*3.2. Assumptions*

When equipment fails during the preventive maintenance period, it is supposed to be shut down for breakdown maintenance. The time and cost of breakdown maintenance are fixed, and the breakdown maintenance will not change the failure rate of the equipment. By adopting maintenance, equipment will return to the state before failure.


#### **4. Modeling of the Maintenance Strategy Optimization**

The multiobjective maintenance model considering equipment energy efficiency under the variable of cost is described as follows: first, the degradation failure model of equipment is developed by using the Weibull distribution and introducing the failure increasing rate factor and the age reduction factor to simulate the degradation process. Furthermore, based on the relationship between the reliability and failure rate, preventive maintenance intervals are calculated, which lays the foundation for the construction of equipment cost model and energy efficiency model. Then, the variable cost model is developed by considering the cost of different maintenance activities of equipment and considering the quality loss of the products produced by the equipment. Third, by calculating the energy consumption of equipment in each state to obtain the total energy consumption and introducing the recovery coefficient to obtain the effective output of equipment, the energy efficiency model of equipment is constructed. Fourth, the decision variables and optimization goals of the proposed model are determined to build a multiobjective decision-making model. Finally, the NSGAII algorithm is selected, and the model is solved based on the algorithm process. The modeling of the maintenance strategy optimization process is shown in Figure 2.

**Figure 2.** Modeling of the maintenance strategy optimization process.

#### *4.1. Degradation Failure Model*

The performance of equipment is continuously degrading as equipment operates. The Weibull distribution is widely used to simulate the cumulative failure analysis of mechanical and electrical equipment. In this paper, the Weibull distribution is used to describe the degradation level of equipment. The failure rate function at running time *t* is expressed as:

$$
\lambda(t) = \frac{\beta}{\alpha} (\frac{t}{\alpha})^{\beta - 1} \tag{1}
$$

where *α*, *β* are the scale parameters and shape parameters of the Weibull distribution, respectively, which are obtained from the historical maintenance data of equipment. As the health rate of equipment cannot be restored to a new state after preventive maintenance, the failure rate increasing factor *bi*(*bi* > 1) and service age decreasing factor *ai*(<sup>0</sup> < *ai* < 1) are introduced to express the change in the failure rate. The failure rate increasing factor indicates that the equipment failure rate at each operating moment will increase after preventive maintenance. The service age decreasing factor indicates that the health state of equipment after preventive maintenance will return to a state between the new state and the state before adopting maintenance. The failure rate expression of equipment after the *i*th preventive maintenance can be obtained:

$$
\lambda\_{i+1} = b\_i \lambda\_i (t + a\_i T\_i) \tag{2}
$$

There is a certain relationship between the reliability of equipment and the failure rate function. When the equipment reaches the preventive maintenance threshold *R*<sup>∗</sup>, preventive maintenance will be carried out. Assuming that *N* times of preventive maintenance are carried out, each preventive maintenance interval can be obtained:

$$e^{-\int\_0^{\bar{\mathcal{I}}\_i} \lambda\_i(t)dt} = \mathcal{R}^\* \quad \mathcal{i} = \{1, 2, \dots, N\} \tag{3}$$

where *Ti* represents the operational time of equipment from *i* − 1th preventive maintenance to *i*th preventive maintenance.

#### *4.2. Variable Cost Model*

In this section, the maintenance cost in each condition is calculated, and the quality loss of the defective products produced is considered. Then, the total cost is divided by the cycle time to obtain the maintenance cost per unit, which is obtained by dividing the total cost by the cycle time *T*. In this paper, the total maintenance cost is mainly composed of preventive maintenance cost, breakdown maintenance cost, replacement cost, and quality loss cost.

#### (1) Preventive maintenance cost

According to the assumption of the model, when the reliability of equipment reaches the threshold of preventive maintenance, preventive maintenance will be carried out. In previous research, the cost of preventive maintenance was fixed. However, the cost of preventive maintenance cannot be fixed due to the deteriorating performance of equipment. When equipment performance significantly deteriorates, it will inevitably cost more to maintain. It means that the cost of preventive maintenance will continue to increase with the degradation of equipment, which is fluctuating. Thus, the cost of *i*th preventive maintenance of equipment is composed of both fixed costs and variable costs, and it is expressed as:

$$\mathcal{C}\_{\mathbf{m}}^{pmi} = \mathcal{C}\_{s}^{pm} + \mathcal{C}\_{vi}^{pm} = \mathcal{C}\_{s}^{pm} + \gamma \lambda\_{i}(T\_{i}) \tag{4}$$

where *Cpms* represents the fixed cost of preventive maintenance, *Cpmvi* represents the variable cost of *i*th preventive maintenance, which is linearly related to the amount of equipment degradation indicated by *<sup>λ</sup>i*(*Ti*), and the correlation coefficient is *γ*.

