Multi-objective Stochastic Paint Optimizer (MOSPO)

Abstract

The single-objective version of stochastic paint optimizer (SPO) is appropriately changed to solve multi-objective optimization problems described as MOSPO. Color theory, the color wheel, and color combination methods are the main concepts of SPO. The SPO will be able to do excellent exploration and exploitation thanks to four simple color combination rules that do not have any internal parameters. Principles like using of fixed-sized external archive make the recommended technique various from the initial single-objective SPO. In addition, to perform multi-objective optimization, the leader selection feature has been added to SPO. The efficiency of recommended multi-objective stochastic paint optimizer (MOSPO) is tested on ten mathematical (CEC-09) and eight multi-objective engineering design problems concerning remarkable precision and uniformity compared to multi-objective particle swarm optimization (MOPSO), multi-objective slap swarm algorithm (MSSA), and multi-objective ant lion optimizer. According to the results of different performance metrics, such as generational distance (GD), inverted generational distance (IGD), maximum spread, and spacing, the proposed algorithm can provide quality Pareto fronts with very competitive results with high convergence.

Appendices

Appendix A: constrained multi-objective test problems used in this paper

1.1 CONSTR

This issue [47] contains two constraints and two design variables that have a convex Pareto front.

$${\text{Minimize: }} f_{1} \left( x \right) = x_{1}$$

(A.1)

$${\text{Minimize:}}\quad f_{2} \left( x \right) = \left( {1 + x_{2} } \right)/x_{1}$$

(A.2)

$${\text{where}}\quad g_{1} \left( x \right) = 6 - \left( {x_{2} + 9x_{1} } \right)$$

(A.3)

$$g_{2} \left( x \right) = 1 + \left( {x_{2} - 9x_{1} } \right)$$

(A.4)

$$0.1 \le x_{1} \le 1, 0 \le x_{2} \le 5$$

1.2 SRN

Srinivas and Deb [48] proposed the following continuous Pareto optimum front for the following issue:

$${\text{Minimize: }}\quad f_{1} \left( x \right) = 2 + \left( {x_{1} {-} 2} \right)^{2} + \left( {x_{2} {-} 1} \right)^{2}$$

(A.5)

$${\text{Minimize:}}\quad f_{2} \left( x \right){ } = 9x_{1} {-} \left( {x_{2} {-} 1} \right)^{2}$$

(A.6)

$${\text{where}}\quad g_{1} \left( x \right){ } = x_{1}^{2} + x_{2}^{2} - 255$$

(A.7)

$$g_{2} \left( x \right) = x_{1} - 3x_{2} + 10$$

(A.8)

$$- 20 \le x_{1} \le 20, - 20 \le x_{2} \le 20$$

1.3 BNH

Binh and Korn [49] provided a this example for the first time as follows:

$${\text{Minimize:}}\quad f_{1} \left( x \right) = 4x_{1}^{2} + 4x_{2}^{2}$$

(A.9)

$${\text{Minimize:}}\quad f_{2} \left( x \right) = (x_{1} - 5)^{2} + (x_{2} - 5)^{2}$$

(A.10)

$${\text{where}}\quad g_{1} \left( x \right) = (x_{1} - 5)^{2} + x_{2}^{2} - 25$$

(A.11)

$$g_{2} \left( x \right) = 7.7 - (x_{1} - 8)^{2} - (x_{2} + 3)^{2}$$

(A.12)

$$0 \le x_{1} \le 5,0 \le x_{2} \le 3$$

1.4 OSY

It was suggested by Osyczka and Kundu [50] that there should be five separate regions. In addition, the following six restrictions and six design variables should be taken into account:

$${\text{Minimize}}:\quad f_{1} \left( x \right) = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} + x_{4}^{2} + x_{5}^{2} + x_{6}^{2}$$

(A.13)

$${\text{Minimize}}:\quad f_{2} \left( x \right) = [25(x_{1} - 2)^{2} + (x_{2} - 1)^{2} + \left( {x_{3} - 1} \right) + (x_{4} - 4)^{2} + \left( {x_{5} - 1)^{2} } \right]$$

(A.14)

$${\text{where}} g_{1} \left( x \right) = 2 - x_{1} - x_{2}$$

(A.15)

$$g_{2} \left( x \right) = - 6 + x_{1} + x_{2}$$

(A.16)

$$g_{3} \left( x \right) = - 2 - x_{1} + x_{2}$$

(A.17)

$$g_{4} \left( x \right) = - 2 + x_{1} - 3x_{2}$$

(A.18)

$$g_{5} \left( x \right) = - 4 + x_{4} + (x_{3} - 3)^{2}$$

(A.19)

$$g_{6} \left( x \right) = 4 - x_{6} - (x_{5} - 3)^{2}$$

(A.20)

$$0 \le x_{1} \le 10, 0 \le x_{2} \le 10, 1 \le x_{3} \le 5$$

(A.21)

$$0 \le x_{4} \le 6, 1 \le x_{5} \le 5, 0 \le x_{6} \le 10$$

Appendix B: constrained multi-objective engineering problems used in this paper

2.1 The four-bar truss design problem

In this 4-bar truss issue [51], there are two objectives as structural volume (${f}_{1}$) and displacement (${f}_{2}$), which are considered to be minimized. This issue has four design variables (${x}_{1}-{x}_{4}$) according to the cross-sectional area of members 1, 2, 3, and 4. This problem may be mathematically formularized as below:

$${\text{Minimize: }} f_{1} \left( x \right) = 200 \times \left( {2 \times x\left( 1 \right) + {\text{sqrt}}\left( {2 \times x\left( 2 \right)} \right) + {\text{sqrt}}\left( {x\left( 3 \right)} \right) + x\left( 4 \right)} \right)$$

