# Parameter Estimation of a Multistate Model for an Aging Piece of Equipment under Condition-Based Maintenance

Authors: Qihong Duan; Xiang Chen; Dengfu Zhao; Zheng Zhao
Journal: Mathematical Problems in Engineering
 (2012)

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## Abstract

We study a multistate model for an aging piece of equipment under condition-based maintenance and apply an expectation maximization algorithm to obtain maximum likelihood estimates of the model parameters. Because of the monitoring discontinuity, we cannot observe any state's duration. The observation consists of the equipment's state at an inspection or right after a repair. Based on a proper construction of stochastic processes involved in the model, calculation of some probabilities and expectations becomes tractable. Using these probabilities and expectations, we can apply an expectation maximization algorithm to estimate the parameters in the model. We carry out simulation studies to test the accuracy and the efficiency of the algorithm.


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## Body

## 1. Introduction

A multistate model may provide more flexibility than a traditional binary state model for modeling equipment conditions. In a multistate model, a piece of equipment is allowed to experience more than two possible states, for example, completely working, partially working, partially failed, and completely failed. Even if every state has an exponential duration distribution, the equipment has a nonexponentially lifetime distribution. Therefore, many authors utilize multistate models for equipment in many disciplines. Examples arise from electric power systems [1], the nuclear science [2], electron devices [3], and so on. Furthermore, although the deterioration process of a system is usually continuous, it is convenient to model it by several discrete states. For example, in [4], a deterioration process is classified into four states: initial, minor deterioration, major deterioration, and failure. Besides, [5, 6] define several discrete states for a system based on different ranges of the hazard rate of the system; that is, a deterioration process is often modeled by a multistate model.On the other hand, maintenance is an indispensable action to keep the reliability level of industrial equipment. To reduce the total maintenance costs, condition-based maintenance is usually implemented. References [7, 8] are good overviews of condition-based maintenance. In this paper, condition-based maintenance involves three key ideas: a routine inspection, a condition indicator, and a repair. Routine inspections are executed with a fixed benchmark interval. In an inspection, we can acquire the current condition indicator for the equipment. An appropriate repair is carried out when the indicator is not less than a warning value. Moreover, we also record the condition indicator right after the equipment is repaired. This kind of condition-based maintenance is recommended by an IEEE standard [9], and there are many types of equipment operating in such mode [10, 11].We assume that every state of the equipment has an exponential duration distribution specified by its parameter. In the paper, we investigate the parameter estimation problem, based on the data obtained from routine inspections and repairs.Such a parameter estimation problem for a multistate model plays an important role in many applications [12, 13]. The problem attracts much attention, and many approaches are proposed. Reference [5] reviews several quite sophisticated empirical approaches. A method of curve fitting is presented in [12] for the probability density function (pdf) of the equipment lifetime. In [14–16], based on proportional hazards models, different approximate methods are proposed to obtain a maximum likelihood estimate (MLE). In these methods, it is assumed that the condition indicators are time independent or the state of the equipment is fixed within two consecutive inspections. Those assumptions are reasonable only for a small benchmark interval or frequent inspections. However, frequent inspections may increase the total costs and may induce a high chance of accidents.There is an alternative method to obtain an MLE. To better illustrate the idea, we start from a complete data set of every state’s durations through continuous monitoring. As it has an exponential duration distribution, the reciprocal of the average of durations of a state is the MLE of its parameter. This simple example gives a clue to the case of condition-based maintenance. Then, given the incomplete data from routine inspections and repairs, we can calculate conditional expectations of every state’s duration,  and we may deduce an MLE. The theoretical foundation of the above discussion is the expectation maximization (EM) algorithms. The EM algorithms are first introduced in [17] to calculate an MLE for incomplete data. Due to its theoretical property, (see, [18, 19] e.g.,), the EM algorithms have good performance in many applications.Recently, a few researchers have tried to use EM algorithms for problems similar to ours. For example, based on a complicated discussion, [20] applied an EM algorithm to a specific 3-state model under devised condition-based maintenance. However, there is no hint in [20] to extend the result to a general multistate model. In general, although an EM algorithm may have good performance, it is difficult to apply an EM algorithm to a specific model. Besides the choice of hidden random variables or stochastic processes for a model, the calculation of involved conditional expectations is often difficult.In this paper, we first propose a proper mathematical framework and a special technique of distributions to simplify the computation of conditional expectations. Then we apply an EM algorithm to obtain MLEs of the model parameters, with the observation of routine inspections and repairs. The paper is organized into seven sections. Section2 establishes a model for our problem. In Section 3, we apply an EM algorithm, which is based on several conditional expectations, to the model. Section 4 establishes formulas of these conditional expectations. Section 5 presents simulation results, and Section 6 presents an application to a real-world dataset. The paper is concluded in Section 7.


