## 5 Application to Manufacturing System Design Problems
### Profit Maximization Problem  
The second problem is a profit maximization problem (PMP) in which total profit is defined as the sum of the profit from the production rate, the cost savings from the reconfiguration of workstations, and reconstruction costs. In this problem both buffer allocation and reconfiguration strategy are determined at the same time.  
\[\begin{split}&\text{Max }TP(\mathbf{Z},\ y_{1},\ y_{2},\ y_{3})\\ &=d_{1}\sum\limits_{l=1}^{3}\{y_{l}\times g_{l}(z_{  
we assumed that four robots can be uninstalled and reused when a station is eliminated. Thus, we set \(d_{3}=0.1\times 4\frac{\text{robots}}{\text{station}}\times\frac{\text{$550,000$}}{ \text{robot}}=\frac{\text{$520,000$}}{\text{station}}\), where 0.1 is the depreciation rate of robots. Finally, \(d_{4}\) is set to \(\text{$55000$}\) per station.  
With these coefficient values, we solved this problem easily using a genetic algorithm supplied from MATLAB(r), because this problem is a nonlinear integer problem and difficult to solve using CPLEX(r). Figure 6 compares the total profits with respect to the reconfiguration strategies after optimal buffer allocations when the product-mix (\(z_{11}\)) is set to 30%. It means that strategy 3 is better than strategies 1 and 2, and it means that \(y_{3}=1\) when the PMP is solved. However, it is difficult to say that strategy 3 is better than strategies 1 or 2 with respect to statistical principles when TB has a large value, because the gaps among strategies are smaller than the confidence intervals. Nevertheless, strategy 3 must be selected if we have to choose only one, and there are any other preferences. Table 11 shows the examples of total profits, production rates obtained by meta-models, and the confidence intervals obtained from simulation experiments when same optimized values of variables are used for simulation models. The cross point between the strategies 1 and 2 (TB = 142) becomes slightly smaller than that of the BAP problem in Section 5.1 (TB = 144). This phenomenon is due to the fact that the profit caused by the production rate is much bigger than the profit obtained from the reconfiguration.  
Other observation is that total profit increases as the increase of TB within the range considered. If the TB increases much more, the total profit will decreases because production rate will not increase anymore, but the investment cost increases as the increase of buffer.  
\begin{table}
\begin{tabular}{c c c c c c c c c c} \hline \hline  & \multicolumn{3}{c}{**Strategy 1**} & \multicolumn{3}{c}{**Strategy 2**} & \multicolumn{3}{c}{**Strategy 3**} \\ \cline{2-10}  & \multicolumn{3}{c}{**Using Meta-Model**} & \multicolumn{3}{c}{**Using Meta-Model**} & \multicolumn{3}{c}{**Simulation**} & \multicolumn{3}{c}{**Using Meta-Model**} & \multicolumn{3}{c}{**Simulation**} \\ \cline{2-10}
**TB** & \multicolumn{3}{c}{**(Optimized)**} & \multicolumn{1}{c}{**Simulation**} & \multicolumn{3}{c}{**(Optimized)**} & \multicolumn{1}{c}{**Simulation**} & \multicolumn{3}{c}{**(Optimized)**} & \multicolumn{1}{c}{**Simulation**} \\ \cline{2-10}  & \multicolumn{3}{c}{**TP**} & \multicolumn{1}{c}{**PR\({}_{1}\)**} & \multicolumn{1}{c}{**95\% C.L.**} & \multicolumn{1}{c}{**TP**} & \multicolumn{1}{c}{**PR\({}_{2}\)**} & \multicolumn{1}{c}{**95\% C.L. of**} & \multicolumn{1}{c}{**TP**} & \multicolumn{1}{c}{**PR\({}_{3}\)**} & \multicolumn{1}{c}{**95\% C.L.**} \\  & \multicolumn{3}{c}{**(\(\times 10^{8}\))**} & \multicolumn{1}{c}{**PR\({}_{1}\)**} & \multicolumn{1}{c}{**(\(\times 10^{8}\))**} & \multicolumn{1}{c}{**PR\({}_{2}\)**} & \multicolumn{1}{c}{**(\(\times 10^{8}\))**} & \multicolumn{1}{c}{**PR\({}_{3}\)**} & \multicolumn{1}{c}{**95\% C.L.**} \\ \hline
70 & 1.2789 & 0.4263 & 0.0013 & 1.2634 & 0.4205 & 0.0017 & 1.2907 & 0.4298 & 0.0020 \\
90 & 1.3665 & 0.4555 & 0.0019 & 1.3529 & 0.4504 & 0.0009 & 1.3791 & 0.4592 & 0.0013 \\
110 & 1.4287 & 0.4762 & 0.0016 & 1.4195 & 0.4725 & 0.0014 & 1.4422 & 0.4803 & 0.0016 \\
130 & 1.4751 & 0.4917 & 0.0027 & 1.4719 & 0.4900 & 0.0015 & 1.4891 & 0