Does KKT works for non-convex problems as well?

Question

I want to make sure that the following claim is correct. Please let me know what you think.

"Let us assume that we have a constrained non-convex and nonlinear minimization problem. The objective function and all the constraints are differentiable. First of all, I can show that LICQ holds in all the points of the space. Then, I apply the first order necessary conditions (KKT) and find all the points satisfy these conditions. My claim is that it is enough to check the value of objective function at these points (that satisfy the KKT conditions) and pick the one which lead to the least objective value. That point (or maybe points) would be the global minimum of the problem."

Edit: Let's assume that the problem has a minimun.

Is anything wrong with above claim? Does non-convexity make any problem for using KKT conditions? BTW, I use KKT conditions from this page.

I think you should read the last paragraph on page 243 of COnvex Optimization book (by Boyd and Vandenberghe). I think from that paragraph one can get the conclusion which you wrote in your post. — Frank Moses, Mar 20 '21 at 10:37

Eff · Accepted Answer · 2017-08-10T19:32:39.410

7

You are correct if there exists an optimum (it is sufficient if e.g. the feasible set is compact). The first order conditions are necessary conditions, hence a global optimum must satisfy them. If you have finitely many points that satisfy the necessary conditions it is enough to check all of them.

However, for convex (and some other) problems, the conditions are also sufficient meaning that any such point satisfying the conditions much necessarily be global optimum.

edited Aug 10 '17 at 19:32

answered Aug 10 '17 at 19:22

Eff

12,989

I think one of the KKT points would guarantee optimality only when the strong duality holds (irrespective of the fact whether it's convex or non-convex). This has clarified in Boyd and Vandenberghe as well. – Akshay Bansal Sep 26 '23 at 00:43

John D · Answer 2 · 2017-08-10T19:34:09.873

3

This claim it's not true. KKT conditions are only necessary for optimality. Example: consider the problem

$min\; f(x)=x^3,$ s.t $\;x\leq 1.$ This problem satisfies LICQ at every point. Furthermore, the problem is unbounded, so no KKT point(x=0 is at least one of them) is a minimum of the function.

$\textbf{EDIT:}$ Even if the function is bounded from below, the statement it is not true. Example:

$min\; \frac{1}{x^2+1},$ s.t $x\leq 0.$

On the other hand, KKT conditions are sufficient for optimality when the objective function and the inequality constraints are convex, and the equality constraints are affine.

edited Aug 10 '17 at 19:34

answered Aug 10 '17 at 19:25

John D

1,862

I know that my objective function is not unbounded. So I guess that case would not happen. Am I right? – Shahram Shahsavari Aug 10 '17 at 19:29
No, even then it may happen that the minimum is not attainable. It would be sufficient that the feasible set is compact. – John D Aug 10 '17 at 19:31
If there is a minimum, should not it satisfy the KKT conditions? – Shahram Shahsavari Aug 10 '17 at 19:33
check the edit. In that case, the function is not unbounded but there is no minimum – John D Aug 10 '17 at 19:35
In that case there is even no KKT point, but you get the idea – John D Aug 10 '17 at 19:36
1

@Magnusseen I agree with your answer. But if it is given that an optimum exists, then it would be enough as my answer adresses, right? – Eff Aug 10 '17 at 19:37
Yes. That is correct – John D Aug 10 '17 at 19:37
Thanks. Let us say that there exist a minimum. Now, is the claim correct? – Shahram Shahsavari Aug 10 '17 at 19:38
@ShahramShahsavari Yes, then the claim is correct as my answer adresses (and as Magnusseen replied to my comment). – Eff Aug 10 '17 at 21:21

Does KKT works for non-convex problems as well?

2 Answers2