Adaptation of the musical composition method for solving constrained optimization problems

Abstract

Many real-world problems may be expressed as nonlinear constrained optimization problems (CNOP). For this kind of problems, the set of constraints specifies the feasible solution space. In the last decades, several algorithms have been proposed and developed for tackling CNOP. In this paper, we present an extension of the “Musical Composition Method” (MMC) for solving constrained optimization problems. MMC was proposed by Mora et al. (Artif Intell Rev 1–15, doi:10.1007/s10462-011-9309-8, 2012a). The MMC is based on a social creativity system used to compose music. We evaluated and analyzed the performance of MMC on 12 CNOP benchmark cases. The experimental results demonstrate that MMC significantly improves the global performances of the other tested metaheuristics on some benchmark functions.

Appendix: G-suite test functions

$$\begin{aligned} \begin{array}{l} \quad \mathbf {G1} \\ \min f(x)=5x_{1}+5_{2}+5x_{3}+5x_{4}-5 \sum _{l=1}^{4} x_{l}^{2}\\ -\sum _{l=5}^{13}x_{l}.\\ \hbox {subject to}\\ 2x_{1}+2x_{2}+x_{10}+x_{11}\le 10\\ -8x_{1}+x_{10}\le 0\nonumber \\ \end{array}\\ \begin{array}{l} -2x_{4}-x_{5}+x_{10}\le 0\\ 2x_{1}+2x_{3}+x_{10}+x_{12} \le 10\\ -8x_{2}+x_{11}\le 0\\ -2x_{6}-x_{7}+x_{11}\le 0\\ 2x_{2}+2x_{3}+x_{11}+x_{11}\le 10\\ -8x_{3}+x_{11}\le 0\\ -2x_{8}-x_{9}+x_{12}\le 0\\ 0\le x_{l} \le 1 \quad \forall l=1,\ldots ,9\\ 0\le x_{l} \le 100 \quad \forall l=10,\ldots ,12\\ 0\le x_{13} \le 1\\ \hbox {where: } f(x^{\star })=-15\\ \hbox {with } x^{\star }=[1,1,1,1,1,1,1,1,1,3,3,3,1]\\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G2}\\ \max f(x)=\left| \frac{\sum _{l=1}^{n} \cos ^{4}(x_{l})-2\prod _{l=1}^{n}\cos ^{2}(x_{l})}{\sqrt{\sum _{l=1}^{n} l*\cos ^{2}(x_{l})}} \right| \\ \hbox {subject to}\\ \prod _{l=1}^{n}x_{l}\ge 0.75\\ \prod _{l=1}^{n}x_{l}\le 0.75n\\ 0\le x_{l} \le 10 \quad \forall l=1,\ldots ,n\\ \hbox {If }n=20 \hbox { then } f(x^{\star })=0.8036 \\ \hbox {If }n=50 \hbox { then } f(x^{\star })=0.83 \\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G3}\\ \max f(x)=(\sqrt{n})^{n}*\prod _{l=1}^{n}x_{l} \\ \hbox {subject to}\\ \sum _{l=1}^{n} x_{l}^{2}=1 0\le x_{l} \le 1\quad \forall l=1,\ldots ,n\\ \hbox {where: } f(x^{\star })=1 \hbox { with } x^{\star }=\left[ \frac{1}{\sqrt{n}}, \frac{1}{\sqrt{n}}, \ldots , \frac{1}{\sqrt{n}}\right] \\ \end{array} \end{aligned}$$

