An enhanced time evolutionary optimization for solving engineering design problems

Abstract

Time evolutionary optimization (TEO) is a novel population-based meta-heuristic optimization algorithm, inspired by natural selection and evolution of creatures over time. Time and the environment are two main factors of evolution at TEO. In this paper, enhanced time evolutionary optimization (ETEO) is presented. ETEO is the new version of TEO which modifies time evolutionary factor and applied population clustering. Population clustering amplified environmental factor to increase the efficiency of ETEO. For this purpose, a memory is used to save some best designs and ETEO can escape from local optimal points. The algorithm was validated by solving several constraint benchmarks and engineering design problems. The comparison results between the proposed algorithm and other metaheuristic methods contain TEO, indicate the ETEO is competitive with them, and in some cases superior to, other available heuristic methods in terms of the efficiency, faster convergence rate, robustness of finding final solution and requires a smaller number of function evaluations for solving constrained engineering problems.

Appendices

Appendix 1

1.1 Constrained problem 1

$$f\left( x \right) = - \left( {\sqrt n } \right)^{n} \cdot \mathop \prod \limits_{i = 1}^{n} x_{i} .$$

Subject to:

$$h\left( x \right) = \mathop \sum \limits_{i = 1}^{n} x_{i}^{2} - 1,$$

$$0 \le x_{i} \le 1\quad i = 1,2, \ldots ,n.$$

1.2 Constrained problem 2

$$\hbox{max} f\left( x \right) = \frac{{sin^{3} \left( {2\pi x_{1} } \right) { \sin }(2\pi x_{2} )}}{{x_{1}^{3} \left( {x_{1} + x_{2} } \right)}}.$$

Subject to:

$$g_{1} \left( x \right) = x_{1}^{2} - x_{2} + 1 \le 0,$$

$$g_{2} \left( x \right) = 1 - x_{1} + \left( {x_{2} - 4} \right)^{2} \le 0,$$

$$0 \le x_{i} \le 1 i = 1,2.$$

1.3 Constrained problem 3

$$f\left( x \right) = 5.3578547x_{3}^{2} + 0.8356891x_{1} x_{5} + 37.293239x_{1} - 40792.141.$$

Subject to:

$$\begin{aligned} & g_{1} \left( x \right) = 85.334407 + 0.0056858x_{2} x_{5} + 0.0006262x_{1} x_{4} - 0.0022053x_{3} x_{5} - 92 \le 0 \\ & g_{2} \left( x \right) = - 85.334407 - 0.0056858x_{2} x_{5} - 0.0006262x_{1} x_{4} - 0.0022053x_{3} x_{5} \le 0 \\ & g_{3} \left( x \right) = 80.51249 + 0.0071317x_{2} x_{5} + 0.0029955x_{1} x_{2} + 0.0021813x_{3}^{2} - 110 \le 0 \\ & g_{4} \left( x \right) = - 80.51249 - 0.0071317x_{2} x_{5} - 0.0029955x_{1} x_{2} - 0.0021813x_{3}^{2} + 90 \le 0 \\ & g_{5} \left( x \right) = 9.300961 + 0.0047026x_{3} x_{5} + 0.0012547x_{1} x_{3} + 0.0019085x_{3} x_{4} - 25 \le 0 \\ & g_{6} \left( x \right) = - 9.300961 - 0.0047026x_{3} x_{5} - 0.0012547x_{1} x_{3} - 0.0019085x_{3} x_{4} + 20 \le 0 \\ & 33 \le x_{2} \le 45 \\ & 27 \le x_{i} \le 45 \quad i = 3,4,5. \\ \end{aligned}$$

1.4 Constrained problem 4

$$f\left( x \right) = \left( {x_{1} - 10} \right)^{2} + 5\left( {x_{2} - 12} \right)^{2} + x_{3}^{4} + 3\left( {x_{4} - 11} \right)^{2} + 10x_{5}^{6} + 7x_{6}^{2} + x_{7}^{4} - 4x_{6} x_{7} - 10x_{6} - 8x_{7} ,$$

