Reliability Analysis of Multi-state System from Time Response

Abstract

In this paper, firstly, it is discrete the engine output and constructed the Markov random process with the discrete state and continuous time. Secondly, we constructed a model of the multi-state Markov model and determined the transition intensity. Thirdly, we computed the expected of the engine capacity deficiency and the forced outage by the method.

1 Introduction

In the engineering practice, most systems are acted on their works under the kinds of their units, which units is limited, additional, the state of these units are diverse, so a system is shown a multi-state by these reasons, that such the system is called a multi-state system (MSS) (Levitin and Lisnianski 2001). The MSS is widely used in the reliability evaluation of the power systems, engine systems and electronic products systems (Mosleh 1991). The reliability of the MSS have been studied, the basic research methods were provided, but these studies have some limit and is start. In addition, because of the complexity of the MSS, the studying has some difficulties, therefore, the articles about the reliability of the MSS is little for now. Recently, the multi-state reliability of the engine is assessed by using the two-state system (Fang et al. 2014), but the assessment result of the engine is error if the reliability of the engine unit has special, or at least limited. The contents of Markov chain and semi-Markov chain become rich in research, theory is quite perfect, gradually. The reliability of the system was studied by using the Markov chains (MC) and semi-Markov Chain (SMC) (Trivedi 2001, Zhang and Wu 2012), reliability of system, availability of system, average operating time of system may be obtained. People studied the reliability of the of the coal-fired generating to be used a single Markov model (Hamoud and Yiu 2015, Jmamal et al. 2015). The reliability problems of the system with the discrete state (DS) and the time continuous (TC) was studied by using the Markov model (MM) with DS and TC (Shnurkov and Ivanov 2015, Atchade 2014). The DS and TC Markov models have been used to assess the reliability of other systems (Fang et al. 2013). The reliability of the multi-state (MS) elements generator system were studied by using the DS and TC Markov model, The result that was predicted has a good effect for the short-term forecasting.

In this paper, we studied the MS random process of the engine unit according to the DS and TC MM and the SMM from the actual reliability of the engine system, we established the multi-state Markov model (MSMM) and accessed the MS the reliability of the engine from time response using the method. Finally, the method is shown by using an engineering practical case.

2 The Multi-state Markov Model

$T$ is the Service deadline of an engine unit, we used the interval $[0,e]$ to indicate the output energy of the engine, $e$ is the engine’s maximum output energy in the time $t$. It is shown that the random process is continuous time and the continuous state, we used the discrete state to substituted it whom is noted as $E(t)$.

• (1) We noted two special state of the engine as 1, N they are corresponding to $e_{1} = 0$ the engine is failed complete and $e_{N} = e$ that the engine is outputted normally, respectively.

• (2) We divided the interval $[0,e]$ to $N - 2$ sub-interval, the every sub-interval’s length is $\Delta e = \frac{e}{N - 2}$.

• (3) If $E(t) \in ((i - 1)\Delta e,(i - 1)\Delta e],i = 2, \cdots ,N - 1$, we noted the state of $E(t)$ is $i(i = 2,3, \cdots N - 1)$ in $t$, $g_{i}$ is its output energy.

• (4) The work energy $g_{i}$ in state $i$ is the average of the interval $((i - 1)\Delta e,(i - 1)\Delta e]$.

Through about quantitative processing, the continuous state and the continuous of the original random process $E(t)$ is translated to the DS and TC of the random process $E_{D} (t)$, the random process $E_{D} (t)$ has $N$($N = 1,2, \cdots ,N$) different outputting level $g_{i}$, while the random process $E_{D} (t)$ is expressed by using the Markov stochastic process (MSP), its transition stats is shown as Fig. 1.

Fig. 1.

The generation unit is established to MSMM

It is shown as Fig. 1, $a_{i,j}$ is the state $i$ translating to the state $j$, $a_{N - 1,\,N - 2}$ is the state $N - 1$ translating to the state $N - 2$, the other and so on. Each state $i$ are corresponding to the output energy $g_{i}$ of the engine. $T_{i}^{m}$ is the mth sojourn time at the state $i$ of the unit. $\{ T_{i}^{(1)} ,T_{i}^{(2)} , \cdots ,T_{i}^{{(k_{i} )}} \}$ is the sample of the unit service period. $k_{ij}$ is the unit number from state $i$ translating to the state $j$, $k_{i}$ is the sojourn number in the state $i$ of the unit. We can obtained the transfer intensity $a_{i,j}$ using the real number that will be observed from the discrete state and time continuous stochastic process.

