# Smart Topology Optimization Using Adaptive Neighborhood Simulated Annealing
## 5 Results
### Cantilever Problem  
To verify the proposed method, the classic problem of TO for the cantilever and MBB beam was selected. The cantilever beam is a design domain fixed on one end and a force applied on the other end as shown in Figure 1.  
The TO was applied to the cantilever beam with \(60\times 30\) elements with a constraint on the volume fraction. The number of elements to discrete the design domain should be high enough to give a good resolution of the final design. In this case, the number of elements was selected based on the similar cases in literature [2]. In other cases, this could be selected simply by considering the minimum required resolution of the final design which depends on the complexity of the expected final design. In order to determine the proper value for \(m\), the optimization process for the cantilever beam problem started with \(m=10\). This is a very small value, the motivation was to find a value smaller than \(60\times 30\times 40\) = 72,000, where \(30\times 60\) is the number of variables and 40 is the commonly used multiplier. Figure 2 shows the plot of the SA converged objective function value for the optimum solution of a 0.5-volume fraction versus the value of \(m\). As shown in Figure 2, the objective function (compliance) starts to converge to the optimum when \(m=100\). By increasing the value of \(m\) until 1000, the converged objective  
Figure 1: Schematic of design domain and loading condition for cantilever beam. Figure 2: SA converged objective function value versus value of \(m\) for the cantilever beam problem with volume fraction 0.5 (\(m\) is the number of objective function evaluations to reach thermal equilibrium in a specific temperature). function value will not have any considerable changes and it is already very close to the optimum solution. Similarly, for other volume fractions, the convergence has the same behavior. Since the computational cost for this problem is not too high, \(m=1000\) was used in the simulations. In more complicated problems, a reasonable value of \(m\) can be determined using a similar approach. The frozen state is defined as the temperature where no new candidate is accepted. It is necessary to determine the initial temperature value. In an attempt to stop the SA before reaching the frozen state and to define the initial temperature, the following study was executed. The number of rejected and accepted new candidates per temperature were computed. It resulted in the graph shown in Figure 3. In the SA, the number of accepted solutions must be as high as possible in the exploration phase. From Figure 3, at a temperature near 100, SA has an acceptance ratio of about 60%. The initial temperature \(T_{0}=100\) is reasonable, and allows the SA to have enough domain exploration. The final temperature was selected as \(T_{end}=0.001\) in which the number of accepted solutions is already small. The real frozen state cannot be reached, because of the checker board problem. The problem of minimizing compliance in the cantilever beam of Figure 1 is solved with the gradient-based method from the literature [6] and the method proposed herein. The results are presented in Table 1 for different volume fractions. The parameters used for the SA in the proposed method are \(T_{0}=100\) and \(T_{end}=0.001\) for initial and final temperatures, \(m=1000\) to reach the thermal equilibrium, and \(\alpha=0.85\) for the cooling factor. The compliance of optimized cantilever beam using the proposed method agrees very well with the values from the literature. Figure 4 shows the optimized topology of cantilever beam for different volume fractions. The shapes from the method proposed herein are similar to the results from the literature. \begin{table}
\begin{tabular}{c c c} \hline
**Volume Fraction** & **Compliance from [6]** & **Compliance from the Method Proposed Herein** \\ \hline
0.5 & 70.41 & 73.71 \\
0.6 & 60.43 & 61.42 \\
0.7 & 53.95 & 54.97 \\
0.8 & 49.68 & 50.21 \\
0.9 & 46.84 & 47.10 \\ \hline \end{tabular}
\end{table}