## 2. Models

In this section, we propose the mathematical framework for our problem. In the following Model1, we present basic concepts of a new multistate model for an aging piece of equipment under condition-based maintenance. Besides, specific hidden random variables are proposed in Model 1.Model 1.
In the multistate model, a piece of equipment withm discrete states under condition-based maintenance can be described by a predefined warning value W≤m and two families of random variables Ti=(Ti1,…,Tim-1) and Ri=(Ri1,…,Rim), i=0,1,…. The random variable Tij has an exponential distribution specified by a parameter θj, j=1,…,m-1.The state space of Rij is {1,…,m}. Finally, random variables Tij and Rij are independent.In Model1, for both Tij and Rij, the subscript i indicates a time index, and the superscript j presents the state of the equipment at the moment. Concretely, we assume that the state of the equipment is j at a time instance i and a repair takes place, and Rij is the state of the equipment right after the repair. And Tij denotes the residual duration of state j after a time instance i, given that there is no maintenance again. As introduced in Section 1, the duration of state j has an exponential distribution, and θj is its parameter. Due to the memoryless property of an exponential distribution, the distribution of the residual duration is also exponential. Here and in the following, we assume that the benchmark interval is 1 for simplicity of argument.To introduce some derivational concepts to describe the behavior of the equipment clearly, we introduce several families of auxiliary random variables. These random variables,Yi, Mi, and Ii, are recursively defined. First, M0=1, I0=0.Fori≥1,(2.1)Yi=max⁡{s∣∑j=Mi-1s-1Ti-1j≤1},
where Yi that  indicates the state of the equipment at a time instance i. In fact, assume that the state of the equipment is Mi-1 at a time instance i-1, Yi is the maximal number s such that the total duration of states Mi-1,…, and s-1 is less than 1. In other words, between two consecutive inspections at time indexes i-1 and i, the equipment goes through states Mi-1,…,s-1.Fork≥1,(2.2)Ik=min⁡{i∣i>Ik-1,Yi≥W},
where Ik indicates the time index when the kth repair takes place.Moreover,(2.3)MIk=RIkYIk,
and for Ik-1<i<Ik,(2.4)Mi=Yi.
Here Mi denotes the state of the equipment right after a time instance i, whether there is maintenance or not. In fact, if Yi<W and no repair takes place, we have Mi=Yi. If Yi≥W, there is a proper repair, and Mi denotes the state of the equipment right after the repair, as showed in (2.3).