(B.1)

$${\text{Minimize: }} f_{2} \left( x \right) = 0.01 \times \left( {\frac{2}{x\left( 1 \right)}} \right) + \left( {\frac{{2 \times {\text{sqrt}}\left( 2 \right)}}{x\left( 2 \right)}} \right) - \left( {\left( {2 \times {\text{sqrt}}\left( 2 \right)} \right)/x\left( 3 \right)} \right) + \left( {2/x\left( 1 \right)} \right))$$

(B.2)

$$1 \le x_{1} \le 3, 1.4142 \le x_{2} \le 3$$

$$1.4142 \le x_{3} \le 3, 1 \le x_{4} \le 3$$

2.2 The welded beam design problem

Ray and Liew [52] first offered this welded beam issue with two objectives, namely the fabrication cost (${f}_{1}$) and beam deflection (${f}_{2}$), which is to be reduced and four constraints. This issue has four numbers of variables, named weld thickness (${x}_{1}$), clamped bar length (${x}_{2}$), bar height (${x}_{3}$), and bar thickness (${x}_{4}$) as follows:

$${\text{Minimize: }} f_{1} \left( x \right) = 1.10471 \times x\left( 1 \right)^{2} \times x\left( 2 \right) + 0.04811 \times x\left( 3 \right) \times x\left( 4 \right) \times \left( {14 + x\left( 2 \right)} \right)$$

(B.3)

$${\text{Minimize: }} f_{2} \left( x \right) = 65856000/\left( {30 \times 10^{6} \times x\left( 4 \right) \times x\left( 3 \right)^{3} } \right)$$

(B.4)

$${\text{where }} g_{1} \left( x \right) = \tau - 13600$$

(B.5)

$$g_{2} \left( x \right) = \sigma - 30000$$

(B.6)

$$g_{3} \left( x \right) = x\left( 1 \right) - x\left( 4 \right)$$

(B.7)

$$g_{4} \left( x \right) = 6000 - P$$

(B.8)

$$0.125 \le x_{1} \le 5, 0.1 \le x_{2} \le 10$$

$$0.1 \le x_{3} \le 10, 0.125 \le x_{4} \le 5$$

$${\text{where }} q = 6000*\left( {14 + \frac{x\left( 2 \right)}{2}} \right);\;D = {\text{sqrt}}\left( {\frac{{x(2)^{2} }}{4} + \frac{{x\left( 1 \right) + x\left( 3 \right))^{2} }}{4}} \right)$$

(B.9)

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

(B.10)

$$\alpha = \frac{6000}{{{\text{sqrt}}\left( 2 \right)*x\left( 1 \right)*x\left( 2 \right)}}$$

(B.11)

$$\beta = Q*\frac{D}{J}$$

(B.12)

2.3 Disk brake design problem

This disk brake design issue has figured by Ray and Liew [52] with two objectives, namely stopping time (${f}_{1}$) and brake mass (${f}_{2}$) to minimize and five constraints for a disk brake. This problem contains five numbers of variables: disk inner radius (${x}_{1}$), outer disk radius (${x}_{2}$), engaging force (${x}_{3}$), and friction surfaces number (${x}_{4}$). The following equations may represent this problem:

$${\text{Minimize: }} f_{1} \left( x \right) = 4.9 \times \left( {10} \right)^{{\left( { - 5} \right)}} \times \left( {x\left( 2 \right)^{\left( 2 \right)} - x\left( 1 \right)^{\left( 2 \right)} } \right) \times \left( {x\left( 4 \right) - 1} \right)$$

(B.13)

$${\text{Minimize: }} f_{2} \left( x \right) = \left( {9.82 \times \left( {10} \right)^{\left( 6 \right)} } \right)) \times \left( {x\left( 2 \right))^{\left( 2 \right)} - x\left( 1 \right)^{\left( 2 \right)} } \right))/\left( {\left( {x\left( 2 \right))^{\left( 3 \right)} - x\left( 1 \right)^{\left( 3 \right)} } \right) \times x\left( 4 \right) \times x\left( 3 \right)} \right)$$

(B.14)

$$g_{1} \left( x \right) = 20 + x\left( 1 \right) - x\left( 2 \right)$$

(B.15)

$$g_{2} \left( x \right) = 2.5 + \left( {x\left( 4 \right) + 1} \right) - 30$$

(B.16)

$$g_{3} \left( x \right) = \left( {x\left( 3 \right)} \right)/\left( {3.14 \times \left( {x\left( 2 \right)^{2} - x\left( 1 \right)^{2} } \right)^{2} } \right) - 0.4$$

