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Binary Artificial Electric Field Algorithm

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Abstract

Artificial Electric Field Algorithm (AEFA) is one of the recent population-based optimization techniques, it is inspired by the electrostatic force theory. This article aims to design a novel binary version of the AEFA scheme to improve the performance of the original AEFA scheme in discrete space. The popular S-shaped and V-shaped functions are used to design the binary versions of the AEFA. The efficiency and the optimization ability of the proposed binary versions of the AEFA are studied theoretically as well as experimentally. An extensive experimental study is performed to understand the performance of the proposed schemes. A set of 24 benchmark problems are solved using binary versions of AEFA the experimental results are compared with nine state-of-the-art algorithms. The running time complexity and Wilcoxon’s signed-rank statistical test are also conducted to judge the proposed algorithms. In addition to the experimental studies, theoretical analysis is also carried out which suggests the convergence scenario of the proposed schemes. These studies suggest that the designed binary versions of the AEFA are very efficient and competent in addressing difficult optimization problems.

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Acknowledgements

The authors are thankful to Dr B. R. Ambedkar National Institute of Technology Jalandhar for the necessary support to this research. The first author is thankful to the Ministry of Education Government of India for providing financial support to carry out this work.

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Authors and Affiliations

  1. Department of Mathematics, Dr. B.R. Ambedkar National Institute of Technology Jalandhar, Jalandhar, Punjab, 144011, India

    Dikshit Chauhan & Anupam Yadav

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  1. Dikshit Chauhan
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  2. Anupam Yadav
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Correspondence to Anupam Yadav.

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A Appendix

A Appendix

1.1 Illustration of binary versions of the AEFA

To illustrate implementing the procedure of binary versions of the AEFA, an illustrative example is presented for the sphere function. The mathematical expression of sphere function is mentioned in Eq. (41).

$$\begin{aligned} F(X)=\sum _{i=1}^{D}X^{2} \end{aligned}$$
(41)

For the sake of convenience and easy to understand the population size \(N=3\) and dimension \(D=8\) are considered. \(vel_{i}^{d}\) and \(X_{i}^{d}\), \(i=1,2,3; d=1,2,3,4,5,6,7,8\) are representing the velocity and position of the candidate solution.

  1. Iteration.1.

    Initialization of position and velocity The initial position of candidate solution are randomly initialized between \([-100,100]\)

    $$\begin{aligned}&X_{1}=[65.4169, -15.9417, 12.3398, -72.7920, 29.2953, \nonumber \\&86.7142, 98.6398, 18.4448],\quad F(X_{1})= 2.8432e+04\nonumber \\&X_{2}=[61.2647, 67.7288, -4.6856, -15.3621, -34.6034, \nonumber \\&6.1326, 38.8467, 1.3461],\quad F(X_{2})=1.1344e+04\nonumber \\&X_{3}=[ 76.2411, -82.2269, 48.7849, 9.1527, 20.5085,\nonumber \\&70.3099, 96.2467, 23.2951],\quad F(X_{3})= 3.0208e+04 \end{aligned}$$

    The initialization of velocity is randomly chosen

    $$\begin{aligned} vel_{1}&=[ 0.0155, 0.2967, -0.0296, 0.6658, \nonumber \\&-0.0948, -0.0644, 0.4611, 0.3427]\nonumber \\ vel_{2}&=[ 0.2253, 0.6273, 0.0912, 0.0932, \nonumber \\&0.1454, 0.2951, 0.5124, 0.1867]\nonumber \\ vel_{3}&=[ 0.1771, 0.6126, -0.1322, -0.1151, \nonumber \\&-0.1293, -0.0313, 0.1487, -0.0340]\nonumber \end{aligned}$$

    Calculate the probability values by using transfer function for \(vel_{1}\), \(vel_{2}\) and \(vel_{3}\), we get

    $$\begin{aligned} S_{1}=[0.4961,0.4439,0.4558] \end{aligned}$$
    (42)

    The fitness values of all agents \(X_{1}\), \(X_{2}\) and \(X_{3}\) are 2.8432e+04, 1.1344e+04 and 3.0208e+04 for iteration one. The best fitness value for the first iteration is 1.1344e+04 that is the best value of 2nd agent. In the next iteration, the values of position switch 0 or 1 according to the random variable that lies within [0, 1] and the probability value of \(S_{1}\). If the value of the random variable is lesser than or equal to the \(S_{1}\) value then position changes their values by 1 otherwise it changes their values by 0.

  2. Iteration.2.
    $$\begin{aligned} X_{1}&=[1,1,0,0,0,0,0,0],\quad F(X_{1})= 2\nonumber \\ X_{2}&=[1,0,0,0,0,0,0,0],\quad F(X_{2})=1\nonumber \\ X_{3}&=[0,1,0,0,0,0,0,0],\quad F(X_{3})= 1\nonumber \end{aligned}$$
    $$\begin{aligned} vel_{1}&=[ -0.0001, 0.1115, -0.0223, 0.2375, \nonumber \\&\quad -0.0588, -0.0230, 0.0397, 0.1105]\nonumber \\ vel_{2}&=[ 0.1610, 0.2564, 0.0356, 0.0784, \nonumber \\&\quad 0.0915, 0.0627, 0.0169, 0.1238]\nonumber \\ vel_{3}&=[0.1078, 0.2033, -0.0998, -0.0720, \nonumber \\&\quad -0.0254 , -0.0011 , 0.1399, -0.0047 ]\nonumber \\\ S_{1}&=[0.5000,0.4598,0.4730] \end{aligned}$$
    (43)

    The fitness values of all agents \(X_{1}\), \(X_{2}\) and \(X_{3}\) are 2, 1 and 1 for iteration two. The best fitness value for the second iteration is 1 that is the best value of 2nd and 3rd agents.