## 3. EM Algorithm

We now apply the EM algorithm to estimate the parameterθj, j=1,…,m-1, given n pairs of observations yi and mi-1, i=1,2,…,n.The EM algorithm, introduced in [17], is an iterative solution to MLE. Each iteration consists of two steps. In the Expectation step, the following functions Q(θ′∣θ) are sought:(3.1)Q(θ′∣θ)=Eθ[log⁡L(θ′)∣Yi=yi,Mi-1=mi-1,i=1,…,n],
where parameters θ,θ′∈R+m-1. θ′ stands for parameters to be estimated in the current iteration, and θ denotes the parameters estimated in the previous iteration. As indicated in [17], the expectation in (3.1) is taken over the hidden random variable Ti and Ri and L(θ′) is the joint pdf of Ti and Ri, with the parameter θ′. Estimates of the parameters θ′ are obtained in the maximization step by solving the following optimization:(3.2)max⁡θ′Q(θ′∣θ).
The estimation formulas are established by setting(3.3)∂Q(θ′∣θ)∂θj′=0,j=1,2,…,m-1.As the pdf of the repairsRi has nothing to do with θ′ and the joint pdf of Ti, i=1,2,…,n has the following form:(3.4)∏i=1n∏j=1m-1θjexp⁡(-θjTij),(3.3) leads to an unique solution θ′=ℛ(θ). Here ℛ(θ)=(ℛ1(θ),…,ℛm-1(θ)) and for j=1,…,m-1,(3.5)Rj(θ)=n(∑i=1nEθ[Tij∣Yi=yi,Mi-1=mi-1,i=1,…,n])-1.Based on the theory of EM algorithms, (see, [17–19] e.g.), we propose the following algorithm to calculate the MLE of θj, j=1,…,m-1.Algorithm 3.1.
Step 1.
Lett=0, and choose an initial value θ0∈R+m-1.Step 2.
Setθt+1=ℛ(θt).Step 3.
Stop and outputθt if the stop criterion is satisfied.Step 4.
Sett+1→t and go to Step 2.


