Hi Anjani, I think what you are asking for is to achieve both objectives by 50% relative to their isolated optimum, correct? How do you know that this is possible? It might not be possible to find a solution like that. This is also why I like the other approach more: Solve for the first objective, then add it as a constraint to a second version of the problem focusing on the second objective. E.g. you could say objective function value no. 1 should achieve a value equal to 50% of the isolated optimum.

]]>Hi Anjani, just because alpha is equal to 50% does not mean the relative objective achievement of both objectives has to be the same. And also the solution will not necessarily be the “middle” of the two isolated objectives. For exampe f1(x1,x2) = 2×1 + 2×2 is objective 1; f2(x1,x2) = 3×1-4×2 is objective 2. Constraint it x1,x2 <= 10 and x1,x2>=0. Optimal solution to f1 is x1=10,x2=10 with f1=40; optimal solution to f2 is x1=10,x2=0 with f2 = 30. If we combine the normalized objective functions f1/40 and f2/30 with alpha=50% then the optimal solution is x1=10,x2=0. In that case f1 is achieved by 50%, f2 is achieved by 100%.

]]>Hi Linnart, Sorry for the confusion. Its only Max not (Max – min). Thanks for correcting.

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