The multi-objective optimization problem settings are as defined below:

  • Objective 1: Minimize $f_1(X_1,X_2)=C_0+C_1(1/X_1)+C_2(X_2/X_1)$

  • Objective 2: Minimize $f_2(X_1,X_2)=D_0+D_1X_1+D_2X_2+D_3(X_2/X_1)+D_4(X_2^2/X_1)$

  • Objective 3: Minimize $f_3(X_1, X_2)=E_1(X_2/X_1)$

  • Constraint 1: $ 0 \le X_1 \le A_1$

  • Constraint 2: $ 0 \le X_2 \le A_2$

  • Constraint 3: $ k_1X_1 + k_2X_2 \le A_3$

  • Constraint 4: $ k_3X_1 \le k_4X_2$

Here, $C_0, C_1,C_2,D_0,D_1,D_2,D_3,D_4,E_1,A_1,A_2, A_3, k_1,k_2,k_3,k_4$ are positive constant numbers. $X_1$ and $X_2$ are decision variables that can be real/integer.

Evolutionary algorithms like NSGA-II are considered mainstream algorithms to solve a non-linear multi-objective optimization algorithm (Pareto optimization).

  1. Other than metaheuristics like NSGA-II, PSO, and SMPSO, are there other exact methods that can be applied to solve Pareto optimization problems (i.e., identifying Pareto optimal solutions)?

  2. Which is better to solve the problem illustrated above NSGA-II or exact methods? Does the non-linearity of the problem play a significant role in the selection of solution algorithms?

  • 1
    $\begingroup$ Since you mentioned "Pareto optimization", should we assume that you are looking for Pareto optimal solutions rather than optimizing a weighted combination of objectives or applying preemptive priorities? If so, are you looking for one Pareto-efficient solution or seeking to explore the Pareto frontier? $\endgroup$
    – prubin
    Jul 20, 2023 at 17:33
  • $\begingroup$ Thank you. I am looking for solutions lying in the Pareto front. But can we derive one Pareto-efficient solution? $\endgroup$
    – vp_050
    Jul 20, 2023 at 17:39

1 Answer 1


Observe that all three objective functions are increasing functions of $X_2.$ For fixed $X_1,$ you want the smallest feasible value of $X_2.$ Constraint 4 is the only one that puts upward pressure on $X_2.$ So for fixed $X_1$ you want $X_2 = \frac{k_3}{k_4} X_1.$ Substitute that into constraint 3 to get $$X_1 \le \frac{k_4}{k_1 k_4 + k_2 k_3} A_3.$$ Combining constraints 2 and 4, we also have $X_1 \le \frac{k_4}{k_3} A_2.$

Rewrite your objective functions, substituting out $X_2.$ The first is now linear in $\frac{1}{X_1},$ the second is linear in $X_1$ and the third is a constant. The domain of $X_1$ is from 0 to $$\min(A_1, \frac{k_4}{k_3} A_2,\frac{k_4}{k_1 k_4 + k_2 k_3} A_3).$$ Since $f_1$ is strictly decreasing and $f_2$ is strictly increasing in $X_1,$ every point in the revised domain of $X_1$ is Pareto efficient.


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