Hybridizing sine–cosine algorithm with harmony search strategy for optimization design problems

Abstract

Recent developments designate the quick growing of optimization meta-heuristics in the domain of optimization. Sine–cosine optimizer is a stochastic technique that generates various preliminary random research agents global optimal solutions and involves them to fluctuate toward or outwards the superior global optima solution utilizing a mathematical model based on sine and cosine trigonometry functions. Nevertheless, standard SCA provides insufficient global optima results on complex dimension functions illustrating poor convergence rate. The search process of the SCA method holds various shortcomings such as slow convergence, weak balance amid exploration and exploitation, and inefficiency in convergence. To overcome these shortcomings in this work, we are trying to present a new heuristic approach based on merging the features of SCA (sine–cosine approach) with a HS (harmony search approach) known as HSCAHS algorithm. The existing approach integrates the merits of the sine–cosine algorithm and HS algorithm to reduce demerits, like the trapping in local optima and unbalanced exploitation. The new approach presents own work performance in two different stages; firstly, the sine–cosine algorithm starts the explore procedure to augment exploration capability. Secondly, harmony search part starts its search from SCA finds so far to augment the exploitation tendencies. Hence, hybrid approach can find best possible solution in which least time improves the exploitation and exploration. Hence, the newly existing hybrid approach can be quickly convergent, statistically sound and more robust. The capability of the hybrid approach is verified by applying it on eighteen tested benchmark, economic dispatch, three-bar truss design, rolling element bearing design, multiple disc clutch brake design, speed reducer design and planetary gear train design problems. The experimental solutions reveal that the hybrid approach is able to finding the superior quality of optimal goal (or solution) of the optimize functions in most cases in comparison with others.

Appendices

Appendix

Benchmark test suites

Appendix

A: Welded beam design problem

Consider $\vec {x}=\left[ {{x}_{1}}{{x}_{2}}{{x}_{3}}{{x}_{4}} \right] =\left[ h\,l\,t\,b \right]$

• $\text {Minimize } f\left( {\vec {x}} \right) =1.10471x_{1}^{2}{{x}_{2}}+0.04811{{x}_{3}}{{x}_{4}}\left( 14.0+{{x}_{2}} \right)$

• Subject to:

• ${{g}_{1}}\left( {\vec {x}} \right) =\tau \left( {\vec {x}} \right) -13600\le 0$

• ${{g}_{2}}\left( {\vec {x}} \right) =\sigma \left( {\vec {x}} \right) -30000\le 0$

• ${{g}_{3}}\left( {\vec {x}} \right) ={{x}_{1}}-{{x}_{4}}\le 0$

• ${{g}_{4}}\left( {\vec {x}} \right) =0.10471\left( x_{1}^{2} \right) +0.04811{{x}_{3}}{{x}_{4}}\left( 14+{{x}_{2}} \right) -5.0\le 0$

• ${{g}_{6}}\left( {\vec {x}} \right) =\delta \left( {\vec {x}} \right) -0.25\le 0$

• ${{g}_{7}}\left( {\vec {x}} \right) =6000-{{p}_{c}}\left( {\vec {x}} \right) \le 0$

• where

• $\tau \left( {\vec {x}} \right) =\sqrt{\left( {{\tau }'} \right) +\left( 2{\tau }'{\tau }'' \right) \frac{{{x}_{2}}}{2R}+{{\left( {{\tau }''} \right) }^{2}}}$

• ${\tau }'=\frac{6000}{\sqrt{2}{{x}_{1}}{{x}_{2}}}$

• ${\tau }''=\frac{MR}{J}$

• $M=6000\left( 14+\frac{{{x}_{2}}}{2} \right)$

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

• $j=2\left\{ {{x}_{1}}{{x}_{2}}\sqrt{2}\left[ \frac{x_{2}^{2}}{12}+{{\left( \frac{{{x}_{1}}+{{x}_{3}}}{2} \right) }^{2}} \right] \right\}$

