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A deteriorating cold standby repairable system with priority in use
European Journal of Operational Research 183 (2007) 278–295 www.elsevier.com/locate/ejor
Stochastics and Statistics
A deteriorating cold standby repairable system with priority in use Yuan Lin Zhang *, Guan Jun Wang Department of Mathematics, Southeast University, Nanjing 210018, China Received 18 November 2005; accepted 27 September 2006 Available online 18 December 2006
Abstract In this paper, a cold standby repairable system consisting of two dissimilar components and one repairman is studied. In this system, it is assumed that the working time distributions and the repair time distributions of the two components are both exponential and component 1 is given priority in use. After repair, component 2 is ‘‘as good as new’’ while component 1 follows a geometric process repair. Under these assumptions, using the geometric process and a supplementary variable technique, some important reliability indices such as the system availability, reliability, mean time to first failure (MTTFF), rate of occurrence of failure (ROCOF) and the idle probability of the repairman are derived. A numerical example for the system reliability R(t) is given. And it is considered that a repair-replacement policy based on the working age T of component 1 under which the system is replaced when the working age of component 1 reaches T. Our problem is to determine an optimal policy T * such that the long-run average cost per unit time of the system is minimized. The explicit expression for the long-run average cost per unit time of the system is evaluated, and the corresponding optimal replacement policy T * can be found analytically or numerically. Another numerical example for replacement model is also given. 2006 Elsevier B.V. All rights reserved. Keywords: Geometric process; Renewal process; Supplementary variable; Generalized Markov process; Priority; Replacement policy
1. Introduction In order to improve the reliability or raise the availability and hence reduce the loss, a two-component redundant system is often employed. Note that such a system is one of the typical systems employed in reliability engineering. Under the conditions that the working time and the repair time of the components in the system both have exponential or general distributions, by using Markov process theory or Markov renewal process theory, some important reliability indices of the system have been derived in the previous literature (see e.g. Barlow and Proschan [1] and Ascher and Feingold [2] for further details). Later, a priority rule for repair or use of a component has been introduced. Nakagawa and Osaki [3] assumed that both the working time and the repair time of the priority component have a general distribution while both the working time *
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0377-2217/$ - see front matter 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2006.09.075
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and the repair time of the non-priority component have an exponential distribution. They obtained some interesting reliability indices of the system using Markov renewal theory. See also Osaki [4] and Buzacott [5] for generalizations. In all of these studies, they assumed that a system (or a component) after repair is ‘‘as good as new’’. This is a perfect repair model. However, this assumption is not always true. In practice, most repairable systems are deteriorative because of the ageing effect and the accumulative wear. Barlow and Hunter [6] first presented a minimal repair model in which the minimal repair does not renew the age of the system. Brown and Proschan [7] first reported an imperfect repair model in which the repair is perfect repair with probability p or minimal with probability 1 p. For a deteriorating repairable system, it is quite reasonable to assume that the successive working times of the system after repair will become shorter and shorter while the consecutive repair times of the system after failure will become longer and longer. Ultimately, it cannot work any longer, neither can it be repaired. For such a stochastic phenomenon, Lam [8,9] first introduced a geometric process repair model to approach it. Under this model, he studied two kinds of replacement policy for a one-component repairable system with one repairman (called a simple repairable system), one based on the working age T of the system and the other based on the failure number N of the system. The explicit expressions of the long-run average cost per unit time under these two kinds of policy are respectively calculated. Finkelstein [10] presented a general repair model based on a scale transformation after each repair to generalize Lam’s work. Zhang [11] generalized Lam’s work by a bivariate replacement policy (T, N) under which the system is replaced at the working age T or at the time of the Nth failure, whichever occurs first. Many replacement policies have been done by Stadje and Zuckerman [12], Stanley [13], Zhang et al. [14,15], Zhang [16], Lam and Zhang [17], Zhang [18], Wang and Zhang [19] and others under the geometric process repair model. However, the above various research is related to the simple repairable system. For multi-component systems, Zhang and Wu [20] first reported a two-component series repairable system under the geometric process repair model. When the working time of component 1 in the system follows the exponential distribution while that of component 2 and the repair times of both components follow general distributions, they derived some reliability indices of the system. Lam and Zhang [21] provided a more in depth analysis of the system studied by Zhang and Wu [20] under the assumptions that the working and repair times of both components all follow the exponential distributions. Lam [22] investigated a two-component parallel repairable system assuming that all working and repair times follow the exponential distributions and one component after repair is ‘‘as good as new’’ while the other is not. The system ROCOF is determined through the Monte Carlo numerical method. Lam and Zhang [23] studied a similar model, more reliability indices of the system are obtained by using the Laplace-transform technique. Lam [24] reported a maintenance model for two-unit redundant system with one repairman. Under this model, he studied two kinds of replacement policy, respectively based on the number of failure and the working age for two units. The long-run average cost per unit time for each kind of replacement policy is derived. Zhang [25] applied the geometric process repair model to a two-component cold standby repairable system with one repairman. He also assumed that each component after repair is not ‘‘as good as new’’. Under this assumption, by using a geometric process, he studied a replacement policy N based on the number of repairs of component 1. The problem is to determine an optimal replacement policy N* such that the long-run expected reward per unit time is maximized. In practical applications, a two-component cold standby repairable system with one repairman and priority in use is often used. For example, in the operating room of a hospital, a patient on the operating table has to discontinue his/her operation as soon as the power source is cut (i.e. the power station failures). Usually, there is a standby power station (e.g. a storage battery) in the operating room. Thus, the power station (e.g. write as component 1) and the storage battery (e.g. write as component 2) form a cold standby repairable lighting system. Obviously, it is reasonable to assume that the power station has priority in use due to the operating cost of the power station is cheaper than the operating cost of the storage battery, and the storage battery after repair is as good as new due to its used time being smaller than the power station, and the repair of the storage battery is also convenient. Besides a cold standby lighting system in a hospital, some similar examples can be found from Lam [24]. The purpose of this paper is to apply the geometric process repair model to a two-component cold standby repairable system with one repairman. Now we may assume that the component 2 after repair is ‘‘as good as new’’ while component 1 follows the geometric process repair, but component 1 has priority in use.
