Keywords

1 Introduction

During last years an intensive attention to the aging and degradation models for technical and biological objects has been attracted. Some wearing and aging models were in focus of many investigators in the framework of shock and damage models. An excellent review and contribution to the earlier papers devoted to the topic one can be found in [4, 16, 18]. The aging and degradation models suppose the study of systems with gradual failures, for which a large number of multi-state reliability models have been elaborated, see e.g. [17, 19].

In this paper, we consider a degradation process in the system with gradual and instantaneous failures proposed in [3]. The main purpose of the work is to develop estimation a small failure probability, in which case crude Monte Carlo simulation turns out to be time-consuming. Following [3], we use the regenerative structure of the basic degradation process, but unlike [3], now we allow the basic process to be non-Markovian, in which case a speed-up simulation, the so-called splitting method, is applied to estimate the failure probability with a given accuracy in an acceptable simulation time. The standard splitting algorithm [5, 7, 12] accelerates the rare event (failure) by means of the splitting each trajectory of the process, upon reaching the predefined thresholds, onto a few independent copies.

The main contribution of this work is to apply a regenerative variant of the splitting, developed in [1, 2], to estimate the failure probability in the non-Markov analogue of the system considered in  [3]. This approach also allows to avoid inversion of the Laplace transform of a convolution of the (degradation) phases. This inversion becomes highly non-accurate even for a moderate number of the exponential degradation phases when we deals with asymptotic reliability function, see Sect. 2.4. The numerical results demonstrate a significant gain in simulation time of the speed-up simulation over the crude (standard) Monte-Carlo simulation. Moreover, the proposed approach also allows to effectively estimate the average length of regeneration cycle with and without failure, the mean (unconditional) cycle length and also to calculate the asymptotic reliability function.

It is worth mentioning that, by the structure of the basic process, the splitting method turns out to be highly adaptive to estimate the required performance measures in the model under consideration. It is in a strong contrast to the application of this method to majority of other models. The main advantage of our model is that each path of degradation process can only increases, while in other models any path can, typically, both increases and decreases, significantly increasing simulation time. Finally, regenerative splitting provides confidence estimation of the required (small) failure probability with a given accuracy.

The paper is organized as follows. In Sect. 2.1, the basic degradation process is described in detail, including description its regenerative structure. In Sect. 2.2, the expressions for the mean regeneration cycle, with and without failure, are deduced. In Sect. 2.3, we focus on i.i.d. exponential degradation phases, in which case analytical results for main parameters are obtained. In Sect. 2.4, the asymptotic reliability function is briefly discussed. In Sect. 3, a brief introduction to the splitting method is given. Finally, simulation results and their comparison with the analytical results, where exist, are given in Sect. 4. We conclude our analysis with Sect. 5.

2 Main Results

2.1 Degradation Process

In this section, we describe the degradation of the system by a random process \(X:=\{X(t)\}_{t\ge 0}\), with a finite state space \(E=\{0, 1, \dots , L, \dots , M, \dots , K; F\}\) describing the degradation stage of the system. In other words, \(X(t)=i\) means that the the system is in the ith stage at instant t. As we show, the states (stages) \(L,\,K,\,M\) and F play a specific role. We assume that the process starts in state \(X(0)=0\) and crosses (step-by-step) \(K-1\) intermediate degradation states. Upon reaching the state M, the following two events are possible. The process X visits the state K, where preventive repair is performed during random time \(U_K\) with a given distribution. Afterwards the process returns to stage L. Otherwise, if during a random time V, the process is still at some intermediate stage \(j\in [M,\,K)\), then an instantaneous failure occurs. As a result, the process X jumps to a complete failure state F. Then the system is unworkable during a random repair time \(U_F\), with a given distribution. After repair, the process returns to initial state 0.

This description is motivated by the application of the controllable model based on recovery policy defined by two parameters (L, K) [3]. In this paper, in particular, the control principles have been illustrated by means of a number of real examples, such as corrosion of the protective covering, damage process due to the fatigue crack growth, wear of a plane bearing, and so on.