Subject to:

$$\begin{aligned} & g_{1\left( x \right)} = 127 - 2x_{1}^{2} - 3x_{2}^{4} - x_{3} - 4x_{4}^{2} - 5x_{5} \ge 0 \\ & g_{2\left( x \right)} = 282 - 7x_{1} - 3x_{2} - 10x_{3}^{2} - x_{4} + x_{5} \ge 0 \\ & g_{3\left( x \right)} = 196 - 23x_{1} - x_{2}^{2} - 6x_{6}^{2} + 8x_{7} \ge 0 \\ & g_{4\left( x \right)} = - 4x_{1}^{2} - x_{2}^{2} + 3x_{{1x_{2} }} - 2x_{3}^{2} - 5x_{6} + 11x_{7} \ge 0 \\ & - 10 \le x_{i} \le 10\quad i = 1,2, \ldots ,7. \\ \end{aligned}$$

Appendix 2

2.1 Three-bar truss design problem

$$f\left( x \right) = \left( {2\sqrt {\left( 2 \right)x_{1} } + x_{2} } \right) \times l.$$

Subject to

$$\begin{aligned} & g_{1} \left( x \right) = \frac{{\sqrt 2 x_{1} + x_{2} }}{{\sqrt 2 x_{1}^{2} + 2x_{1} x_{2} }}P - \sigma \le 0 \\ & g_{2} \left( x \right) = \frac{{x_{2} }}{{\sqrt 2 x_{1}^{2} + 2x_{1} x_{2} }}P - \sigma \le 0 \\ & g_{3} \left( x \right) = \frac{1}{{\sqrt 2 x_{2} + x_{1} }}P - \sigma \le 0 \\ & 0 \le x_{i} \le \quad i = 1,2 \\ & l = 100\;{\text{cm,}}\,P = 2\;\frac{\text{kN}}{{{\text{cm}}^{2} }}, \sigma = 2\;\frac{\text{kN}}{{{\text{cm}}^{2} }} \\ \end{aligned}$$

2.2 Pressure vessel design problem

$$f\left( x \right) = 0.6224x_{1} x_{3} x_{4} + 1.7781x_{2} x_{3}^{2} + 3.1661x_{1}^{2} x_{4} + 19.84 x_{1}^{2} x_{3} .$$

Subject to:

$$\begin{aligned} & g_{1} \left( x \right) = - x_{1} + 0.0193x_{3} \le 0, \\ & g_{2} \left( x \right) = - x_{2} + 0.00954x_{3} \le 0, \\ & g_{3} \left( x \right) = - \pi x_{3}^{2} x_{4} - \frac{{4\pi x_{3}^{3} }}{3} + 1296,000 \le 0, \\ & g_{4} \left( x \right) = x_{4} - 240 \le 0, \\ & 0 \le x_{i} \le 100\quad i = 1,2, \\ & 10 \le x_{i} \le 100\quad i = 3,4. \\ \end{aligned}$$

2.3 Speed reducer design problem

$$f\left( x \right) = 0.7854x_{1} x_{2}^{2} (3.3333x_{3}^{2} + 14.9334x_{3} - 43.0934) - 1.508x_{1} (x_{6}^{2} + x_{7}^{2} ) + 7.4777(x_{6}^{3} + x_{7}^{3} ) + 0.7854(x_{4} x_{6}^{2} + x_{5} x_{7}^{2} ).$$

Subject to:

$$\begin{aligned} & g_{1} \left( x \right) = \frac{27}{{x_{1} x_{2}^{2} x_{3} }} - 1 \le 0 \\ & g_{2} \left( x \right) = \frac{397.5}{{x_{1} x_{2}^{2} x_{3}^{2} }} - 1 \le 0 \\ & g_{3} \left( x \right) = \frac{{1.93x_{4}^{3} }}{{x_{2} x_{6}^{4} x_{3} }} - 1 \le 0 \\ & g_{4} \left( x \right) = \frac{{1.93x_{5}^{3} }}{{x_{2} x_{7}^{4} x_{3} }} - 1 \le 0 \\ & g_{5} \left( x \right) = \frac{{\left[ {\left( {745x_{4} /x_{2} x_{3} } \right)^{2} + 16.9 \times 10^{6} } \right]^{0.5} }}{{110x_{6}^{3} }} - 1 \le 0 \\ & g_{6} \left( x \right) = \frac{{\left[ {\left( {745x_{5} /x_{2} x_{3} } \right)^{2} + 157.5 \times 10^{6} } \right]^{0.5} }}{{85x_{7}^{3} }} - 1 \le 0 \\ & g_{7} \left( x \right) = \frac{{x_{2} x_{3} }}{40} - 1 \le 0 \\ & g_{8} \left( x \right) = \frac{{5x_{2} }}{{x_{1} }} - 1 \le 0 \\ & g_{9} \left( x \right) = \frac{{x_{1} }}{{12x_{2} }} - 1 \le 0 \\ & g_{10} \left( x \right) = \frac{{1.5x_{6} + 1.9}}{{x_{4} }} - 1 \le 0 \\ & g_{11} \left( x \right) = \frac{{1.1x_{7} + 1.9}}{{x_{5} }} - 1 \le 0 \\ \end{aligned}$$

