Improved particle swarm optimization based approach for bilevel programming problem-an application on supply chain model

Abstract

Bilevel programming problem (BLPP) is a NP-hard problem and very difficult to be resolved by using the classical method. This paper attempts to develop an efficient method based on improved bilevel particle swarm optimization algorithm (IBPSO) to deal with BLPP. The improved algorithm adopts dynamic self-adapting inertia weight and Cauchy distribution to ensure global search ability and faster convergence speed of IBPSO. IBPSO employs two improved PSO as a main program and a subprogram respectively. According to the interactive iterations of two improved PSO algorithms, IBPSO can solve BLPP without some assumptions of BLPP, such as the gradient information of the objective functions, the convexity of constraint regions, and so on. Twelve bilevel problems in the literatures are employed to illustrate the performance of IBPSO and some comparisons are also given. The results demonstrate that the proposed algorithm IBPSO exhibits higher accuracy than other algorithms. Then IBPSO is adopted to solve two supply chain models proposed in this paper, and some features of the proposed bilevel model are given based on the experimental data. The results support the finding that IBPSO is effective in optimizing BLPP.

Appendix

Problem 1:

$$ \begin{gathered} \hbox{max} \;f_{1} = - 2x_{1} + 11x_{2} \;{\text{where}}\;x_{2} \;{\text{solves}} \hfill \\ \hbox{max} \;f_{2} = - x_{1} - 3x_{2} \hfill \\ {\text{s}} . {\text{t}} .\;x_{1} - 2x_{2} \le 4,\;2x_{1} - x_{2} \le 24,\;3x_{1} + 4x_{2} \le 96,x_{1} + 7x_{2} \le 126 \hfill \\ \quad \; - 4x_{1} + 5x_{2} \le 65,\;x_{1} + 4x_{2} \ge 8,\;x_{1} \ge 0,\;x_{2} \ge 0 \hfill \\ \end{gathered} $$

Problem 2:

$$ \begin{gathered} \hbox{max} \;f_{1} = x_{2} \;{\text{where}}\;x_{2} \;{\text{solves}} \hfill \\ \hbox{max} \;f_{2} = - x_{2} \hfill \\ {\text{s}} . {\text{t}} .\; - x_{1} - 2x_{2} \le 10,\;x_{1} - 2x_{2} \le 6,\;2x_{1} - x_{2} \le 21 \hfill \\ \quad \;x_{1} + 2x_{2} \le 38,\; - x_{1} + 2x_{2} \le 18,\;x_{1} \ge 0,\;x_{2} \ge 0 \hfill \\ \end{gathered} $$

Problem 3:

$$ \begin{gathered} \hbox{max} \;f_{1} = x_{1} + 3x_{2} \;{\text{where}}\;x_{2} \;{\text{solves}} \hfill \\ \hbox{max} \;f_{2} = - x_{2} \hfill \\ {\text{s}} . {\text{t}} .\; - x_{1} + x_{2} \le 3,\;x_{1} + 2x_{2} \le 12 \hfill \\ \quad \;4x_{1} - x_{2} \le 12,\;x_{1} \ge 0,\;x_{2} \ge 0 \hfill \\ \end{gathered} $$

Problem 4:

$$ \begin{gathered} \hbox{max} \;f_{1} = 8x_{1} + 4x_{2} - 4y_{1} + 40y_{2} + 4y_{3} \;{\text{where}}\;y\;{\text{solves}} \hfill \\ \hbox{max} \;f_{2} = - x_{1} - 2x_{2} - y_{1} - y_{2} - 2y_{3} \hfill \\ {\text{s}} . {\text{t}} .\;y_{1} - y_{2} - y_{3} \ge - 1 \hfill \\ \quad \, - 2x_{1} + y_{1} - 2y_{2} + 0.5y_{3} \ge - 1 \hfill \\ \quad \, - 2x_{2} - 2y_{1} + y_{2} + 0.5y_{3} \ge - 1 \hfill \\ \quad \;x_{1} ,x_{2} ,y_{1} ,y_{2} ,y_{3} \ge 0 \hfill \\ \end{gathered} $$

Problem 5:

$$ \begin{gathered} \hbox{min} \;f_{1} = - x_{1}^{2} - 3x_{2} - 4y_{1} + y_{2}^{2} \hfill \\ {\text{s}} . {\text{t}} .\;x_{1}^{2} + 2x_{2} \le 4,\;x_{1} \ge 0,\;x_{2} \ge 0 \hfill \\ {\text{where}}\;y\;{\text{solves}} \hfill \\ \hbox{min} \;f_{2} = 2x_{1}^{2} + y_{1}^{2} - 5y_{2} \hfill \\ {\text{s}} . {\text{t}} .\;x_{1}^{2} - 2x_{1} + x_{2}^{2} - 2y_{1} + y_{2} \ge - 3 \hfill \\ \quad \;x_{2} + 3y_{1} - 4y_{2} \ge - 4,\;y_{1} \ge 0,\;y_{2} \ge 0 \hfill \\ \end{gathered} $$