For the further analysis, we need the following notations from  [3]. Denote by \(T_i\) the transition time from state i to \(i+1\), \(i\in E\setminus \{K,F\}\). We assume \(\{T_i\}\) to be independent, and call the corresponding degradation process homogeneous, if \(\{T_i\}\) are independent, identically distributed (i.i.d.). Namely this case is studying in the present paper. Note that

$$\begin{aligned} S_{i\,j}:=\sum _{k=i}^{j-1}T_k,\, \,\,\;\;0 \le i \le K-1,\, j>i, \end{aligned}$$

is the transition time from state j to state i. We define the following distribution functions:

$$\begin{aligned}&F_i(t)=\mathbb {P}(T_i \le t) ;\;\; F_V(t)=\mathbb {P}(V \le t);\end{aligned}$$
(1)
$$\begin{aligned}&F_{i\,j}(t)=\mathbb {P}(S_{i\,j}\le t) = F_{i,\,j-1}*F_j(t) = \int _{0^-}^{t}F_{i,\,j-1}(t-v)dF_j(v); \end{aligned}$$
(2)
$$\begin{aligned}&F_{i,\,i+1}(t) = F_i(t),\;\;\; F_{i\,i}(t)\equiv 0, \end{aligned}$$
(3)

where \(*\) means convolution.

Fig. 1.
figure 1

Realization of the degradation process with two types of cycle

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2.2 Basic Relations for General Case

By construction the process X has a regenerative structure. We will consider the regeneration points \(\{\tau _n\}\) as the return times to state M. Denote by \(Z_k,\,k\ge 1\), the transition instants of the process X. Then the regeneration points are recursively defined as

$$\begin{aligned} \tau _{n+1} =\{Z_i > \tau _n:\, X(Z_i^+)=M\},\,\,n\ge 0,\,\,\tau _0:=0. \end{aligned}$$

and \(Y_k= \tau _{k+1} - \tau _k,\,k\ge 1\), are the i.i.d. lengths of the regeneration cycles. As it is seen in Fig. 1, after the process X hits state M, an instantaneous failure may happen during period V, otherwise, a preventive repair occurs during time \(S_{MK}\). Thus, there are two types of the regeneration cycles of the process X.

Denote by \(Y_F = V+U_F+S_{0M},\) the length of the (generic) regeneration cycle with failure, and \(Y_{NF} = S_{MK}+U_{KL}+S_{LM}\) the length of cycle ended by the preventive repair. Thus, an unconditional (generic) length of a regenerative cycle Y is

$$\begin{aligned}&Y= Y_F\cdot I_{\{V \le S_{MK}\}} + Y_{NF}\cdot I_{\{S_{MK} < V\}}, \end{aligned}$$
(4)

where I denotes indicator function. Now we calculate the main performance measures of the stationary process X. To find the mean regeneration cycle length \(\mathbb {E}[Y]\), consider the probability of a failure during regeneration cycle,

$$\begin{aligned} p_F = \mathbb {P}(S_{MK} \ge V)= \int _{0}^{\infty }(1 - F_{MK}(t))dF_V(t). \end{aligned}$$
(5)

By definition (4), it immediately follows that

$$\begin{aligned} \mathbb {E}[Y]&= \mathbb {E}[V\cdot I_{\{V\le S_{MK}\}}]+\mathbb {E}[S_{MK}\cdot I_{\{V> S_{MK}\}}]\\&+\,\mathbb {E}[(U_F+S_{0M})\cdot I_{\{V \le S_{MK}\}}] + \mathbb {E}[(U_{KL}+S_{LM})\cdot I_{\{V > S_{MK}\}}. \end{aligned}$$

