## 5 Results and Discussion
### Maintenance Strategies under Different Parameters  
By varying the discount factor based on the availability of maintenance funds and applying it to the dynamic programming algorithm mentioned above, the optimal maintenance strategies for elements in service for 6 years are obtained under different maintenance budgets and different states. The results are shown in Figure 5.  
From Figure 5, it can be observed that, for the same state, the maintenance strategies differ when the discount factor varies. For example, for State 3, when the discount factor ranges from 0.1 to 0.3 (considering only minimizing the current maintenance cost without considering future cost savings), Maintenance Strategy 2 is used to repair the concrete cracks. When the discount factor ranges from 0.4 to 0.6 (which considers not only minimizing the current maintenance cost, but also considers some cost savings in the future), Maintenance Strategies 2 and 4 are used to repair concrete cracks and visible defects. When the discount factor ranges from 0.7 to 1 (which considers both minimizing the current maintenance cost and considering cost savings in the future), Maintenance Strategies 2 and 3 are used to repair concrete cracks and exposed reinforcement corrosion.  
For bridge elements with different service durations, if the actual condition of the elements varies, their maintenance strategies also differ. Figure 6 illustrates the maintenance strategies for elements with a service duration of 2 years.  
Figure 5: Diagram of the maintenance plans under different parameters in the sixth year.  
Figure 6: Diagram of the maintenance plans under different parameters in the second year.  
```
Objective: Find the optimal state and optimal strategy. Input: State distribution vector \(D_{6}\), repair effect matrix \(M_{A_{k}}\), maintenance cost matrix \(R\), initial policy \(\pi(a|s)\), initial value function \(v^{\pi}(s,t)\), discount factor \(\gamma\), and state duration distribution \(T(s,s^{\prime},t^{\prime})=\{T_{1},T_{2},T_{3},T_{4},T_{5}\}\) While iteration < 1000 do \(\Delta<0\); //Policy evaluation for s = 1:5 do/ /5 technical condition levels \(v(s,t)=v^{\pi}(s,t)\) //compute value function \(New\_v(s,t)=\)R(s)+\(D_{T}\cdot\pi(a|s)\cdot T(s,s^{\prime},t^{\prime})\cdot\sum\gamma\cdot v^{\pi}(s^{ \prime},t^{\prime}+t)\) \(\Delta=\max\left(\left|\left|v(s,t)-New_{v(s,t)}\right|,\ \ \Delta\right)\right.\) Until \(\Delta<\theta//\)check if the value function has converged //policy update for s = 1:5 do/ /5 technical condition levels \(old\_policy=\pi(a|s)//\)previous maintenance strategy \(q(s,a,t)=R_{a,s}+\gamma\sum\limits_{s\in S}M_{A_{k}}(s,st)\cdot New\_v(s,t)\) \(new\_policy=max_{a}q(s,a,t)//\) updated maintenance strategy End If old_policy = new_policy do  return \(New\_v(s,t)//\)selecting the optimal value function and optimal strategy else: \(\pi(a|s)=new\_policy//\)return to policy evaluation and repeat the process  End iteration+ = 1 End Output: The optimal strategy.
```  
**Algorithm 1**Pseudo code of the policy iteration al gorithm.  
When comparing Figures 5 and 6, it can be observed that the maintenance strategies for elements with a service duration of 6 years and 2 years differ. Due to the lesser number of defects in elements with a service duration of 2 years, the maintenance strategies for elements in each state of the bridge were as follows: (1) For elements in State 3, execute Maintenance Strategy 4 to repair the visible defects in the concrete. (2) For elements in State 4, execute either Maintenance Strategy 3 or Strategy 4 to repair either the exposed reinforcement corrosion or the visible defects in the concrete. (3) For elements in State 5, execute Maintenance Strategies 2 and 3 to repair the concrete cracks and exposed reinforcement corrosion.