Intelligent Optimization Algorithm for Maintenance Scheme Based on Life Cycle Cost

2.1 Description of the problem

LCC analysis is the basis of engineering projects and scheme optimization, this paper took the reinforced concrete bridge deck as the research object, the initial reliability was set to be β=8.0, the degradation start time was set as T₁=15, and the minimum reliability index was β^*=4.2. Considering the optimization calculation of LCC, three basic maintenance scenarios were set up through GA, respectively are: take no maintenance measure/preventive maintenance measures only, take major maintenance measures, and take major maintenance measures and preventive maintenance measures. The expected target and output were the optimal cost and service life under the constraints of the three scenarios.

The calculation formula for the LCC of the example engineering project over the entire life cycle [16] is as follows:

$L C C(T)=C_{c}+C_{I N}(T)+C_{M}(T)+C_{R}(T)+C_{F}(T)$     (1)

where, C_c is the initial cost; C_IN(T) is the test cost; C_M(T) is the routine maintenance cost; C_R(T) is the maintenance cost and the loss cost caused by the maintenance; C_F(T) is the failure cost, namely the loss caused by failure.

2.2 Determination of parameters

2.2.1 Initial cost C_c

Since the initial cost won’t affect the choice of scheme, to simplify the process of comparative analysis, C_c was set to be a constant.

2.2.2 Routine test maintenance cost C_IN,M(t)

To facilitate computation, the routine test cost C_IN(T) and the routine maintenance cost C_M(T) are generally combined together as the routine test & maintenance cost C_IN,M(t), which is calculated in years.

$C_{I N, M}(t)=0.02(1+0.05 t) C_{c},\left(t<T_{1}\right)$     (2)

$C_{I N, M}(t)=0.02\left[1+0.05 t+0.5 \frac{\left[\alpha_{1}+\alpha^{'}\left(t-T_{1}\right)\right]\left(t-T_{1}\right)^{0.8}}{\beta^{*}}\right] C_{c},\left(t \geq T_{1}\right)$     (3)

where, T₁ is the start time of initial degradation, its value took 15 in this paper; α₁is the degradation rate at the start time of initial degradation, its value took 0.01 in this paper; α' is the increase coefficient of the degradation rate, it’s a variate [16].

2.2.3 Maintenance cost

According to LCC theory, the maintenance cost is divided into two parts, namely the preventive maintenance cost C_R,PM and the reinforcement maintenance cost C_R,EM. Considering the actual situation of the example, the two parts of the maintenance cost are:

$C_{R, P M}=0.1 C_{C}+C_{C} \times\left(\frac{\Delta A}{0.05}\right)^{2}$     (4)

$C_{R, E M}=C_{R, E M}(\Delta \beta)$     (5)

where, $\Delta A$  is the difference in the degradation rate before and after the maintenance; $\Delta \beta$ is the difference in the reliability index before and after the maintenance.

2.2.4 Structural failure loss C_F(t)

Structural failure loss is usually related to the failure level, the classification method of failure level adopted in this paper is shown in Table 1, and the corresponding calculation formula for failure loss is shown as Formula 7.

Table 1. Reliability indexes of failure levels

Structural failure loss C_F(t) can be calculated as follows:

$C_{F}(t)=\sum_{i=1}^{5}\left\{p_{f}^{i}(t) \times C_{c} \times k^{i}_{C F 1}\left(1+k_{F}^{i}\right)\right\}$     (6)

where, $p_{f}^{i}(t)$ is the failure probability of the structure, and its calculation formula [16] is:

$p_{f}=\frac{1}{2}\left(1+\alpha_{1} \beta+\alpha_{2} \beta^{2}+\alpha_{3} \beta^{3}+\alpha_{4} \beta^{4}+\alpha_{5} \beta^{5}+\alpha_{6} \beta^{6}\right)^{-16}$     (7)

where,

α₁=0.049867374; α₂=0.0211410061; α₃=0.0032776263;

α₄=0.00003800036; α₅=0.0000488906; α₆=0.000005383.

2.2.5 The overall objective of the example project

The overall objective of the example project took account both economy and safety issues, that is, achieve maximum expected service life while the structure satisfies the constrain of safe operation; or achieve minimum expected project cost under the condition that the benefits of the project can hardly be estimated accurately.

