Stochastic and Statistics
MUT of a one out of two system with preventive maintenance

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Abstract

We are interested in the MUT (Mean Up Time) of a one out of two system in cold standby with preventive maintenance: a preventive maintenance occurs when the working unit reaches a given age. We study in details the stationary distribution of the Markov chain describing the state of the system at the beginning of its working periods. We give exact analytical formulas from which we derive a way to compute the MUT and we compare the results with those of Smith and Decker [M.A.J. Smith, R. Decker, Preventive maintenance in a 1 out of n system: The uptime, downtime and costs, European Journal of Operational Research 99 (1997) 565–583] which are based on approximations. We also investigate some discontinuities problems.

Introduction

In a system, single items (which can be components or subsystems) may be important for the system’s ability to fulfill its mission. In such a situation, to ensure high system performances, it is necessary to introduce redundancies. From an idealistic point of view, the greater is the number of redundancies, the greater are performances. But from a practical point of view, it is often impossible to have many redundancies, because for instance of their costs or of the space they take. Therefore it is interesting to have accurate evaluations of the system performances, even in the case of few redundancies.

A realistic modelization of the items’ behavior is also necessary. Assuming constant failure and repair rates allows to consider sophisticated dependancies between the items, using Markov models [2, Chapter 5], [3, Chapter 9], [12, Chapter 8]. But it is well known that constant rates are unrealistic, specially constant repair rates. In the case of items with constant failure rates and general repair rates, some redundancy problems are solved with queuing theory methods since practical queues are M/G (Markovian inputs, general services) [8, Chapter 5], [13, Chapter 8]. In the case of general models (dependant components with general failure and repair rates), some formulas can be obtained [5] but they do not provide tractable computational methods. Indeed, the results are obtained by adding supplementary variables in order to obtain a Markovian model and to apply general results to such models. But the number of supplementary variables and/or their nature (continuous variables instead of discrete variables) often make the problem untractable from a computational point of view. The introduction of preventive maintenance makes the problems still more complex, even if some particular cases can be solved [6].

In [14], Scarf underlines that “case studies have (often) been motivated by the need to find an application for a particular model, rather than by the solution of the problem of interest to the engineer or manager”. The motivation of this paper is exactly the converse: the model is a simple and widely studied one but the question to be solved leads to mathematical developments which could not have been thought up at the beginning of the study and which are interesting in themselves.

We are interested in a two units system in cold standby with preventive maintenance. Such systems have been studied for a long time under various assumptions and purpose. In [11], Chapter 7 is devoted to such models. Usually constant failure rates are assumed: for instance, in [7] the units have also constant repair rates but some dependancies are introduced, in [10] hidden failures of the standby item are considered, in [15] maximum operation times are introduced. In [18], [19], the author assumes general failure rates and considers implicitly a system of two units in cold standby but focuses on the behavior of the standby unit under several preventive maintenance policies.

Few people consider the Mean Up Time (MUT) of the system i.e. the asymptotic mean duration of a working period of the system, probably because of the difficulty of its computation. It cannot be deduced from the steady states of the initial model describing the system when this model is not a classical Markovian one. In [17] a Markovian model is considered. In [16], Smith and Decker study a one out of n system which is preventively maintained. They suggest an approximation for the MUT which would be true if the successive beginning times of up periods were renewal times. They give numerical results for n = 2, 3, 4.

In this paper, we consider the same model as in [16] with n = 2, but we do not make any approximation. Let us describe it. The system consists in two units in cold standby with preventive maintenance. The system works if one unit works. Both units are not working simultaneously. A preventive maintenance is performed on the working unit when it reaches age a. Then it is maintained and instantaneously replaced by the other unit if this unit is not failed. The preventive maintenance takes Rp units of time. If the working unit fails before age a, a corrective maintenance is performed and it is instantaneously replaced by the other unit if this unit is not failed. The corrective maintenance takes Rc (⩾Rp) units of time. At the end of a preventive or corrective maintenance, the unit is considered new, it becomes a waiting unit if the other unit is working. Otherwise it starts instantaneously and becomes the working unit. A unit which is waiting cannot fail. We assume general failure rates.

We know how to compute the Mean Time To Failure (MTTF), i.e. the mean duration of the first working period of the system and we have developed the method for n = 2, 3 see [4] for an abstract. We have exhibited discontinuity points for the MTTF seen as a function of the replacement age a. These discontinuities occur at the points a = Rp and a = Rc when n = 2 and at the points a = Rp/2 and a = Rc/2 when n = 3.

The aim of this paper is to compute the MUT of the system when n = 2 and to detect discontinuity points. For this purpose, we add supplementary variables to the model in order to have a Markov process. Then we study carefully the stationary distribution π of the Markov chain describing the residual maintenance duration of the failed unit at the beginning of the system’s working periods. We prove that it is the sum of Dirac measures and of a measure which is absolutely continuous. We then provide formulas to compute them.