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Furthermore, we assume that the working time and the repair time of both components are exponentially distributed. Under these assumptions, by using the geometric process and a supplementary variable technique we not only obtain some important reliability indices such as availability, reliability, MTTFF, ROCOF, and the idle probability of the repairman, but also determine the replacement policy based on the working age T of component 1. The problem is to choose an optimal replacement policy T* such that the long-run average cost per unit time is minimized. The explicit expression for the long-run average cost per unit time of the system is evaluated, and the corresponding optimal replacement policy T* can be found analytically or numerically. Finally, two numerical examples are respectively given for the system reliability R(t) and the system replacement policy T. For further reference, we first state the definitions of stochastic order and a geometric process. Definition 1. Given two random variables n and g, n is said to be stochastically larger than g or g is stochastically smaller than n, if P ðn > aÞ P P ðg > aÞ;
for all real a;
denoted by n P stg or g 6 stn (see e.g. Ross [26]). Furthermore, we say that a stochastic process {Xn, n = 1, 2,. . .} is stochastically decreasing if Xn P st Xn+1 and stochastically increasing if Xn 6 st Xn+1 for all n = 1, 2,. . . Definition 2. Assume that {nn, n = 1, 2,. . .} is a sequence of independent non-negative random variables. If the distribution function of nn is Fn(t) = F(an1t), n = 1, 2, . . . , and if a is a positive constant, then {nn, n = 1, 2,. . .} is called a geometric process, and a is the ratio of the geometric process (see e.g. Lam [8] and Zhang [11] for further details). Obviously, if a > 1, then {nn, n = 1, 2,. . .} is stochastically decreasing. If 0 < a < 1, then {nn, n = 1, 2,. . .} is stochastically increasing. If a = 1, then the geometric process is the sequence of interarrival times of a renewal process. 2. Model We study a two-component cold standby repairable system with one repairman and priority in use by making the following assumptions: Assumption 1. Initially, the two components are both new, and component 1 is in working state while component 2 is in cold standby state. Assumption 2. Assume that component 2 after repair is as good as new while component 1 follows a geometric process repair. The repair rule is ‘‘first-in-first-out’’. If a component fails during the repair of the other, it must wait for repair and the system is down. Component 1 has priority in use. Assumption 3. Assume that the time interval between the completion of the (n 1)th repair and the completion of the nth repair of component i is called the nth cycle of component i, i = 1, 2; n = 1, 2,. . . ðiÞ Let X ðiÞ n and Y n be, respectively, the working time and the repair time of component i in the nth cycle, i = 1, ðiÞ 2; n = 1, 2,. . . We assume that the distributions of X ðiÞ n and Y n ði ¼ 1; 2Þ are, respectively, n1 tÞ ¼ 1 expðan1 k1 tÞ; F ð1Þ n ðtÞ ¼ F ða
Gnð1Þ ðtÞ ¼ Gðbn1 tÞ ¼ 1 expðbn1 l1 tÞ; F ð2Þ ðtÞ ¼ 1 expðk2 tÞ; Gð2Þ ðtÞ ¼ 1 expðl2 tÞ; where t P 0, a P 1, 0 < b 6 1, ki > 0, li > 0; i = 1, 2.
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ðiÞ Assumption 4. Assume that X ðiÞ n and Y n ; i ¼ 1; 2; n ¼ 1; 2; . . . are independent.
Obviously, such a model for a two-component cold standby repairable system with one repairman and priority in use based on these assumptions above takes the deterioration into account, it is more realistic than the model in which two components after repair are both ‘‘as good as new’’. 3. System analysis Let {N(t), t P 0} be a stochastic process characterized by the following mutually exclusive events: {N(t) = 0}: {N(t) = 1}: {N(t) = 2}: {N(t) = 3}: {N(t) = 4}:
Component Component Component Component Component
1 1 2 1 2
is is is is is
working, component 2 is working, component 2 is working, component 1 is under repair, component under repair, component
in cold standby at time t. under repair at time t. under repair at time t. 2 is waiting for repair at time t. 1 is waiting for repair at time t.
Then {N(t), t P 0} is a stochastic process with state space X = {0, 1, 2, 3, 4}. The set of working states is W = {0, 1, 2} and the set of failure states is F = {3, 4}. According to the model assumptions, {N(t), t P 0} is not a Markov process. However, it can be extended to a generalized Markov process by introducing a supplementary variable. Let S(t) be the number of the cycle of component 1 at time t. Then {N(t), S(t), t P 0} constitutes a generalized (i.e. two-dimensional) Markov process. The state marginal probabilities of the system at time t are defined by pjk ðtÞ ¼ P fN ðtÞ ¼ j; SðtÞ ¼ kg;
j 2 X;
k ¼ 1; 2; . . .
According to the model assumptions and the supplementary variable technique, we can obtain the following differential equations for the system. By straightforward probability arguments, for example, we have p0k ðt þ DtÞ ¼ pfN ðt þ DtÞ ¼ 0; Sðt þ DtÞ ¼ kg ¼ p0k ðtÞð1 ak1 k1 DtÞ þ p1k ðtÞl2 Dt þ p2 letting Dt tend to zero, we can get d k1 þ a k1 p0k ðtÞ ¼ l2 p1k ðtÞ þ bk2 l1 p2 dt
k1 ðtÞ
k1 ðtÞb
k2
l1 Dt þ oðDtÞ;
ðk P 2Þ:
In the same way, we have d k1 þ a k1 þ l2 p1k ðtÞ ¼ bk2 l1 p3 k1 ðtÞ ðk P 2Þ; dt d þ bk1 l1 þ k2 p2k ðtÞ ¼ ak1 k1 p0k ðtÞ þ l2 p4k ðtÞ ðk P 2Þ; dt d þ bk1 l1 p3k ðtÞ ¼ k2 p2k ðtÞ ðk P 2Þ; dt d þ l2 p4k ðtÞ ¼ ak1 k1 p1k ðtÞ ðk P 2Þ; dt d þ k1 p01 ðtÞ ¼ 0; dt p11 ðtÞ ¼ 0; d þ k2 þ l1 p21 ðtÞ ¼ k1 p01 ðtÞ þ l2 p41 ðtÞ; dt d þ l1 p31 ðtÞ ¼ k2 p21 ðtÞ; dt p41 ðtÞ ¼ 0:
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The initial conditions are p01 ð0Þ ¼ 1;
p0k ð0Þ ¼ 0;
ðk P 2Þ;
pjk ð0Þ ¼ 0;
j ¼ 1; 2; 3; 4;
k ¼ 1; 2; . . .