3 The Transfer Intensity of the Engine Unit Will be Determined

$E_{D} (t)$ is a DS and TC Markov process, we neglected the translation time of the unit state from $i$ $j(i \ne j)$, the transfer moment only is interested, so $E_{D} (t)$ is considered as a DS and time discrete (TD) random process, it is noted as $E_{Di} (n),n = 0,1,2, \cdots$, its transition probability of one step is noted as $\pi_{ij} ,i,j = 1,2, \cdots N$.

The unit from state $i$ translating to the state $j(i \ne j)$ whose The cumulative probability distribution function is as follow.

$$ Q_{ij} (t) = 1 - e^{{ - a_{ij} t}} $$

(1)

The probability of the first transfer of the unit state from $i$ to $j$ before the moment $t$ is noted as $H_{ij} (t),i,j = 1, \cdots ,N$, the random process $E_{D} (t)$ whose the nuclear matrix $H(t)$ of will be consisted by all $H_{ij} (t)$.

$$ H_{ik} (t) = \int\limits_{0}^{t} {[1 - Q_{i1} (u)] \cdots [1 - Q_{ik - 1} (u)]} [1 - Q_{ik + 1} (u)] \cdots [1 - Q_{ik - 1} (u)]dQ_{ik} (u)] $$

(2)

From Eqs. (1) and (2), it is obtained as follow.

$$ H_{ik} (t) = \frac{{a_{ik} }}{{\sum\limits_{j = 1}^{n} {a_{ij} } }}[1 - e^{{ - \sum\limits_{j = 1}^{N} {a_{ij} t} }} $$

(3)

The unit sojourn time $T_{i}$ whose cumulative probability distribution function in the state $i$ is as follow.

$$ Q_{i} (t) = \sum\limits_{k = 1}^{N} {H_{ik} (t) = 1 - } e^{{ - \sum\limits_{j = 1}^{N} {a_{ij} t} }} $$

(4)

From Eq. (4), the $T_{i}$ is obeyed to the exponential distribution, $Q_{i} (t)$ mean is follow.

$$ T_{imean} = \frac{1}{{\sum\limits_{j = 1}^{n} {c_{ij} } }} = \frac{1}{A} $$

(5)

Where $A = \sum\limits_{j = 1}^{n} {a_{ij} }$.

The mean that is obtained by using the observation samples is noted as follow.

$$ \hat{T}_{imean} = \frac{{\sum\limits_{j = 1}^{{k_{i} }} {T_{i}^{(j)} } }}{{k_{i} }} $$

(6)

It is obtained by using Eqs. (5) and (6) as follow.

$$ \hat{A} = \frac{1}{{\hat{T}_{imean} }} = \frac{{k_{i} }}{{\sum\limits_{j = 1}^{{k_{i} }} {T_{i}^{(j)} } }} $$

(7)

We can obtain the transition probability of one step to use the embedded Markov random processing.

$$ \pi_{ij} = \mathop {\lim }\limits_{t \to \infty } H_{ij} (t) $$

(8)

On the formula was simplified as follow.

$$ \pi_{ij} = \frac{{c_{ik} }}{{\sum\limits_{j = 1}^{N} {c_{ij} } }} $$

(9)

It is obtained by using the above equation.

$$ a_{ik} = \pi_{ij} \sum\limits_{j = 1}^{N} {a_{ij} } $$

(10)

The single transition intensity in the state $i$ can be obtained by using Eq. (10).