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$$\begin{aligned}&\begin{array}{l} \quad \mathbf {G4}\\ \min f(x)=5.3578547x_{3}^{2}+0.8356891x_{1}x_{5} \\ + 37.293239x_{1}-40792.141\\ \hbox {subject to}\\ 85.334407+0.0056858x_{2}x_{5}+0.0006262x_{l}x_{4}\\ -0.0022053x_{3}x_{5}\ge 0\\ 85.334407+0.0056858x_{2}x_{5}+0.0006262x_{l}x_{4}\\ -0.0022053x_{3}x_{5}\le 92\\ 80.51249 + 0.0071317x_{2}x_{5} + 0.0029955x_{l}x_{2}\\ +0.0021813x_{3}^{2}\ge 90\\ 80.51249 + 0.0071317x_{2}x_{5} + 0.0029955x_{l}x_{2}\\ +0.0021813x_{3}^{2}\le 110\\ 9.300961 + 0.0047026x_{3}x_{5} + 0.00l2547x_{l}x_{3} \\ +0.0019085x_{3}x_{4}\ge 20\nonumber \\ \end{array}\\&\begin{array}{l} 9.300961 + 0.0047026x_{3}x_{5} + 0.00l2547x_{l}x_{3} \\ +0.0019085x_{3}x_{4}\le 25\\ 78\le x_{1} \le 102\\ 33\le x_{2} \le 45\\ 27\le x_{l} \le 45 \quad \forall l=3,4,5\\ \hbox {where: } f(x^{\star })=-30,665.5 \\ \hbox { with } x^{\star }=[78, 33, 29.995, 45, 36.776 ]\\ \end{array} \end{aligned}$$

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$$\begin{aligned}&\begin{array}{l} \quad \mathbf {G5}\\ \min f(x)=3x_{1}+0.000001x_{1}^{2}+2x_{2}+ \frac{0.000002}{3x_{2}^{3}}\\ \hbox {subject to}\\ x_{4}-x_{3}+0.55\ge 0\\ x_{3}-x_{4}+0.55\ge 0\\ 1{,}000 \sin (-x_{3}-0.25)+1,000\sin (-x_{4}-0.25)+894.8\\ -x_{1}=0\nonumber \\ \end{array} \\&\begin{array}{l} 1000 \sin (x_{3}-0.25)+ 1{,}000\sin (x_{3}-x_{4}-0.25)+ 894.8\\ \!-x_{2} \!=\! 0\\ 1{,}000 \sin (x_{4}-0.25)+ 1{,}000\sin (x_{4}-x_{3}-0.25)\\ +1{,}294. 8 = 0\\ 0\le x_{l} \le 1,200 \quad \forall l=1,2\\ -0.55\le x_{i} \le 0.55 \quad \forall l=3,4\\ \hbox {where: } f(x^{\star })=5{,}126.4981\\ \hbox {with } x^{\star }\!=\![679.9453,1{{,}}026.067,0.1188764,-0.3962336]\\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G6}\\ \min f(x)=(x_{1}-10)^{3}+(x_{2}-20)^{3}\\ \hbox {subject to}\\ (x_{1}-5)^{2}+(x_{2}-5)^{2}-100\ge 0\\ -(x_{1}-6)^{2}-(x_{2}-5)^{2}+82.81\ge 0\\ 13\le x_{1} \le 100\\ 0\le x_{2} \le 100 \\ \hbox {where: } f(x^{\star })=-6961 .81381\\ \hbox {with } x^{\star }=[14.095, 0.84296]\\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G7}\\ \min f(x)=x_{1}^{2}+x_{2}^{2}+x_{1}x_{2}-14x-{1}-16x_{2}\\ +(x_{3}-10)^{2} +4(x{4}-5)^{2} +(x_{5}-3)^{2}+2(x_{6}-1)^{2}\\ +5x_{7}^{2}+7(x_{8}-11)^{2} +2(x_{9}-10)^{2} + (x_{10}-7)^{2}+45\\ \hbox {subject to}\\ 105-4x_{1}-5x_{2}+3x_{7}-9x_{8}\ge 0\\ -3(x_{1}-2)^{2}-4(x_{2}-3)^{2}-2x_{3}^{2}+7x_{4}+120\ge 0\\ -10x-{1}+8x_{2}+17x_{7}-2x_{8}\ge 0\\ -x_{1}^{2}-2(x_{2}-2)^{2}+2x_{1}x_{2}-14x_{5} +6x_{6}\ge 0\\ 8x_{1}-2x_{2}-5x_{9}+2x_{10}+12\ge 0\\ -5x_{1}^{2}-8x_{2}-(x_{3}-6)^{2}+2x_{4}+40\ge 0\\ 3x_{1}-6x_{2}-12(x_{9}-8)^{2}+7x_{10} \ge 0\\ -0.5(x_{1}-8)^{2}-2(x_{2}-4)^{2}-3x_{5}^{2} + x_{6} +30 \ge 0\\ -10\le x_{l} \le 10 \hbox { }\forall l=1,\ldots ,10\\ \hbox {where: } f(x^{\star })=24.3062091 \\ \hbox {with } x^{\star }=[2.171996,2.363683,8.773926,5.095984,\\ 0.99065048,1.430574,1.321644,9.828726,8.280092,\\ 8.375927]\\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G8}\\ \max f(x)=\frac{\sin ^{3}(2\pi x_{1})*\sin (2\pi x_{2})}{x_{1}^{3}*(x_{1}+x_{2})}\\ \hbox {subject to}\\ x_{1}^{2}-x_{2}+1\le 0\\ 1-x_{1}-(x_{2}-4)^2\le 0\\ 0\le x_{l} \le 10 \quad \forall l=1,2\\ \hbox {where: } f(x^{\star })=0.1 \\ \end{array} \end{aligned}$$