According to the theory of Markov process, it is easy to see that above equations are actually Kolmogorov forward equations. Based on the property (Lipschitz condition) of differential equation, the solution will exist and is unique. Obviously, a numerical method should be applicable for the solution of the system of differential equations above. Here, we adopt the method of Laplace transform to determine Laplace transform solution of the system of differential equations. In Section 4.6, a numerical example for the system reliability is given to illustrate the numerical method. R1 Denote the Laplace transform of Pjk(t) by pjk ðsÞ ¼ 0 est pjk ðtÞdt. The Laplace transforms of the above differential equations are respectively given by ðs þ ak1 k1 Þp0k ðsÞ ¼ l2 p1k ðsÞ þ bk2 l1 p2 ðs þ ðs þ ðs þ ðs þ
k1 ðsÞ ðk P 2Þ; k2 a k1 þ l2 Þp1k ðsÞ ¼ b l1 p3 k1 ðsÞ ðk P 2Þ; bk1 l1 þ k2 Þp2k ðsÞ ¼ ak1 k1 p0k ðsÞ þ l2 p4k ðsÞ ðk P 2Þ; bk1 l1 Þp3k ðsÞ ¼ k2 p2k ðsÞ ðk P 2Þ; l2 Þp4k ðsÞ ¼ ak1 k1 p1k ðsÞ ðk P 2Þ; k1 Þp01 ðsÞ ¼ 1;
ð1Þ
k1
ð2Þ ð3Þ ð4Þ ð5Þ
ðs þ p11 ðsÞ ¼ 0;
ð6Þ ð7Þ
ðs þ k2 þ l1 Þp21 ðsÞ ¼ k1 p01 ðsÞ; ðs þ l1 Þp31 ðsÞ ¼ k2 p21 ðsÞ;
ð8Þ ð9Þ
p41 ¼ 0:
ð10Þ
According to Eqs. (4) and (2), respectively, we obtain s þ bk2 l1 p3 k1 ðsÞ ðk P 2Þ; k2 s þ ak1 k1 þ l2 p1k ðsÞ ðk P 2Þ: p3 k1 ðsÞ ¼ bk2 l1
p2
k1 ðsÞ
¼
Therefore, p2
k1 ðsÞ
¼
ðs þ ak1 k1 þ l2 Þðs þ bk2 l1 Þ p1k ðsÞ bk2 l1 k2
ðk P 2Þ:
ð11Þ
Substituting Eq. (11) into Eq. (1) entails that ðs þ ak1 k1 Þp0k ðsÞ ¼
k2 l2 þ ðs þ ak1 k1 þ l2 Þðs þ bk2 l1 Þ p1k ðsÞ ðk P 2Þ: k2
ð12Þ
Hence p0
k1 ðsÞ
¼
k2 l2 þ ðs þ ak2 k1 þ l2 Þðs þ bk3 l1 Þ p1 k2 ðs þ ak2 k1 Þ
k1 ðsÞ
ðk P 3Þ:
ð13Þ
Substitution of Eqs. (5) into Eq. (3) yields ðs þ bk1 l1 þ k2 Þp2k ðsÞ ¼ ak1 k1 p0k ðsÞ þ
ak1 k1 l2 p ðsÞ s þ l2 1k
ðk P 2Þ:
Or, ðs þ bk2 l1 þ k2 Þp2
k1 ðsÞ
¼ ak2 k1 p0
k1 ðsÞ
þ
ak2 k1 l2 p s þ l2 1
k1 ðsÞ
ðk P 3Þ:
ð14Þ
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283
Substitution of Eqs. (11) and (13) into Eq. (14) yields k2
p1k ðsÞ ¼
k1 l1 ðabÞ f ðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ p1 ðs þ l2 Þðs þ ak2 k1 Þðs þ bk2 l1 Þðs þ bk2 l1 þ k2 Þðs þ ak1 k1 þ l2 Þ
k1 ðsÞ
¼ Ak p12 ðsÞ
ðk P 3Þ; ð15Þ
where f ðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ ¼ ðs þ l2 Þ½k2 l2 þ ðs þ bk3 l1 Þðs þ ak2 k1 þ l2 Þ þ k2 l2 ðs þ ak2 k1 Þ; k Y k1 l1 ðabÞm2 f ðs; a; b; k1 ; k2 ; l1 ; l2 ; mÞ Ak ¼ : m2 k Þðs þ bm2 l Þðs þ bm2 l þ k Þðs þ am1 k þ l Þ 1 2 1 1 1 2 m¼3 ðs þ l2 Þðs þ a According to Eq. (15), Eqs. (5) and (12) become, respectively, ak1 k1 Ak p12 ðsÞ ðk P 3Þ; p4k ðsÞ ¼ s þ l2 p0k ðsÞ ¼
ð16Þ
k2 l2 þ ðs þ bk2 l1 Þðs þ ak1 k1 þ l2 Þ Ak p12 ðsÞ ðk P 3Þ: k2 ðs þ ak1 k1 Þ
ð17Þ
Substituting Eqs. (16) and (17) into Eq. (3) yields p2k ðsÞ ¼ gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞAk p12 ðsÞ ðk P 3Þ;
ð18Þ
where gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ ¼
ak1 k1 f ðs; a; b; k1 ; k2 ; l1 ; l2 ; k þ 1Þ : k2 ðs þ l2 Þðs þ ak1 k1 Þðs þ bk1 l1 þ k2 Þ
Finally, using Eqs. (2), (4) and (6)–(10) for k = 2, we have p3k ðsÞ ¼
k2 gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ Ak p12 ðsÞ s þ bk1 l1
p01 ðsÞ ¼
1 ; s þ k1
ðk P 3Þ;
p11 ðsÞ ¼ 0; p41 ðsÞ ¼ 0; p21 ðsÞ ¼
k1 ; ðs þ k1 Þðs þ k2 þ l1 Þ
p31 ðsÞ ¼
k1 k2 ; ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þ
p12 ðsÞ ¼
k1 k2 l1 : ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ
With these explicit expressions obtained from the differential equations and the associated Laplace transforms, we are ready to derive some important reliability indices of the repairable system. 4. Reliability indices 4.1. System availability By the definition, the system availability at time t is given by AðtÞ ¼ P ðthe system is working at time tÞ ¼ P ðN ðtÞ 2 W Þ ¼
1 X k¼1
½p0k ðtÞ þ p1k ðtÞ þ p2k ðtÞ:
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The Laplace-transform of A(t) is A ðsÞ ¼
1 X
½p0k ðsÞ þ p1k ðsÞ þ p2k ðsÞ
k¼1
¼ ½p01 ðsÞ þ p11 ðsÞ þ p21 ðsÞ þ p02 ðsÞ þ p12 ðsÞ þ p22 ðsÞ þ
1 X ½p0k ðsÞ þ p1k ðsÞ þ p2k ðsÞ; k¼3
where, according to (1) for k = 2, (3) for k = 2 and (5) for k = 2 yields p02 ðsÞ ¼
k1 l1 ½k2 l2 þ ðs þ l1 Þðs þ ak1 þ l2 Þ ; ðs þ k1 Þðs þ l1 Þðs þ ak1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ
p22 ðsÞ ¼
ak21 l1 hðs; a; b; k1 ; k2 ; l1 ; l2 Þ ðs þ k1 Þðs þ k2 þ l1 Þðs þ k2 þ bl1 Þðs þ l1 Þðs þ l2 Þðs þ ak1 þ l2 Þ
and where hðs; a; b; k1 ; k2 ; l1 ; l2 Þ ¼ k2 l2 ½ðs þ l2 Þ þ ðs þ ak1 Þðs þ k2 þ bl1 Þ þ ðs þ l1 Þðs þ l2 Þðs þ ak1 þ l2 Þ: Thus, A ðsÞ ¼
1 k1 l1 1þ 1þ s þ k1 s þ k 2 þ l1 ðs þ l1 Þðs þ ak1 þ l2 Þ k2 l2 þ ðs þ l1 Þðs þ ak1 þ l2 Þ ak1 hðs; a; b; k1 ; k2 ; l1 ; l2 Þ k2 þ þ s þ ak1 ðs þ l2 Þðs þ k2 þ bl1 Þ k1 k2 l1 ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ 1 X k2 l2 þ ðs þ bk2 l1 Þðs þ ak1 k1 þ l2 Þ þ gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ Ak : 1þ k2 ðs þ ak1 k1 Þ k¼3
þ
In reliability theory, the system availability is one of the important indices. However, in engineering application, the steady state availability or the limiting availability of the system is more important. Now, using a Tauberian theorem, when a, b 5 1 the steady state availability or the limiting availability of the system is given by A ¼ lim AðtÞ ¼ lim sA ðsÞ ¼ 0; t!þ1
s!0
which is consistent with a physical interpretation. In fact, since component 1 after repair is not as good as new, it implies that the availability tends to 0 as t ! + 1. 4.2. System reliability In order to determine the reliability R(t) of the system, we consider the two-dimensional continuous-time e ðtÞ; SðtÞ; t P 0g. The difference between the Markov processes {N(t), S(t), homogeneous Markov process f N e t P 0} and f N ðtÞ; SðtÞ; t P 0g is that the set of failure states of {N(t), S(t), t P 0} is now the set of absorbing e ðtÞ; SðtÞ; t P 0g. Now, let states of f N e ðtÞ ¼ j; SðtÞ ¼ kg; qjk ðtÞ ¼ P f N
j 2 X;
k ¼ 1; 2; . . . ;
then RðtÞ ¼ P fthe working time of the system > tg ¼
1 X k¼1
½q0k ðtÞ þ q1k ðtÞ þ q2k ðtÞ:
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285
By a similar argument as in Section 3, we obtain the following differential equations: d k1 þ a k1 q0k ðtÞ ¼ l2 q1k ðtÞ þ bk2 l1 q2 k1 ðtÞ ðk P 2Þ; dt d k1 þ a k1 þ l2 q1k ðtÞ ¼ 0 ðk P 2Þ; dt d k1 þ k2 þ b l1 q2k ðtÞ ¼ ak1 k1 q0k ðtÞ ðk P 2Þ; dt d þ k1 q01 ðtÞ ¼ 0; dt q11 ðtÞ ¼ 0; d þ k2 þ l1 q21 ðtÞ ¼ k1 q01 ðtÞ: dt The initial conditions are q01 ð0Þ ¼ 1;
q0k ð0Þ ¼ 0; k ¼ 2; 3; . . . ;
qjk ð0Þ ¼ 0;
j ¼ 1; 2; k ¼ 1; 2; . . .