Further, by the between \(U_F\), \(S_{0M}\), \(I_{\{V \le S_{MK}\}}\), and also by independence between \(U_{KL}\), \(S_{LM}\), \(I_{\{V> S_{MK}\}}\), we obtain

$$\begin{aligned} \mathbb {E}[Y]&= \mathbb {E}[\min \{V,S_{MK}\}] + (\mathbb {E}[U_F]+\mathbb {E}[S_{0M}])\mathbb {P}(V\le S_{MK})\\&+\,(\mathbb {E}[U_{KL}]+\mathbb {E}[S_{LM}])\mathbb {P}(V> S_{MK})\nonumber \\&= \mathbb {E}[\min \{V,S_{MK}\}] + (\mathbb {E}[U_F]+\mathbb {E}[S_{0M}])p_F \nonumber \\&+\,(\mathbb {E}[U_{KL}]+\mathbb {E}[S_{LM}])(1-p_F). \nonumber \end{aligned}$$
(6)

Also by (3),

$$\begin{aligned}&\mathbb {E}[\min \{V,S_{MK}\}]=\int _{0}^{\infty }(1-F_V(t))(1 - F_{MK}(t))dt, \\&F_{MK}(t) = \int _{0}^{t}F_{M,K-1}(t-v)dF_K(v). \end{aligned}$$

Denote \(T_F\) the length of cycle up to an instantaneous failure, provided it happens, and assume that the density \(F^{'}_{V}(x)=f_V(x)\) exists. The following result is immediate:

$$\begin{aligned} \mathbb {E}[T_F]= & {} \mathbb {E}[V|V\le S_{MK}] = \int _0^{\infty }\frac{y \,f_V(y)\,\mathbb {P}(S_{MK}\ge y)}{\mathbb {P}(V\le S_{MK})}\,dy \nonumber \\= & {} \frac{1}{p_F}\int _0^{\infty }y f_V(y)(1 - F_{MK}(y))dy. \end{aligned}$$
(7)

Now we calculate the length of a regeneration cycle with failure. By the independence between \(U_F\), \(S_{0M}\), \(I_{\{V \le S_{MK}\}}\), and by (4):

$$\begin{aligned} \mathbb {E}[Y_F] = \mathbb {E}[Y | V \le S_{MK}] = \mathbb {E}[T_F] + \mathbb {E}[U_F] + \mathbb {E}[S_{0M}]. \end{aligned}$$
(8)

Assume that the density \(F^{\,'}_{S_{MK}}(x):=f_S(x)\) exists, and find the mean length of the regeneration cycle without failure. Using (3), (4) and the independence between \(U_{KL}\), \(S_{LM}\), \(I_{\{V > S_{MK}\}}\), we obtain

$$\begin{aligned} \mathbb {E}[Y_{NF}]= & {} \mathbb {E}[Y | S_{MK}< V] = \mathbb {E}[(S_{MK}+U_{kl}+S_{LM}) | S_{MK}<V] \nonumber \\= & {} \frac{1}{1 - p_F}\int _0^{\infty }y \,f_{S}(y)\,(1 - F_V(y))\,dy + \mathbb {E}[U_{KL}] + \mathbb {E}[S_{LM}]. \end{aligned}$$
(9)

These performance measures include convolutions, and can be explicitly calculated in a few special cases only. Moreover, solution normally includes Laplace-Stieltjes transform, and to obtain explicit expression, inversion of the transform is required. However a low accuracy is the weakest point of this numerical approach. Nevertheless the model with i.i.d. exponential degradation stages allows an explicit solution. Below we use it to verify the accuracy and efficiency of the splitting simulation method, which can be applied to more general setting, where analytical solution is unavailable.

2.3 The Exponential Degradation Stages

Recall that we consider the homogeneous case, with i.i.d. exponential \(\{T_i\}\). It allows to find \(\mathbb {E}[T_F]\), \(\mathbb {E}[Y_{NF}]\) analytically, provided V and \(U_{KL}\) are exponential as well.

