A hybrid TLNNABC algorithm for reliability optimization and engineering design problems

Abstract

The paper aims to present a new TLNNABC hybrid algorithm to solve reliability and engineering design optimization problems. In this algorithm, the structure of the artificial bee colony (ABC) algorithm has been improved by incorporating the features of the neural network algorithm (NNA) and teaching-learning based optimization (TLBO). In the standard ABC, the onlooker bees apply the same searching method as the employed bees, which causes slow convergence and also restricts its practical application of solving optimization problems. In view of this inadequacy and resulting in a better balance between exploration and exploitation, searching procedures for employed bees and onlooker bees of the conventional ABC are renovated based on NNA and improved TLBO algorithms respectively and a new hybrid algorithm called TLNNABC has been developed in this paper. In TLNNABC, for the employed bee phase, NNA is used to increase the population diversity. However, the improved teaching learning-based optimization is embedded in the onlooker bee phase. In this context, a new search operator is introduced which increases the exploitation capability of the algorithm to operate, and a probabilistic selection strategy, which helps to determine whether to apply the original or the new search operator to construct a new solution. Finally, the performance of the proposed TLNNABC algorithm has been demonstrated by the well-known benchmark problems related to reliability optimization, structural engineering design problems, and 23 unconstrained benchmark functions and finally compared with several existing algorithms. Experimental results show that the proposed algorithm is very effective and achieves superior performance than the other algorithms.

Appendix 1

1.1 1.1 The welded beam design problem

The objective of this problem is to minimize the fabricating cost of the welded beam subject to constraints on shear stress ($\tau$), bending stress in the beam ($\sigma$), end deflection of the beam ($\delta$), buckling load on the bar ($P_c$) and side constraints. In the problem, there are four design variables $h(x_1), l(x_2), t(x_3)$, and $b(x_4$) as shown in Fig. 4a. The problem can be mathematically formulated as follows:

$$\begin{array}{ll} \text {Minimize} & f(x)=1.10471x_1^2 x_2+0.04811x_3 x_4 (14.0+x_2) \\ \\ \text {subject to} & g_1 (x)= \tau (x)-\tau _{\text {max}} \le 0 ; \qquad g_5 (x)=0.125-x_1 \le 0 \\ \\ & {} g_2 (x)= \sigma (x)-\sigma _{\text {max}} \le 0 ; \qquad g_6 (x)=\delta (x)-\delta _{\text {max}} \\ \\ & g_3 (x)=x_1-x_4 \le 0 ; \qquad g_7(x)= P-P_c (x) \le 0 \\ \\ & {} g_4 (x)=0.1047x_1^2+0.04811x_3 x_4 (14.0+x_2 )-5.0 \le 0 \\ \\ \text {where} & {} \tau (x) = \sqrt{(\tau ')^2 +2 \tau ' \tau '' \frac{x_2}{2R}+\tau ''^2}, \ \ \tau ' = \frac{P}{\sqrt{2}x_1x_2}, \ \ \tau ''=\frac{MR}{J}, \ \ M = P\left( L+\frac{x_2}{2}\right) ,\\ \\ & {} R =\sqrt{\frac{x_2^2}{4}+ \left( \frac{x_1+x_3}{2} \right) ^2}, \ \ J = 2 \left\{ \sqrt{2} x_1x_2 \left[ \sqrt{\frac{x_2^2}{12} +\left( \frac{x_1+x_3}{2} \right) ^2}\right] \right\} , \ \ \sigma (x) = \frac{6PL}{x_4x_3^2}, \ \ \delta (x) = \frac{4PL^3}{Ex_4x_3^3}, \\ \\ & {} P_c (x) = \frac{4.103E\sqrt{\frac{x_3^2x_4^6}{36}}}{L^2} \left( 1- \frac{x_3}{2L} \sqrt{\frac{E}{4G}}\right) , \ \ P = 6000\, {\rm lb},\ \ \ L=14in, E =30\times 10^{06} \, {\rm psi}, \\ \\ & {} G= 12\times 10^{06}\, {\rm psi}. \ \ \ \tau _{\text {max}} = 136,000 \, {\rm psi}, \ \ \sigma (x) = 30,000\, {\rm psi}, \ \ \delta _{\text {max}} = 0.25\, {\rm in.} ,\\ \\ & {} 0.1 \le x_1 \le 2.0, \ \ 0.1 \le x_2 \le 10.0, \ \ 0.1 \le x_3 \le 10.0, \ \ 0.1 \le x_4 \le 2.0 \end{array}$$

1.2 1.2 Tension/compression spring design problem

The main objective of this problem is to minimize the weight of the tension/compression spring, subject to shear stress, surge frequency and minimum deflection constraints. The structural diagram of this problem is shown in Fig. 4b. The major design variables for this problem are the wire diameter $d(x_1)$, the mean coil diameter $D(x_2)$, and the number of active coils $P(x_3)$. The mathematical formulation of this problem can be described as follows:

$$\begin{aligned} \begin{array}{ll} \text {Minimize} &{} f(x)=(x_3+2) x_2 x_1^2 \\ \text {subject to} &{} g_1(x) = 1 -\frac{x_2^3x_3}{71785x_1^4}\le 0 \\ &{} g_2(x) = \frac{4x_2^2-x_1 x_2}{12566(x_1^3 x_2-x_1^4)} +\frac{1}{5108x_1^2}-1 \le 0 \\ &{} g_3(x) = 1 -\frac{140.45x_1}{x_2^2x_3}\le 0 \\ &{} g_4(x) = \frac{x_1+x_2}{1.5}-1\le 0 \\ \text {where} &{} 0.05 \le x_1 \le 2.0, \ \ 0.25 \le x_2 \le 1.3, \ \ 2 \le x_3 \le 15.0 \end{array} \end{aligned}$$

