0
$\begingroup$

In multiobjective optimization, by using exact methods, we need to find the set of efficient solutions in the decision space or the set of non-dominated solutions in the criteria space. So, we must provide a set of several solutions for the decision maker (which can be 100, 1000 or even more) which is not efficient for the decision maker. Another approach was proposed, is to fix a new objective function which we must optimize on the set of efficient solutions. This approach provide one efficient solutions depending on the added objective function. This approach is also biaised directly by the added objective function and is not necessary giving a good solution if the added solution is not good enough or it deberatly miss some directions.

My question: Is there another approach where we can provide a subset of the efficient solutions set (we can say 3 to 5 solutions max) that take in consideration all the criteria but without providing all the efficient set.

Propositions:

  1. Using metaheuristics to get efficient solutions: This is not a good approach because we will not be sure if the provided solution is efficient.
  2. Using exact approach to get 3 to 5 efficent solutions then stop: This is a possible alternative but we will never know if is there another efficient solution which may be better for the decision maker.

What I search is an approach that provide a subset of the efficient solutions but it will take the other efficient solutions in consideration, and this subset can be seen as an representative of the efficicent solutions set.

$\endgroup$

1 Answer 1

1
$\begingroup$

Years ago two of my colleagues proposed an interactive approach. (This was for LPs, but should generalize to MIPs.) It requires reasonably fast solution time for the model. The gist was to optimize a weighted combination of objective functions, show the solution to the decision maker, and ask if they were happy with it. If the expressed a desire for a better outcome on some criterion, you asked them what the would be willing to trade (for instance, allow an extra 30 minutes of production time to save $100 in production cost), used that information to adjust the objective weights, solved again, and asked the decision maker about the new solution. The process repeated until the decision maker was satisfied (or exhausted, which I suspect would be the more frequent outcome).

$\endgroup$
6
  • $\begingroup$ Thanks for your response, this is what we call the interactive methods like STEM method. But those methods have a big inconvenient, the decision maker can accept a solution where he may could wait to improve some objectives to get better. What i search is a principle or an idea that has a mathematics background which when we represent the subsolutions set then I can ignore the others. I could say like a gravity center of all the solutions, but not one but a few solutions that can be considered like that. $\endgroup$ Commented Mar 14 at 21:15
  • $\begingroup$ Any sort of "center" of the solutions will be Pareto-inefficient. $\endgroup$
    – prubin
    Commented Mar 14 at 22:07
  • $\begingroup$ One of my colleague propose to take the ideal point as center, then search for the solutions with the smallest distance to this point, the idea guide directly to the supported efficient solutions (or Pareto solutions of rank 1), but i search maybe another distance with another point or something like that which has a signinificant meaning in mathematics. $\endgroup$ Commented Mar 14 at 22:29
  • $\begingroup$ Also, a center of gravity must be among the efficient solutions that bring to him the other efficient solutions, or may be create clusters among the efficient solutions then choose one solution from each cluster like a location problem, but i do not have a clear idea or a method that can do that. $\endgroup$ Commented Mar 14 at 23:12
  • $\begingroup$ If you optimize each objective individually and then create a vector in objective space consisting of the optimal values, that is sometimes called the "utopia point". You can then create a weighted distance ($L_1$, $L_2$ or $L_\infty$ metric) and optimize that to get an efficient solution. Changes the weights (or maybe the metric) and repeat a few times. $\endgroup$
    – prubin
    Commented Mar 15 at 3:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.