Assuming that the maintenance time for each preventive maintenance is *tpmm* , the total cost and total time of preventive maintenance in the cycle is therefore expressed as:

$$T\mathbb{C}\_{\mathfrak{m}}^{\text{pw}} = \sum\_{i=1}^{N} \mathbb{C}\_{\mathfrak{m}}^{\text{pwi}} = \sum\_{i=1}^{N} \mathbb{C}\_{s}^{\text{pw}} + \gamma \lambda\_{i}(T\_{i}) \tag{5}$$

$$T\_{\mathbf{m}}^{pm} = \mathbf{N} \times t\_{\mathbf{m}}^{pm} \tag{6}$$

#### (2) Breakdown maintenance cost

According to assumption 1, the total cost of breakdown maintenance is equal to the number of failures multiplied by each breakdown maintenance cost. The number of failures in a preventive maintenance cycle can be calculated from Equation (3), which is expressed as:

$$N\_c^i = \int\_0^{T\_i} \lambda\_i(t)dt = -\ln R^\* \tag{7}$$

The time and cost of each breakdown maintenance are *tpmm* , *Ccmm* , respectively. The total time and total cost of breakdown maintenance can be obtained according to Equation (7):

$$TC\_m^{$$

$$T\_{\rm m}^{cm} = t\_{\rm m}^{cm} \times \left( (N+1) \times (-\ln R^\*) \right) \tag{9}$$

#### (3) Quality loss cost

Generally, equipment will produce a certain amount of substandard products during the production process. Most of these substandard products can be caused by equipment designed so that it cannot be reduced, and some are caused by equipment degradation. The cost of quality loss is the loss of inferior products that cannot be sold properly due to quality problems. With the degradation of equipment, the product quality will continue to degrade. At this time, the number of defective products will continue to increase, resulting in a particular cost. According to past sales data, the revenue of each product can be measured, and the cost of quality loss can be measured by the original sales revenue of defective products. Thus, it is necessary to calculate the number of defective products closely related to the defective product rate. According to Assumption 5, the defective rate of equipment varies, and it can be expressed as:

$$p(\lambda\_i(t)) = p\_0 + \mu[1 - e^{-\sigma\lambda\_i(t)^\theta}] \tag{10}$$

where *p*0 represents the defective rate in the new state of equipment,*μ* represents the boundary of quality deterioration, and *σ* and *θ* are constants [36].

Assuming that the loss cost of a single product is *C<sup>d</sup> v*, and the production rate is *v*, the quality loss cost of equipment in the period *T* is:

$$T\mathbb{C}\_d = \mathbb{C}\_{dv} \times \sum\_{i=1}^{N+1} T\_i \times v \times \overline{p}\_i \tag{11}$$

In Equation (11), *pi* represents the average defective rate of products produced by equipment during *i* − 1th and *i*th preventive maintenance intervals. The calculation equation of the average defective rate is as follows:

$$\overline{p}\_{i} = \frac{p(\lambda\_{i}(0)) + p(\lambda\_{i}(T\_{i}))}{2} \tag{12}$$

According to Assumption 3, replacement cost and time are *Cr* and *Tr*, respectively. Currently, the total cost of equipment can be obtained. The next cycle time needs to be calculated. As is shown in Figure 1, the period *T* from the new state of equipment to the next replacement is composed of preventive maintenance time, equipment normal operation time, breakdown maintenance time, and replacement time. Finally, based on the above analysis, the variable cost model can be developed expressed by the maintenance cost per unit obtained by dividing the total cost during the cycle time *T*. The expression is as follows:

$$\begin{array}{rcl}ETC &=& \frac{T\_{\text{w}}^{\text{pw}} + TC\_{\text{m}}^{\text{nc}} + C\_{\text{r}} + T\_{\text{d}}}{\sum\limits\_{i=1}^{N}T\_{i} + T\_{\text{m}}^{\text{pw}} + T\_{\text{m}}^{\text{c}} + T\_{\text{r}}}\\ &=& \frac{\sum\limits\_{i=1}^{N}C\_{\text{s}}^{\text{pw}} + \gamma\lambda\_{i}(T\_{i}) + C\_{\text{m}}^{\text{c}m} \times ((N+1)\times(-\ln R^{\*})) + C\_{\text{r}} + \sum\limits\_{i=1}^{N+1}C\_{\text{d}v} \times v \times \overline{\mathbb{P}\_{i} \times T\_{i}}}{\sum\limits\_{i=1}^{N+1}T\_{i} + N \times t\_{\text{w}}^{\text{pw}} + T\_{\text{r}} + t\_{\text{w}}^{\text{cm}} \times ((N+1) \times (-\ln R^{\*}))}\end{array} \tag{13}$$

