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I have following constraint in my optimization problem:

$0 \leq t \leq \frac{S}{B \ R(p)}$

where $t,p$ are optimization variables and $S,B$ are positive constants. $R(p)$ is a concave function which is in the denominator, is there any way to make this constraint convex?

Thanks in advance!

Syed
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1 Answers1

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From your constraint, it's clear $R$ is positive as well. Then,we have $$\frac1{R(p)}=\max_{\lambda>0}\left(2\lambda-\lambda^2R(p)\right)$$ and since $R$ is concave, that's the maximum of a set of convex functions in $p$, i.e. it's convex (for a proof, look here).

  • Thank you for your answer. Now, when I try to implement it in the cvx it gives error of Invalid constraint: {real affine} <= {convex} Do you have any idea to overcome this problem in cvx? thanks in advance – Syed Sep 23 '17 at 03:10
  • Or maybe to convert the above mentioned constraint as a concave expression, then it can be solved by cvx – Syed Sep 23 '17 at 04:09
  • @Syed Sorry, I misunderstood your question: I showed $1/R(p)$ is a convex function, but you want a "convex constraint" satisfying the rather stringent conditions in your specific software, and couldn't be bothered to explain that. Good luck with your question. –  Sep 23 '17 at 21:29