## 4. Conditional Expectations

To calculate the conditional expectations involved in (3.5), we propose some lemmas and theorems in the following.Lemma 4.1.
Assume that random variablesU1,…,Ur are independent, Ui has an exponential distribution specified by a parameter λi, and, for i≠j, we have λi≠λj. Then, the summation S=∑i=1rUi has a pdf g(x;λ1,…,λr). Here,
(4.1)g(x;λ1,…,λr)=∑i=1rciλiexp⁡(-λix)
for x≥0. ci is a function for arguments λ1,…,λr, defined by
(4.2)ci(λ1,…,λr)=∏j=1rλjλi∏j=1,j≠ir(λj-λi).
The above denominator is λi when r=1.The proof of Lemma4.1 is based on the properties of polynomials and is presented in Appendix A. Further, we can calculate a useful conditional expectation and some basic probabilities based on Lemma 4.1.Lemma 4.2.
Under the assumption and notations in Lemma4.1, we have the conditional expectation E(Ui∣S=x)=bi(x). Here, for x≥0,
(4.3)bi(x)=1g(x;λ1,…,λr)∑j=1,j≠ircjλj(-xe-λix+e-λjx-e-λix(λi-λj)).Proof.
LetSi=S-Ui, then Si and Ui are independent. Based on Lemma 4.1, the pdf fSi(x) of Si has a form
(4.4)fSi(x)=∑j=1,j≠ircj,iλjexp⁡(-λjx)
for x≥0, where
(4.5)cj,i=∏l=1rλlλiλj∏l=1,l≠i,jr(λl-λj).
Letf(u,x) be the joint pdf of Ui and S. Since Si and Ui are independent and S=Ui+Si, we have the following:(4.6)f(u,x)=fUi(u)⋅fSi(x-u).
Hence, the conditional pdf of Ui with respect to S is
(4.7)fUi|S(u∣x)=fUi(u)⋅fSi(x-u)g(x;λ1,…,λr).
In the sequel, it follows from the expression of fSi(x) that
(4.8)E(Ui∣S=x)=1g(x;λ1,…,λr)∫0xufUi(u)fSi(x-u)du=λig(x;λ1,…,λr)∑j=1,j≠ircj,iλj(xe-λixλj-λi+e-λjx-e-λix(λj-λi)2).
The result follows from the above equation.Lemma 4.3.
Suppose that a random variableV has an exponential distribution specified by a parameter φ, random variables U1,…,Ur and V are independent, and φ≠λi for 1≤i≤r. Under the assumption and notations of Lemma 4.1, we have P(S≤1)=p(λ1,…,λr) and P(S≤1,S+V>1)=p(λ1,…,λr;φ). Here two functions p(λ1,…,λr) and p(λ1,…,λr,φ) have the following form:
(4.9)p(λ1,…,λr)=∑i=1rci(1-e-λi),p(λ1,…,λr;φ)=∑i=1rciλie-λi-e-φφ-λi.Proof.
By computing the definite integral of the pdf ofS on the interval [0,1], we can obtain the expression of P(S≤1).
It follows from the independence amongU1,…,Ur and V that S=∑i=1rUi and V are independent. Therefore,
(4.10)P(S≤1,S+V>1)=∬Dg(x;λ1,…,λr)fV(v)dxdv.
Here and in the following, D={(x,v)|x≤1,x+v>1}. Besides, fV(x) is the pdf of V. In the sequel,
(4.11)P(S≤1,S+V>1)=∑i=1r∫01ciλie-λix(∫1-x∞φe-φvdv)dx
follows from the above integral.Then we can calculate several auxiliary important conditional expectations.Theorem 4.4.
Under the assumption and notations of Lemma4.3, we have
(4.12)E(Ui∣S<1)=gi(λ1,…,λr)p(λ1,…,λr),E[Ui∣S≤1,S+V>1]=gi(λ1,…,λr;φ)p(λ1,…,λr;φ),E[V∣S≤1,S+V>1]=h(λ1,…,λr;φ).
Here, gi(λ1)=(1/λ1)(1-e-λ1)-e-λ1 and, for r>1, gi(λ1,…,λr) has the following form:
