1
$\begingroup$

I have the following optimization problem: \begin{align}\max_x&\quad\log_2(1+|a+bx|^2+cx^2)\\\text{s.t.}&\quad0\le x\le1\\&\quad(1-x^2)\ge\text{constant}\end{align} where $a$ and $b$ are two complex numbers, and $c$ is a real positive number.

Does anyone have any idea regarding how this objective function can be converted into a convex one?

Is it true to say that the optimal value of $x$ is equal to $\sqrt{1-\text{constant}}$?

Improve this question
$\endgroup$
1
  • $\begingroup$ Just pointing out that since you are maximising, maybe you want to get a concave function instead? $\endgroup$
    – Nikos Kazazakis
    Apr 14 at 10:43
6
$\begingroup$

You are maximizing a convex quadratic (the monotonic log is irrelevant) so the maximum is attained at the border, i.e. either $0$ or $\min(1,\sqrt{1-\text{constant}})$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Thank you for your prompt reply, I have another question! what if we have the following constraint $log_2(1+|a+b*x|^2+c*x^2)\geq constant2$ as well? Do we get the previous solution still? $\endgroup$
    – Shayan zargari
    Apr 15 at 12:47
  • $\begingroup$ If you pick the maximum, and it does not satisfy that constraint, then the problem is obviously not solvable. $\endgroup$
    – Johan Löfberg
    Apr 15 at 12:50
  • $\begingroup$ So, I cannot derive a close-form solution! Shall I resort to SCA approach? $\endgroup$
    – Shayan zargari
    Apr 15 at 13:02
  • 1
    $\begingroup$ I have already given you the closed-form solution, and the added constraint does not change anything as it either will be a redundant constraint (the maximal solution satisfies it) or render the problem infeasible (the maximal solution does not satisfy it) $\endgroup$
    – Johan Löfberg
    Apr 15 at 13:05
  • $\begingroup$ I get it, Thank you so much! $\endgroup$
    – Shayan zargari
    Apr 15 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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