where

$$\begin{aligned} & 2.6 \le x_{1} \le 3.6, 0.7 \le x_{2} \le 0.8, 17 \le x_{3} \le 28, 7.3 \le x_{4} \le 8.3 \\ & 7.3 \le x_{5} \le 8.3, 2.9 \le x_{6} \le 3.9, 5.0 \le x_{7} \le 5.5. \\ \end{aligned}$$

2.4 Tension/compression spring design problem

$$f\left( x \right) = (x_{3} + 2)x_{2} x_{1}^{2} .$$

Subject to:

$$\begin{aligned} & g_{1} \left( x \right) = 1 - \frac{{x_{2}^{3} x_{3} }}{{71.785x_{1}^{4} }} \le 0 \\ & g_{2} \left( x \right) = 4x_{2}^{2} - \frac{{x_{1} x_{2} }}{{12.566\left( {x_{2}^{{}} x_{1}^{3} - x_{1}^{4} } \right)}} + \frac{1}{{5108x_{1}^{2} }} - 1 \le 0 \\ & g_{3} \left( x \right) = 1 - \frac{{140.45x_{1} }}{{x_{2}^{2} x_{3} }} \le 0 \\ & g_{4} \left( x \right) = \frac{{x_{1} + x_{2} }}{1.5} - 1 \le 0 \\ & 0.05 \le x_{1} \le 2.00 \\ & 0.25 \le x_{2} \le 1.30 \\ & 2.00 \le x_{3} \le 15.00 \\ \end{aligned}$$

2.5 Welded beam design problem

$$f\left( x \right) = 1.10471x_{1}^{2} x_{2} + 0.04811x_{3} x_{4} (14 + x_{2} ).$$

Subject to:

$$\begin{aligned} & g_{1} \left( x \right) = \tau \left( x \right) - \tau_{ \text{max} } (x) \le 0 \\ & g_{2} \left( x \right) = \sigma \left( x \right) - \sigma_{ \text{max} } \left( x \right) \le 0 \\ & g_{3} \left( x \right) = x_{1} - x_{4} \le 0 \\ & g_{4} \left( x \right) = 0.10471x_{1}^{2} + 0.04811x_{3} x_{4} \left( {14 + x_{2} } \right) - 5 \le 0 \\ & g_{5} \left( x \right) = 0.125 - x_{1} \le 0 \\ & g_{6} \left( x \right) = \delta (x) - \delta_{\text{max} } \le 0 \\ & g_{7} \left( x \right) = P - P_{\text{c}} \left( x \right) \le 0 \\ & 0.1 \le x_{i} \le 2,\quad i = 1, 4 \\ & 0.1 \le x_{i} \le 10,\quad i = 2, 3. \\ \end{aligned}$$

where

$$\tau \left( x \right) = \sqrt {(\tau^{'} )^{2} + 2\tau^{'} \tau^{''} \frac{{x_{2} }}{2R} + (\tau^{''} )^{2} } ,$$

$$\tau^{'} = \frac{P}{{\sqrt 2 x_{1} x_{2} }}$$

$$\tau^{''} = \frac{\text{MR}}{J}$$

$$M = 6000\left( {14 + \frac{{x_{2} }}{2}} \right)$$

$$R = \sqrt {\frac{{x_{2}^{2} + \left( {x_{1} + x_{3} } \right)^{2} }}{4}}$$

$$J = 2\left( {\sqrt 2 x_{1} x_{2} \left( {\frac{{x_{2}^{2} }}{12}} \right) + \left( {\frac{{\left( {x_{1} + x_{3} } \right)^{2} }}{4}} \right)} \right)$$

$$\sigma \left( x \right) = \frac{504000}{{x_{4} x_{3}^{2} }}$$

$$\delta \left( x \right) = \frac{201952}{{x_{4} x_{3}^{3} }}$$

$$P_{c} \left( x \right) = 4.013E\left( {1 - 0.028234x_{3} } \right)\frac{{x_{3} x_{4}^{3} }}{{6L^{2} }}$$

$$L = 14,E = 30 \times 10^{6}$$