(4.13)∑j=1,j≠ircjλj(e-λiλi(1+1λi-1λj-λi)+e-λjλj(λj-λi)-1λi2+1λjλi).
Besides,
(4.14)gi(λ1;φ)=λ1(φ-λ1)((1-1φ-λ1)e-λ1+e-φφ-λ1).
And, for r>1, gi(λ1,…,λr;φ) has the following form:
(4.15)∑j=1,j≠ircjλj(-e-λiφ-λi-e-φ-e-λi(φ-λi)2+1λi-λj(e-λj-e-φφ-λj-e-λi-e-φφ-λi)).
Moreover, h(λ1,…,λr;φ) has the following form:
(4.16)1+1φ+1p(λ1,…,λr,φ)∑j=1rcjλjφ-λj(e-λj-e-φφ-λj-e-λj).Proof.
In the following, we calculate three expectations respectively.Case 1.
E[V∣S≤1,S+V>1]. We have the following:
(4.17)E[V∣S≤1,S+V>1]=∬DvfS(x)fV(v)dxdvP(S≤1,S+V>1).
Here fS(x) and fV(v) are pdfs of random variables S and V, respectively. We have fS(x)=g(x;λ1,…,λr). The denominator is presented in Lemma 4.3. Similarly to the proof of Lemma 4.3, the above numerator is
(4.18)∬DvfS(x)fV(v)dxdv=∫01∑i=1rciλie-λix(∫1-x∞φve-φvdv)dx=∑i=1rciλi(1+1φ-λi+1φ)e-λi-e-φφ-λi-∑i=1rciλie-λiφ-λi.
Simplifying the above expression, we can obtain that  E[V∣S]=h(λ1,…,λr;φ).Case 2.
We haveE[Ui∣S≤1]. When r=1, it follows from S=U1 that we have
(4.19)E(U1∣U1≤1)=1P(U1≤1)∫01λ1xe-λ1xdx=1λ1-e-λ11-e-λ1.
When r>1, according to the total expectation formula, we have
(4.20)E(Ui∣S≤1)=E[E(Ui∣S)∣S≤1].
Then it follows from Lemma 4.2 and (4.20) that
(4.21)E[Ui∣S≤1]⋅p(λ1,…,λr)=∫01bi(x)fS(x)dx,gi(λ1,…,λr) is the above definite integral.Case 3.
We haveE[Ui∣S≤1,S+V>1]. When r=1, it is obvious that
(4.22)E(U1∣U1≤1,U1+V>1)=∬Dxλ1e-λ1xφe-φvdvdxP(U1≤1,U1+V>1),
and the result follows.
Whenr>1, similarly to the discussion of Case 2, we have(4.23)E[Ui∣S≤1,S+V>1]=E[E(Ui∣S,V)∣S≤1,S+V>1].
As U1,…,Ur and V are independent, and S is a function of U1,…,Ur, we have E(Ui∣S,V)=E(Ui∣S). Hence, it follows from Lemma 4.2 that
(4.24)E[Ui∣S≤1,S+V>1].p(λ1,…,λr,φ)=∫01∫1-x∞φbi(x)fx(x)e-φvdvdx=∫01∑j=1,j≠ircj,iλje-φ(-xe(φ-λi)x+(e(φ-λj)x-e(φ-λi)x)λj-λi)dx.
The result follows from the above equation.Now we turn to the following Lemma4.5 based on the theories of measure and stochastic processes. Please see Appendix B for a proof of Lemma 4.5.Lemma 4.5.
For random variables involved in Model1, we have the following:
(4.25)E[Tkj∣Mi-1=mi-1,Yi=yi,i=1,…,n]=E[Tkj∣Mk=mk,Yk+1=yk+1].
Here, 0≤k≤n-1 and 1≤j≤m-1.Finally, we arrive at conditional expectations in (3.5), which is necessary to Algorithm 3.1.Theorem 4.6.
We assume that there is a guessθ=(θ1,…,θm-1) for the parameter vector of Model 1 for which θi≠θj if i≠j. Let exkj=Eθ[Tkj∣Yi=yi,Mi-1=mk-1,i=1,…,n] for 1≤j≤m-1 and 0≤k≤n-1. When yk+1=mk, we have the following:
(4.26)exkj={1θj,j≠yk+1,1+1θyj,j=yk+1.
When yk+1=m, we have
(4.27)exkj={1θj,j<mk,gj-mk+1(θmk,…,θm-1)p(θmk,…,θm-1),mk≤j≤m-1.
Finally, when yk+1≠m and yk+1≠mk, we have
(4.28)exkj={1θj,j<mkorj>yk+1,gj-mk+1(θmk,…,θyk+1-1;θyk+1)p(θmk,…,θyk+1-1;θyk+1),mk≤j<yk+1,h(θmk,…,θyk+1-1;θyk+1),j=yk+1.Proof.
Except some trivial cases, this result follows from Theorem4.4 and Lemma 4.5.