More precisely, under technical assumptions, we prove thatMUT=Ez(M0)π(dz),(Corollary 7) where Ez(M0) is given in Proposition 1. The measure π is a probability distribution on [0, Rc] which can be writtenπ(dz)=k=1n1ykδRc-k(a+Rp)+xn11{2Rcn2(a+Rp)}δn1(a+Rp)-Rc+τaδa+π1(z)dz,where the yk’s, xn1, and τa are given in Proposition 13 and Lemma16 and n1 is characterized by (n1  1)(a + Rp) < Rc  n1(a + Rp) (Theorem 23). Moreover, the density π1 satisfies(1-F¯2(a)1{a<z<Rp})π1(z)=k0F¯k(a)g(u,z+k(a+Rp))+1{z<Rp}k1F¯k(a)g(u,k(a+Rp)-z)π1(u)du+k0=1n1yg(Rc-(a+Rp),z+k(a+Rp))+τag(a,z+k(a+Rp))+xn11{2Rcn2(a+Rp)}g(n1(a+Rp)-Rc,z+k(a+Rp))+1{z<Rp}k1=1n1yg(Rc-(a+Rp),k(a+Rp)-z)+1{z<Rp}τag(a,k(a+Rp)-z)+1{z<Rp}xn11{2Rcn2(a+Rp)}g(n1(a+Rp)-Rc,k(a+Rp)-z),whereg(z,y)=1{yRc}1{y<a+Rc-z}f(y-Rc+z)+α(z)1{y<a}f(y)+1{y<a+Rc-Rp}β(z)f(y-Rc+Rp)and f is the density function of the items’ lifetime when no preventive maintenance is performed. The way to compute α(z) and β(z) is given in Section 5.1.

When a < Rc, the probability distribution π has Dirac components and the density π1 has discontinuity points. When Rc  a, the probability measure π is absolutely continuous and its p.d.f. π1 is continuous.

Numerical examples are given.

Section snippets

Characterization of the MUT

Let X be the intrinsic lifetime of a unit i.e. the lifetime of a unit when no preventive maintenance is performed. Let X1 = min(X, a) be its lifetime when it is preventively maintained at age a. Let us assume that X has a probability density function, namely f. Let F be the probability distribution function of X and F¯ its survival function:F(t)=0tf(x)dx,F¯=1-F.The function F¯ is the reliability function of a unit which is not maintained.

In the following, we assume that:

  • Rp  Rc, which is a quite

Stationary distribution of the entrance time in a working period

To compute the stationary distribution π associated with the transition function p, we have to solve the following equation:π(A)=p(z,A)π(dz).

Let us recall that (2) implies that p(z, ·), and therefore π, are concentrated on [0, Rc]. We will use this remark in all the following without recalling it again.

The absolutely continuous component of the stationary distribution

In this section, we do not separate the cases Rp=Rc and Rp < Rc since the results of Section 3.1 can be seen as those of Section 3.2 with Rp = Rc.

For numerical computation, it is impossible to work with equations which are true almost everywhere as Eq. (23). We will show that we can omit this restriction.

To simplify the notations, we extend the function g to R2, taking it equal to zero outside [0, Rc] × [0, Rc]. In the same way, we extend π1 to R by taking π1 equal to zero outside [0, Rc].

Lemma 26

For almost

Computations of α and β

First we have to compute α(z) and β(z). LetQ(z,Rp)=Pz(Zσ1=Rp,σ1T1),Q(z,Rc)=Pz(Zσ1=Rc,σ1T1).Lemma 2 givesQ(z,Rp)=P(zX1,X1=a)=1{za}F¯(a),Q(z,Rc)=P(zX1<a)=1{z<a}(F¯(z)-F¯(a)).By using the Markov property at time σ1, it can be easily seen thatα(z)=A1Q(z,Rp)+B1Q(z,Rc),withA1=n2PRp(Zσ1{Rp,Rc},σ1T1,,Zσn-3{Rp,Rc},σn-3T1,σn-2=Rc,σn-2T1)=1+n4PRp(Zσ1{Rp,Rc},σ1T1,,Zσn-3=Rp,σn-3T1)Q(Rp,Rc)+n4PRp(Zσ1{Rp,Rc},σ1T1,,Zσn-3=Rc,σn-3T1)Q(Rc,Rc)=Q(Rp,Rc)+β(Rp)Q(Rp,Rc)+α(Rp)Q(Rc,Rc).SimilarlyB

Conclusion

Our aim was to compute the MUT of a one out of two system in cold standby, a unit being preventively maintained when it reaches age a and the duration of a preventive (respectively corrective) maintenance being Rp (respectively Rc).

We also wanted to see if the MUT is not a continuous function of a at the points a = Rp and a = Rc, as the MTTF.

For this purpose, we have studied the stationary probability distribution of the Markov chain, taking its values in [0, Rc], which describes the residual

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