I have two objective functions say f1 where I have to minimize (X+Y) and another function f2 where I have to maximize (A-B). The two functions are conflicting.

I need to convert them into a minimization problem involving one single objective function using the weighed sum approach where the final form would become f3 = Min {uf1 (+ or-) vf2} where u+v=1.

I am not sure what would be the sign before vf2 ? I wonder if it should be f3 = Min {uf1 + vf2} or f3 = Min {uf1 - v*f2}.

Thanks in advance.

  • $\begingroup$ Why not simply multiply the maximization problem by -1 (making it a minimization problem), then add? $\endgroup$ – NietzscheanAI Jun 2 '17 at 16:05
  • $\begingroup$ I was confused by this definition : Max f(x) <=> -Min -f(x), which sound weird at some point and confused me. So f3 = Min {uf1 + vf2} is that what you meant @NietzscheanAI? Thanks :) $\endgroup$ – user2567806 Jun 2 '17 at 16:09
  • $\begingroup$ This depends entirely on what you're trying to achieve. I don't think this question can be answered in any generality. $\endgroup$ – David Richerby Jun 3 '17 at 12:38

To turn a maximisation problem into a minimisation one (or vice versa), simply multiply the value by -1.

Hence you want w1 * f1 - w2 * f2, for some appropriate choice of weights w1, w2 (which could indeed both be 1).

Unless there's some other requirement you haven't mentioned, there's no specific need for w1 and w2 to add up to 1: it depends on the relative importance of f1 and f2, which is problem-specific.

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  • $\begingroup$ thanks for your answer. By the way, for the sum of w1 and w2 to be 1 is the weighted sum approach that states that. Regards. $\endgroup$ – user2567806 Jun 5 '17 at 17:50

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