Taking Laplace-transforms on both sides of the above differential equations and using the initial conditions, we obtain the following for k P 2 q0k ðsÞ ¼
bk2 l1 q s þ ak1 k1 2
k1 ðsÞ; k2
¼
k1 l1 ðabÞ q0 ðs þ ak1 k1 Þðs þ bk2 l1 þ k2 Þ
k1 ðsÞ;
¼ Bk q01 ðsÞ; ak1 k1 q0k ðsÞ s þ bk1 l1 þ k2 ak1 k1 Bk q01 ðsÞ; ¼ s þ bk1 l1 þ k2
q2k ðsÞ ¼
where Bk ¼
k Y m¼2
ðk1 l1 ÞðabÞm2 ; ðs þ am1 k1 Þðs þ bm2 l1 þ k2 Þ
q01 ðsÞ ¼
1 : s þ k1
Clearly q1k ðsÞ ¼ 0
ðk P 1Þ; k1 : q21 ðsÞ ¼ ðs þ k1 Þðs þ k2 þ l1 Þ
Therefore, the Laplace transform of R(t) is given by R ðsÞ ¼
1 1 X X ½q0k ðsÞ þ q2k ðsÞ ¼ ½q01 ðsÞ þ q21 ðsÞ þ ½q0k ðsÞ þ q2k ðsÞ k¼1
1 ¼ s þ k1
"
k1 1þ s þ k 2 þ l1
k¼2
1 X 1þ þ k¼2
# ak1 k1 Bk : s þ bk1 l1 þ k2
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4.3. System MTTFF The system MTTFF is one of the most important reliability indices for a repairable system. In particular, it will produce the calamity result as soon as the system fails. Based on the system reliability and a property of the Laplace transform, the system MTTFF is given by X Y Z þ1 1 k 1 1 1 ak1 bm2 l1 MTTFF ¼ RðtÞdt ¼ lim R ðsÞ ¼ þ þ k1 : þ s!0 k 1 k 2 þ l1 k1 b l1 þ k2 m¼2 aðbm2 l1 þ k2 Þ 0 k¼2 Let Ck ¼
Y k 1 ak1 bm2 l1 þ k1 ; k1 b l1 þ k2 m¼2 aðbm2 l1 þ k2 Þ
P1 then k¼2 C k is a positive series under k1, k2, l1, a and b are the positive P1 constants. And when a P 1, 0 < b 6 1, according to the ratio test of the positive series, it is easy to see k¼2 C k < 1. Thus, the result of the system MTTFF is finite. Obviously, if any component in the system is always no failure, the system MTTFF is infinite. 4.4. System ROCOF Let Nf(t) be the number of failures of a system that have occurred by time t, then the mean number of failures of the system in (0, t] is given by M f ðtÞ ¼ E½N f ðtÞ: Its derivative mf(t) is called the rate of occurrence of failure (ROCOF) or the failure frequency of the system at time t. Hence Z t mf ðxÞdx: M f ðtÞ ¼ 0
Obviously, if mf(t) is increasing, the mean number of failures that have occurred in (s, s + t] (where s is positive constant) is larger than the mean number in (0, t], and hence the system is deteriorating. On the other hand, if mf(t) is decreasing, the system is improving. Therefore, the ROCOF is one of the important indices in reliability theory, in particular in the study of a deteriorating or improving system. According to Lam [22], we have mf ðtÞ ¼
1 X X
pjk ðtÞajlk ;
j2W ;l2F k¼1
where ajlk is transition rate from state j to l in kth cycle. In view of the model assumptions, we have a03k = a04k = a13k = a24k = 0, a14k = ak1k1, a23k = k2. Therefore, mf ðtÞ ¼
1 X
½p0k ðtÞða03k þ a04k Þ þ p1k ðtÞða13k þ a14k Þ þ p2k ðtÞða23k þ a24k Þ ¼
k¼1
1 X
½ak1 k1 p1k ðtÞ þ k2 p2k ðtÞ:
k¼1
The Laplace transform of mf(t) is given by mf ðsÞ ¼
1 1 X X ½ak1 k1 p1k ðsÞ þ k2 p2k ðsÞ ¼ ½k1 p11 ðsÞ þ k2 p21 ðsÞ þ ½ak1 k1 p1k ðsÞ þ k2 p2k ðsÞ k¼1
k¼2
1 X k1 k2 k1 k2 l1 ½ak1 k1 þ gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞk2 þ Ak : ¼ ðs þ k1 Þðs þ k2 þ l1 Þ k¼2 ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
287
4.5. The idle time probability of the repairman Clearly, the repairman will be idle if and only if one component is functioning while the other component is in cold standby. Thus, the idle time probability of the repairman at time t is given by 1 X p0k ðtÞ: IðtÞ ¼ P fN ðtÞ ¼ 0g ¼ k¼1
The Laplace-transform of I(t) is I ðsÞ ¼
1 X
p0k ðsÞ ¼ ½p01 ðsÞ þ p02 ðsÞ þ
k¼1
1 X
p0k ðsÞ
k¼3
1 k1 l1 k2 l2 1þ 1þ ¼ s þ k1 ðs þ ak1 Þðs þ k2 þ l1 Þ ðs þ l1 Þðs þ ak1 þ l2 Þ 1 k2 X k1 l1 ½k2 l2 þ ðs þ b l1 Þðs þ ak1 k1 þ l2 Þ Ak : þ ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þðs þ ak1 k1 Þ k¼3 Using a Tauberian theorem, we can obtain the idle probability of the repairman in the steady- state as follows: I ¼ lim IðtÞ ¼ lim sI ðsÞ ¼ 0; t!þ1
s!0
i.e. the steady-state idle time probability of the repairman tends to 0. In fact, this conclusion is also in agreement with our intuition. Because component 1 after repair is not as good as new and its consecutive repair times after failure are stochastically increasing, it will be unrepairable finally. This implies that the idle time probability of the repairman will tends to zero as t ! + 1. In particular, if a = b = 1, ki = k, li = l, i = 1, 2 the system reduces to a two-identical-component cold standby repairable system with one repairman and without priority rule. This is the classical case which has been studied by e.g., Barlow and Proschan [1] and Ascher and Feingold [2]. 4.6. A numerical example for the system reliability R(t) Based on these results above, we can find that it is very difficult to obtain the transient results of the reliability indices for the system proposed in this paper. We can only obtain the steady-state results of some reliability indices of the system. Moreover, the method by using the inverse Laplace transform to get the transient results of the reliability indices is also very tedious. Thus, in engineering applications, a numerical method based on the Runge–Kutta method is often adopted. Here, an approximative solution of R(t) is given to illustrate the numerical method. Let a = 1.1, b = 0.9, k1 = 0.01, k2 = 0.1, l1 = 0.1, l2 = 0.5. According to the differential equations and the initial conditions about qjk(t) in Section 4.2, when k = 1, 2, 3; t = 10, 20, 30, 40, 50, we can obtain these approximative solutions of qjk(t), j = 0, 2 (see Table 1).
