## 4 Case Study
### Element Maintenance Decision Model
#### 4.2.6 Definition of the Initial Strategy  
Using the policy iteration method obtained from the dynamic programming algorithm, we solved the sequential decision making problem. First, we defined the maintenance strategy, denoted as strategy \(\pi(a|s)\), in this study. Strategy A is defined as the proportion  
\begin{table}
\begin{tabular}{c c c c c c c c} \hline \hline
**Element** & \multicolumn{6}{c}{**Proportion Of Each State**} \\
**Age** & **1** & **2** & **3** & **4** & **5** & **1–2** & **1–3** & **1–4** \\ \hline
0 & 100.00\% & 0.00\% & 0.00\% & 0.00\% & 0.00\% & 100.00\% & 100.00\% \\
1 & 66.67\% & 33.33\% & 0.00\% & 0.00\% & 0.00\% & 100.00\% & 100.00\% \\
2 & 36.84\% & 28.77\% & 34.39\% & 0.00\% & 0.00\% & 65.61\% & 100.00\% & 100.00\% \\
3 & 29.82\% & 22.11\% & 16.84\% & 31.23\% & 0.00\% & 51.93\% & 68.77\% & 100.00\% \\
4 & 20.00\% & 7.37\% & 31.93\% & 35.79\% & 4.91\% & 27.37\% & 59.30\% & 95.09\% \\
5 & 19.30\% & 10.18\% & 28.42\% & 36.49\% & 5.61\% & 29.47\% & 57.89\% & 94.39\% \\
6 & 18.60\% & 4.91\% & 30.18\% & 39.30\% & 7.02\% & 23.51\% & 53.68\% & 92.98\% \\
7 & 18.60\% & 10.18\% & 20.00\% & 43.51\% & 7.72\% & 28.77\% & 48.77\% & 92.28\% \\
8 & 18.25\% & 2.81\% & 24.56\% & 46.32\% & 8.07\% & 21.05\% & 45.61\% & 91.93\% \\
9 & 17.54\% & 4.21\% & 20.35\% & 47.72\% & 10.18\% & 21.75\% & 42.11\% & 89.82\% \\
10 & 14.39\% & 2.81\% & 18.25\% & 52.28\% & 12.28\% & 17.19\% & 35.44\% & 87.72\% \\
11 & 13.68\% & 1.75\% & 19.65\% & 50.18\% & 14.74\% & 15.44\% & 35.09\% & 85.26\% \\ \hline \hline \end{tabular} Note: States 1–2 refers to State 1 and State 2; States 1–3 refers to State 1, State 2, and State 3; and States 1–4 refers to State 1, State 2, State 3, and State 4.  
\end{table}
Table 6: The proportion of box girder members in each state.  
\begin{table}
\begin{tabular}{c c c} \hline \hline
**State** & **Weibull Distribution Parameters** \\  & **a\({}_{1}\)** & **b\({}_{1}\)** \\ \hline State 1 & 0.3870 & 0.5745 \\ States 1–2 & 0.2091 & 0.999 \\ States 1–3 & 0.1040 & 1.1127 \\ States 1–4 & 0.0261 & 1.5009 \\ \hline \hline \end{tabular}
\end{table}
Table 7: Weber distribution parameters.  
of elements in each state that will be maintained using four maintenance actions every year in the future.  
\[\pi(a\mid s)=P[A_{t}=a\mid S_{t}=s]=\begin{bmatrix}a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24}\\ a_{31}&a_{32}&a_{33}&a_{34}\\ a_{41}&a_{42}&a_{43}&a_{44}\\ a_{51}&a_{52}&a_{53}&a_{54}\end{bmatrix}, \tag{25}\]  
where the element in the \(i\)th row and \(k\)th column represents the proportion of bridge elements in state \(i\) that will be maintained using maintenance measure \(A_{k-1}\). For example, the element \(a_{12}\) in the maintenance strategy matrix represents the percentage of elements in State \(1\) that will be maintained using maintenance action \(A_{1}\). The element \(a_{23}\) represents the percentage of elements in State \(2\) that will be maintained using maintenance action \(A_{2}\).  
Before starting the iterative computation, an initial strategy needs to be arbitrarily defined as the input for the algorithm.  
\[\pi(a\mid s)=P[A_{t}=a\mid S_{t}=s]=\begin{bmatrix}0.25&0.25&0.25&0.25\\ 0.25&0.25&0.25&0.25\\ 0.25&0.25&0.25&0.25\\ 0.25&0.25&0.25&0.25\\ 0.25&0.25&0.25&0.25\end{bmatrix} \tag{26}\]