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I have an optimization problem of the form:

\begin{align} \begin{cases} x_2 \rightarrow \min, \\ \text{subject to:} \\ f_1(x) \leq 0, \\ f_2(x) \leq 0, \end{cases} \end{align}

with $x= (x_1,x_2)^T$ as the optimization variable.

Here, $f_1(\cdot)$ is a convex function. But $f_2(\cdot)$ is a concave function.

Explicitly, $f_2(x)$ is of the form: \begin{align} f_2(x) = 2(D-x_1-2R)+\beta \frac{D^2-x_1^2+2DR-2Rx_1}{4R}-x_2, \end{align} with $D,R,\beta$ constant.

How to convert $f_2(x)$ to a convex function? Is there any method that do such a modification?

Thanks in advance.

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  • $\begingroup$ Hey Tina, I edited your question, maybe you could make sure that the form of $f_2$ is still correct. $\endgroup$ – k1next Jul 8 '15 at 6:00
  • $\begingroup$ yes.It is still correct.Thanks.. $\endgroup$ – Tina Jul 8 '15 at 6:17
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No, you can't. If $f_2$ was convex, any local optimum will be a global optimum. This is not true if $f_2$ is concave: there may be many local optima. So these are two quite different types of problem, and in general problems with non-convex constraints are much harder. There is no way to "transform" your way out of it.

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