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Enhanced harmony search algorithm with circular region perturbation for global optimization problems

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Abstract

To improve the searching effectiveness of the harmony search (HS) algorithm, an enhanced harmony search algorithm with circular region perturbation (EHS_CRP) is proposed in this paper. In the EHS_CRP algorithm, a global and local dimension selection strategy is designed to accelerate the search speed of the algorithm. A selection learning operator based on the global and local mean level is proposed to improve the balance between exploration and exploitation. Circular region perturbation is employed to avoid the algorithm stagnation and get a better exploration region. To assess performance, the proposed algorithm is compared with 10 state-of-the-art swarm intelligent approaches in a large set of global optimization problems. The simulation results confirm that EHS_CRP has a significant advantage in terms of accuracy, convergence speed, stability and robustness. Moreover, EHS_CRP performs better than other tested methods in engineering design optimization problems. Thus, the EHS_CRP algorithm is a viable and reliable alternative for some difficult and multidimensional real-world problems.

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Acknowledgements

The authors are grateful to the editor and the anonymous referees for their constructive comments and recommendations, which have helped to improve this paper significantly. The authors would also like to express their sincere thanks to P. N. Suganthan for the useful information about meta-heuristic algorithm and optimization problems on their home webpages. This work is supported by Guangzhou Science and Technology Plan Project (201804010299), National Nature Science Foundation of China (Grant No. 61806058, 51505092, 51775122), Guangdong natural science foundation project (2018A030310063) and 2017 undergraduate innovation training program of Guangzhou University (201711078068, CX2017070). Guangzhou university talent launch program (2700050326).

All the authors have no conflict of interest. This article does not contain any studies with human participants performed by any of the authors.

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Authors and Affiliations

  1. School of Mechanical and Electric Engineering, Guangzhou University, Guangzhou, 510006, China

    Wenqiang Wu, Haibin Ouyang & Chunliang Zhang

  2. Operations Research Department, Institute of Statistical Studies and Research, Cairo University, Giza, 12613, Egypt

    Ali Wagdy Mohamed

  3. Graduate School of Business and Law, RMIT University, Melbourne, 3000, Australia

    Steven Li

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Appendix 1

Appendix 1

1.1 Problem 1 Welded beam design

$$ {\displaystyle \begin{array}{l}f(x)=1.1047{x}_1^2{x}_2+0.04811{x}_3{x}_4\left(14.0+{x}_2\right)\\ {}s.t.\left\{\begin{array}{l}{g}_1(x)=\tau (x)-{\tau}_{\mathrm{max}}\le 0\\ {}{g}_2(x)=\sigma (x)-{\sigma}_{\mathrm{max}}\le 0\\ {}{g}_3(x)={x}_1-{x}_4\le 0\\ {}{g}_4(x)=0.1047{x}_1^2{x}_2+0.04811{x}_3{x}_4\left(14.0+{x}_2\right)-5\le 0\\ {}{g}_5(x)=0.125-{x}_1\le 0\\ {}{g}_6(x)=\theta (x)-{\theta}_{\mathrm{max}}\le 0\\ {}{g}_7(x)=P- Pc(x)\le 0\end{array}\right.\\ {}0.125\le {x}_1\le 10,\kern0.5em 0.1\le {x}_2\le 10,0.1\le {x}_3\le 10,0.1\le {x}_4\le 5\end{array}} $$
$$ {\displaystyle \begin{array}{c}\tau (x)=\sqrt{\tau^{\hbox{'}2}+2{\tau}^{\hbox{'}}{\tau}^{"}\frac{x2}{2R}+{\tau}^{"2}},{\tau}^{\hbox{'}}=P/\left(\sqrt{2}{x}_1{x}_2\right),{\tau}^{"}= MR/J,M=P\left(L+{x}_2/2\right),\\ {}R=0.5\sqrt{{x_2}^2+{\left({x}_1+{x}_3\right)}^2},J=2\left\{\sqrt{2}{x}_1{x}_2\left[\frac{{x_2}^2}{4}+{\left(\frac{x_1+{x}_3}{2}\right)}^2\right]\right\},\kern0.5em \sigma (x)=6 PL/\left({x}_4{x_3}^2\right),\\ {}\theta (x)=6{PL}^3/\left({Ex}_3^2{x}_4\right), Pc(x)=\frac{4.013E\sqrt{x_3^2{x}_4^6/36}}{L}\left[1-\frac{x^3}{2L}\sqrt{\frac{E}{4G}}\right],\\ {}P=6000.L=14,E=30\times {10}^6,G=12\times {10}^6,{\tau}_{\mathrm{max}}=30600,{\sigma}_{\mathrm{max}}=30000,{\theta}_{\mathrm{max}}=0.25\end{array}} $$

Where h(x1) is the weld thickness, l(x2) is the welded joint length, t(x3) is the width of the beam, b(x4) is the thickness of the beam. The problem is subject to five constraints including shear stress (τ), bending stress (s), buckling load (Pc), deflection (δ) and geometric constraints. E is young’s modulus of bar stock, G is shear modulus of bar stock, P is the loading condition, L is overhang length of the beam.

