# How can we choose the right weight to solve multi-objective problem using weighted sum method?

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

• Welcome to OR SE. What do you mean by "right weight"?
– prubin
Jul 22, 2022 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 Jul 22, 2022 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 Bot
Jul 22, 2022 at 19:23

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.

• Yes, of course, the weighted sum method gives only one solution. But the problem is how to choose the weight? Jul 24, 2022 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. Jul 24, 2022 at 12:21