The overall objective of the example project can be expressed as:

$\operatorname{Max}\{E[U(T)]-E[L C C(T)]\}$ or $\operatorname{Min} E[L C C(T)]$     (8)

subject to $\beta \geq \beta^{*}$ or $p_{s} \geq p_{s}^{*}$     (9)

where, $E[U(T)]$ is the sum of expected benefits within the life cycle $(T)$ of the project; $\beta$ is the structural reliability index (within service life); $\beta^{*}$ is the limit of target reliability index (the lowest target reliability index during the trial period); $p_{s}$ is the structural reliability (within service life); $p_{s}^{*}$ is the limit of target reliability (the lowest target reliability during the trial period)

In this paper, the overall objective of the example project was set as a constraint for the optimization of the scheme. The scheme optimization was conducted from two aspects of “satisfying a certain service life and calculating the minimum cost” and “satisfying a certain cost and calculating the longest service life”.

According to the engineering project example in [16] we can know that, under the condition of no maintenance measure had been applied, the service life of concrete bridge deck is 70 years. From two perspectives, this paper set three different maintenance scenarios and conducted comparative analysis from the vertical and horizontal directions, as shown in Table 2. With MATLAB R2013a software as the operating platform, LCC theory as the basis, Formula (1) as the objective function, the overall objective of the example project as the constraint condition, a GA program was written and compiled to optimize the schemes that satisfied the conditions and find out the optimal solution. The parameters of the evolutionary algorithm are shown in the schemes listed in sections 3.1 to 3.3.

Table 2. Reliability indexes of failure levels

4.1 Optimal scheme for no maintenance measure and preventive maintenance measures only

4.1.1 Control the service life and calculate the cost

5.png

Figure 5. Trend of optimal point

6.png

Figure 6. Trend of average value of optimal point

The optimization goal is to calculate the scheme with the minimum cost under the constraint condition that the service life is 100 years. In terms of Scenario 1, the evolution parameters were set as follows: population size 100, number of iterations 100 generations, crossover rate 0.4, mutation rate 0.1. Targeting at the optimization goal, the maintenance scheme was optimized by the GA program, and the data results and analysis are shown as follows:

(1) Calculation results:

After the optimization operations, the specific cost and optimization process of each maintenance scheme are shown in Table 3, and the convergence diagrams are shown in Figure 5 and Figure 6

(2) Result analysis

Table 3 shows the optimization process of the schemes under Scenario 1 by GA, the total cost of the initial scheme was 9.50E + 03C_C, and the scheme was “no measure taken”; after the optimization operation, it reached optimum in the 13^th generation and converged to the end; the total cost after optimization was 7.856C_C, and the scheme was “maintain every 10 years”. Figure 5 shows the trend of the individual with the optimal fitness value and Figure 6 shows the trend of the average fitness value of the optimal solution. It can be seen that the algorithm had good convergence characteristics and high solution efficiency. The calculation results were compared with the existing examples [16], and its rationality had been proved. Since the GA had found the optimal solution quickly and accurately, its high efficiency had been proved as well.

Table 3. Data summary of schemes

4.1.2 Control the cost and calculate the service life

The optimization goal is to calculate the scheme with the longest service life under the constraint condition of a given total maintenance cost. In this paper, the initial cost C_c was set to 1 million yuan, the limit of total maintenance cost was set to 7 million yuan, and the constraint condition was Formula (9). According to the reliability degradation model [16] and Formula (1), a GA optimization program was written and compiled using MATLAB to calculate the service life under the Scenario 1.

(1) Calculation results:

Table 4 Service life of maintenance schemes

(2) Result analysis

According to the results in Table 4, it can be seen that the service life was the longest 93 years when the scheme was “maintain every 10 years”, and the corresponding reliability value was 5.1282. It can be seen from the results that the shorter the interval, the longer the service life extended, this result is consistent with the existing research conclusions [16], which proved the rationality of the algorithm.

4.2 Optimal scheme for major maintenance measures

4.2.1 Control the service life and calculate the cost

The optimization goal is to calculate the scheme with the minimum cost under the constraint condition that the service life is 100 years. In terms of Scenario 2, the evolution parameters were set as follows: population size 100, number of iterations 100 generations, crossover rate 0.4, mutation rate 0.1. Targeting at the optimization goal, the maintenance scheme was optimized by the GA program, and the data results and analysis are shown as follows:

(1) Calculation results:

After the optimization operations, the specific cost and optimization process of each maintenance scheme are shown in Table 5, and the convergence diagrams are shown in Figure 7 and Figure 8.