#### *4.3. Energy Efficiency Model*

The key to achieving carbon neutrality lies in energy conservation and emission reduction. In order to achieve energy conservation and emission reduction, it is necessary to explore the problem of excessive energy consumption in the process of product production and equipment maintenance and to improve energy efficiency. Energy efficiency is an important indicator to measure the input–output ratio of the manufacturing industry. The energy efficiency in the manufacturing process is expressed as the ratio between the total output capacity and the total input energy [37]. Thus, in order to obtain the energy efficiency of equipment, the total energy consumption and the total effective output of equipment need to be calculated. The first step is to find out the energy consumed by equipment. It is common knowledge that the energy consumption of equipment in different states varies. According to the difference, the state of equipment can be divided into normal operation state, standby state, warm-up state, power-on state, and power-off state. Since the time of power-on and power-off is very short, the energy consumption is negligible. The energy consumption of equipment is shown in Figure 3.

The energy consumption considered in this paper includes two parts, the energy consumption of equipment and the energy consumption of maintenance. The energy consumption of equipment includes the energy consumption of the normal operation, the standby energy consumption, and the warm-up energy consumption. The maintenance energy consumption is mainly the energy consumption of three maintenance activities. The various energy consumptions are calculated below.

**Figure 3.** Different energy consumption states of equipment.

#### (1) Operation energy consumption of equipment

The energy consumption of equipment during operation increases with degradation, which is linearly related to the failure rate. The energy consumption per unit before *i*th preventive maintenance is expressed as:

$$X\_i(t) = X\_0 + \omega \lambda\_i(t) \tag{14}$$

where *X*0 represents the energy consumption of equipment at the initial stage, and *ω* represents the linear relationship between the variation of equipment energy consumption and equipment failure rate. Thus, the total operation energy consumption of equipment in period *T* is expressed as:

$$E\_o = \sum\_{i=1}^{N+1} \int\_0^{T\_i} X\_i(t)dt = \sum\_{i=1}^{N+1} \int\_0^{T\_i} X\_0 + \omega \lambda\_i(t)dt\tag{15}$$

#### (2) Standby energy consumption of equipment

According to Assumption 2, equipment is on standby while preventive maintenance is being performed. The standby time of equipment can be measured by the preventive maintenance time. Assuming that the standby energy consumption per unit of equipment is *E*iv, the total standby energy consumption can be expressed as:

$$E\_{\rm i} = E\_{\rm iv} \times T\_{\rm m}^{\rm pm} = E\_{\rm iv} \times \mathcal{N} \times t\_{\rm m}^{\rm pm} \tag{16}$$

#### (3) Warm-up energy consumption of equipment

According to Assumption 4, the equipment needs to go through a warm-up time after it is turned on. The equipment needs to be shut down for maintenance and replacement. Assuming that the energy required to warm up equipment once is *Ewv*, then the total warm-up energy consumption of equipment is expressed as:

$$E\_{\rm av} = E\_{\rm uvv} \times \sum\_{i=1}^{N+1} N\_{\rm c}^i = E\_{\rm uvv} \times \left( (N+1) \times (-\ln R^\*) \right) \tag{17}$$

#### (4) Equipment maintenance energy consumption

When equipment performs maintenance activities, it needs to consume other energy such as electricity. Assuming that the energy consumption of each breakdown maintenance and preventive maintenance is *Epv*, and the energy consumption of equipment replacement is *Epr*, then the maintenance energy consumption of equipment in the period *T* is:

$$E\_p = E\_{pv} \times \left( N + (N+1) \times (-\ln R) \right) + E\_{pr} \tag{18}$$

The above equation has calculated the total energy consumption of equipment, and then the total effective output of equipment needs to be obtained. The total effective output of equipment includes qualified products and defective products that can be recycled. With the deterioration of equipment, the defective rate is increasing, and the recovery factor will change correspondingly. The recovery factor *τi* is introduced to describe the gradual decrease in the amount of recovery. Thus, the final effective output of equipment is obtained by subtracting the number of defective products that cannot be recovered and is expressed as:

$$Y = \sum\_{i=1}^{N+1} T\_i \times v - \sum\_{i=1}^{N+1} v \times \overline{p}\_i \times T\_i \times (1 - \tau^i) \tag{19}$$