## 5. Simulation Results

In this section, we test the efficiency and accuracy of the EM algorithm, based on simulation data.At first, we investigate the necessity of the condition “θi≠θj for i≠j” in Theorem 4.6. We can find similar conditions for many theorems and lemmas in Section 4, and there is the root in Lemmas 4.1 and 4.2. Although these conditions make our analysis clear and convenient, they appear to be obstacles for an application.Now we turn to the assumption of Lemma4.2. Assume that random variables U1, U2, and U3 are independent, and Ui has an exponential distribution specified by a parameter θi. Based on Lemma 4.2, we can compute the conditional expectation of U1 conditional on the summation S=∑i=13Ui to be any value, for example, 1.Letθ3=4 and(5.1)B(θ1,θ2)=E[U1∣S=1].
We draw the surface of B(θ1,θ2) in Figure 1. Figure 1 shows that although B(θ1,θ2) does not exist if θ1=θ2, θ1=θ3, or θ2=θ3, but by filling proper values, B(θ1,θ2) is continuous. That is, B(θ1,θ2) has some removable discontinuities. The result is obvious from the viewpoint of engineering; why expectations in Section 4 have “jumps" for any special cases?  Similar conclusion can be obtained for the assumption of Lemma 4.1 from the viewpoint of mathematics. In fact, the pdf of Ui is an analytical function on θi>0. As a convolution, the pdf of S=U1+U2+U3 is also an analytical function on θi>0, i=1,2,3. And any discontinuity is removable for an expression of such a function.Figure 1
B(θ1,θ2) has some removable discontinuities.Similar discussions can be applied to other theorems and lemmas in Section4. In a word, these conditions may not trouble us for any application. For a numerical implement of the EM algorithm, if parameters become the same by accident, a small perturbation can fix the problem.Secondly, we present a method to simulate the behavior of a 4-state model for an aging piece of equipment under condition-based maintenance, based on some preset parameters. The simulation can provide us a test data set.Without maintenance, the equipment would run through states 1, 2, 3 orderly to a failure state 4. With maintenance, the straight path to a failure is regularly deflected by inspections and maintenance. In an inspection, if we find that the current state is 1 or 2, no repair is applied. If the state is 3, an appropriate repair is carried out. In this case, the state of the equipment right after the repair is a random variable, which is 1 with probability 0.1, is 2 with probability 0.3, and is 3 with probability 0.6. If the state of the equipment is 4, we must replace it with a new piece of equipment, and the state right after this replacement is 1. The duration of states 1, 2, and 3 have exponential distributions specified by parameters 0.3, 0.29, and 0.5. These parameters are suggested by [21].Now we turn to five experiments in which we test the efficiency and accuracy of the EM algorithm, based on simulation data provided by the above method.In the first experiment, we study the relationship between convergence error and the number of iterations. As EM algorithms are recursive, this experiment will provide us with an intuitive stop criterion for Algorithm3.1. For a data set from 100 inspections, Figure 2 shows that all curves of the estimated parameters θ1, θ2, and θ3 converge while the number of iterations increases. We can conclude from Figure 2 that for 100 or more iterations, we can obtain reasonable results.Figure 2
Relationship between convergence error and the number of iterations.In the second experiment, we run the EM algorithm for different initial values. Figure3 draws the estimated parameter pairs (θ1,θ2) for 100 different initial values which are chosen randomly from 0.001 to 15. We can see that the EM algorithm converges to the same result for a great range of initial values.Figure 3
Estimated parameters for different initial values.In the third experiment, we run the EM algorithm for different sample sizes. Given the number of inspectionsI = 20, 40, 100, 200, 400, and 1000, we run the simulating equipment and obtain 6 data sets. Then based on these data sets, we estimate parameters θ1, θ2, and θ3, respectively. For the EM algorithm, the initial values of estimated parameters are θ1=θ2=θ3=0.1, and the number of iterations is 100. Table 1 shows the estimated parameters for different data sets. Of course, the estimated parameters are nonmonotonic for different numbers of inspections, which is inherent to such a random problem.Table 1
Estimated parameters with different numbers of inspections.
Iθ1θ2θ3200.40890.27000.7639400.37540.31230.65811000.33420.27780.58872000.32710.30400.54144000.32150.28310.535410000.30750.29830.5256In the fourth experiment, we repeat the third experiment 100 times to obtain the standard errors of each estimated parameter. Table2 shows the standard errors of the estimated parameters. Again, all cases demonstrate a decrease in the standard errors, when the number of inspections increases. We can conclude from Tables 1 and 2 that, for 100 or more inspections, we can obtain reasonable estimated parameters with an acceptable standard error.Table 2
Standard errors of estimated parameters.
Iθ1θ2θ3200.09020.06320.2337400.04620.05820.16811000.03540.02800.09632000.02650.02630.07404000.01780.01640.053210000.01290.01140.0406Finally, in the fifth experiment, we compare our method with another method.As indicated in Section1, [14–16] used an assumption that the state of the equipment is fixed during two consecutive inspections for their problems. Under this assumption, we can establish an MLE for every parameter. For n pairs of observations Yi and Mi-1, i=1,2,…,n, we write V1=0, S1=M0, and for k≥1,(5.2)Vk=min⁡{i≥Vk-1∣Mi≠SkorMi≠Yi},Sk=MVk.
Here Vk records the moment when the state of the equipment is changed and Sk records the corresponding new state. Let 𝒩kj be the number of the set {l≤k∣Sl=j}. Then, under the assumption of [14–16],(5.3)XNkjj=Vk+1-Vk
records the 𝒩kjth duration of the state j of the equipment. Finally, let the total number of changing K be the maximal subscript of Vk. Then, we have the following MLE:(5.4)θ̂j=NKj(∑l=1NKjXlj)-1.On the other hand, although we assume that, in our method, the benchmark interval is 1 for simplicity of argument, it is not too difficult to implement our method for other benchmark intervals. We run the simulating equipment for 400 time units and obtain inspection data sets for different benchmark intervals, respectively.Figure4 shows the estimated parameter θ1 of two methods. We can conclude that for frequent inspections, (the corresponding benchmark interval is less than 0.2) two methods have similar behaviors. However, we can find that two methods show a striking contrast when the benchmark interval is not less than 0.2. As indicated in [21], the latter is more important for practical situations. As our method is stable for all cases, it is more practical.Figure 4
Estimated parameters of two methods for different benchmark intervals.