We can obtain the transition probability of single step of the embedded Markov chain (EMC) using the unit work energy which can be observed in the random process.

$$ \hat{\pi }_{ik} = \frac{{k_{ik} }}{{k_{i} }} $$

(11)

We can compute the transition intensity by using Eqs. (7), (10) and (11).

$$ \hat{a}_{ik} = \pi_{ik} \hat{C} = \frac{{k_{ik} }}{{k_{i} }}\frac{1}{{\hat{T}_{imean} }} = \frac{{k_{ik} }}{{\sum\limits_{j = 1}^{{k_{i} }} {T_{i}^{(j)} } }},i,k = 1, \cdots ,N $$

(12)

For the MSM system that has $N$ states, it has as follow.

$$ \sum\limits_{j = 1}^{N} {a_{ij} } = 0 $$

(13)

So:

$$ \hat{a}_{ii} = - \sum\limits_{\begin{subarray}{l} j = 1 \\ i \ne j \end{subarray} }^{N} {a_{ij} } $$

(14)

Above all, there is an algorithm for model of the MSM system that has $N$ states.

• Step 1. The system will be quantitative processed by using the method in the 2nd part content.

• Step 2. The unit in the every state $i$ total sojourn time is computed.

• $$ T_{{\sum\limits_{i} {} }} = \sum\limits_{m = 1}^{{k_{i} }} {T_{i}^{(m)} } $$

  (15)

• Step 3. The transition intensity of the state from $i$ to state $j$ can be computed by using Eqs. (16) and (17).

• $$ \hat{a}_{ij} = \frac{{k_{ij} }}{{T_{{\sum\limits_{i} {} }} }},i \ne j. $$

  (16)
• $$ \hat{a}_{ii} = - \sum\limits_{\begin{subarray}{l} j = 1 \\ i \ne j \end{subarray} }^{N} {a_{ij} } $$

  (17)

4 The MSMM and the Engine Short Time Reliability

4.1 The Transition Intensity Will be Computed

Commonly, a fuel engine output power of is 288 kW, its service deadline is $T = 5$ years, the Markov will be established by using the above algorithm.

$e_{1} = 0\;\text{kW}$, $e_{4} = 288\;\text{kW}$ can be computed by using the method.

$$ \Delta e = \frac{288}{4 - 2} = 144\;\text{kW} $$

(18)

Then the output energy of the engine can be divided to two intervals as $[0,144]\;\text{kW}$, $[144,288]\;\text{kW}$.

The remainder two output level is $e_{2} = 123\;\text{kW}$, $e_{3} = 241\;\text{kW}$, respectively. The state transition intensity $\hat{a}_{ij}$ from state $i$ to the state $j$ as follow.

$$ \left[ {\begin{array}{*{20}c} { - 0.0933} & {0.0800} & {0.0133} & 0 \\ {0.0294} & { - 0.3823} & {0.3235} & {0.0294} \\ 0 & {0.0288} & { - 0.3846} & {0.3558} \\ {0.0002} & {0.0001} & {0.0007} & {0.0010} \\ \end{array} } \right] $$

(19)

The number of the transition and the sojourn time is shown in Table 1.

Table 1. The transition number, residence time

4.2 MSS Model of Four-State

The state transition for four state of the MM is as the follow figure. The state 1, 2, 3 and 4 steady state probability is respectively shown as follows (Fig. 2).

Fig. 2.

Engine unite is divided to four-state MSS

$$ P_{1} = 0.0018,P_{2} = 0.0008,P_{3} = 0.0025,P_{4} = 0.9949 $$

(20)

The probability $P_{i} (t),(i = 1,2,3,4)$ $i$ is computed after the differential equations (21a)–(21d) is solved.

$$ \frac{{dP_{1} (t)}}{dt} = - (a_{12} + a_{13} + a_{14} )P_{1} (t) + a_{21} P_{2} (t) + a_{31} P_{3} (t) + a_{41} P_{4} (t) $$

(21a)

$$ \frac{{dP_{2} (t)}}{dt} = a_{12} P_{1} (t) - (a_{21} + a_{23} + a_{24} )P_{2} (t) + a_{32} P_{3} (t) + a_{42} P_{4} (t) $$

(21b)

$$ \frac{{dP_{3} (t)}}{dt} = a_{13} P_{1} (t) + a_{23} P_{2} (t) - (a_{31} + a_{32} + a_{34} )P_{3} (t) + a_{43} P_{4} (t) $$

(21c)

$$ \frac{{dP_{4} (t)}}{dt} = a_{14} P_{1} (t) + a_{12} P_{2} (t) + a_{13} P_{3} (t) - (a_{41} + a_{42} + a_{43} )P_{4} (t) $$

(21d)

The probability of the state $i$ can be obtained by solving the differential equation (21) under the special initial conditions in any time, the reliability of the engine can be computed too.

