## 3 Research Methods
### Differential Evolution Algorithm (DE)
The best solutions among the top 50 percent are \(X_{2}^{1}\), \(X_{3}^{1}\), \(X_{6}^{1}\), \(X_{7}^{1}\) found from \[pX_{i}^{G}:\]5 Because it is the first iteration, the local optimal solution is equivalent to the solution itself, i.e., \(pX_{i}^{G}=X_{i}^{G}\). Then, the mutation solution \(V_{i}^{1}\) of this solution is calculated according to Equation (36). Taking \(X_{7}^{1}\) as an example, assuming that the three random integers \(r_{1}\), \(r_{2}\), \(r_{3}\) are 1, 3, and 8 respectively and mutation factor F = 0.5, then \(v_{7,1,1}^{1}=36.89+0.5\times(29.92-35.34)=34.18\) and the rest of the solutions for variables \(j\) are analogous. The results are shown in \[pX_{i}^{G}:\]6
3. After obtaining all the mutation solutions in Table 4, mate a new solution \(U_{i}^{1}\) according to Equation (37). Take \(U_{2}^{1}\) as an example. Let the evolution parameter \(CR=0.75\), then generate random numbers \(rand(t,j)\) and \(rnb(i,t)\). Assuming \(rand(1,1)=0.85\), \(rnb(2,1)=1\), although \(rand(1)=0.85>CR\), the new solution variable \(u_{2,1,1}^{1}\) must accept the mutation solution \(v_{2,1,1}^{1}\) in the first solution variable of time period 1 because \(rnb(2)=1\). When reaching the second solution variable of time period 1, \(rand(1,2)=0.52\) and \(rnb(2,1)=1\), the new solution variable \(u_{2,1,2}^{1}=v_{2,1,2}^{1}\) at this time because \(rand(1,2)=0.52<CR\). Additionally, for the first solution variable of time period 2, \(rand(2,1)=0.83\) and \(rnb(2,2)=2\), the new solution variable is its own solution, which means \(u_{2,2,1}^{1}=x_{2,2,1}^{1}\), because \(rand(2,1)=0.83>CR\) and \(rnb(2,2)=2\
eq 1\). If the last solution variable \(rand(2,2)=0.29\) and \(rnb(2,2)=2\), then \(u_{2,2,2}^{1}=v_{2,2,2}^{1}\). After the solution variables of solutions of this set are all mated, enter the solutions of the next set \(U_{3}^{1}\), and randomly generate \(rand(t,j)\) and \(rnb(i,t)\). The results are shown in \[pX_{i}^{G}:\]7
4. Compare the pros and cons of \(F(U_{i}^{1})\) and \(F(X_{i}^{1})\) to determine the retained solution \(X_{i}^{1}\) of this generation according to Equation (38). If \(F(U_{i}^{1})<F(X_{i}^{1})\), then the solution of this generation is \(X_{i}^{1}=U_{i}^{1}\); if \(F(U_{i}^{1})>F(X_{i}^{1})\), and the solution of this generation remains unchanged \(X_{i}^{1}=X_{i}^{1}\). Then, combine this result with the other 50% of the solutions that did not enter the local search. The results are shown in \[pX_{i}^{G}:\]8 Finally, compare it with the best solution in the global region as shown in Tables 7 and 8. Then, enter the update of the next generation starting algorithm, i.e., the local search of the DE algorithm is completed. \begin{table}
\begin{tabular}{c c c c c c} \hline \hline  & \(x^{1}_{i,1,1}\) & \(x^{1}_{i,2}\) & \(x^{1}_{i,2,1}\) & \(x^{1}_{i,2,2}\) & \(F(\mathbf{X^{1}_{i}})\) \\ \hline \(X^{2}_{1}\) & 36.89 & 24.32 & 15.88 & 45.61 & 168.59 \\ \hline \(X^{2}_{2}\) & 32.09 & 23.13 & 12.36 & 47.46 & 131.69 \\ \hline \(X^{2}_{2}\) & 30.39 & 30.29 & 10.34 & 51.26 & 145.01 \\ \hline \(X^{2}_{4}\) & 39.52 & 23.24 & 17.61 & 47.57 & 159.38 \\ \hline \(X^{2}_{5}\) & 41.69 & 18.97 & 21.66 & 58.35 & 181.67 \\ \hline \(X^{2}_{6}\) & **28.18** & **22.68** & **23.12** & **52.83** & **153.26** \\ \hline \(X^{2}_{7}\) & 33.41 & 24.98 & 14.95 & 43.34 & 142.24 \\ \hline \(X^{2}_{8}\) & 35.34 & 19.55 & 19.53 & 55.98 & 171.83 \\ \hline \hline \end{tabular}
\end{table} 
Table 6: Updating the solutions of this generation \(X^{1}_{i}\).