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I have a multi-objective problem with three objectives F1, F2, and F3. the problem was formulated as a weighted sum. Now I didn't know how I can choose the right weight for my problem

in the image I showed my experimentation with different weights

charafeddine
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  • Welcome to OR SE. What do you mean by "right weight"? – prubin Jul 22 '22 at 16:12
  • the weight that gives us better solution, Because i don't want to find the Pareto front, i want to find only one solution from the Pareto front – charafeddine Jul 22 '22 at 16:14
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jul 22 '22 at 19:23

2 Answers2

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I'm going to assume here that bigger is better for all criteria. Let $x_{a,c}$ be the score for alternative $a$ on criterion $c,$ where $a\in \lbrace 1,\dots,A\rbrace$ and $c\in\lbrace 1,\dots,C\rbrace.$ Any easy way to find a single Pareto-efficient point is to generate a random vector $w\in (0,1)^C$ and compute the score $s_a = \sum_{c=1}^C w_c x_{a,c}$ for each alternative $a.$ The alternative with the highest score is automatically efficient, since any alternative that dominated it would have a higher score.

prubin
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As far as I know, solving the multi-objective optimization by the weighted sum method should give one of the solutions that already exists on the Pareto non-dominance solution frontier. Let us make a simple example. Suppose, there is a multi-objective problem with only two objects. Named $Z_1$ and $Z_2$. Now, solving this by a weighted sum and again by $\epsilon$-constraints. The results are as follows:

Weighted sum method

  • $Z_1 = 2800$ and $Z_2 = 17.9$ within the arbitrary weights for each objective.

$\epsilon$-constraint method with an $\epsilon = 0.00001$

  • The first solution is: $Z_1 = 2790$ and $Z_2 = 20$.
  • The second solution is: $Z_1 = 2800$ and $Z_2 = 17.9$.

As you can see, the first method gives the second non-dominance solution. Also, I am not aware of if, there exists any contradiction to that. I hope this will be helpful.

A.Omidi
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  • Yes, of course, the weighted sum method gives only one solution. But the problem is how to choose the weight? – charafeddine Jul 24 '22 at 11:05
  • @charafeddine, please see this link. As I check for some problems with an arbitrary weight the solution (every single objective) is a same. Also, as I said, do not see any contradiction to that. – A.Omidi Jul 24 '22 at 12:21