Table 1 Some results of qjk(t) q01(t) q21(t) q02(t) q22(t) q03(t) q23(t)
t = 10
t = 20
t = 30
t = 40
t = 50
0.8187 0.0281 0.1159 0.0037 0.0061 1.9176 · 105
0.6703 0.0231 0.2074 0.0077 0.0287 2.0088 · 104
0.5488 0.0189 0.2601 0.0101 0.0588 6.5363 · 104
0.4493 0.0155 0.2854 0.0113 0.0893 0.0014
0.3679 0.0127 0.2919 0.0117 0.1162 0.0023
288
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
Thus, we can further obtain these approximative solutions of R(t), i.e. RðtÞ ¼
1 3 X X ½q0k ðtÞ þ q2k ðtÞ ½q0k ðtÞ þ q2k ðtÞ ¼ q01 ðtÞ þ q21 ðtÞ þ q02 ðtÞ þ q22 ðtÞ þ q03 ðtÞ þ q23 ðtÞ: k¼1
k¼1
When t = 10, 20, 30, 40, 50, the results of R(t) are, respectively, Rð10Þ 0:9725;
Rð20Þ 0:9374;
Rð30Þ 0:8974;
Rð40Þ 0:8522;
Rð50Þ 0:8027:
Clearly, we can see that the value of the system reliability R(t) is decreasing when t increases. For a deteriorating system, the observation is consistent with our intuition. 5. Replacement model under policy T 5.1. Long-run average cost under policy T In this section, we consider a repair-replacement policy based on the working age T of component 1 which is other than the policy N in Zhang [25]. The system will be replaced by a new one when the working age of component 1 reaches T. Our objective is to determine an optimal replacement policy T* such that the long-run average cost per unit time of the system is minimized. To do this, based on Section 2 we add the following assumptions. Assumption 5. Assume that a component in the system is replaced sometime by a new and identical one, and the replacement time is negligible. Assumption 6. Assume that a component in the system cannot produce the working reward during cold standby, and no cost is incurred during waiting for repair. Assumption 7. Assume that the repair cost rate of component i is cðiÞ r ; i ¼ 1; 2 while the working reward rate of two components is same cw. And the replacement cost of the system is C. Let s1 be the first replacement time of the system, and sn(n P 2) be the time between the (n 1)th replacement and the nth replacement of the system under policy T. Obviously, {s1, s2,. . .} forms a renewal process, and the interarrival time between two consecutive replacements is called a renewal cycle. Let C(T) be the long-run average cost per unit time of the system under policy T. Thus, according to renewal reward theorem (see, for example Ross [26]), we have CðT Þ ¼
the expected cost incurred in a renewal cycle : the expected length of a renewal cycle
ð19Þ
Because component 1 has priority in use, component 1 has only the working state, the repair state and the waiting for repair state. Thus, when the working age of component 1 reaches T, then component 2 is either under a cold standby state or under a repair state (see Fig. 1). Let W be the length of a renewal cycle of the system under the policy T, then W ¼T þ
K X j¼1
ð1Þ
Yj þ
K X ð2Þ ð1Þ Y j1 X j vn j¼2
ð2Þ
ð1Þ
o vn
Y j1 X j >0
ð1Þ
ð2Þ
o;
Y j1 X j1 >0
where, the second term and the third term are respectively the length of repair time and the length of time of waiting for repair of component 1 before the working age of component 1 reaches T; v is a indicator function such that 1; If event A occurs vA ¼ 0; If event A does not occur:
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
289
Fig. 1. A possible course of the system.
K is the failure number of component 1 before the working age of component 1 reaches T. Obviously, K is a random variable. Hence
EW ¼ T þ E
K X
2
!
þ E4
ð1Þ Yj
j¼1
K X
ð2Þ Y j1
ð1Þ Xj
vn
j¼2
3 ð2Þ
o vn
ð1Þ
Y j1 X j >0
ð1Þ
ð2Þ
o 5:
ð20Þ
Y j1 X j1 >0
To evaluate the long-run average cost per unit time under the policy T, first of all, we determine the probability mass function of K. Since ( P fK P kg ¼ P
k X
) ð1Þ Xj
ð1Þ
¼ H k ðT Þ;
j¼1 ð1Þ
where H k is the distribution of
Pk
ð1Þ j¼1 X j ,
therefore ð1Þ
ð1Þ
P fK ¼ kg ¼ P fK P kg P fK P k þ 1g ¼ H k ðT Þ H kþ1 ðT Þ:
ð21Þ ð2Þ
ð1Þ
According to Assumption 3 and the definition of the convolution, the distribution functions of Y j1 X j ð2Þ ð1Þ and X j Y j are, respectively,
Uj ðuÞ ¼ Gð2Þ ðuÞ 1 F ð1Þ ðaj1 uÞ ;
Wj ðvÞ ¼ F ð2Þ ðvÞ 1 Gð1Þ ðbj1 vÞ : Therefore 2
ð2Þ ð1Þ E4 Y j1 X j vn
3 ð2Þ
ð1Þ
o vn
Y j1 X j >0
ð1Þ
ð2Þ
o 5 ¼ Wj1 ð0Þ
Y j1 X j1 >0
Z
1
u dUj ðuÞ:
ð22Þ
0
By using the property of the conditional expectation and Eq. (21), we have the following results:
E
K X
"
! ð1Þ Xj
¼E E
j¼1
K X
!# ð1Þ X j jK
j¼1
¼
1 k X X k¼1
j¼1
1 aj1 k1
!
¼
1 X k¼1
E
k X
! ð1Þ Xj
P ðK ¼ kÞ
j¼1
1 h i 1 X 1 ð1Þ ð1Þ ð1Þ H ðT Þ < 1: H k ðT Þ H kþ1 ðT Þ ¼ k1 k¼1 ak1 k
ð23Þ
290
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
Similarly K X
E
! ð2Þ Xj
¼
j¼1
1 1 X ð1Þ H ðT Þ; k2 k¼1 k
ð24Þ
!
