Multi- Objective optimization: how to prove :An optimal solution of the max-ordering problem is weakly efficient but not necessarily efficient.

Asked Apr 18 '19 at 23:10

Active Apr 19 '19 at 05:01

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For multi-objective optimization problem. A method to find weakly efficient is using max-ordering, which is $min_{x\in X}\ max_{i=1,...n}f_i(x)$.

How to prove that:An optimal solution $x^o$ of the max-ordering problem $min_{x\in X}\ max_{i=1,...n}f_i(x)$ is weakly efficient.

I have try to prove by contradiction. Firstly suppose that there is a point $x^c$ that $f_j(x^c) < f_j(x^o)$ . When the $f_j(x^c) = max_{i=1,...n}f_i(x)$. It is contradict with the suppose.

However, when the $f_j(x^c) \neq max_{i=1,...n}f_i(x)$ . There is seems not contradiction.

Where is the problem of my proof?

edited Apr 19 '19 at 05:01

asked Apr 18 '19 at 23:10

Alan

Please provide some more information about what you have tried, where you got stuck, and what your specific questions are. This feels a bit like you are asking us to solve a homework problem for you. – LarrySnyder610 Apr 19 '19 at 00:25
Thanks for your advice very much. I have updated my question. Can I have your further advice? Thank you. – Alan Apr 19 '19 at 00:56
I would try coming up with a counterexample, i.e., a specific instance of the optimization problem, with actual numbers, and show that $x^o$ is weakly efficient but not efficient. – LarrySnyder610 Apr 19 '19 at 01:20

Multi- Objective optimization: how to prove :An optimal solution of the max-ordering problem is weakly efficient but not necessarily efficient.

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