Theorem 1

For the i.i.d. exponential \(\{T_i\}\) with parameter \(\lambda \) and exponential V with parameter \(\nu \),

$$\begin{aligned} \mathbb {E}[T_F]&= \frac{1}{\nu p_F}\Big [1-\frac{\lambda +(K-M+1)\nu }{\lambda +\nu }\Big (\frac{\lambda }{\lambda +\nu }\Big )^{K-M}\Big ], \end{aligned}$$
(10)

where

$$\begin{aligned} p_F = 1-\Big (\frac{\lambda }{\lambda +\nu }\Big )^{K-M}. \end{aligned}$$
(11)

Proof

In this case \(S_{MK}\) has Erlang distribution (Erlang (\(\lambda ,\,K-M\)))

$$\begin{aligned} F_{MK} (x) = 1 - e^{-\lambda y}\sum _{i=0}^{K-M-1}\frac{(\lambda y)^i}{i!}. \end{aligned}$$
(12)

Then it follows from (7) that

$$\begin{aligned} \mathbb {E}[T_F]= & {} \frac{\nu }{p_F}\int _0^{\infty }ye^{-\nu y} e^{-\lambda y}\sum _{i=0}^{K-M-1}\frac{(\lambda y)^i}{i!}dy \nonumber \\= & {} \frac{\nu }{p_F} \sum _{i=0}^{K-M-1}\frac{\lambda ^i}{i!}\int _0^{\infty }y^{i+1}e^{-(\lambda + \nu ) y}dy. \end{aligned}$$
(13)

Because, see [13],

$$\begin{aligned} \displaystyle \int _0^{\infty }y^{n}e^{-\mu y}dy = n!\mu ^{-n-1}, \end{aligned}$$
(14)

then we obtain from (13)

$$\begin{aligned}&\mathbb {E}[T_F]= \frac{\nu }{p_F (\lambda + \nu )^2} \sum _{i=0}^{K-M-1}\Big (\frac{\lambda }{\lambda +\nu }\Big )^i (i+1). \end{aligned}$$

It is known [13] that,

$$\begin{aligned} \sum _{i=0}^{n-1}(a+ir)\rho ^a = \frac{a-(a+(n-1)r)\rho ^n}{1-\rho } + \frac{r\rho (1-\rho ^{n-1})}{(1-\rho )^2}. \end{aligned}$$
(15)

Taking \(n=K-M, \,a=1, \,r=1\), we arrive to

$$\begin{aligned} \mathbb {E}[T_F]= & {} \frac{1}{p_F}\Big [\frac{1}{\nu } - \Big (\frac{\lambda }{\lambda +\nu }\Big )^{K-M}\Big [\frac{(K-M)\nu +\lambda +\nu }{\nu (\lambda + \nu )}\Big ]\Big ]. \end{aligned}$$
(16)

Because V is exponential, it follows from (5) and (12) the followinf equality

$$\begin{aligned} p_F= & {} \nu \int _{0}^{\infty }e^{-(\lambda + \nu ) y}\sum _{i=0}^{K-M-1}\frac{(\lambda y)^i}{i!}dy =\frac{\lambda }{\lambda + \nu }\sum _{i=0}^{K-M-1}\Big (\frac{\lambda }{\lambda +\nu }\Big )^{i}, \end{aligned}$$
(17)

which is equivalent to (11).

Theorem 2

For the i.i.d. exponential \(\{T_i\}\) with parameter \( \lambda \) and exponential \(U_{KL}\) with parameter \(\mu \),

$$\begin{aligned}&\mathbb {E}[Y_{NF}]=\frac{K-M}{(1-p_F)(\lambda +\nu )}\Big (\frac{\lambda }{\lambda +\nu }\Big )^{K-M} +\frac{1}{\mu }+\frac{M-L}{\lambda }, \end{aligned}$$
(18)

where

$$\begin{aligned} p_F = 1-\Big (\frac{\lambda }{\lambda +\nu }\Big )^{K-M}. \end{aligned}$$
(19)

Proof

Since \(S_{LM}\) is Erlang \((\lambda ,\,M-L)\) and \(U_{KL}\) is exponential, then

$$\begin{aligned}&\mathbb {E}[U_{KL}] + \mathbb {E}[S_{LM}] = \frac{1}{\mu }+\frac{M-L}{\lambda }. \end{aligned}$$
(20)

As above, using (9) and (14), we obtain

$$\begin{aligned}&\int _0^{\infty }y \frac{f_{S}(y)(1 - F_V(y))}{1 - p_F}dy \nonumber \\&= \int _{0}^{\infty }y^{K-M}\frac{\lambda ^{K-M}}{(K-M-1)!}e^{-(\lambda + \nu )y}dy =\frac{K-M}{\lambda + \nu }\Big (\frac{\lambda }{\lambda +\nu }\Big )^{K-M}. \end{aligned}$$
(21)

By (9), (20) and (21), now (18) follows immediately.