1 1 X 1 ð1Þ H k ðT Þ; k1 l b 1 j¼1 k¼1 " # Z K 1 X X ð2Þ ð1Þ ð1Þ E X j Y j vfX ð2Þ Y ð1Þ >0g ¼ F j ðT Þ K X
E
ð1Þ Yj
j
j¼1
2 K X ð2Þ E4 Y j vn j¼1
E
T
ð26Þ
1 X ð1Þ o5 ¼ 1 Wk1 ð0ÞH k ðT Þ; l2 k¼1
Wk1 ð0Þ
Z
ð27Þ
1
3 ð2Þ
o vn
ð1Þ
Y j1 X j >0
ð1Þ
ð2Þ
Y j1 X j1 >0
o5
ð1Þ u dUk ðuÞ H k ðT Þ;
ð28Þ
0
k¼2
"
v dWj ðvÞ;
0
j¼1
ð1Þ ð2Þ Y j1 X j1 >0
j¼2 1 X
j
1
3
2 K X ð2Þ ð1Þ E4 Y j1 X j vn
¼
ð25Þ
¼
K X
#
! ð1Þ
vfY ð1Þ X ð2Þ >0g ¼
Xj
K
j¼1
K
1 X
ð1Þ
ð1Þ
Wk ð0Þ H k ðT Þ H kþ1 ðT Þ
T
k¼1
k X j¼1
! 1 : k1 aj1
ð29Þ
Substituting (25) and (28) into (20), we have Z 1 1 1 X 1 X 1 ð1Þ ð1Þ H ðT Þ þ Wk1 ð0Þ u dUk ðuÞ H k ðT Þ: EW ¼ T þ l1 k¼1 bk1 k 0 k¼2
ð30Þ
The total work time of the system in a renewal cycle is U ¼T þ
K X
K K n o X X ð2Þ ð1Þ ð2Þ ð2Þ ð1Þ min X j ; Y j Xj X j Y j v ¼Tþ
j¼1
j¼1
ð2Þ
ð1Þ
:
X j Y j >0
j¼1
According to the expressions (24) and (26), we have EU ¼ T þ
Z 1 1 1 X 1 X ð1Þ ð1Þ H k ðT Þ H k ðT Þ v dWk ðvÞ: k2 k¼1 0 k¼1
ð31Þ
The total repair time of the system in a renewal cycle is V ¼ V 1 þ V 2; where V1 and V2 denotes, respectively, the repair times of the components 1 and 2 in a renewal cycle, i.e. V1 ¼
K X
ð1Þ
Yj ;
j¼1
V2 ¼
K 1 X j¼1
ð2Þ Y j vfY ð1Þ X ð2Þ >0g j
j
þ
T
K X j¼1
! ð1Þ Xj
v
ð1Þ
ð2Þ
:
Y K X K >0
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
291
According to the expressions (25), (27) and (29), we have EV ¼ EV 1 þ EV 2 1 1 1 X 1 1 X ð1Þ ð1Þ ¼ H k ðT Þ þ Wk ð0ÞH k ðT Þ k1 l1 k¼1 b l2 k¼1 þ
1 X
Wk ð0Þ
ð1Þ H k ðT Þ
ð1Þ H kþ1 ðT Þ
T
k¼1
k X j¼1
! 1 : k1 aj1
ð32Þ
Therefore, substituting the expressions (30)–(32) into the expression (19), then the long-run average cost per unit time of the system under policy T is given by ð2Þ cð1Þ r EV 1 þ cr EV 2 þ C cw EU EW h P i ð1Þ 1 crð1Þ l11 1 H ðT Þ k1 k k¼1 b ¼
R 1 ð1Þ P1 P1 1 ð1Þ 1 T þ l1 k¼1 bk1 H k ðT Þ þ k¼2 Wk1 ð0Þ 0 u dUk ðuÞ H k ðT Þ h P i P1 Pk ð1Þ ð1Þ ð1Þ 1 1 1 cð2Þ W ð0ÞH ðT Þ þ W ð0Þ H ðT Þ H ðT Þ T k k j1 k k kþ1 r k¼1 k¼1 j¼1 k1 a l2 þ
R 1 ð1Þ P1 1 ð1Þ P1 1 T þ l1 k¼1 bk1 H k ðT Þ þ k¼2 Wk1 ð0Þ 0 u dUk ðuÞ H k ðT Þ h i R1 P1 ð1Þ P ð1Þ cw T þ k12 1 H ðT Þ H ðT Þ u dW ðvÞ C k k¼1 k k¼1 k 0 :
R P P 1 ð1Þ ð1Þ 1 1 1 T þ l11 k¼1 bk1 H k ðT Þ þ k¼2 Wk1 ð0Þ 0 u dUk ðuÞ H k ðT Þ
CðT Þ ¼
ð33Þ
Then the following result will show a sufficient condition that the optimal replacement policy T* can be found by minimizing C(T). To do this, we need a lemma as follows. P ð1Þ a 1 Lemma 1. If t P a1 F 1 ðbÞ; a > 1; 0 < b < 1, then 1 where F1 is the inverse function of the k¼1 bk H k ðtÞ ¼ P1, ð1Þ ð1Þ ð1Þ k distribution function of X 1 , and H k is the distribution function of j¼1 X j . (Its proof can be found from Lam [8].) a TheoremP1. Let d0 ¼ a1 F 1 ðbÞ, where a > 1, 0 < b < l, F1 is the inverse function of the distribution function of ð1Þ ð1Þ 1 X 1 . If k¼1 H k ðT Þ < 1 when T P d0, then the optimal replacement policy T* must be found by minimizing C(T) in (0, d0). ð2Þ
ð1Þ
ð2Þ
ð1Þ
Proof. According to the distribution functions of Y j1 X j and X j Y j , we have Z Z Z þ1 ð2Þ ð1Þ
ðy xÞdGðyÞdF ðak1 xÞ u dUk ðuÞ ¼ E Y k1 X k v ð2Þ ð1Þ ¼ 0
¼
Z
þ1
Z
0
Y k1 X k >0
D1
þ1
ðy xÞdGðyÞ dF ðak1 xÞ 6
Z
þ1
Z
0
x
þ1 0
1 ð2Þ y dGðyÞ dF ðak1 xÞ ¼ EY k1 ¼ ; l2
where D1 = {(x,y)|x > 0,y > 0 and y x > 0}. Based on Lemma 1 and the condition in Theorem 1, when T P d0 we can get Z 1 1 1 1 X X 1 ð1Þ 1 X ð1Þ ð1Þ Wk1 ð0Þ u dUk ðuÞ H k ðT Þ ¼ Wk1 ð0Þ H k ðT Þ 6 H k ðT Þ < 1: l l 0 2 2 k¼2 k¼2 k¼2 Similarly, Z 1
v dWk ðvÞ ¼ E
0
¼
ð2Þ ð1Þ X k Y k v
Z þ1 Z 0
þ1
ð2Þ
ð1Þ
X k Y k >0
Z Z
¼ ðx yÞdF ðxÞdGðbk1 yÞ
ðx yÞdF ðxÞ dGðb y
ð34Þ
D2 k1
Z
þ1
Z
yÞ 6 0
0
þ1
ð2Þ
x dF ðxÞ dGðbk1 yÞ ¼ EX k ¼
1 ; k2
292
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
where D2 = {(x,y)|x > 0,y > 0 and x y > 0}. Thus 1 Z þ1 1 X 1 X ð1Þ ð1Þ v dWk ðvÞ H k ðT Þ 6 H k ðT Þ < 1 k 2 0 k¼1 k¼1 and 1 X
k X ð1Þ ð1Þ Wk ð0Þ H k ðT Þ H kþ1 ðT Þ T
k¼1
j¼1
1 k1 aj1
! 6
ð35Þ
1 X k¼1
ð1Þ H k ðT Þ
ð1Þ H kþ1 ðT Þ
T
k X j¼1
1 k1 aj1
!