  3. Iteration.3.
    $$\begin{aligned} X_{1}&=[1,0,0,1,0,0,0,0],\quad F(X_{1})= 2\nonumber \\ X_{2}&=[1,0,0,0,0,0,0,0],\quad F(X_{2})=1\nonumber \\ X_{3}&=[0,0,0,0,0,0,0,0],\quad F(X_{3})= 0\nonumber \\ vel_{1}&=[ -0.0001, 0.0510, -0.0140, 0.1306, \nonumber \\&\quad -0.0467, -0.0053, 0.0132, 0.0180]\nonumber \\ vel_{2}&=[ 0.1107, 0.1418, 0.0268, 0.0563, \nonumber \\&\quad 0.0560, 0.0300, 0.0101, 0.0913]\nonumber \\ vel_{3}&=[ 0.0229, 0.0500 , -0.0037 , -0.0227 ,\nonumber \\&\quad -0.0089 , -0.0002 , 0.0651 , -0.0030 ]\nonumber \\ S_{1}&=[0.5000,0.4723,0.4942] \end{aligned}$$
    (44)

    The fitness values of all agents \(X_{1}\), \(X_{2}\) and \(X_{3}\) are 2, 1 and 0 for iteration three. The best fitness value for the third iteration is 0 that is the best value of 3rd agent.

  4. Iteration.4.
    $$\begin{aligned} X_{1}&=[0,1,0,0,0,0,0,0],\quad F(X_{1})= 1\nonumber \\ X_{2}&=[0,0,0,0,0,0,0,0],\quad F(X_{2})=0\nonumber \\ X_{3}&=[0,0,0,0,0,0,0,0],\quad F(X_{3})= 0\nonumber \\ vel_{1}&=[ -0.0001, 0.0478, -0.0112, 0.0557, \nonumber \\&\quad -0.0296, -0.0023, 0.0009, 0.0051]\nonumber \\ vel_{2}&=[0.0554, 0.0107, 0.0133, 0.0032, \nonumber \\&\quad 0.0393, 0.0087, 0.0023, 0.0565]\nonumber \\ vel_{3}&=[0.0192, 0.0204, -0.0011, -0.0208, \nonumber \\&\quad -0.0067, -0.0002, 0.0509, -0.0012]\nonumber \\ S_{1}&=[0.5000,0.4861,0.4952] \end{aligned}$$
    (45)

    The fitness values of all agents \(X_{1}\), \(X_{2}\) and \(X_{3}\) are 1, 0 and 0 for iteration four. The best fitness value for the fourth iteration is 0 that is the best value of 2nd and 3rd agents.

  5. Iteration.5.
    $$\begin{aligned} X_{1}&=[1,0,0,0,0,0,0,0],\quad F(X_{1})= 1\nonumber \\ X_{2}&=[0,0,0,0,0,0,0,0],\quad F(X_{2})=0\nonumber \\ X_{3}&=[0,0,0,0,0,0,0,0],\quad F(X_{3})= 0\nonumber \\ vel_{1}&=[ -0.0000, 0.0033, -0.0107, 0.0095,\nonumber \\&\quad -0.0294, -0.0015, 0.0005, 0.0018]\nonumber \\ vel_{2}&=[ 0.0433, 0.0069, 0.0123, 0.0008, \nonumber \\&\quad 0.0366, 0.0085, 0.0015, 0.0249]\nonumber \\ vel_{3}&=[0.0047, 0.0075, -0.0003, -0.0047, \nonumber \\&\quad -0.0006, -0.0001, 0.0390, -0.0000]\nonumber \\ S_{1}&=[0.5000,0.4891,0.9882] \end{aligned}$$
    (46)

    The fitness values of all agents \(X_{1}\), \(X_{2}\) and \(X_{3}\) are 1, 0 and 0 for iteration five. The best fitness value for the fifth iteration is 0 that is the best value of 2nd and 3rd agents.

  6. Iteration.6.
    $$\begin{aligned} X_{1}&=[0,0,0,0,0,0,0,0],\quad F(X_{1})= 0\nonumber \\ X_{2}&=[0,0,0,0,0,0,0,0],\quad F(X_{2})=0\nonumber \\ X_{3}&=[0,0,0,0,0,0,0,0],\quad F(X_{3})= 0\nonumber \end{aligned}$$

    The fitness values of all agents \(X_{1}\), \(X_{2}\) and \(X_{3}\) are 0, 0 and 0 for iteration six. The best fitness value for the sixth iteration is 1 that is the best value of all three agents. Since all agent has a fitness value 0 so no improvement is needed.

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Chauhan, D., Yadav, A. Binary Artificial Electric Field Algorithm. Evol. Intel. 16, 1155–1183 (2023). https://doi.org/10.1007/s12065-022-00726-x

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