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

• $\delta \left( {\vec {x}} \right) =\frac{65856000}{\left( 30\times {{10}^{6}} \right) {{x}_{4}}x_{3}^{3}}$

• ${{p}_{c}}\left( {\vec {x}} \right) =\frac{4.013\left( 30\times {{10}^{6}} \right) \sqrt{\frac{x_{3}^{2}x_{4}^{6}}{36}}}{196}\left( 1-\frac{{{x}_{3}}\sqrt{\frac{30\times {{10}^{6}}}{4\left( 12\times {{10}^{6}} \right) }}}{28} \right)$

• $\text {with } 0.1\le {{x}_{1}},{{x}_{4}}\le 2.0\,and\,0.1\le {{x}_{2}},{{x}_{3}}\le 10.0$

B: Tension/compression spring design problem

Consider:

• $\vec {x}=\left[ {{x}_{1}}{{x}_{2}}{{x}_{3}} \right] =\left[ d\,D\,N \right]$

• $\text {Minimize } f\left( {\vec {x}} \right) =\left( {{x}_{3}}+2 \right) {{x}_{2}}x_{1}^{2}$

• $\text {subject to: }$

• ${{g}_{1}}\left( {\vec {x}} \right) =1-\frac{x_{2}^{3}{{x}_{3}}}{71785x_{1}^{4}}\le 0$

• ${{g}_{2}}\left( {\vec {x}} \right) =\frac{4x_{2}^{2}-{{x}_{1}}{{x}_{2}}}{12566\left( {{x}_{2}}x_{1}^{3}-x_{1}^{4} \right) }+\frac{1}{5108x_{1}^{2}}-1\le 0$

• ${{g}_{3}}\left( {\vec {x}} \right) =1-\frac{140.45{{x}_{1}}}{x_{2}^{2}{{x}_{3}}}\le 0$

• ${{g}_{4}}\left( {\vec {x}} \right) =\frac{{{x}_{1}}+{{x}_{2}}}{1.5}-1\le 0$

• $\text {with } 0.05\le {{x}_{1}}\le 2.0,0.25\le {{x}_{2}}\le 1.3,and\,2.0\le {{x}_{3}}\le 15.0$

C: Rolling element bearing design problem

$\text {Maximum}[C_d(X)] ={\left\{ \begin{array}{ll}max(-f_cz^{2/3}D_b^{1.8}) &{} if D_b =\le 25.5mm 0\\ max(-3.647f_cz^{2/3}D_b^{1.4}) &{} if D_b =\le 25.5mm \end{array}\right. }$

• $\text {Subject to: }$

• $g_1(x)=\frac{\phi _0}{2\sin ^{-1}(D_b/D_m)}-Z+1\ge 0,$

• $g_2(x)=2D_b-K_{D_{min}}(D-d)\ge 0,$

• $g_3(x)=K_{D_{max}}(D-d)-2D_b\ge 0,$

• $g_4(x)=\zeta B_w-D_b\ge 0,$

• $g_5(x)=D_m-0.5(D+d)\ge 0,$

• $g_6(x)=(0.5+e)(D+d)-D_m\ge 0,$

• $g_7(x)=0.5(D-D_m-D_b)-\zeta D_b\ge 0,$

• $g_8(x)=f_1\ge 0.515,$

• $g_9(x)=f_0\ge 0.515,$

• where

• $f_c=37.91[1+[1.04(\frac{1-\gamma }{1+\gamma })^{1.72}] (\frac{f_i(2f_0-1)}{f_i-1})^{0.41}]^{10/3}]^{-0.3}$

• $\gamma = \frac{D_b\cos \alpha }{D_m},$

• $f_1=\frac{r_1}{D_b}$

• $\phi _0=2\pi -20 \cos ^{-1}[\frac{(D-d)/2-3(T/4)^2+[D/2-(T/4-D_b)^2]-[d/2+(T/4)]^2]}{[2(D-d/2-3(T/4)][(T/4)-D_b]}]$