1 1 X X ð1Þ ð1Þ ð1Þ 6 T H k ðT Þ H kþ1 ðT Þ 6 T H k ðT Þ k¼1
k¼1
< 1:
ð36Þ
According to Lemma 1, the condition in Theorem 1 and Eqs. (34)–(36), when T P d0, we have CðT Þ crð1Þ : Because the total lifetime of the cold standby repairable system is limited, the minimum of the long-run average cost per unit time under the policy T exists. Thus, under the condition of theorem, the optimal replacement policy T * can be found by minimizing C(T) in (0, d0), because for T P d0, we have CðT Þ crð1Þ . Clearly, T * can be found analytically or numerically. h 5.2. A numerical example In this section, we provide an example to illustrate the theoretical results. ð1Þ ð2Þ ð1Þ ð2Þ Now, we denote the probability density functions of X j ; X j ; Y j and Y j , respectively, by fj(t), f(t), gj(t) ð2Þ ð1Þ and g(t) for j = 1, 2,. . . According to Assumption 3, then the probability density functions of Y j1 X j and ð2Þ ð1Þ X j Y j are, respectively, 8 j1 < aj1 k1 l2 el2 u ; u P 0; a k1 þl2 /j ðuÞ ¼ gðuÞ fj ðuÞ ¼ j1 k l j1 a : 1 2 ea k1 u ; u < 0; aj1 k1 þl 8 j1 2 2 l1 < b kj1 ek2 v ; v P 0; k2 þb l1 wj ðvÞ ¼ f ðvÞ gj ðvÞ ¼ : bj1 k2 l1 ebj1 l1 v ; v < 0: k þbj1 l 2
Therefore Z 1
u dUk ðuÞ ¼
0
Z
1
Z Z
1
1
u/k ðuÞdu ¼
0
ak1 k1 ; l2 ðak1 k1 þ l2 Þ
ð37Þ
1
bk1 l1 ; ð38Þ k2 ðk2 þ bk1 l1 Þ 0 0 k2 ð2Þ ð1Þ Wk ð0Þ ¼ P X j Y j < 0 ¼ : ð39Þ k2 þ bk1 l1 Pk ð1Þ ð1Þ To determine the distribution H k of j¼1 X j , the following lemma is essential, see e.g., Chiang [27] for the proof. v dWk ðvÞ ¼
vwk ðvÞdv ¼
Lemma 2. Assume that {Xi, i = 1, 2,. . .} is a sequence of independent random variables,P and Xi has the exponential distribution Exp(ki) with ki 5 kj for i 5 j. Then n the probability density function of ni¼1 X i is given by hn ðxÞ ¼ ð1Þ
n1
k1 . . . kn
n X
eki x kj Þ
Pnj¼1 ðki i¼1 j6¼i
for
x P 0;
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
293
and 0 otherwise. In particular, according to Assumption 3, 4 and ki = ai1k, the density function of this paper is given by n1
hnð1Þ ðxÞ ¼ ð1Þ
a
nðn1Þ 2
k
n X
ea
ð1Þ
Z
T
ð1Þ
hk ðxÞdx ¼ ð1Þ
ð1Þ j¼1 X j
in
i1 k
Pnj¼1 ðai1 i¼1 j6¼i
x aj1 Þ
and 0 otherwise. Thus, the distribution function of H k ðT Þ ¼
Pn
k1 kðk1Þ=2
a
0
for Pk
k X i¼1
t P 0;
ð1Þ j¼1 X j
is given by i1
1 ea kT : i1 a Pkj¼1 ðai1 aj1 Þ
ð40Þ
j6¼i
Thus, substituting the expressions (37)–(40) into the expression (33), then ð2Þ cð1Þ r EV 1 þ cr EV 2 þ C cw EU EW h P i ð1Þ 1 1 crð1Þ l11 k¼1 bk1 H k ðT Þ ¼ P1 P ð1Þ ð1Þ ak1 k1 k2 1 T þ l11 1 k¼1 bk1 H k ðT Þ þ k¼2 ðk2 þbk2 l1 Þðak1 k1 þl2 Þl2 H k ðT Þ h P i P1 Pk ð1Þ ð1Þ ð1Þ 1 k2 k2 1 cð2Þ T j¼1 k1 a1j1 k¼1 k2 þbk1 l1 H k ðT Þ þ k¼1 k2 þbk1 l1 H k ðT Þ H kþ1 ðT Þ r l2 þ P P1 ð1Þ ð1Þ ak1 k1 k2 1 T þ l11 1 k¼1 bk1 H k ðT Þ þ k¼2 ðk2 þbk2 l1 Þðak1 k1 þl2 Þl2 H k ðT Þ h i P1 ð1Þ cw T þ k¼1 k þb1k1 l H k ðT Þ C 2 1 ; P1 1 ð1Þ P1 k1 k ð1Þ 1 1 2 T þ l1 k¼1 bk1 H k ðT Þ þ k¼2 ðk þbk2 al Þðakk1 H ðT Þ k k þl Þl
CðT Þ ¼
2
ð1Þ
1
1
2
ð41Þ
2
ð1Þ
where H k ðT Þ and H kþ1 ðT Þ are both determined by the expression (40). There are a = 1.1, b = 0.9, k1 = 0.01, k2 = 0.1, l1 = 0.1, l2 = 0.5 in Section 4.6. Now, we further assume that crð1Þ ¼ 10; cð2Þ r ¼ 8; cw ¼ 9 and C = 2500. Substituting the above values into the expression (41) with the expression (40) and overpassing numerical calculation, we can obtain some results presented in Table 2 and Fig. 2. It is easy to find that C(500) = 1.7933 is the minimum of the long-run average cost per unit time of the system. In other words, the optimal policy is T * = 500, and we should replace the system when the working age of component 1 reaches 500. Table 2 or Fig. 2 indicates that the optimal policy T * is unique. However, we should point out that T is a continuous variate. To produce the plot of C(T) versus T as quickly as possible, we have to select 50 units as the step size of T. Thus, the above results in Table 2 and Fig. 2 are obtained under the condition of 50 units as the step size of T.
Table 2 Some results obtained using expressions (40) and (41) T
C(T)
T
C(T)
T
C(T)
5 · 50 6 · 50 7 · 50 8 · 50 9 · 50 10 · 50 11 · 50
1.2117 0.0624 0.9180 1.4828 1.7924 1.7933 1.3490
12 · 50 13 · 50 14 · 50 15 · 50 16 · 50 17 · 50 18 · 50
0.3394 1.1587 2.8354 4.3618 5.5755 6.4696 7.1046
19 · 50 20 · 50 21 · 50 22 · 50 23 · 50 24 · 50 25 · 50
7.5491 7.8586 8.0732 8.2204 8.3192 8.3825 8.4195
294
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4
3
C(T)
2
1
0
1
2
250
300
350
400
450
500 T
550
600
650
700
750
Fig. 2. The plot of C(T) against T.
6. Concluding remarks In this paper, we introduce the model for a two-component cold standby repairable system with one repairman and use priority. Assume that component 1 after repair is not ‘‘as good as new’’ such that the successive working times of component 1 form a decreasing geometric process and the consecutive repair times of component 1 form a increasing geometric process. Obviously, the model in this paper is more realistic than the model in which two components after repair are both ‘‘as good as new’’. In this model, we study not only some important reliability indices of the system but also a replacement policy of the system when the working time distributions and the repair time distributions of the two components are both exponential. Thus, this paper has definite theoretical interest and potential practical application. However, based on our analyses and the examples above, we have the following remarks: (1) Let N(t) be the state of the system at time t. It is clear from model assumptions that {N(t), t P 0} is not a Markov chain. However, it can be extended as a two-dimensional Markov process by introducing a supplementary variable. To obtain the system reliability indices, we need to determine the state marginal probabilities of the system at time t. Accordingly, we can derive the system of differential equations about Pjk(t), j 2 X, k = 1, 2,. . . Finally, the Laplace transform results of reliability indices of the system and the steady-state results of some reliability indices of the system are obtained. (2) We can see that the method by using the inverse Laplace transform to get the transient results of the reliability indices is very tedious, and is also very difficult. But, the Laplace transform results of reliability indices of the system is discommodious for practical application. Thus, in engineering, a numerical method based on the Runge–Kutta method is often adopted. In Section 4.6, an approximative solution of P1 R(t) is given to illustrate the numerical method. Note that the expression of RðtÞ ¼ k¼1 ½q0k ðtÞ þ q2k ðtÞ is an infinite series. However, we can select the values of k and t according to the practical request. (3) In this paper, we consider a replacement policy based on the working age T of component 1. An optimal replacement policy T * for minimizing C(T) is determined. And under some conditions, we can prove that the optimal replacement policy T * must be found by minimizing C(T) in (0, d0). In fact, when a = 1.1, b = 0.9, we can compute
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
d0 ¼
295
a F 1 ðbÞ 2532:8436: a1
Obviously, the optimal policy T * = 500 2 (0, d0) = (0, 2532.8463). It is consistent with the theoretical analysis in replacement model. (4) From Table 2 and Fig. 2 in Section 5.2, we should point out that the optimal solution is obtained under certain conditions. Note that the working age T of component 1 is a continuous variable. To find optimal solution of the objective function C(T) as quickly as possible, we have to select the step sizes of T. Thus, the optimal solution in this paper can be obtained under certain conditions, i.e. the step sizes of T is 50 units (see Table 2 or Fig. 2). Clearly, different step sizes of T will obtain different optimal solution. Because the total life of the system is finite, we can find a corresponding optimal solution.