For this setting (and exponential \(U_F\) with parameter \(\mu _F\)), the average unconditional regeneration cycle length is

$$\begin{aligned} \mathbb {E}[Y]= & {} \frac{1}{\nu }\Big (1-\Big (\frac{\lambda }{\lambda +\nu }\Big )^{K-M}\Big ) \nonumber \\+ & {} \Big (\frac{1}{\mu }+\frac{M-L}{\lambda }\Big )(1-p_F) + \Big (\frac{1}{\mu _F}+\frac{M}{\lambda }\Big ) p_F, \end{aligned}$$
(22)

where \(p_F\) satisfies (11). Finally note that, by (10),

$$\begin{aligned}&\mathbb {E}[\min \{V,S_{MK}\}] = \int _{0}^{\infty }e^{-(\lambda + \nu ) y}\sum _{i=0}^{K-M-1}\frac{(\lambda y)^i}{i!}dy = \frac{p_F}{\nu }. \end{aligned}$$
(23)

2.4 Asymptotic Reliability Function

The most important reliability descriptor of operation quality is a reliability function

$$\begin{aligned} R(t)=\mathbb {P}[T>t|X(0)=0],\,\,t\ge 0, \end{aligned}$$

where T stands for the life time of the system. In our model, the mean life time can be written as

$$\begin{aligned} \mathbb {E}[T]=\mathbb {E}[Y_{NF}](\mathbb {E}[N]-1)+\mathbb {E}[T_F], \end{aligned}$$

where \(\mathbb {E}[N]:=1/p_F\) is the mean number of cycles until complete failure. This number follows geometric distribution with parameter \(1-p_F\). For exponential (pure Markovian) reliability model, the classical evaluation method of R(t) includes solution of the Kolmogorov differential equations for the state probabilities

$$\begin{aligned} \pi _i(t)=\mathbb {P}[\hat{X}(t)=i|\hat{X}(0)=0], \end{aligned}$$

of an auxiliary Markov process \(\{\hat{X}(t),\,t\ge 0\}\) with an absorption state F. In this case,

$$\begin{aligned} R(t)=\sum _{x\in E\setminus \{F\}}\pi _i(t). \end{aligned}$$

Another way is to evaluate the density function

$$\begin{aligned} r_x(t)=\frac{1}{dt}\mathbb {P}[T\in [t,t+dt)|\hat{X}(0)=x] \end{aligned}$$

of the remaining life time given the initial state is x, see [3], in which case the reliability function becomes,

$$\begin{aligned} R(t)=1-\int _0^t r_0(u)du,\,\,\,t\ge 0. \end{aligned}$$

In both methods the results are obtained in terms of Laplace transforms (LT). However, these results in general are not applicable, provided a (natural) assumption that the failure is a rare event. In this case the required inversion of corresponding LT is hardly available, while numerical methods lead to highly incorrect result. Alternatively, it is possible to use the asymptotic of reliability function R(t). This analysis is based on the renewal theory and the limit properties of regenerative processes, see, for instance, [14, 15, 20, 21]. The main result for our model can be formulated as the following limit statement provided failure becomes rarer [21]:

$$\begin{aligned} R\Big (\frac{\mathbb {E}[Y_{NF}]}{p_F} \, t\Big )= \mathbb {P}\Big (T\ge \frac{\mathbb {E}[Y_{NF}]}{p_F} \, t\Big ) \rightarrow e^{-t},\,\,\,p_F\rightarrow 0, \end{aligned}$$
(24)

uniformly in t. It can be rewritten as

$$\begin{aligned} R(t) \rightarrow e^{-t\,\frac{p_F}{\mathbb {E}[Y_{NF}]}},\,\,\,p_F\rightarrow 0. \end{aligned}$$
(25)

