Past present future: a new human-based algorithm for stochastic optimization

Abstract

Past present future (PPF) is a new stochastic optimization algorithm inspired by the phenomena of the way an individual learns from a successful person in society. PPF is based on the concept of “future improvement of a person’s life depending on his/her past experience and present work.” The influence of successful persons also affects the improvement of the future life of an individual. This work develops a mathematical model for PPF following the above facts. In this new algorithm, the population is divided into subpopulations and a switching mechanism is followed among the subpopulations to track the change in optimal positions of an individual thereby accelerating the convergence rate. In addition, this switching mechanism also prevents pre-mature convergence. PPF was found to possess low computational complexity with fast convergence characteristics. The proposed PPF is compared with 41 up-to-date meta-heuristic algorithms taking an extensive set of benchmark functions to verify the efficiency. In addition, five classical engineering design problems are simulated to estimate the efficacy of the PPF algorithm in optimizing engineering problems. The results confirm the superior performance of the proposed algorithm to get the optimal solution with less iteration and have shown the best competitive performance compared to all other algorithms.

Appendices

Appendix A

See Tables 22, 23 and 24.

Table 22 Unimodal benchmark functions

Table 23 Multimodal benchmark functions

Table 24 Fixed-dimensional multimodal benchmark functions

Appendix B

2.1 The welded beam design problem

$$ {\text{Min}}\;f\left( {x_{1} ,x_{2} ,x_{3} ,x_{4} } \right) = 1.10471x_{1}^{2} x_{2} + 0.04811x_{3} x_{4} \left( {14.0 + x_{2} } \right) $$

Subject to:

$$ \begin{aligned} & g_{1} \left( X \right) = x_{1} - x_{4} \le 0 \\ & g_{2} \left( X \right) = \delta - 0.25 \le 0 \\ & g_{3} \left( X \right) = \tau - 13600 \le 0 \\ & g_{4} \left( X \right) = \rho - 30000 \le 0 \\ & g_{5} \left( X \right) = 0.10471x_{1}^{2} + 0.04811x_{3} x_{4} \left( {14.0 + x_{2} } \right) - 5 \le 0 \\ & g_{7} \left( X \right) = 6000 - F \le 0 \\ \end{aligned} $$

where the variables satisfy $0.1 \le x_{1}$, $x_{4} \le 2.0$ and $0.1 \le x_{2}$, $x_{3} \le 10$.

$$ \begin{aligned} & \rho = 50,400/x_{3}^{2} x_{4} \\ & Q = 6000(14 + x_{2} /2) \\ & D = \frac{1}{2}\sqrt {x_{2}^{2} + \left( {x_{1} + x_{3} } \right)^{2} } \beta = \frac{QD}{J} \\ & J = \sqrt 2 x_{1} x_{2} (x_{2}^{2} /6 + \left( {x_{1} + x_{3} } \right)^{2} /2) \\ & \delta = 65,856/3000x_{3}^{3} x_{4} \alpha = 6000/\left( {\sqrt 2 x_{1} x_{2} } \right) \\ & \tau = \sqrt {\alpha^{2} + \frac{{\alpha \beta x_{2} }}{D} + \beta^{2} } \\ & F = 0.61423 \times 10^{6} \frac{{x_{4}^{3} x_{3} }}{6}\left( {1 - \frac{{x_{3} \sqrt {30/48} }}{28}} \right) \\ \end{aligned} $$

2.2 Tension/compression spring design problem

${\text{Min}}\;f\left( {x_{1} ,x_{2} ,x_{3} } \right) = \left( {x_{3} + 2} \right)x_{1}^{2} x_{2}$.

Subject to:

$$ \begin{aligned} & g_{1} \left( X \right) = 1 - \frac{{x_{2}^{2} x_{3} }}{{71785x_{1}^{4} }} \le 0 \\ & g_{2} \left( X \right) = \frac{{x_{2} \left( {4x_{2} - x_{1} } \right)}}{{12566x_{1}^{3} \left( {x_{2} - x_{1} } \right)}} + \frac{1}{{5108x_{1}^{2} }} - 1 \le 0 \\ & g_{3} \left( X \right) = 1 - \frac{{140.45x_{1} }}{{x_{2}^{2} x_{3} }} \le 0 \\ & g_{4} \left( X \right) = \frac{{2\left( {x_{2} + x_{1} } \right)}}{3} - 1 \le 0 \\ \end{aligned} $$

where the variables satisfy $0.05 \le x_{1} \le 2$, $0.25 \le x_{2} \le 1.3$ and $2 \le x_{3} \le 15.$

2.3 Gear train design problem

$$ \begin{aligned} & {\text{Min}}\;f\left( {x_{1} ,x_{2} ,x_{3} ,x_{4} } \right) = \left( {\frac{1}{6.931} - \frac{{x_{3} x_{2} }}{{x_{1} x_{4} }}} \right)^{2} \\ & {\text{Variable}}\;{\text{range}}\;12 \le x_{1} ,x_{2} ,x_{3} ,x_{4} \le 60 \\ \end{aligned} $$

2.4 The three-bar truss design problem

${\text{Min}}\;f\left( {x_{1} ,x_{2} } \right) = \left( {2\sqrt 2 x_{1} + x_{2} } \right)*l$

Subject to:

$$ \begin{aligned} & g_{1} \left( X \right) = \frac{{\sqrt 2 x_{1} + x_{2} }}{{\sqrt 2 x_{1}^{2} + 2x_{1} x_{2} }}P - \sigma \le 0 \\ & g_{2} \left( X \right) = \frac{{x_{2} }}{{\sqrt 2 x_{1}^{2} + 2x_{1} x_{2} }}P - \sigma \le 0 \\ & g_{3} \left( X \right) = \frac{1}{{\sqrt 2 x_{1}^{2} + 2x_{1} x_{2} }}P - \sigma \le 0 \\ \end{aligned} $$

Variable range $0 \le x_{1} ,x_{2} \le 1$

Where l = 100 cm. P = 2 KN/cm² $\sigma = 2\;{\text{KN/cm}}^{2}$.

2.5 Cantilever beam design problem

${\text{Min}}\;f\left( {x_{1} ,x_{2} ,x_{3} ,x_{4} ,x_{5} } \right) = 0.6224\left( {x_{1} + x_{2} + x_{3} + x_{4} + x_{5} } \right).$

Subject to:

$$ g_{1} \left( X \right) = \frac{61}{{x_{1}^{3} }} + \frac{27}{{x_{2}^{3} }} + \frac{19}{{x_{3}^{3} }} + \frac{7}{{x_{4}^{3} }} + \frac{1}{{x_{5}^{3} }} - 1 \le 0 $$

Variable range $0.01 \le x_{1} ,x_{2} ,x_{3} ,x_{4} ,x_{5} \le 100$.