A Novel U-Shaped Transfer Function for Binary Particle Swarm Optimisation

Abstract

Particle swarm optimisation (PSO), one of the most elegant algorithms in the field of nature-inspired optimisation, has many variants for solving different types of problems. One of these variants is binary particle swarm optimisation (BPSO), which is suitable for solving combinatorial optimisation problems. A main component of BPSO is the transfer function that maps continuous velocity values to probability values which in turn are used to update particle positions. Transfer function has a significant impact on the performance of BPSO algorithm. This paper proposes a novel transfer function with tunable parameters that allows different U-shaped transfer functions. For evaluating the proposed transfer functions, a set of benchmark functions and 0/1 knapsack problems are employed. The results show that the U-shaped transfer functions can significantly improve the performance of BPSO. It is also shown that the BPSO algorithms equipped with U-shaped transfer functions provide superior results compared to the existing transfer functions in the literature.

Appendix

1.1 Unimodal Test Functions

$$ \begin{aligned}&f1(\varvec{x}) = \sum \limits _{i=1}^{n} x_i^2 \\&f2(\varvec{x}) = \sum _{i=1}^{n} | x_i | + \prod _{i=1}^{n} | x_i |\\&f3(\varvec{x}) = \sum _{i=1}^{n} \left( \sum _{j=1}^{i} x_j \right) ^2\\&f4(\varvec{x}) = \max _i \{ \left| x_i \right| , 1 \le i \le n \} \\&f5(\varvec{x}) = \sum _{i=1}^{n-1} \left( 100(x_{i+1}-x_i^2)^2 -(x_i - 1)^2 \right) \\&f6(\varvec{x}) = \sum _{i=1}^{n} ( x_i + 0.5)^2 \end{aligned} $$

1.2 Multi-modal Test Functions

$$\begin{aligned}&f7(\varvec{x}) = \sum _{i=1}^{n} ix_i^4 + \text {random}[0,1)\\&f8(\varvec{x}) = \sum _{i=1}^{n} \left( -x_i \sin \left( \sqrt{|x_i|}\right) \right) \\&f9(\varvec{x}) = \sum _{i=1}^{n} \left( x_i^2 - 10\cos (2 \pi x_i)+ 10 \right) \\&f10(\varvec{x}) = -20 e^{-0.2\sqrt{\frac{1}{n} \sum _{i=1}^{n}x_i^2}} -e^{\frac{1}{n} \sum _{i=1}^{n} \cos (2 \pi x_i)} + 20 + e\\&f11(\varvec{x}) = \frac{1}{4000} \sum _{i=1}^{n}x_i^2 - \prod _{i=1}^{n} \cos \left( \frac{x_i}{\sqrt{i}}\right) +1 \\&f12(\varvec{x}) = \frac{\pi }{n} \{ 10\sin (\pi y_1) + \sum _{i=1}^{n-1} (y_i-1)^2 \left[ 1+10 \sin ^2(\pi y_{i+1}) \right] \\&\qquad \qquad \quad + (y_n -1)^2 \} + \sum _{i=1}^{n} u(x_i,10,100,4)\\&y_i = 1 + \frac{x_i+1}{4} \end{aligned}$$

$$\begin{aligned} u(x_i,a,k,m) = {\left\{ \begin{array}{ll} k(x_i - a)^m &{} x_i > a \\ 0 &{} -a<x_i<a \\ k(-x_i - a)^m &{} x_i < -a \end{array}\right. } \end{aligned}$$

See Table 5 for details of the test functions.

Table 5 Details of the test functions