1.2 Problem 2 Speed reducer design

$$ {\displaystyle \begin{array}{l}\begin{array}{c}f(x)=0.7854{x}_1{x}_2^2\left(3.3333{x}_3^2+14.9334{x}_3-43.0934\right)-\\ {}1.506{x}_1\left({x}_6^2+{x}_7^2\right)+7.4777\left({x}_6^3+{x}_7^3\right)+0.7854\left({x}_4{x}_6^2+{x}_5{x}_7^2\right)\end{array}\\ {}s.t\left\{\begin{array}{l}{g}_1(x)=\frac{27}{x_1{x}_2^2{x}_3}-1\le 0,{g}_2(x)=\frac{397.5}{x_1{x}_2^2{x}_3^2}-1\le 0,{g}_3(x)=\frac{1.93{x}_4^3}{x_2{x}_3{x}_6^4}-1\le 0,\\ {}{g}_4(x)=\frac{1.93{x}_5^3}{x_2{x}_3{x}_7^4}-1\le 0,{g}_5(x)=\frac{\sqrt{{\left(\frac{745{x}_4}{x_2{x}_3}\right)}^2+16.9\cdot {10}^6}}{110{x}_6^3}-1\le 0\\ {}{g}_6(x)=\frac{\sqrt{{\left(\frac{745{x}_4}{x_2{x}_3}\right)}^2+157.5\cdot {10}^6}}{85{x}_7^3}-1\le 0\\ {}{g}_7(x)={x}_2{x}_3-40\le 0,{g}_8(x)=-\frac{x_1}{x_2}+5\le 0,{g}_9(x)=\frac{x_1}{x_2}-12\le 0,\\ {}{g}_{10}(x)=\frac{1.5{x}_6+1.9}{x_4}-1\le 0,{g}_{11}(x)=\frac{1.5{x}_7+1.9}{x_5}-1\le 0\\ {}2.6\le {x}_1\le 3.6,0.7\le {x}_2\le 0.8,17\le {x}_3\le 28,7.3\le {x}_4\le 8.3\\ {}7.3\le {x}_5\le 8.3,2.9\le {x}_6\le 3.9,0.5\le {x}_6\le 5.5\end{array}\right.\end{array}} $$

Here, the face width b(x1), module of teeth m(x2), number of teeth in the pinion z(x3), length of the first shaft between bearings l1(x4), length of the second shaft between bearings l2(x5) and the diameter of the first shaft d1(x6) and second shaft d2(x7). Note that the third variable x3 (number of teeth) is of integer value while all other variables are continuous.

1.3 Problem 3 Complex (bridge) system reliability design

$$ {\displaystyle \begin{array}{l}\max\ f\left(\mathbf{r},\mathbf{n}\right)={R}_1{R}_2+{R}_3{R}_4+{R}_1{R}_4{R}_5+{R}_2{R}_3{R}_5-{R}_1{R}_2{R}_3{R}_4-{R}_1{R}_2{R}_3{R}_5\\ {}\kern5.25em -{R}_1{R}_2{R}_4{R}_5-{R}_1{R}_3{R}_4{R}_5-{R}_2{R}_3{R}_4{R}_5+2{R}_1{R}_2{R}_3{R}_4{R}_5\\ {}s.t.\left\{\begin{array}{l}{g}_1\left(\mathbf{r},\mathbf{n}\right)=\sum \limits_{i=1}^m{w}_i{v}_i^2{n}_i^2-V\le 0,\\ {}{g}_2\left(\mathbf{r},\mathbf{n}\right)=\sum \limits_{i=1}^m{\alpha}_i{\left(-\frac{1000}{\ln \left({r}_i\right)}\right)}^{\beta_i}\left[{n}_i+\exp \left(0.25{n}_i\right)\right]-C\le 0,\\ {}{g}_3\left(\mathbf{r},\mathbf{n}\right)=\sum \limits_{i=1}^m{w}_i{n}_i\exp \left(0.5{n}_i\right)-W\le 0,\\ {}0\le {r}_i\le 1,{n}_i\in {Z}^{+},1\le i\le m.\end{array}\right.\end{array}} $$