In other words, the reliability function is asymptotically exponential, and the mean life time becomes (approximately)

$$\begin{aligned} \mathbb {E} [T]=\int _0^\infty R(t)dt=\frac{\mathbb {E}[Y_{NF}]}{p_F}. \end{aligned}$$
(26)

This result is intuitive because, as we mention above, the number of cycles up to the (first) failure is geometric with parameter \(1-p_F\).

To determine the reliability function for moderate \(p_F\), neither analytic formulas nor asymptotic relations can be effectively used. To overcome this difficulty, a speed-up simulation (splitting) method is applied below, to effectively estimate the reliability function for arbitrary \(p_F\).

3 Splitting Method for Degradation Process

In this section, we briefly describe the so-called splitting method developed for the accelerated estimation of rare events, see [5,6,7, 11, 12]. We start the splitting upon the process X reaches stage M, after which an instantaneous failure becomes possible, see Fig. 1.

First, we estimate \(p_F\) and cycle parameters \(\mathbb {E}[Y]\), \(\mathbb {E}[Y_{NF}]\), \(\mathbb {E}[Y_F]\) by means of a combination of splitting and regenerative simulation. Then we compare obtained estimates with result of Crude Monte-Carlo simulation, and with analytical results, when available.

Consider a modification of the standard splitting, which starts at the regeneration instants \(\tau _k\), when the process reaches stage M, see Fig. 1 The hitting each next degradation stage i corresponds to the ith splitting threshold in the standard splitting, when \(R_i\) copies of the random variables (r.v.) \(T_i\) is realised, \(M \le i \le K-1\). (We take \(R_M = 1\)). Recall that all new i degradation processes evolve independently. Then each original path generates \(D=R_M\cdots R_{K-1}\) (dependent) regeneration cycles called group of cycles. The dependence is generated by the same pre-history of realizations of \(S_{MK}\) before splitting point at each stage. Thus, we obtain D realizations of \(S_{MK}\) for one group of cycles. The cycles belonging to different groups are independent by construction. The total number of groups is denoted by \(R_{M-1}\). The total number of the failures in the ith group is

$$\begin{aligned} A_i=\sum _{j=(i-1)\cdot D+1}^{i\cdot D}I^{(j)}, i=1, \dots , R_{M-1}, \end{aligned}$$

where indicator \(I^{(j)}=1\) for the cycle with failure (\(I^{(j)}=0\), otherwise), and the groups are i.i.d. The sequence \(\{I^{(j)}, j \ge 1\}\) is discrete regenerative with constant cycle length \(\beta _i=D\) and regeneration instants \(i\cdot D\), \(i \in [1, R_{M-1}]\). In particular, the unbiased point estimator \(\hat{p}_F\) of \(p_F\) is strongly consistent: as \(R_{M-1} \rightarrow \infty \), w.p.1

$$\begin{aligned} \hat{p}_F = \frac{\sum _{j=1}^{R_{M-1}}A_j}{R_{M-1}\cdot D} \rightarrow \frac{\mathbb {E}\sum _{j=1}^{D}I^{(j)}}{D} = p_F. \end{aligned}$$
(27)

To construct a confidence interval for the required parameters the regenerative approach is also applicable [8,9,10]. It follows from [9] that the asymptotic \(100(1-\delta )\%\) confidence interval for \(p_F\) is

$$\begin{aligned} \Big [\hat{p}_F \pm \frac{z(\delta )\sqrt{v_n}}{\sqrt{n}} \Big ], \end{aligned}$$
(28)

where \(z(\delta )\) is defined by \(\mathbb {P}[N(0,1) \le z(\delta )] = 1 - \delta /2\), N is standard normal variable, and

$$\begin{aligned} v_n = \frac{n^{-1}\sum _{i=1}^{n}[A_i - \hat{p_F}D]^2}{D^2} \end{aligned}$$
(29)

is a weakly consistent estimator of \(\sigma ^2 = \mathbb {E}[A_1 - p_FD]^2/D^2\), that is

$$\begin{aligned} v_n\Rightarrow \sigma ^2,\,\,\,n \rightarrow \infty , \end{aligned}$$

in probability.

