Is there a solution to find the global optimum of a convex function , subject to non convex constraints?

For example:

$\min x_1^2 + x_2^2$

subject to

$x_2 + x_1^2 \geq 2$

$x_2 - x_1^2 \leq 3$


It's only an example; I am looking for a general method to find the optimal point in any convex function with non-convex constraints

  • $\begingroup$ What inequality constraints define the feasible region? $\endgroup$ – Rodrigo de Azevedo Feb 20 '17 at 22:58

The two constraints here are actually both convex, which you can see by drawing out the shape and observing that circles (and lunes) are convex. Alternatively, you know that $x_1^2 + x_2^2$ is a convex function, so $(x_1 - 1)^2 + (x_2 - 1)^2$ is a convex function, and hence the inequality $(x_1 - 1)^2 + (x_2 - 1)^2 \leq 3$ is a convex region.

Also by drawing the regions out, you can see that the origin is included in the feasible set, and $(0, 0)$ clearly minimises $x_1^2 + x_2^2$, so the minimum is $0$.

  • $\begingroup$ I think i have chosen an incorrect example for my question, in general have can we solve a convext function subject to non-convex constraints? $\endgroup$ – Hojjatollah Bakhtiyari Kiya Feb 20 '17 at 20:23
  • $\begingroup$ @B.Mehta I disagree, the feasible region is not convex: If you draw a line segment between the bottom left "corner" and bottom right "corner", it passes through points that are not in the feasible set, therefore it is not convex. $\endgroup$ – Kuifje Feb 20 '17 at 23:20
  • $\begingroup$ @Kuifje The feasible region was originally the intersection of two circles, but has since been edited to a non-convex set, I agree. $\endgroup$ – B. Mehta Feb 20 '17 at 23:33
  • $\begingroup$ @B.Mehta: all righty then ! $\endgroup$ – Kuifje Feb 20 '17 at 23:49
  • $\begingroup$ @B.Mehta I changed the constraints to show that, I am not looking for the current problem's solution. I Only want to know if there exist a solution for non-convex constraints . $\endgroup$ – Hojjatollah Bakhtiyari Kiya Feb 21 '17 at 3:44

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