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### formerly Central European Journal of Mathematics

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# Laplace-Stieltjes transform of the system mean lifetime via geometric process model

Gökhan Gökdere
/ Mehmet Gürcan
Published Online: 2016-06-10 | DOI: https://doi.org/10.1515/math-2016-0034

## Abstract

Operation principle of the engineering systems occupies an important role in the reliability theory. In most of the studies, the reliability function of the system is obtained analytically according to the structure of the system. Also in such studies the mean operating time of the system is calculated. However, the reliability function of some systems, such as repairable system, cannot be easily obtained analytically. In this case, forming Laplace-Stieltjes transform of the system can provide a solution to the problem. In this paper, we have designed a system which consists of two components that can be repairable with the aging property. Firstly, the Laplace-Stieltjes transform of the system is formed. Later, the mean operating time of the system is calculated by means of Laplace-Stieltjes transform. The system’s repair policy is evaluated depending on the geometric process. This property provides the aging of the system. We also provide special systems with different marginal lifetime distributions to illustrate the theoretical results in this study.

MSC 2010: 62K05; 65R10; 90B25

## 1 Introduction

In many engineering systems, cold standby redundancy is an effective way to achieve high system reliability while preserving limited power resources. Cold standby redundancy technique use one or more redundant components that are unpowered, do not consume any energy and do not fail until being activated to replace a faulty online component. Whenever working component fails, then an available cold standby component is immediately powered up to take over the mission task. Some recent works on the research of the cold standby systems are in ([2, 7, 11, 12]).

In perfect repair model, it is assumed that the repair completely restores all properties of failed components. However, this is not always true in real world implementations. In practice, after the repair most repairable components are not “as good as new” because of the stresses. To pay attention to this problem, much work has been done by Brown and Proschan [1], Park [9], Kijima [4], Makis and Jardine [8]. For an imperfect repair model, it is more acceptable to consider that the successive operating times of the component after repair will be even shorter, while the consecutive repair times of the component after its failure will be even longer. For such a stochastic phenomenon, Lam [5, 6] studied a new repair-replacement policy and introduced a geometric process (GP) model. In this model, after the repair the successive operating times of the system are stochastically decreasing, while the consecutive repair times after the failure are stochastically increasing.

The following are the definition of stochastic order and geometric process, respectively, which can be seen from Ross [10] and Lam [6], respectively.

Given X and Y random variables. For all real α, X is said to be stochastically larger than Y or Y is stochastically smaller than X, if $P(X>α)≥P(Y>α).$

Furthermore, a stochastic process {Xn, n = 1, 2, } is called stochastically decreasing if Xnst Xn + 1 and stochastically increasing if Xnst Xn + 1.

Let {Xn, n = 1, 2, …} be a sequence of non-negative independent random variables. Under the condition that α is a positive constant, if the distribution function of Xn is Fn(t) = F(an-1t), t ≥ 0, then {Xn, n = 1, 2, …} is called a geometric process. The positive constant ˛is the ratio of the geometric process.

It is evident that:

• If α > 1, then the geometric process {Xn, n = 1, 2, …} is stochastically decreasing.

• If 0 < α < 1, then the geometric process {Xn, n = 1, 2, …} is stochastically increasing.

• If α = 1, then the geometric process {Xn, n = 1, 2, …} is a renewal process.

In this paper, we study two non-identical components called the component 1 and component 2 and one repairman. Initially, the component 1 begins to operate and the other component is in cold standby state. As soon as the operating component 1 fails, the standby component is switched on immediately and the component 1 is taken for repair. When the repair of the component 1 is completed and the component 2 is still working, the repaired component returns to the standby pool and is once again available to be used as working component. The system breakdown occurs when the working component fails while the other component is under repair.

The organization of this paper is as follows. In Section 2, we give some assumptions concerning failures and repairs that will be useful throughout the paper. In Section 3, we describe the system mean lifetime measured when the component 1 fails. In Section 4, we present Laplace-Stieltjes (LS) transform of the system mean lifetime. In Section 5, we give a Gamma distributed example and a Weibull distributed example to illustrate the theoretical results for the proposed model. In the last section, we summarize what we have done in the article.

## 2 Model assumptions

In this section, some assumptions concerning failures and repairs are given. Initially, a two component cold standby repairable system is set up, in which the component 1 operates while the component 2 is in cold standby state.