7.png

Figure 7. Trend of optimal point

8.png

Figure 8. Trend of average value of optimal point

(2) Result analysis

Table 5. Data summary of schemes

Table 5 shows the optimization process of the schemes under Scenario 2 by GA, the total cost of the initial scheme was 8.5443C_C, and the scheme was “reinforce in the 70^th year”; after optimization operation, it reached optimum in the 6^th generation and converged to the end; after optimization, the total cost was 8.0978C_C, and the scheme was “reinforce in the 50^th year”. Figure 7 shows the trend of the individual with the optimal fitness value and Figure 8 shows the trend of the average fitness value of the optimal solution. The results showed that, under the constraint of 100-year service life, the cost of the “reinforce in the 50^th year” scheme was the lowest.

4.2.2 Control cost and calculate the service life

The initial cost C_c was set to 1 million yuan, the limit of total maintenance cost was set to 7 million yuan, and the constraint condition was Formula (9). According to the reliability degradation model and Formula (1), a GA optimization program was written and compiled using MATLAB to calculate the service life of the schemes under Scenario 2.

(1) Calculation results

The Calculation results is as shown in table 6.

(2) Result analysis

According to the results in Table 6, it can be seen that the service life reached the longest 59 years when the scheme was “reinforce in the 70^th year”, and the corresponding reliability value was 5.5229. From the degradation curve of reliability, we can know that, the longer the service life, for the same interval, the greater the difference of in reliability ($\Delta \beta$), and the longer the extended service life ($\Delta T$) after the reinforcement.

Table 6. Service life of maintenance schemes

4.3 Optimal scheme for major maintenance measures and preventive maintenance measures

4.3.1 Control the service life and calculate the cost

The optimization goal is to calculate the scheme with the minimum cost under the constraint condition that the service life is 100 years. In terms of Scenario 3, under the condition of reinforcement only, the evolution parameters were set as follows: population size 100, number of iterations 100 generations, crossover rate 0.4, mutation rate 0.1. Targeting at the optimization goal, the maintenance schemes were optimized by the GA program, and the data results and analysis are shown as follows:

(1) Calculation results:

After the optimization operations, the specific cost and optimization process of each maintenance scheme are shown in Table 7, and the convergence diagrams are shown in Figure 9 and Figure 10.

Table 7. Data summary of schemes

9.png

Figure 9. Trend of optimal point

10.png

Figure 10. Trend of average value of optimal point

(2) Result analysis

Table 7 shows the optimization process of schemes under Scenario 3 by GA, the total cost of the initial scheme was 8.7388C_C, and the scheme was “reinforce in the 70^th year and conduct no maintenance afterwards”; after optimization operation, it reached optimum in the 6^th generation and converged to the end; the total cost after optimization was 7.8560C_C, and the scheme was “maintain every 10 years”. Figure 9 shows the trend of the individual with the optimal fitness value and Figure 10 shows the trend of the average fitness value of the optimal solution. The results showed that, among the schemes for major maintenance measures and preventive maintenance measures, the optimal scheme that satisfied the constraint was “maintain every 10 years”.

4.3.2 Control cost and calculate the service life

The initial cost C_c was set to 1 million yuan, the limit of total maintenance cost was set to 7 million yuan, and the constraint condition was Formula (9). According to the reliability degradation model and Formula (1), a GA optimization program was written and compiled using MATLAB to calculate the service life of the schemes under Scenario 3.

(1) Calculation results:

Table 8. Service life of maintenance schemes

(2) Result analysis

Under the conditions of initial cost C_c was 1 million yuan, the limit of total maintenance cost was 7 million yuan, and the constraint was Formula (9), according to the results in Table 8, it can be seen that the service life reached the longest 93 years when the scheme was “maintain every 10 years”, and the corresponding reliability value was 5.1282, indicating that among the maintenance schemes that satisfied the constraint, under condition of a given cost, the service life of the scheme “maintain every 10 years” was the longest. This result corresponded to the results in Table 7, which proved the excellency of the scheme and the rationality of the result.