The final energy efficiency model can be obtained and expressed as:

$$\begin{array}{lcl} EE &=& \frac{Y}{E\_{\sigma} + E\_{i} + E\_{w} + E\_{p}} \\ &=& \frac{\sum\_{i=1}^{N+1} T\_{i} \times v - \sum\_{i=1}^{N+1} v \times \overline{p}\_{i} \times T\_{i} \times (1 - \tau^{i})}{\sum\_{i=1}^{N+1} \int\_{0}^{T\_{i}} X\_{0} + \omega \lambda\_{i}(t) dt + E\_{\bar{w}} \times N \times t\_{\mathfrak{m}}^{\mathfrak{m}\mathfrak{n}} + E\_{w\mathbb{C}^{\mathfrak{d}}} \times ((N+1) \times (-\ln \mathbb{R}^{\mathfrak{e}})) + E\_{\mathbb{P}^{3}} \times (N + (N+1) \times (-\ln \mathbb{R})) + E\_{\mathbb{P}^{2}}} \end{array} \tag{20}$$

#### *4.4. Multiobjective Maintenance Model*

In order to achieve a balance between the economic benefits and social benefits of the enterprise, not only maintenance costs but also energy efficiency need to be considered. Thus, this paper aims to minimize the maintenance cost per unit and maximize energy efficiency, using the preventive maintenance threshold and preventive maintenance times as the decision variables. In addition, a multiobjective maintenance model can be constructed. The expression is as follows:

$$D = \begin{cases} \min ET(R^\*, N^\*)\\ \max EE(R^\*, N^\*) \end{cases} \tag{21}$$

#### *4.5. Solution of the Maintenance Strategy Optimization*

NSGAII is a multiobjective genetic algorithm widely used to analyze and solve multiobjective optimization problems due to its advantages of fast solution speed and good convergence of solutions. In this paper, the relationship between energy efficiency and cost per unit needs to be reconciled to satisfy each objective as far as possible. For this reason, the NSGAII algorithm is used to solve the model; the model solution process is shown in Figure 4. Its processes are as follows.

**Figure 4.** NSGAII algorithm processes.


the distance along the direction of the first objective and the distance along the direction of the second objective. The formula is expressed as:

$$i\_d = \left[f\_1(i+1) - f\_1(i-1)\right] + \left[f\_2(i-1) - f\_2(i+1)\right] \tag{22}$$


**Figure 5.** The crowding distance of the ith point.

#### **5. Case Study**

#### *5.1. Data Preparation*

The validity and adaptability of the multiobjective decision model are verified by a case study. In this paper, the production equipment of a manufacturing company is selected for the study. The production equipment produces products at a fixed rate every day. The equipment will produce a small number of defective products that can be recycled to a certain extent. Meanwhile, the defective product rate will increase with equipment degradation, and equipment operation and maintenance consume more energy. Referring to the historical data of the equipment, it can be found that the failure rate of the equipment obeys the Weibull distribution with the shape parameter of 3 and the size parameter of 110, and then referring to the general calculation of comprehensive energy consumption, the following parameters related to maintenance and energy consumption are obtained. As the types of products produced by the equipment will change with customer demand, the defective data of each product varies. In this paper, one of the products is selected, and the initial defective rate is obtained by analyzing the defective data. Other parameters of defective products are determined by referring to the literature [36]. Furthermore, due to the change in customer demand, the production rate of the equipment is not fixed. We

assume that the production rate of the equipment is 100 pieces per day, thus obtaining the total parameter table, as shown in Table 1.


**Table 1.** Value of related parameters.

#### *5.2. Results Analysis*

In order to narrow the search for multiobjective solution sets and ensure the accuracy and convenience of the solution, the maintenance cost per unit and energy efficiency of equipment under different combinations of preventive maintenance thresholds and maintenance times were obtained by numerical simulation.

The results can be seen in Figure 6. When the preventive maintenance threshold is in the range (0.6, 0.8), and the number of preventive maintenance is in the range (0, 6), there is a minimum value of maintenance cost per unit and a maximum value of energy efficiency. In addition, the graph of simulation results shows that the maintenance cost per unit tends to decrease and then increase as the number of maintenance increases. When preventive maintenance is less frequent, the time from the beginning of use to the replacement of equipment is shorter, and the equipment maintenance cost is mainly composed of replacement cost and breakdown maintenance cost, which makes the maintenance cost per unit higher. With the increase in maintenance times, the cycle time will gradually become longer, while the maintenance cost slowly increases, and the maintenance cost per unit shows a downward trend. When the number of maintenance exceeds a certain threshold due to frequent maintenance, the preventive maintenance cost of equipment significantly increases, and the cycle time slowly increases at this time, which makes the maintenance cost per unit show an upward trend.