• $T=D-d-2D_b,$

• $D=160, d=90, B_w=30,$

• $0.5(D+d)\le D_m\le 0.6(D+d),$

• $0.15(D-d)\le D_b\le 0.45(D-d),$

• $4\le Z\le 50, 0.515 \le f_1\le 0.6, 0.515\le f_0 \le 0.6,$

• $0.4 \le K_{D_{min}}\le 0.5, 0.6 \le K_{D_{max}}\le 0.7,$

• $0.3\le \epsilon \le 0.4, 0.02\le e \le 0.1, 0.6\le \xi \le 0.85$

D: Multiple disc clutch brake design problem

• $Minimize f(x) = \pi (r_0^2-r_i^2)(Z+1)\rho t$

• s.t ;

• $g_1(x)=r_0-r_i-\Delta r\geqslant 0$

• $g_2(x)=l_max-(Z+1)(t+\delta )\geqslant 0$

• $g_3(x)=P_max-P_{rz}\geqslant 0$

• $g_4(x)=P_max v_{vrmax}-P_{rz}v_{sr}\geqslant 0$

• $g_5(x)=v_{srmax}-v_{sr}\geqslant 0$

• $g_6(x)=T_max-T\geqslant 0$

• $g_7(x)=M_h-sM_s\geqslant 0$

• $g_8(x)=T\geqslant 0$

• where

• $M_h=\frac{2}{3}\mu FZ\frac{r_0^3-r_i^3}{r_0^2-r_i^2}$

• $P_{rz}=\frac{2}{3}\frac{F}{\pi (r_0^2-r_i^2)}$

• $v_{rz}=\frac{2\pi n (r_0^3-r_i^3)}{90(r_0^3-r_i^3)}$

• $T = \frac{I_z\pi n}{30(M_h-M_f)}$

• $\Delta r=20mm, I_z=55kgm^2, P_max=1MPa, F_max=1000N, T_max=15s,\mu =0.5$

• $s=1.5, M_s=40Nm, M_f=3Nm, n=250r/min, v_{srmax}=10m/s, l_{max}=30mm$

• $60mm\le r_i\le 80mm, 90mm\le r_0\le , 110mm, 1.5mm\le t \le 3mm,600N\le F\le 1000N, 2 \le Z \le 9$

E: Speed reducer design problem

• $Minimize(f(x))=0.7854x_1x_2^2(3.3333x_3^2+14.9334x_3-43.0934)-1.508x_1(x_6^2+x_7^2)+7.4777(x_6^3+x_7^3)+0.7854(x_4x_6^2+x_5x_7^2)$

• s.t;

• $g_1(x)=\frac{27}{x_1x_2^2x_3}-1\le 0$

• $g_2(x)=\frac{397.5}{x_1x_2^2x_3^2}-1\le 0$

• $g_3(x)=\frac{1.93x_4^3}{x_2x_6^4x_3}-1\le 0$

• $g_4(x)=\frac{1.93x_5^3}{x_2x_7^4x_3}-1\le 0$

• $g_5(x)=\frac{[(745x_4/x_2x_3)^2+16.9 \times 10^6]^0.5}{110x_6^3}-1\le 0$

• $g_6(x)=\frac{[(745x_5/x_2x_3)^2+157.5 \times 10^6]^0.5}{85x_7^3}-1\le 0$

• $g_7(x)=x_2x_3/40-1\le 0$

• $g_8(x)=\frac{5x_2}{x_1}-1\le 0$

• $g_9(x)=\frac{x_1}{12x_2}-1\le 0$

• $g_10(x)=\frac{1.5x_6+1.9}{x_4}-1\le 0$

• $g_11(x)=\frac{1.1x_7+1.9}{x_5}-1\le 0$

• where

• $2.6\le x_1\le , 0.7\le x_2\le 0.8, 17\le x_3 \le 28, 7.3\le x_4 \le 8.3, 7.3\le x_5\le 8.3$

• $2.9\le x_6\le 3.9, 5.0\le x_7\le 5.5$