Here, m is the number of subsystems in the system; ni is the number of components in subsystem i; ri is the reliability of each component in subsystem i, qi = 1- ri is the failure probability of each component in subsystem i; Ri(ni) =1- (qi) ni is the reliability of subsystem i, f(r, n) is the system reliability. wi is the weight of each component in subsystem i; vi is the volume of each component in subsystem i, and ci is the cost of each component in subsystem i; Furthermore, V is the upper limit on the sum of the subsystems’ products of volume and weight; C is the upper limit on the cost of the system; W is the upper limit on the weight of the system. The parameters □i and □i are physical features of system components. Constraint g1(r, n) is a combination of weight, redundancy allocation and volume. g2(r, n) is a cost constraint, while g3(r, n) is a weight constraint.

1.4 Problem 4 Overspeed protection system for a gas turbine

$$ {\displaystyle \begin{array}{l}\max\ f\left(r,n\right)=\prod \limits_{i=1}^m\left[1-{\left(1-{r}_i\right)}^{n_i}\right]\\ {}s.t.\left\{\begin{array}{l}{g}_1=\sum \limits_{i=1}^m{v}_i{n}_i^2-V\le 0,\\ {}{g}_2=\sum \limits_{i=1}^mC\left({r}_i\right)\left[{n}_i+\exp \left(0.25{n}_i\right)\right]-C\le 0,\\ {}{g}_3=\sum \limits_{i=1}^m{w}_i{n}_i\exp \left(0.5{n}_i\right)-W\le 0,\\ {}0.5\le {r}_i\le 1-{10}^{-6},{r}_i\in {R}^{+},1\le {n}_i\le 10,{n}_i\in {Z}^{+}.\end{array}\right.\end{array}} $$

Here, ri, ni, vi, and wi refer to the reliability of each component, the number of redundant components, the product of weight and volume per element, and the weight of each component, all in stage i. The term exp.(0.25ni) accounts for the interconnecting hardware. The cost of each component with reliability ri at subsystem i is given by \( C\left({r}_i\right)={\alpha}_i{\left(-\frac{T}{\ln \left({r}_i\right)}\right)}^{\beta_i} \) . The parameters □i and □i are physical features of each component at stage i. T is the operating time during which the component must not fail.

1.5 Problem 5 Convex quadratic reliability design problem

$$ {\displaystyle \begin{array}{l}\max\ R(m)=\prod \limits_{j=1}^{10}\left(1-{\left(1-{p}_j\right)}^{m_j}\right)\\ {}\mathrm{s}.\mathrm{t}.{g}_i(m)=\prod \limits_{j=1}^{10}\left({a}_{ij}{m}_j^2+{c}_{ij}{m}_j\right)\le {b}_i,i=1,2,3,4\end{array}} $$

Here, pj is the reliability of the jth component in a system. All the problem parameters are generated from uniform distributions: pj from [0.80, 0.99], aij from range [0, 10], cij from range [0, 10]. A random set of these coefficients for the case of n = 4 and t = 10 are as below: P = (0.81, 0.93, 0.92, 0.96, 0.99, 0.89, 0.85, 0.83, 0.94, 0.92).

$$ {\displaystyle \begin{array}{l}a=\left(\begin{array}{cccccccccc}2& 7& 3& 0& 5& 6& 9& 4& 8& 1\\ {}4& 9& 2& 7& 1& 0& 8& 3& 5& 6\\ {}5& 1& 7& 4& 3& 6& 0& 9& 8& 2\\ {}8& 3& 5& 6& 9& 7& 2& 4& 0& 1\end{array}\right),c=\left(\begin{array}{cccccccccc}7& 1& 4& 6& 8& 2& 5& 9& 3& 3\\ {}4& 6& 5& 7& 2& 6& 9& 1& 0& 8\\ {}1& 10& 3& 5& 4& 7& 8& 9& 4& 6\\ {}2& 3& 2& 5& 7& 8& 6& 10& 9& 1\end{array}\right)\\ {}b=\left(2.0\times {10}^{13},3.1\times {10}^{12},5.7\times {10}^{13},9.3\times {10}^{12}\right).{m}_j\in \left[1,6\right],j=1,\dots, 10\end{array}} $$

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Wu, W., Ouyang, H., Mohamed, A.W. et al. Enhanced harmony search algorithm with circular region perturbation for global optimization problems. Appl Intell 50, 951–975 (2020). https://doi.org/10.1007/s10489-019-01558-6

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