4 Simulation Results

In general, \(\{T_i\}\) not need to be identical. In particular, independent exponential \(\{T_i\}\) with different parameters \(\lambda _i\) are considered in [3]. Another distributions are not considered in [3] because in this case analytical results are hardly available, if possible. Conversely, in this paper we obtain numerical results for more general setting based on the slitting simulation method. To illustrate this method, we again consider the simplest exponential setting with \(\lambda _i = \lambda .\)

Example 1

First we find the main performance measures (cycle characteristics).

  1. (a)

    Assume the i.i.d. \(\{T_i\} \sim exp(\lambda )\), \(\nu = 2.5\), \(M = 5\), \(K = 17\), \(L = 1\), \(\mu _F = 1.5\), \(\mu = 2\), and let n be the number of simulated regeneration cycles. By (11), we have the following exact values of \(p_F\) for different values of \(\lambda \):

    $$\begin{aligned} \lambda =50:\, p_F = 0.443; \;\lambda =70:\, p_F = 0.343;\nonumber \\ \lambda =90:\, p_F=0.280;\;\,\lambda =110:\,p_F=0.236. \end{aligned}$$
    (30)

    We compare simulation results based on Crude Monte Carlo (MC), and results obtained by regenerative splitting (RS). All tests were executed on a Intel(R) Xeon(R) CPU E5-2630 2.30 GHz processor with 4 GB of RAM, running Linux openSUSE 42.2. We use 50 samples to find (in evident notation) the estimates the variances of \(\hat{p}_F\), respectively, \(Var_{MC}\), and \(Var_{RS}\). The results are given in Table 1. (simulation time \(t_{MC}\) and \(t_{RS}\) is measured in seconds) shows the estimates of \(p_F\).

    Note that both methods are well consistent with analytical expression in (11), while for the same number of cycles n, the RS is much less time-consuming than MC.

    Note that \(Var_{RS}\) is enough large, and \(Var_{RS}>Var_{MC}\), perhaps because \(p_F\) is not small. To construct \(100(1-\delta )\%\) confidence interval for \(p_F\), we use (28), and consider the case when the instantaneous failure is not very rare. We take \(\lambda =110\), in which case exact value \(p_F=0.236\). Figure 2 represents the \(95\%\) confidence intervals and the point RS estimates of \(p_F\).

    Table 1. Comparison of MC and RS estimators for exponential \(T_i\)
    Full size table
    Fig. 2.
    figure 2

    \(95\%\) confidence interval for \(p_F\)

    Full size image
  2. (b)

    Table 2 gives the estimates of the mean cycle length \(\mathbb {E}[Y]\). For both methods the number of cycles with failure, \(n_R\), is close for given number of cycles n. The number \(n_R\) for MC method is given in Table 2. Table 2 shows that both methods provide a good approximation of exact value \(\mathbb {E}[Y]\) in (22).

  3. (c)

    The estimates of \(\mathbb {E}[Y_{NF}]\), the mean length of cycle without failure are given in Table 3. Again, the RS and MC estimates are very close to the exact value given by (18).