## 6. Application

In this section, we apply the EM algorithm to a real-world dataset. The dataset is a 25-year record of inspections and repairs of a power transformer substation under condition-based maintenance. The power transformer substation consists of two power transformers, and it is a cool backup system. Once the primary transformer incurs faults, the secondary transformer replaces it automatically, and it can generally serve the entire load. The benchmark interval of inspections is three months. During an inspection, a transformer is repaired if it incurs faults, and the repaired primary transformer is put into operation. The repair is perfect because we can replace a transformer if necessary.The behavior of the power transformer substation can be described by a 3-state model with a warning valueW=2 and a three-month benchmark interval. Here state 1 implies that the primary transformer works, 2 implies that the secondary transformer works, and 3 implies that two transformers incur faults. In Table 3, we present the record of inspections. Here Y and M are defined by (2.1), (2.3), and(2.4)Table 3
Inspections of a power transformer substation.
Inspection no.01–101112–373839–555657–737475–808182-838485–9899Y—12121212121312M11111111111111—Based on the dataset given in Table3, we apply Algorithm 3.1 and obtain the following MLE θ̂1 and θ̂2 for the primary and the secondary transformers, respectively(6.1)θ̂1=0.0243,θ̂2=0.164.
That is, the mean duration of the primary transformer is 41.2 months, and it is 6.10 months for the secondary transformer.We use the total time on test (TTT) plot to check the efficiency of the result. The TTT plot is obtained by plotting(6.2)G(rn)=(∑i=1rZi:n+(n-r)Zr:n)∑i=1nZi:nagainstr/n. Here Zi:n is the order statistics of the sample. See [22] for a detailed introduction to TTT plot. From the dataset of Table t010207, the total lifetime of two transformers are 22, 81, 54, 54, 21, 9, and 45 months. We apply Monte Carlo simulations, with the MLE parameters θ̂1 and θ̂2, to generate 100 independent total lifetime of two transformers. The TTT plots for the real-world, dataset and the simulated dataset is shown in Figure 5. In Figure 5, the circles represent the real data, and the solid points represent the simulated data with the estimated parameters. Figure 5 shows the efficiency of the result.Figure 5
The TTT plots for real-world dataset and the simulated dataset.


## 7. Conclusion

In this paper, we have addressed the parameter estimation problem for the multistate model for aging equipment under condition-based maintenance. Based on the memoryless property of exponential distributions, we proposed a convenient mathematical framework for the problem. In this framework, the calculation of involved conditional expectations became tractable. Then we applied an EM algorithm to obtain MLE of parameters. A sequence of simulation experiments shows that the estimated parameters converge to the preset value of the parameters, even for a moderate number of inspections. Moreover, the EM algorithm is stable for different length of benchmark intervals. Hence, the algorithm can be recommended for practical applications. It is convenient to extend the algorithm to the situation with random benchmark intervals.In this paper, we assume that the state of the equipment can be directly observed at an inspection. However, it is a difficult task for many types of industrial equipment. In these cases, some other variables may be used to obtain estimates of the states. To establish a full model consisting of equipment, inspections, indirect variables, and investigating parameter estimations, it is challenging and valuable for practical situations.



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*Source: 347675-2012-01-05.xml*