## 5 Case Study
 \tag{17}\]  
where \(m_{0c1}=0\), the reason is that the spare parts of the steering control motor are unavailable; all other coefficients are large integers, which means the related resources are enough for the requirements of the corresponding maintenance tasks. \begin{table}
\begin{tabular}{c c c c} \hline \hline
**Transition** & **Maintenance Task** & **Distribution** & **Parameter (Minutes)** \\ \hline \(t_{1}\) & Hoist the aircraft & Lognormal & \(\mu=1.28\), \(\sigma=1.16\) \\ \(t_{2}\) & Fault diagnosis & Lognormal & \(\mu=0.88\), \(\sigma=1.14\) \\ \(t_{3}\) & Not applicable & Not applicable & Not applicable \\ \(t_{4}\) & Get the spare part of the steering control motor & Exponential & \(\theta=5.53\) \\ \(t_{5}\) & Dismantle hoses and cables & Lognormal & \(\mu=1.12\), \(\sigma=0.29\) \\ \(t_{6}\) & Dismantle connectors (Nuts and screws) & Normal & \(\mu=0.95\), \(\sigma=0.06\) \\ \(t_{7}\) & Replace the steering control motor & Exponential & \(\theta=1.2\) \\ \(t_{8}\) & Install the connectors (Nuts and screws) & Normal & \(\mu=1.35\), \(\sigma=0.06\) \\ \(t_{9}\) & Install hoses and cables & Lognormal & \(\mu=1.43\), \(\sigma=0.24\) \\ \(t_{10}\) & Not applicable & Not applicable & Not applicable \\ \(t_{11}\) & Dismantle the torque link & Normal & \(\mu=0.9\), \(\sigma=0.08\) \\ \(t_{12}\) & Install the new torque link & Normal & \(\mu=1.22\), \(\sigma=0.06\) \\ \(t_{13}\) & Adjustment & Lognormal & \(\mu=0.91\), \(\sigma=0.29\) \\ \hline \hline \end{tabular}
\end{table} 
Table 1: Detailed information of transitions. The ultimate marking of the maintenance process based on SCPN can be expressed as  
\[ \begin{split}& M_{u}(s_{si})=\left\{\begin{array}{cc}c_{s}&i=14\\ 0&i\
eq 14\end{array}\right.\\ & M_{u}(s_{c})=m_{uc1}c_{c1}+m_{uc2}c_{c2}+m_{uc3}c_{c3}+m_{uc4}c_{c4}\\ & M_{u}(s_{r})=m_{ur1}c_{r1}+m_{ur2}c_{r2}+m_{ur3}c_{r3}+m_{ur4}c_{r4}+m_{ur5}c_ {r5}+m_{ur6}c_{r6}\end{split}  \tag{18}\]  
where the coefficients \(m_{uc1}\sim m_{uc4}\) and \(m_{ur1}\sim m_{ur6}\) are unknown, because we do not know what resources may be left after the maintenance process. According to Equations (12) and (16)-(18), we can get two solutions of \(\mathbf{T}_{F}\)  
\[ \begin{split}&\mathbf{T}_{F1}=\{t_{1},t_{2},t_{3},t_{4},t_{5},t_{6},t_{7 },t_{8},t_{9},t_{13}\}\\ &\mathbf{T}_{F2}=\{t_{1},t_{2},t_{10},t_{11},t_{12},t_{13}\}\end{split}  \tag{19}\]  
When transitions in \(\mathbf{T}_{F1}\) are fired, it means the maintenance event expressing the replacement of the steering control motor is conducted. According to Equations (14) and (15), we can get the required quantities of resources, which are one steering control motor, eight washers, four cotter pins, two maintenance staffs, two spanners, two pliers, two screwdrivers, one lifting jack and one diagnostic equipment. When transitions in \(\mathbf{T}_{F2}\) are fired, it means the maintenance event expressing the replacement of the torque link is conducted. In the same way, we determine that one toque link, four washers, two cotter pins, two maintenance staffs, one spanner, one plier, one screwdriver, one lifting jack and one diagnostic equipment are required for this maintenance event. We get 1000 maintenance time samples in terms of the discrete-event simulation proposed in Section 4.2. According to these samples, the mean of the maintenance time is 22.28 min, the standard deviation of the maintenance time is 7.56 min and 89.2% of the maintenance time samples are less than half an hour, which is the up limit of the line maintenance for civil aircraft. The elapsed time of our discrete-event simulation procedure running in MATLAB is less than one second. The computer used was Intel (TM) i5-6500 four Core Processor 3.2 GHz, 8 GB of RAM. We have also developed a GRASP based method [1] to obtain the maintenance time samples. In terms of these samples, the mean of the maintenance time is 21.6 min, the standard deviation of the maintenance time is 7.39 min and 90.5% of the maintenance time samples are less than half an hour.