Table 2. Comparison of MC and RS estimators for \(\mathbb {E}[Y]\)
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Table 3. Comparison of MC and RS estimators for \(\mathbb {E}[Y_{NF}]\)
Full size table
Fig. 3.
figure 3

(a) \(R_a(t)\) for \(p_F=0.236\); (b) \(R_a(t)\) for \(p_F=0.000118\)

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Example 2

To calculate the reliability function R(t) we instead use the asymptotic relation (25)

$$\begin{aligned} R_a(t):= e^{-D_F\, t}, \end{aligned}$$

where \(D_F= {p_F}/\mathbb {E}[Y_{NF}]\), assuming that a failure within cycle is to be rare event. We use both methods to obtain the estimate \(\hat{D}_F= \hat{p}_F/\hat{\mathbb {E}}[Y_{NF}]\), and use approximation (25) in the case of not rare failure, \(p_F=0.236\), \(\lambda =110\). Other parameters are taken as in Example 1. In Fig. 3(a) presents plots of function \(R_a(t)\) with parameter \(\hat{D}_F\) obtained by MC, RS, and \(D_F\) with exact values \(p_F\) and \(\mathbb {E}[Y_{NF}]\). Now we take parameters \(\nu =0.001\), \(\lambda =110\), \(\mu =2\), \(\mu _F=1.5\), \(K=18\), \(L=1\), \(M=5\). Then \(p_F=0.000118\) by (11). In this case we are unable to calculate function R(t) analytically, because of a problem to calculate Laplace transform. However, estimates obtained by MC and RS simulation are available and given by Fig. 3(b). In some cases, the value of function R(t) at a given point t is required. This problem can be solved by means of simulation as well. For instant, it is difficult to calculate the value R(30000) numerically, while both simulation methods gives satisfactory closeness:

$$\begin{aligned} R_a(30000)[MC] = 0.0043,\,\,\,\, R_a(30000)[RS] = 0.0037. \end{aligned}$$
(31)

In both examples, the proximity of estimated obtained by both methods is observed. Moreover, the first example demonstrates a high consistency between the exact value of \(D_F\) and the estimates obtained by both simulation methods. It is rather surprising because in this case the probability is not small. It is worth mentioning that, for the i.i.d exponential \(\{T_i\}\), the exact value R(t) is available by the numerical inversion of the following Laplace transform expression of R(t):

$$\begin{aligned} \tilde{R}(s)=\frac{\frac{\nu }{\lambda +\nu +s}\Big (\frac{\lambda }{\lambda +s}\Big )^M\sum _{i=0}^{K-M-1}\Big (\frac{\lambda }{\lambda +\nu +s}\Big )^i}{1-\frac{\mu }{\mu +s}\Big (\frac{\lambda }{\lambda +\nu +s}\Big )^{K-M}\Big (\frac{\lambda }{\lambda +s}\Big )^{M-1}}. \end{aligned}$$
(32)

In particular, it gives \(R(30000)=0.0044425\), which is consistent with (31). For the non-identical \(\{T_i\}\) (even exponential) the Laplace transform inversion becomes labour-consuming and highly inaccurate as the number of phases increases.

Table 4. Simulation results for \(T_i\) Weibull
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Example 3

Finally, we consider the i.i.d. \(\{T_i\}\) with light-tailed Weibull distribution

$$\begin{aligned} F_T(x) = 1- e^{-3x^4}, x \ge 0, \end{aligned}$$

implying \(\mathbb {E}[T]=0.6886\). Assume that time \(V=1/\nu \), that is constant. In this case it seems quite hard (if possible) to find the probability \(p_F\) analytically, while MC and RS estimates are available being close. At that, the RS method allows to reduce simulation time significantly. In particular, the estimation of probability of order \(10^{-7}\) by MC simulation is extremely long and often gives no results in an acceptable time (see empty cells in Table 4).

5 Conclusion

In this paper, we apply a speed-up simulation technique, called multilevel regenerative splitting, to estimate a small failure probability related to a degradation process. The dynamics of the degradation process is highly adapted to the application of this method. The key idea of the method is to treat the group of the degradation process paths, obtained by the splitting and having the same root, as a regenerative cycle. This approach leads to a modified splitting procedure to preserve properties of the estimates obtained by the standard splitting. This approach allows to compensate a lack or inefficiency of analytical methods discussed in [3]. In particular, simulation can be applied far outside exponential models. As experiments show the RS method gives a significant reduction of simulation time in comparison with the MC simulation. Both methods provide a good approximation of the analytical results, when they are available.