According to the probability of the state, the engine stability probability when its state is the state $i$ can be obtained by using the follow equation.

$$ p_{i} = \mathop {\lim }\limits_{t \to \infty } P_{i} (t),(i = 1,2,3,4) $$

(22)

4.3 The Engine Reliability from Time Response

For the engine system, the evaluation for the rate of the work outage (ROW) of the engine is an important. ROW is a whole failure probability for the engine, that is to say the engine work energy is zero, its function about time that is stopped at the state one as follow.

$$ ROW(t) = P_{1} (t) $$

(23)

Of course, the result of ROW(t) depend on the initial conditions.

It is clearly, the initial conditions can be preset as follows.

$$ P_{1} (0) = 0,P_{2} (0) = 0,P_{3} (0) = 0,P_{4} (4) = 1 $$

(24)

The result computed is shown in the Fig. 3. It is shown that the engine will be gone to stability after 81 h, the stability probability at the state 1 is as follow in this time.

Fig. 3.

ROW(t) under the initial condition (24) in the 1000 time

$$ p_{1} = \mathop {\lim }\limits_{t \to \infty } P_{1} (t) = 0.0018 $$

(25)

If the initial conditions can be preset as follows.

$$ P_{1} (0) = 0,P_{2} (0) = 0,P_{3} (0) = 1,P_{4} (4) = 0 $$

(26)

ROW(t) is shown as Fig. 4.

Fig. 4.

ROW(t) under the initial condition (26) in the 1000 time

If the initial conditions can be preset as follows.

$$ P_{1} (0) = 0,P_{2} (0) = 1,P_{3} (0) = 0,P_{4} (4) = 0 $$

(27)

ROW(t) is shown as Fig. 5.

Fig. 5.

ROW(t) under the initial condition (27) in the 1000 time

If the initial conditions can be preset as follows.

$$ P_{1} (0) = 1,P_{2} (0) = 0,P_{3} (0) = 0,P_{4} (4) = 0 $$

(28)

ROW(t) is shown as Fig. 6.

Fig. 6.

ROW(t) under the initial condition (28) in the 1000 time

It is shown in Figs. 3, 4, 5 and 6, the ROW of the engine is stabilized 0.0018 under the initial conditions (24), (26), (27) and (28). It is also shown in the four figures, the maximum for ROW under the initial conditions (26)–(28) is larger more than the maximum for ROW under the initial conditions (24), it is because the engine will be more happened malfunction. Obviously, the engine has been turn to the fault state while it is begun to run under the Eq. (27), it is confirmed to the actual situation.

When $W = 200\;\text{kW}$ at the 1000th h, the state would transits to the state 2, the work energy could not reach to 123 kW, it is to say, that the deficiency capacity (DC) will produce.

$$ DC2 = (W - 123) = 76\;\text{kW} $$

(29)

If the state become to the state 1, the work energy can not reach the required, it is to say, that the DC) will produce as follow.

$$ DC1 = (W - 0) = 200\;\text{kW} $$

(30)

The expected deficiency capacity (EDC) is too a function from time response, it is shown as follow.

$$ EDC(t) = P_{2} (t)DC2 + P_{1} (t)DC1 $$

(31)

The EDC is shown in Figs. 7, 8, 9 and 10 under the initial condition (24), (26), (27) and (28).

Fig. 7.

EDC(t)4 under the initial condition (24) in the 1000 time

Fig. 8.

EDC(t)3 under the initial condition (26) in the 1000 time

Fig. 9.

EDC(t)2 under the initial condition (26) in the 1000 time

Fig. 10.

EDC(t)1 under the initial condition (28) in the 1000 time

5 Conclusion

For the MS engine system, we discrete the engine system and quantization processed, we constructed the multi-state Markov model. We given the algorithm of transition intensities according to the observed data. The ROW and the EDC of the engine changing over time can be calculate by the method. This method can be applied to calculate the generator system failure rate changes from time response and the output of the energy deficit expectations. Finally, it is shown that the method is feasibility, effectiveness and practicability by using an engineering practical example.