I have a combinatorial optimization problem where there are three objectives F1, F2, and F3 to be minimized. The problem was formulated as a weighted sum where F=alphaF1+betaF2+gamma*F3. My question is how we choose the weights alpha, beta, and gamma. I tested someone randomly chosen 1. Do you think the weights I choose are logical? from the result in the picture we can see that 0.3 0.35 0.35 is the best combination with that can i use only this combination for all my rest experimentation?
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2$\begingroup$ The F3 values are a magnitude smaller than F1 and F2. Therefore, they'll contribute less to the weighted objective. Is that what you want? More generally, the weights should reflect the relative importance of the objectives. Having a look at the concept of "Pareto frontier" would be relevant here. $\endgroup$– Solver MaxCommented Apr 2, 2022 at 1:41
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1$\begingroup$ How do you solve your problem? I ask because the solution corresponding to $(0.15,0.7,0.15)$ is dominated by the solution corresponding to weights $(0.3,0.35,0.35)$. This cannot happen if you solve the problem to optimality $\endgroup$– SuneCommented Apr 2, 2022 at 9:29
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$\begingroup$ @Sune Good catch! $\endgroup$– prubin ♦Commented Apr 2, 2022 at 15:59
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2 Answers
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Do you think the weights I choose are logical?
No, I don't think they are logical, but that is because you have chosen them randomly (according to your description). Hence, there is not really any logic behind the choice of weights. What is good, however, is that you have strictly positive weights in all you scenarios. This will ensure that the resulting solutions are Pareto optimal if you solve the weighted sum problems to optimality. From your results shown in your attached picture we can see that you do not solve the problems to optimality, as there are dominated outcomes among your results.
From the result in the picture we can see that 0.3 0.35 0.35 is the best combination
Generally, you cannot say that one solution (set of weights) is better than another in a multi-objective optimization problem, as it depends on the preferences of the decision maker. You could argue that it is the entire idea behind multi-objective optimization: Generate (a subset of) the set of non-dominated outcomes and then let the decision maker decide which of the Pareto optimal solutions corresponding to non-dominated outcomes they find more appealing.
with that can i use only this combination for all my rest experimentation?
If you settle for one set of weights and stick to that set, I would argue that you do not really solve a multi-objective optimization problem: You have mixed apples and oranges in one bowl and you are left with the problem of interpreting the objective function. On the other hand, if you solve your problem for a small set of weight combinations, say $\{w^1,...,w^K\}$ with $w^k_i>0$, you will get at most $K$ different Pareto optimal solutions you can choose from based on your or your decision maker's preferences. And then I would say that you are dealing with multi-objective optimization.
To get familiar with the theory of multicriteria/multi-objective optimization I would refer to:
- Multicriteria Optimization by Matthias Ehrgott. You may look into chapter 3 for information on the weighted sum method.
- For more theory on how to chose the weights such that you can generate all the efficient extreme points of the set of non-dominated solutions you may look into "weight set decomposition". For the three-objective case you can start with An inner approximation method to compute the weight set decomposition of a triobjective mixed-integer problem by P. Halffmann, T. Dietz, A. Przybylski, and S. Ruzika.
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In the weighted sum approach, we scale our set of goals into a single goal by multiplying each of our objectives by a user-supplied weight. This method is one of the most widely used approaches.
The weight given to an objective is normally assigned to be proportional to the objective’s relative importance in our problem. However, we also need to make sure that we account for the magnitude of our objective functions. We resolve this issue by scaling all of our objective functions to the same magnitude which is called normalization. For more details, please see:
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1$\begingroup$ Can you correct your first link? It is not working. $\endgroup$ Commented Feb 5, 2023 at 5:58
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