(B.17)

$$g_{4} \left( x \right) = \left( {2.22 \times \left( {10} \right)^{{\left( { - 3} \right)}} \times x\left( 3 \right) \times \left( {x\left( 2 \right)^{3} - x\left( 1 \right)^{3} } \right)} \right)/\left( {\left( {x\left( 2 \right)^{2} - x\left( 1 \right)^{2} } \right)^{2} } \right) - 1$$

(B.18)

$$g_{5} \left( x \right) = 900 - \left( {2.66 \times \left( {10} \right)^{{\left( { - 2} \right)}} \times x\left( 3 \right) \times x\left( 4 \right) \times \left( {x\left( 2 \right)^{3} - x\left( 1 \right)^{3} } \right)} \right)/\left( {\left( {x\left( 2 \right)^{2} - x\left( 1 \right)^{2} } \right)^{2} } \right)$$

(B.19)

$$55 \le x_{1} \le 80, 75 \le x_{2} \le 110$$

$$1000 \le x_{3} \le 3000, 2 \le x_{4} \le 20$$

2.4 Speed reducer design problem

This issue [51, 53] contains two objectives such as weight (${f}_{1}$) and stress (${f}_{2}$), which are to be minimized. The problem may represent with a diagram as given in Fig. 10. Also, the problem has eleven constraints with seven numbers of design variables such as width of gear face (${x}_{1}$), teeth module (${x}_{2}$), pinion teeth number (${x}_{3}$ numeral variable), a distance between of bearings 1 (${x}_{4}$), a distance of bearings 2 (${x}_{5}$), shaft 1 diameter (${x}_{6}$), and shaft 2 diameter (${x}_{7}$). The equations may clearly represent this problem as below:

$${\text{Minimize: }} f_{1} \left( x \right) = 0.7854 \times x\left( 1 \right) \times x(2)^{2} \times (3.3333 \times x\left( {3)^{2} + 14.9334 \times x\left( 3 \right)} \right) \ldots$$

$$- 43.0934) - 1.508 \times x\left( 1 \right) \times (x(6)^{2} + x(7)^{2}$$

(B.20)

$${\text{Minimize}}:{ }f_{2} \left( x \right) = (({\text{sqrt}}\left( {\left( {\left( {745*x\left( 4 \right)} \right)/x\left( 2 \right)*x\left( 3 \right)} \right))^{2} + 19.9e6} \right)/\left( {0.1*x\left( {6)^{3} } \right)} \right)$$

(B.21)

$${\text{where }} g_{1} \left( x \right) = 27/(x\left( 1 \right) \times x\left( {2)^{2} \times x\left( 3 \right)} \right) - 1$$

(B.22)

$$g_{2} \left( x \right) = 397.5/(x\left( 1 \right) \times x(2)^{2} \times x\left( {3)^{2} } \right) - 1$$

(B.23)

$$g_{3} \left( x \right) = (1.93 \times (x\left( {4)^{3} } \right)/(x\left( 2 \right) \times x\left( 3 \right) \times x\left( {6)^{4} } \right) - 1$$

(B.24)

$$g_{4} \left( x \right) = (1.93 \times (x\left( {5)^{3} } \right)/(x\left( 2 \right) \times x\left( 3 \right) \times x\left( {7)^{4} } \right) - 1$$

(B.25)

$$g_{5} \left( x \right) = (({\text{sqrt}}\left( {\left( {745 \times x\left( 4 \right)} \right)/x\left( 2 \right) \times x\left( {3)))^{2} + 16.9e6} \right)} \right)/\left( {110 \times x\left( {6)^{3} } \right)} \right) - 1$$

(B.26)

$$g_{6} \left( x \right) = (({\text{sqrt}}\left( {\left( {745 \times x\left( 4 \right)} \right)/x\left( 2 \right) \times x\left( {3)))^{2} + 157.5e6} \right)} \right)/\left( {85 \times x\left( {7)^{3} } \right)} \right) - 1$$

(B.27)

$$g_{7} \left( x \right) = \left( {\left( {x\left( 2 \right) \times x\left( 3 \right)} \right)/40} \right) {-} 1$$

(B.28)

$$\tau = {\text{sqrt}}\left( {\alpha^{2} + 2 \times \alpha \times \beta \times \frac{x\left( 2 \right)}{{2 \times D}} + \beta^{2} } \right)$$

(B.29)

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

(B.30)

$${\text{tmpf}} = 4.013 \times \frac{{30 \times 10^{6} }}{196}$$

(B.31)

$$P = {\text{tmpf}} \times {\text{sqrt}}\left( {x(3)^{2} \times \frac{{x(4)^{6} }}{36}} \right) \times \left( {1 - x\left( 3 \right) \times \frac{{{\text{sqrt}}\left( {\frac{30}{{48}}} \right)}}{28}} \right)$$

(B.32)