# A Network-Based Method for the EMU Train High-Level Maintenance Planning Problem
## 5 Empirical Case Study and Sensitivity Analysis
### Empirical Case Study  
Shanghai railway owns 91 CRH1 series and CRH380D EMU train-sets, which including 24 sets of CRH1B (16-cars), 19 sets of CRH1E (8-cars) and 48 sets of CRH380D (8-cars). Among these trains, 11 sets of CRH1B, eight sets of CRH1E, and 41 sets of CRH380D needed maintaining during the planning horizon. The start date of the planning horizon is 18 August 2016, denoted by number 1, and the end date is 1 February 2018, so the length of the planning horizon is 533 days, i.e., \(T=533\). The maintenance regulations for EMU train-sets are listed in Table 1. Table 4 shows the initial condition of train-sets needed maintaining at the beginning of the planning horizon, including the ID, average daily operating mileage, type, initial cumulative mileage, maintenance level, maintenance level service time, the interval for delivering train-sets to the workshop, and the ETA and the deadline for HM. The eta of maintenance is calculated by the formula as follows:  
\[\text{eta}_{e}=[(L_{e}^{ideal}-l_{e}^{0})/\overline{l_{e}}]+1 \tag{19}\]  
Similarly,  
\[\text{deadline}_{e}=[(L_{e}^{upper}-l_{e}^{0})/\overline{l_{e}}]+1 \tag{20}\]In order to calculate conveniently in this empirical case, a simple formula should be followed:  
\[L_{e}^{min}=L_{e}^{upper}-p\overline{l_{e}} \tag{21}\]  
where \(p\) is the modified parameter for guaranteeing the feasible solution. According to Table 1, we set \(p=80\) (see Section 5.2), approximately, after multiple trials. If \(L_{e}^{min}<0\), then \(L_{e}^{min}=0\). Moreover, if many of train-set needed maintenance at the same time, we can increase the value of \(p\) to enlarge the solution space to disperse the start date of HM under the limited maintenance capacity. During the planning horizon, there exists four long travel rushes, including two spring festival, one spring season and one summer holiday. All maintenance work should be finished at the 532 th day. According to practice, \(b_{1}(a_{1,k})\) is valued as follows:  
\[b_{1}(a_{1,k})=\left\{ \begin{array}{l}12\;if\;k\in[149,189)\\ 107\;if\;k\in[189,318)\\ 108\;if\;k\in[318,380)\\ 115\;if\;k=533\\ 105\;otherwise\end{array} \right. \tag{22}\]  
The unit of measurement of \(b_{1}(a_{1,k})\) and \(b_{i}\) (\(i=2,\,3,\,4\)) is a standard set (a train-set includes eight cars), and the 16-car train-set is equal to two standard sets. All of the train-sets undergo third-level maintenance at the depot, and undergo fourth- and fifth-level maintenance at the plant. The maximal number of standard sets that can be maintained simultaneously at the depot for the third-level is seven, i.e., \(b_{2}=6\). The maintenance capacity of the plants is sufficient. We set the simultaneous maintenance capacity for fourth-level and fifth-level as large positive numbers, e.g., \(b_{3}+b_{4}=20\). Meanwhile, the receiving capability per day for depot and plant are both limited, e.g., \(N_{2}=1\) and \(N_{3}+N_{4}=2\). The unit of measurement of \(N_{i}\) (\(i=2,\,3,\,4\)) is the natural set. \begin{table}
\begin{tabular}{c c c c c c c c} \hline
**e** & \(\overline{l_{e}}\) (**km**) & \(l_{e}^{0}\) (**km**) & **g** & \(\mu\) (d)** & \(Interval\) (d) & \(eta_{e}\) & \(deadline_{e}\) \\ \hline  
\hline \end{tabular}
\end{table} 
Table 4: Initial conditions of all train-sets. Based on the data given above, model I and model II were solved, respectively, and both of them obtained the optimal solution. The termination rules for two models were set up with a solver relative gap of 0 in Spyder. One of the detailed results are shown in Table 5, the first column is the identifier (ID) of the train-set; the second column is the start date of HM denoted by \(d_{e}\), i.e., the date of delivering train-set to the workshop. \begin{table}
\begin{tabular}{c c c c c c c c} \hline
**e** & \(\overline{I_{e}}\) **(km)** & \(\overline{I_{e}^{0}}\) **(km)** & **g** & \(\mu\) **(d)** & _Interval_ **(d)** & \(eta_{e}\) & _deadline_\({}_{e}\) \\ \hline  
\hline \end{tabular}
\end{table}