Failure time of the component k is the random variable ${X}_{k}^{\left(n\right)}$ with the distribution function $FXk(n)(t)=Fk(mkt),$

where k = 1, 2; t ≥ 0, ${m}_{k}=\left\{\begin{array}{ll}{a}^{n},\hfill & k=1\hfill \\ {a}^{n-1},\hfill & k=2\hfill \end{array}$, and a > 1 is the ratio of the GP for component k. The finite mean of ${X}_{k}^{\left(n\right)}$ is $E{X}_{k}^{\left(n\right)}=\frac{{T}_{k}}{{m}_{k}}$.

Repair time of component k is the random variable ${Y}_{k}^{\left(n\right)}$ with the distribution function $GYk(n)(t)=Gk(bn−1t),$

where k = 1, 2; t ≥ 0. The finite mean of ${Y}_{k}^{\left(n\right)}$is $E{Y}_{k}^{\left(n\right)}=\frac{1}{{b}^{n-1}{\mu }_{k}}$, where 0 < b < 1 is the ratio of the GP for component k.

Let${X}_{k}^{\left(1\right)},{X}_{k}^{\left(2\right)},...$ be successive working times and ${Y}_{k}^{\left(1\right)},{Y}_{k}^{\left(2\right)},...$ be successive repair times of the component k, (k = 1, 2). It is supposed that all these random variables are independent.

## 3 The mean lifetime of the system

In this section, we have studied a system consisting of two components which can be repaired. There are some difficulties to obtain the reliability function of a repairable system due to the repair time of the faulty components. When any component of the system breaks down and needs repair, it increases the operating time of the system. In such conditions the repaired component’s performance is lower than the new component. Apart from that, the repairing time increases after each break down and repair. Based on the above discussion, our proposed model for repairable system is defined as below.

Let τ12(a,b) denote the system mean lifetime under the assumption that at t = 0, the component 1 begins to work and the component 2 is inactive (on standby). Let τ(a,b) denote the system mean lifetime measured from the moment when the component 1 fails. It is evident that τ12(a,b) = X1 + τ(a,b), where X1 and τ(a,b) are independent random variables. Also, X1 is the random variable with the distribution function FX1 with finite mean EX1 = T1.

The system described above fails when for some n ≥1 either event

$An={X2(j)>Y1(j),X1(j)>Y2(j),j=1,2,....,n−1;X^2(n)≤Y1(n)}$

or event

$Bn={X2(j)>Y1(j),X1(j)>Y2(j),j=1,2,....,n−1;X2(n)>Y1(n),X^1(n)≤Y1(n)}$

occurs. Here ${\stackrel{^}{X}}_{1}^{\left(n\right)}$ and ${\stackrel{^}{X}}_{2}^{\left(n\right)}$ are random variables which have the same distribution as ${X}_{1}^{\left(n\right)}$ and ${X}_{2}^{\left(n\right)}$ given that ${X}_{1}^{\left(n\right)}<{Y}_{2}^{\left(n\right)}$ and ${X}_{2}^{\left(n\right)}<{Y}_{1}^{\left(n\right)}$, respectively. A possible process for events An and Bn are shown in Figure 1 and 2, respectively as follows

Fig. 1

A diagram for event An in cycle n.

Fig. 2

A diagram for event Bn in cycle n.

Let

$αik(n)=P{Xk(n)>Yi(n)}=∫0∞GY1(n)(t)dFXk(n)(t)$(1)

where i = 1, 2; k = 1, 2; ik. Note that, α12(n) is the probability that after the n-th repair the working component 2 does not fail while the component 1 is under (n + 1)-th repair, and α21(n) is the probability that after the (n + 1)-th repair, the working component 1 does not fail while the component 2 is under n-th repair. Then, using w(1), the probability that the system breakdown in the cycle n is obtained as

$α=P{(X2(n)≤Y1(n))∪[(X2(n)>Y1(n))]∩(X1(n)≤Y2(n))}=(1−α12(n))+[α12(n)(1−α21(n))]=1−α12(n)α21(n)$(2)