Problem 6:

$$ \begin{gathered} \hbox{min} \;f_{1} = x^{2} + \left( {y - 10} \right)^{2} \hfill \\ {\text{s}} . {\text{t}} .\;x + 2y - 6 \le 0,\; - x \le 0 \hfill \\ {\text{where}}\;y\;{\text{solves}} \hfill \\ \hbox{min} \;f_{2} = x^{3} - 2y^{3} + x - 2y - x^{2} \hfill \\ {\text{s}} . {\text{t}} .\; - x + 2y - 3 \le 0,\; - y \le 0 \hfill \\ \end{gathered} $$

Problem 7:

$$ \begin{gathered} \hbox{min} \;f_{1} = \left( {x - 5} \right)^{4} + \left( {2y + 1} \right)^{4} \hfill \\ {\text{s}} . {\text{t}} .\;x + y - 4 \le 0,\; - x \le 0 \hfill \\ {\text{where}}\;y\;{\text{solves}} \hfill \\ \hbox{min} \;f_{2} = e^{ - x + y} + x^{2} + 2xy + y^{2} + 2x + 6y \hfill \\ {\text{s}} . {\text{t}} .\; - x + y - 2 \le 0,\; - y \le 0 \hfill \\ \end{gathered} $$

Problem 8:

$$ \begin{gathered} \hbox{min} \;f_{1} = \left( {x_{1} - y_{2} } \right)^{4} + \left( {y_{1} - 1} \right)^{2} + \;\left( {y_{1} - y_{2} } \right)^{2} \hfill \\ {\text{s}} . {\text{t}} .\; - x_{1} \le 0 \hfill \\ {\text{where}}\;y\;{\text{solves}} \hfill \\ \hbox{min} \;f_{2} = 2x_{1} + e^{{y_{1} }} + y_{1}^{2} + 4y_{1} + 2y_{2}^{2} - 6y_{2} \hfill \\ {\text{s}} . {\text{t}} .\;6x_{1} + 2y_{1}^{2} + e^{{y_{2} }} - 15 \le 0,\; - y_{1} \le 0,\;y_{1} - 4 \le 0 \hfill \\ \quad \;5x_{1} + y_{1}^{4} - y_{2} - 25 \le 0,\; - y_{2} \le 0,\;y_{2} - 2 \le 0 \hfill \\ \end{gathered} $$

Problem 9–12:

$$ \begin{gathered} \mathop {\hbox{min} }\limits_{x} \;f_{1} = r\left( {x_{1}^{2} + x_{2}^{2} } \right) - 3y_{1} - 4y_{2} + 0.5\left( {y_{1}^{2} + y_{2}^{2} } \right) \hfill \\ \mathop {\hbox{min} }\limits_{y} f_{2} = 0.5\left[ {y_{1} \quad y_{2} } \right]H\left[ {y_{1} \quad y_{2} } \right]^{T} - b\left( x \right)^{T} \left[ {y_{1} \quad y_{2} } \right]^{T} \hfill \\ \quad \;{\text{s}} . {\text{t}} .\; - 0.333y_{1} + y_{2} - 2 \le 0,\;y_{1} - 0.333y_{2} - 2 \le 0,\;y_{1} \ge 0,\;y_{2} \ge 0 \hfill \\ \end{gathered} $$

Problem 9:

$$ r = 0.1,\;H = \left[ {\begin{array}{*{20}c} 1 & { - 2} \\ { - 2} & 5 \\ \end{array} } \right],\;b\left( x \right) = \left[ \begin{gathered} x_{1} \hfill \\ x_{2} \hfill \\ \end{gathered} \right] $$

Problem 10:

$$ r = 0.1,\;H\;{\text{and}}\;b\left( x \right)\;{\text{are}}\;{\text{the}}\;{\text{same}}\;{\text{as}}\;{\text{in}}\;{\text{Problem}}\; 9 $$

Problem 11:

$$ r = 0.1,\;H = \left[ \begin{gathered} 1\quad 3 \hfill \\ 3\quad 10 \hfill \\ \end{gathered} \right],\;b\left( x \right) = \left[ \begin{gathered} - 1\quad 2 \hfill \\ 3\quad - 3 \hfill \\ \end{gathered} \right]\,\left[ \begin{gathered} x_{1} \hfill \\ x_{2} \hfill \\ \end{gathered} \right] $$

Problem 12:

$$ r = 0.1,\;H\;{\text{is}}\;{\text{the}}\;{\text{same}}\;{\text{as}}\;{\text{in}}\;{\text{Problem}}\; 1 1,\;b\left( x \right) = \left[ \begin{gathered} x_{1} \hfill \\ x_{2} \hfill \\ \end{gathered} \right] $$