1)\ge y$ $\endgroup$ – Cesareo May 29 '18 at 16:29
  • $\begingroup$ Ok, I found a lot of problem such as $f(x,y) = \frac{\min\left\{ \log_2 \left( 1+x\right), \log_2 \left( 1+y\right)\right\}}{1+x+y}$ which were transformed into $f(x,y) = \frac{z}{1+x+y}$ s.t. $ \log_2 \left( 1+x\right) \geq z$ and $ \log_2 \left( 1+y\right) \geq z$ so I thought that the same transformation can be applied in my case. Do you have any idea how I can solve my problem? $\endgroup$ – user3648211 May 29 '18 at 16:36
  • $\begingroup$ I am sorry. I didn't have noticed that the feasible region was defined also by $y > 0$ and $x > 0$. With those additional restrictions, into the feasible region, $f(x,y)$ is convex. $\endgroup$ – Cesareo May 29 '18 at 16:45
  • $\begingroup$ Great, It is a good news that I can apply such transformation to my optimization problem. Thanks a lot. $\endgroup$ – user3648211 May 30 '18 at 10:04

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