Thus using (1) and (2), the probabilities of events An and Bn are found to be

$P{An}=∏j=1n−1(1−α(j))α¯21(n),$

and

$P{Bn}=∏j=1n−1(1−α(j))α12(n)α¯21(n),$

where ${\overline{\alpha }}_{12}\left(n\right)=1-{\alpha }_{12}\left(n\right)$ and ${\overline{\alpha }}_{21}\left(n\right)=1-{\alpha }_{21}\left(n\right)$. For the system mean lifetime measured from X1 we have

$P{τ(a,b)=ζ1+ζ2+...+ζn−1+X^2(n)}=∏j=1n−1(1−α(j))α¯12(n),$(3)$P{τ(a,b)=ζ1+ζ2+...+ζn−1+X2(n)+X^1(n)}=∏j=1n−1(1−α(j))α12(n)α¯12(n),$(4)

where ${\zeta }_{j}={X}_{2}^{\left(j\right)}+{X}_{1}^{\left(j\right)}$, j = 1, 2, …, n – 1 and for n = 1 the sum and product vanishes.

## 4 Construction of LS transform of the system mean lifetime

In this section, we obtain LS transform of the system mean lifetime. The system consists of two components. Each component can be repairable and has aging property. We introduce the following notations;

$qik(p)(s)=∫0∞e−stGYi(p)(t)dFXk(p)(t),$(5)$q¯ik(p)(s)=∫0∞e−stG¯Yi(p)(t)dFXk(p)(t)=fXk(p)(s)−qik(p)−qik(p)(s),$(6)$fXk(p)(s)=∫0∞e−stdFXk(p)(t),$(7)

where i = 1, 2; k = 1, 2; ik; p ≥ 1 and ${\overline{G}}_{{Y}_{i}\left(p\right)}\left(t\right)=1-{G}_{{Y}_{i}\left(p\right)}\left(t\right)$. Using the formula of total expectation and (3)-(4) we have

$Ee−sτ(a,b)=∑n=1∞(Ee−s(ζ1+ζ2+...+ζn−1+X^2(n))∏j=1n−1(1−α(j))α¯12(n)+Ee−s(ζ1+ζ2+...+ζn−1+X2(n)+X^2(n))∏j=1n−1(1−α(j))α12(n)α¯12(n)).$(8)

Let ${X}_{k}^{\left(j\right)}$ be independent random variables. For $P\left\{{X}_{k}^{\left(j\right)}{Y}_{i}^{\left(j\right)}\right\}$, we have $Ee−sXk(j)=qik(j)(s)αik(j).$(9)

Proof. Using the definition $P\left\{{X}_{k}^{\left(j\right)}{Y}_{i}^{\left(j\right)}\right\}$, we have $Ee−sXk(j)=∫0∞e−stdP{Xk(j)Yi(j)}=1P{Xk(j)>Yi(j)}∫0∞e−stdP{Xk(j)Yi(j)}.$

From (1), it follows that $Ee−sXk(j)=1αik(j)∫0∞e−std[∫0∞P{Xk(j)Yi(j),Xk(j)=uj}dFXk(j)(uj)]=1αik(j)∫0∞e−std[P{ujYi(j)}dFXk(j)(uj)+∫t∞P{ujYi(j)}dFXk(j)(uj)].$

The right hand-side of the above equality tends to zero as uj < t. Hence it takes the form $Ee−sXk(j)=1αik(j)∫0∞e−std[∫0tP{uj>Yi(j)}dFXk(j)(uj)]=1αik(j)∫0∞e−std[∫0tGYi(j)(uj)dFXk(j)(uj)]=1αik(j)∫0∞e−stGYi(j)(t)dFXk(j)(t)$

From this and (5) we get the proof. □

Let ${\stackrel{^}{X}}_{k}^{\left(n\right)}$ be independent random variable. For $P\left\{{X}_{k}^{\left(n\right)}, we have $Ee−sX^k(n)=∫0∞e−stdP{Xk(n)(10)

Proof. It can be obtained similarly as Lemma 4.1. □

Taking into consideration the Assumption 2.3 and substituting (9) and (10) in (8) we find that

$Ee−sτ(a,b)=∑n=1∞[q¯12(n)(s)+q12(n)(s)q¯21(n)(s)]∏j=1n−1[q12(j)(s)q21(j)(s)].$

In view of τ12(a, b) = X1 +τ(a, b) and ${f}_{{X}_{1}}\left(s\right)=E{e}^{-s{X}_{1}}=\underset{0}{\overset{\infty }{\int }}{e}^{-st}d{F}_{{X}_{1}}\left(t\right)$, the LS transform of the system 0mean lifetime τ12(a,b)is obtained as