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$$\begin{aligned}&\begin{array}{l} \quad \mathbf {G9}\nonumber \\ \min f(x)=(x_{1}-10)^{2}+5(x_{2}-12)^2+x_{3}^{4}+3(x_{4}-11)^{2}\nonumber \\ +10x_{5}^{6}+7x_{7}^{4} -4x_{6}x_{7}-10x_{6}-8x_{7}\nonumber \\ \end{array}\\&\begin{array}{l} \hbox {subject to}\\ 127-x_{1}^{2}-3x_{2}^{4}-x_{3}-4x_{4}^{2}-5x_{5}\ge 0\\ 196-23x_{1}-x_{2}^{2}-2x_{6}^{2}+8x_{1}\ge 0\\ 282-7x_{1}-3x_{2}-10x_{3}^{2}-x_{4}+x_{5}\ge 0\\ -4x_{1}^{2}-x_{2}^{2}+3x_{1}x_{2}-5x_{6}+11x_{7}\ge 0\\ -10\le x_{l} \le 10 \quad \forall l=1,\ldots ,7\\ \hbox {where: } f(x^{\star })=680.6300873\\ \hbox {with }x^{\star }=[2.330499,1.951372,-0.4775414\\ 4.365726,-0.6244870,1.038131,1.594227]\\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G10}\\ \min f(x)=x_{1}+x_{2}+x_{3}\\ \hbox {subject to}\\ 1-0.0025(x_{4}+x_{6})\ge 0\\ 1-0.01(x_{8}-x_{5})\ge 0\\ x_{2}x_{7}-1{,}250x_{5}-x_{2}x_{7}\ge 0\\ 1-0.0025(x_{5}+x_{7}-x_{4})\ge 0\\ x_{1}x_{6}-833.33252x_{4}-100x_{1}+83,333.333\ge 0\\ x_{3}x_{8}-1{,}250,000-x_{3}x_{5}+2,500x_{5}\ge 0\\ 100\le x_{1} \le 10,000 \\ 1{,}000\le x_{l} \le 10{,}000\quad \forall l=2,3\\ 10\le x_{l} \le 1,000\quad \forall l=4,\ldots ,8\\ \hbox {where: } f(x^{\star })=7{,}049.330923 \\ \hbox {with }x^{\star }=[579.3167,1,359.943,5,110.071,\\ 182.074,295.5985,217.9799,286.4162,395.5979]\\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G11}\\ \min f(x)=x_{1}^{2}+(x_{2}-1)^{2}\\ \hbox {subject to}\\ x_{2}-x_{1}^{2}=0\\ -1\le x_{l} \le 1 \quad \forall l=1,2\\ \hbox {where: } f(x^{\star })=0.75000455 \\ \hbox {with }x^{\star }=[\pm 0.70711,0.5]\\ \end{array} \end{aligned}$$

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$$\begin{aligned} \begin{array}{l} \quad \mathbf {G12}\\ \max f(x)=\frac{100-(x_{1}-5)^{2}-(x_{2}-5)^{2}-(x_{3}-5)^{2}}{100}\\ \hbox {subject to}\\ (x_{1}-\rho _{1})^{2}+(x_{2}-\rho _{2})^{2}+(x_{3}-\rho _{3})^{2}\le 0.25\\ \hbox {for: } \rho _{s}=1,3,5,7,9 \forall s=1,2,3\\ 0\le x_{l} \le 10 \forall l=1,2,3\\ \hbox {where: } f(x^{\star })=1 \\ \hbox {with }x^{\star }=[5,5,5]\\ \end{array} \end{aligned}$$

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