1.3 1.3 Pressure vessel design problem

The objective of this problem is to minimize the total cost, including the cost of material, forming, and welding. Figure 4c describes the structure of the pressure vessel problem in which a cylindrical vessel is capped at both ends by hemispherical heads. It is a mixed discrete-continuous constrained optimization problem with four design variables: the thickness of the pressure vessel $T_s (x_1)$, the thickness of the head $T_h(x_2)$, the inner radius of the vessel $R(x_3)$, and the length of the cylindrical component $L(x_4)$. $T_s$ and $T_h$ are the available thicknesses of rolled steel plates, which are integer multiples of 0.0625 in., and R and L are continuous variables. The problem can be mathematically formulated as follows:

$$\begin{aligned} \begin{array}{ll} \text {Minimize} &{} f(x)=0.6224x_1 x_3 x_4+1.7781x_2 x_3^2 +3.1661x_1^2 x_4+19.84x_1^2 x_3 \\ \text {subject to} &{} g_1 (x)=-x_1+0.0193x_3 \le 0 \\ &{} g_2 (x)=-x_2+0.00954x_3 \le 0 \\ &{} g_3 (x)=- \pi x_3^2 x_4- \frac{4}{3} \pi x_3^2+1296000 \le 0 \\ &{} g_4 (x)=x_4-240 \le 0 \\ \text {where} &{} 1 \times 0.0625\le x_1, x_2\le 99 \times 0.0625, \ \ \ 10\le x_3, x_4\le 200 \end{array} \end{aligned}$$

1.4 1.4 Speed reducer design problem

The design of the speed reducer problem is shown in Fig. 4d, which consists of seven decision variables namely, face width (b), module of teeth (m), number of teeth on pinion (z), length of shaft 1 between bearing $(l_1)$, length of shaft 2 between bearing $(l_2)$, diameter of shaft 1 $(d_1)$ and diameter of shaft 2 $(d_2)$. The optimization model by considering the design variables $X=(b,m,z,l_1,l_2,d_1,d_2) = (x_1,x_2$, $x_3$, $x_4$, $x_5,x_6,x_7)$, is defined as

$$\begin{aligned} \begin{array}{ll} \text {Minimize} &{} f(x)=0.7854x_1 x_2^2 (3.3333x_3^2+14.9334x_3-43.0934) -1.508x_1 (x_6^2+x_7^2 )+7.4777(x_6^3+x_7^3 )+0.7854(x_4 x_6^2+x_5 x_7^2)\\ \text {subject to} &{} g_1 (x) = \frac{27}{x_1 x_2^2 x_3}-1 \le 0; \qquad g_7(x) = \frac{x_2x_3}{40}-1 \le 0 \\ &{} g_2 (x)= \frac{397.5}{x_1 x_2^2 x_3^2}-1 \le 0 ; \qquad g_8(x) = \frac{5x_2}{x_1}-1 \le 0 \\ &{} g_3 (x)= \frac{1.93 x_4^3}{x_2 x_3 x_6^4}-1 \le 0 ; \qquad g_9(x) = \frac{x_1}{12x_2}-1 \le 0 \\ &{} g_4 (x)= \frac{1.93x_5^3}{x_2 x_3x_7^4}-1 \le 0 ; \qquad g_{10}(x) = \frac{1.56x_6+1.9}{x_4}-1 \le 0 \\ &{} g_5 (x)= \frac{\sqrt{(\frac{745x_4}{x_2x_3})^2 +16.9 \times 10^6}}{110x_6^3}-1 \le 0 ; \qquad g_{11}(x) = \frac{1.1x_7+1.9}{x_5}-1 \le 0 \\ &{} g_6 (x)= \frac{\sqrt{(\frac{745x_5}{x_2x_3})^2 +157.5 \times 10^6}}{85x_7^3}-1 \le 0 \\ \text {where} &{} 2.6 \le x_1 \le 3.6, \ \ 0.7 \le x_2 \le 0.8, \ \ 17 \le x_3 \le 28, \ \ 7.3 \le x_4 \le 8.3, 7.3 \le x_5 \le 8.3, \ \ \ 2.9 \le x_6 \le 3.9, \ \ 5.0 \le x_7 \le 5.5 \end{array} \end{aligned}$$

1.5 1.5 The three-bar truss design problem

This design optimization problem was firstly revealed by the author [42]. According to this, it is desired that three bars are placed as shown in Fig. 4e and aim is to minimize the weight of bars. The mathematical formulation of the problem, with decision variables $X=(A_1, A_2)=(x_1, x_2)$, is stated as

$$\begin{aligned} \begin{array}{ll} \text {Minimize} &{} f(x)=(2\sqrt{2} x_1+x_2 )l \\ \text {subject to} &{} g_1(x) = \frac{\sqrt{2}x_1+x_2}{\sqrt{2}x_1^2+2x_1x_2} p -\sigma \le 0 \\ &{} g_2(x) = \frac{x_2}{\sqrt{2}x_1^2+2x_1x_2} p -\sigma \le 0 \\ &{} g_3(x) = \frac{1}{\sqrt{2}x_2+x_1} p -\sigma \le 0 \\ \text {where} &{} 0 \le x_1 \le 1,\ \ \ 0 \le x_2 \le 1 \end{array} \end{aligned}$$

where $l =100 \, {\text {cm}}$, ${P} = 2\, {\text {KN}}/{\text {cm}}^{2}$, $\sigma = 2\, {\text {KN}}/{\text {cm}}^2$