$Ee−sτ12(a,b)=fX1(s)(∑n=1∞[q¯12(n)(s)+q12(n)(s)q¯21(n)(s)]∏j=1n−1[q12(j)(s)q21(j)(s)]).$(11)

With the LS transform, the mean lifetime of the system τ12(a, b) is expressed in terms of the integrals qik(p)(s) and ${\overline{q}}_{ik\left(p\right)}\left(s\right)$, i = 1, 2; k = 1, 2; ik. The LS transform of the system mean lifetime given by the equation (11) is required to calculate the expected value of the system’s mean lifetime.

We will use the following special function and integral in the next section to obtain explicit expression of the Laplace-Stieltjes transform of the system mean lifetime for a Weibull distributional example. The function is the error function defined by (Equation 8.250.1 in [3])

$Φ(x)=2π∫0xe−t2dt.$(12)

The integral is

$∫0∞e−x24β−γxdx=πβeβγ2[1−Φ(γβ)]$(13)

where β > 0 (Equation 3.322.2 in [3]).

## 5 Numerical examples

In probabilistic design it is common to use parametric statistical models to illustrate the theoretical results for the proposed model. In this section, we apply our model to a Weibull and Gamma distributional examples.

## 5.1 A Weibull distributional example

Assume that the working time of the component k yields exponential distribution while the repair time of the component k yields Weibull distribution, i.e.

$FX1(t)=1−eλ1t,FX1(n)(t)=F1(ant)=1−eλ1ant,FX2(n)(t)=F2(an−1t)=1−eλ1an−1t,GYk(n)(t)=Gk(bn−1t)=1−e(μkbn−1t)βk,$

where t ≥ 0, a > 1, 0 < b < 1, μk > 0, βk > 0; k = 1, 2. In order to obtain the explicit expression of the LS transform of the system mean lifetime expediently, let β1 = β2 = 2, we can get

$fX1(s)=∫0∞e−std(1−e−λ1t)=λ1s+λ1q12(n)(s)=∫0∞e−st(1−e−(μ1bn−1t)2)d(1−e−λ2an−1t)=λ2an−1s+λ2an−1−πλ2an−12μ1bn−1e(An)2[1−Φ(An)]q21(n)(s)=∫0∞e−st(1−e−(μ2bn−1t)2)d(1−e−λ1ant)=λ1ans+λ1an−πλ1an2μ2bn−1e(Bn)2[1−Φ(Bn)]q¯12(n)(s)=∫0∞e−st(e−(μ1bn−1t)2)d(1−e−λ2an−1t)=πλ2an−12μ1bn−1e(An)2[1−Φ(An)]q¯21(n)(s)=∫0∞e−st(e−(μ2bn−1t)2)d(1−e−λ1ant)=πλ1an2μ2bn−1e(Bn)2[1−Φ(Bn)]$

where ${A}_{n}=\frac{s+{\lambda }_{2}{a}^{n-1}}{2{\mu }_{1}{b}^{n-1}}$ and ${B}_{n}=\frac{s+{\lambda }_{1}{a}^{n}}{2{\mu }_{2}{b}^{n-1}}$. In above, we also used equation (12) and (13). Substituting the above equations into (11), we can get the explicit expression of the LS transform of the system mean lifetime as follows

$Ee−sτ12(a,b)=λ1s+λ1[limN→∞∑n=1N{(πλ2an−12μ1bn−1e(An)2[1−Φ(An)]) +(λ2an−1s+λ2an−1−πλ2an−12μ1bn−1e(An)2[1−Φ(An)])(πλ1an2μ2bn−1e(Bn)2[1−Φ(Bn)])} ×∏j=1n−1(λ2aj−1s+λ2aj−1−πλ2aj−12μ1bj−1e(Aj)2[1−Φ(Aj)])(πλ1aj2μ2bj−1e(Bj)2[1−Φ(Bj)])]$(14)

When the above series is convergent, the remaining terms after certain N are neglected in the numerical calculations. From (14) and $E{\tau }_{12}\left(a,b\right)=-\frac{d}{ds}E{e}^{-s{\tau }_{12}\left(a,b\right)}\left(0\right)$ we can get the mean time to failure of the system for Weibull distributional example. Table 1 contains some numerical results for selected values of the parameters λ1, λ2, a, b, μ1, μ2 and N. We have used the programming language Wolfram Mathematica 9 for differentiating (14) and calculations.

