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
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$\begingroup$ Welcome to OR SE. What do you mean by "right weight"? $\endgroup$– prubin ♦Commented Jul 22, 2022 at 16:12
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$\begingroup$ 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 $\endgroup$– charafeddineCommented Jul 22, 2022 at 16:14
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$\begingroup$ 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. $\endgroup$– Community BotCommented Jul 22, 2022 at 19:23
2 Answers
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.
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.
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$\begingroup$ Yes, of course, the weighted sum method gives only one solution. But the problem is how to choose the weight? $\endgroup$ Commented Jul 24, 2022 at 11:05
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