## 3 The Proposed Approach
### Numerical Examples  
The experiments performed considered different scenarios regarding the availability of the failure data of systems, including situations where the number of failures is small, by grouping similar subsystems data that enable suitable adjustments of theoretical models. For the first scenario, the example from [PERSON] et al. [21] was used to provide maintenance data for a central cooler in a nuclear power plant. The example considers five repairable systems observed for 600 days, preventive maintenance carried out at times 150, 250, 500, and 600 (days), with severities equal to 40%, 50%, 90%, and 20%, respectively (simultaneous maintenance in all five systems). A total of 93 failures were observed in this example, randomly distributed among the five systems, as graphically illustrated in Figure 3.  
In this first case example, preventive maintenance costs were defined in two scenarios: (a) varying the PM cost as a function of intervention level (Table 1); (b) fixed preventive maintenance cost of USD 4000, regardless of intervention level. The corresponding results of this example allow one to verify that the optimization will return a result close to the expected. As for a fixed cost, the optimal known solution is to perform maintenance as comprehensively as possible, which results in the best improvement factor in the system condition. In the second case example, a scenario with three identical components subject to independent maintenance actions was considered. In this case, the intervention levels were fixed at 20%, 70%, and 100% with the respective preventive maintenance costs, according to Table 2. The corrective maintenance cost was set to be USD 15,000, with a planning horizon of 24 months (730 days). The failure data of the three components are presented in Table 3. \begin{table}
\begin{tabular}{c c} \hline
**Intervention Level** & **MP Cost (USD)** \\ \hline
20\% & 2000 \\
40\% & 4000 \\
50\% & 9000 \\
90\% & 12,000 \\ \hline \end{tabular}
\end{table} 
Table 1: Preventive maintenance cost as a function of intervention level for first case example, scenario (a). Scenario (b) considers a constant preventive maintenance cost, equal to USD 4000. \begin{table}
\begin{tabular}{c c} \hline
**Intervention Level** & **PM Cost (USD)** \\ \hline
20\% & 500 \\
70\% & 2000 \\
100\% & 3000 \\ \hline \end{tabular}
\end{table} 
Table 2: PM costs for the second case example. Figure 3: Graphic representation of first case example data, which considers five similar systems subject to synchronized preventive maintenance actions. The markers in the form of an x on the horizontal lines indicate the times of the failures in each system. In the third case example, a scenario with two components and independent maintenance actions was considered. In this case, the first of the two observed systems did not have failures, and the intervention levels were always 100%. Thus, both the corrective maintenance cost and preventive maintenance cost were considered constant and defined as USD 17,000 and USD 4000, respectively. The planning horizon of the maintenance policy was 48 months (1460 days). The failure data of this example are presented in Table 4.  
For the fourth case example, four identical components received preventive maintenance with an intervention level of 50% or 100%, with respective costs of USD 2000 and USD 3000. The corrective maintenance cost was USD 15,000, with a planning horizon of 24 months (730 days). The data of this case are presented in Table 5.