Table 1

Mean time to failure of the system for Weibull distributional example.

## 5.2 A Gamma distributional example

Assume that the working time of the component k yields exponential distribution while the repair time of the component k yields Gamma distribution, i.e.

$FX1(t)=1−eλ1t,FX1(n)(t)=F1(ant)=1−eλ1ant,FX2(n)(t)=F2(an−1t)=1−eλ2an−1t,GYk(n)(t)=Gk(bn−1t)=1−∑i=01(μkbn−1t)i!e−μkbn−1t.$

where t ≥ 0, a > 1, 0 < b < 1, λk > 0, μk > 0; k = 1, 2. In the following, we will compute several integrations so that the expressions of ${f}_{{X}_{1}}\left(s\right),{q}_{12\left(n\right)}\left(s\right),{q}_{21\left(n\right)}\left(s\right),{\overline{q}}_{12\left(n\right)}\left(s\right),{\overline{q}}_{21\left(n\right)}\left(s\right)$ can be derived, respectively.

$∫0∞e−std(1−e−λ1t)=λ1s+λ1∫0∞e−st(1−∑i=01(μ1bn−1t)ii!e−μ1bn−1t)d(1−e−λ2an−1t)=λ2(μ1)2an−1b2n−2(s+λ2an−1)(s+μ1bn−1+λ2an−1)2,∫0∞e−st(1−∑i=01(μ2bn−1t)ii!e−μ2bn−1t)d(1−e−λ1ant)=λ1(μ2)2anb2n−2(s+λ1an)(s+μ2bn−1+λ1an)2,∫0∞e−st(∑i=01(μ1bn−1t)ii!e−μ1bn−1t)d(1−e−λ2an−1t)=λ2an−1(1+s+μ1bn−1+λ2an−1)(s+μ1bn−1+λ2an−1)2,∫0∞e−st(∑i=01(μ2bn−1t)ii!e−μ2bn−1t)d(1−e−λ1ant)=λ1an(1+s+μ2bn−1+λ1an)(s+μ2bn−1+λ1an)2.$

Here, we used suitable transformations and simplifications. Now, substituting the above equations into (11), we can get the explicit expression of the LS transform of the system mean lifetime as follows

$Ee−sτ12(a,b)=λ1s+λ1[limN→∞∑n=1N{(λ2an−1(1+s+μ1bn−1+λ2an−1(s+μ1bn−1+λ2an−1)2) +(λ1λ2(μ1)2a2n−1b2n−2(1+s+μ2bn−1+λ1an)(s+λ2an−1)(s+μ1bn−1+λ2an−1)2(s+μ2bn−1+λ1an)2) ×∏j=1n−1λ1λ2(μ1)2(μ2)2a2n−1b4n−4(s+λ2an−1)(s+λ1an)(s+μ1bn−1+λ2sn−1)2(s+μ2bn−1+λ1an)2].$(15)

When the above series is convergent, the remaining terms after certain Nare neglected in the numerical calculations. From (15) and $E{\tau }_{12}\left(a,b\right)=-\frac{d}{ds}E{e}^{-s{\tau }_{12}\left(a,b\right)}\left(0\right)$ we can get the mean time to failure of the system for Gamma distributional example. Table 2 contains some numerical results for selected values of the parameters λ1, λ2, a, b, μ1, μ2 and N. We have used the programming language Wolfram Mathematica 9 for differentiating (15) and calculations.

Table 2

Mean time to failure of the system for Gamma distributional example.

## 6 Conclusion

In this article, we have studied a repairable aging cold standby system with a single standby unit when the components are independent. We have presented Laplace-Stieltjes transform of the system mean lifetime and compute its expected value for Weibull distributional example and Gamma distributional example.

## Acknowledgement

The authors would like to express gratitude to the referees for their valuable suggestions and comments that led to some improvements in this article.

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## About the article

Accepted: 2016-05-04

Published Online: 2016-06-10

Published in Print: 2016-01-01

Citation Information: Open Mathematics, ISSN (Online) 2391-5455,

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