**Figure 6.** Maintenance cost per unit and energy efficiency under different threshold combinations.

Similarly, energy efficiency tends to increase and then decrease as the number of maintenance increases. When the number of preventive maintenance times is small, the energy consumption of equipment maintenance is mainly composed of replacement energy consumption and operation energy consumption, which makes the increase in equipment output exceed the increase in energy consumption, and the energy efficiency shows an upward trend. When the number of maintenance times exceeds a certain threshold, the

cycle will gradually become longer as the number of maintenance increases. At this time, the effective output of the equipment slowly increases, but the energy consumption rapidly increases due to frequent preventive maintenance and breakdown maintenance, which makes the energy efficiency show a downward trend.

The optimal solution for the single objective can be obtained by conducting a simulation in the interval of the preventive maintenance thresholds of (0.6, 0.8) and the preventive maintenance times of (0, 6), as shown in Tables 2 and 3. The simulation results show that when the preventive maintenance threshold is 0.77 and the number of maintenance visits is 4, the lowest maintenance cost per unit is achieved at 25.81566. When the preventive maintenance threshold is 0.73 and the number of maintenance visits is 2, the highest equipment energy efficiency is achieved at 0.542660.

Thus, in this paper, we set the range of preventive maintenance threshold as (0.6, 0.8), the range of maintenance times as (2, 4), the number of individuals in the population as 100, the crossover rate as 0.9, the variation rate as 0.1, and the maximum number of iterations as 200. The above parameters were input into the algorithm of NSGAII, and the following results were obtained by a simulation using python, as shown in Figure 7. By comparing them, it is found that when *N* = 2, the energy efficiency of the equipment is the highest, but the maintenance cost per unit of equipment is also high. When *N* = 3, the energy efficiency of the equipment is lower than when *N* = 2, and the maintenance cost per unit of equipment is reduced more. When *N* = 4, the energy efficiency of the equipment is the lowest, but the maintenance cost per unit of equipment is not significantly reduced. Therefore, the comprehensive analysis yields that the energy efficiency and maintenance cost per unit of equipment is generally better for different maintenance thresholds at *N* = 3, so the Pareto curve at *N* = 3 is the final set of Pareto solutions for the model.

**Table 2.** Simulation results for maintenance costs per unit.



**Table 3.** Simulation results for energy efficiency.

**Figure 7.** Pareto curve at different *N*.

#### *5.3. Results Comparison*

Then, in order to verify the superiority of the model, a comparison with a single target is required. By comparing the results of the single objective decision with the multiobjective decision, we can see that when only the cost of maintenance is considered, the optimal solution is obtained with an objective value of (25.8157, 0.5423). It is clear that improvements in energy efficiency need to be made. When only energy efficiency is considered, the optimal solution obtained corresponds to an objective value of (26.5679, 0.5427), a decision that is clearly not optimal in terms of maintenance costs. In comparison, the compromise solution is (25.8622, 0.5426), where the energy efficiency is not much different from the single objective and the maintenance cost per unit is better, thus showing that the integrated consideration of maintenance cost and energy efficiency can help enterprises to achieve the goals of energy conservation and emission reduction.

#### *5.4. Sensitivity Analysis*

Finally, in order to analyze the relationship between the objective function and the parameters, this paper selects the maintenance cost parameters for sensitivity analysis by changing one maintenance cost parameter, keeping the other parameters constant, and observing the sensitivity of the objective function to the maintenance cost parameter. In this paper, the fixed cost of preventive maintenance, breakdown maintenance, and replacement cost affect the objective function. The range of parameter variation is −50~+50%. The analysis results obtained are shown in Table 4.

**Table 4.** Sensitivity analysis results.


According to the sensitivity analysis results, the maintenance cost per unit is more sensitive to the fixed cost of preventive maintenance and the replacement cost, and it varies positively with both parameters. When the fixed cost of preventive maintenance and the replacement cost decrease, the change in cost is more obvious than when they increase, which means that the maintenance costs per unit can be reduced by reducing the fixed costs of preventive maintenance and replacement costs when making maintenance decisions. In addition, the maintenance costs per unit and energy efficiency are not sensitive to breakdown maintenance cost, and changes in the fixed cost of preventive maintenance, breakdown maintenance cost, and replacement cost do not have a significant impact on changes in energy efficiency.