Acknowledgement The authors are grateful for the Editor’s and the referee’s valuable comments and suggestions, which have considerably improved the presentation of the paper. References [1] R.E. Barlow, F. Proschan, Mathematical Theory of Reliability, Wiley, New York, 1965. [2] H. Ascher, H. Feingold, Repairable Systems Reliability, Marcel Dekker, New York, 1984. [3] T. Nakagawa, S. Osaki, Stochastic behaviour of a two-unit priority standby redundant system with repair, Microelectronics Reliability 14 (3) (1975) 309–313. [4] S. Osaki, Reliability analysis of a two-unit standby redundant system with priority, Canadian Journal of Operations Research Society 8 (1970) 60–62. [5] J.A. Buzacott, Availability of priority standby redundant systems, IEEE Transactions of Reliability R-20 (1971) 60–63. [6] R.E. Barlow, L.C. Hunter, Optimum preventive maintenance policy, Operations Research 8 (1960) 90–100. [7] M. Brown, F. Proschan, Imperfect repair, Journal of Applied Probability 20 (1983) 851–859. [8] Y. Lam, Geometric processes and replacement problem, Acta Mathematica Applied Sinica 4 (4) (1988) 366–377. [9] Y. Lam, A note on the optimal replacement problem, Advances in Applied Probability 20 (1988) 479–482. [10] M.S. Finkelstein, A scale model of general repair, Microelectronics Reliability 33 (1993) 41–44. [11] Y.L. Zhang, A bivariate optimal replacement policy for a repairable system, Journal of Applied Probability 31 (1994) 1123–1127. [12] W. Stadje, D. Zuckerman, Optimal strategies for some repair replacement models, Advance in Applied Probability 22 (1990) 641–656. [13] A.D.J. Stanley, On geometric processes and repair replacement problems, Microelectronics Reliability 33 (1993) 489–491. [14] Y.L. Zhang, R.C.M. Yam, M.J. Zuo, Optimal replacement policy for a deteriorating production system with preventive maintenance, International Journal of System Science 32 (10) (2001) 1193–1198. [15] Y.L. Zhang, R.C.M. Yam, M.J. Zuo, Optimal replacement policy for a multistate repairable system, Journal of Operations Research Society 53 (2002) 336–341. [16] Y.L. Zhang, A geometric-process repair-model with good-as-new preventive repair, IEEE Transactions Reliability R-51 (2) (2002) 223–228. [17] Y. Lam, Y.L. Zhang, A geometric-process maintenance model for a deteriorating system under a random environment, IEEE Transactions Reliability R-52 (1) (2003) 83–89. [18] Y.L. Zhang, An optimal replacement policy for a three-state repairable system with a monotone process model, IEEE Transactions Reliability R-53 (2004) 452–457. [19] G.J. Wang, Y.L. Zhang, A shock model with two-type failures and optimal replacement policy, International Journal of System Science 36 (4) (2005) 209–214. [20] Y.L. Zhang, S.M. Wu, The reliability analysis of a two-unit series repairable system without being repaired ‘‘as good as new’’. In: Proceedings of the Fourth Symposium on Reliability Mathematics, Guilin, China, 1992, pp. 464–469. [21] Y. Lam, Y.L. Zhang, Analysis of a two-component series system with a geometric process model, Naval Research Logistics 43 (2003) 491–502. [22] Y. Lam, Calculating the rate of occurrence of failures for continuous-time Markov chains with application to a two-component parallel system, Journal of Operations Research Society 45 (1995) 528–536. [23] Y. Lam, Y.L. Zhang, Analysis of a parallel system with two different units, Acta Mathematica Applied Sinica 12 (1996) 408–417. [24] Y. Lam, A maintenance model for two-unit redundant system, Microelectronics Reliability 37 (3) (1997) 497–504. [25] Y.L. Zhang, An optimal geometric process model for a cold standby repairable system, Reliability Engineering and Systems Safety 63 (1999) 107–110. [26] S.M. Ross, Stochastic Processes, second ed., Wiley, New York, 1996. [27] C.L. Chiang, An Introduction to Stochastic Processes and Their Applications, Robert E. Krieger Publishing Co., New York, 1980.
Stochastics and Statistics
A deteriorating cold standby repairable system with priority in use Yuan Lin Zhang *, Guan Jun Wang Department of Mathematics, Southeast University, Nanjing 210018, China Received 18 November 2005; accepted 27 September 2006 Available online 18 December 2006
Abstract In this paper, a cold standby repairable system consisting of two dissimilar components and one repairman is studied. In this system, it is assumed that the working time distributions and the repair time distributions of the two components are both exponential and component 1 is given priority in use. After repair, component 2 is ‘‘as good as new’’ while component 1 follows a geometric process repair. Under these assumptions, using the geometric process and a supplementary variable technique, some important reliability indices such as the system availability, reliability, mean time to first failure (MTTFF), rate of occurrence of failure (ROCOF) and the idle probability of the repairman are derived. A numerical example for the system reliability R(t) is given. And it is considered that a repair-replacement policy based on the working age T of component 1 under which the system is replaced when the working age of component 1 reaches T. Our problem is to determine an optimal policy T * such that the long-run average cost per unit time of the system is minimized. The explicit expression for the long-run average cost per unit time of the system is evaluated, and the corresponding optimal replacement policy T * can be found analytically or numerically. Another numerical example for replacement model is also given. 2006 Elsevier B.V. All rights reserved. Keywords: Geometric process; Renewal process; Supplementary variable; Generalized Markov process; Priority; Replacement policy
1. Introduction In order to improve the reliability or raise the availability and hence reduce the loss, a two-component redundant system is often employed. Note that such a system is one of the typical systems employed in reliability engineering. Under the conditions that the working time and the repair time of the components in the system both have exponential or general distributions, by using Markov process theory or Markov renewal process theory, some important reliability indices of the system have been derived in the previous literature (see e.g. Barlow and Proschan [1] and Ascher and Feingold [2] for further details). Later, a priority rule for repair or use of a component has been introduced. Nakagawa and Osaki [3] assumed that both the working time and the repair time of the priority component have a general distribution while both the working time *
Corresponding author. Tel.: +86 25 8379 3841. E-mail address: [email protected] (Y.L. Zhang).
0377-2217/$ - see front matter 2006 Elsevier B.V. All rights reserved. doi:10.1016/j.ejor.2006.09.075
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279
and the repair time of the non-priority component have an exponential distribution. They obtained some interesting reliability indices of the system using Markov renewal theory. See also Osaki [4] and Buzacott [5] for generalizations. In all of these studies, they assumed that a system (or a component) after repair is ‘‘as good as new’’. This is a perfect repair model. However, this assumption is not always true. In practice, most repairable systems are deteriorative because of the ageing effect and the accumulative wear. Barlow and Hunter [6] first presented a minimal repair model in which the minimal repair does not renew the age of the system. Brown and Proschan [7] first reported an imperfect repair model in which the repair is perfect repair with probability p or minimal with probability 1 p. For a deteriorating repairable system, it is quite reasonable to assume that the successive working times of the system after repair will become shorter and shorter while the consecutive repair times of the system after failure will become longer and longer. Ultimately, it cannot work any longer, neither can it be repaired. For such a stochastic phenomenon, Lam [8,9] first introduced a geometric process repair model to approach it. Under this model, he studied two kinds of replacement policy for a one-component repairable system with one repairman (called a simple repairable system), one based on the working age T of the system and the other based on the failure number N of the system. The explicit expressions of the long-run average cost per unit time under these two kinds of policy are respectively calculated. Finkelstein [10] presented a general repair model based on a scale transformation after each repair to generalize Lam’s work. Zhang [11] generalized Lam’s work by a bivariate replacement policy (T, N) under which the system is replaced at the working age T or at the time of the Nth failure, whichever occurs first. Many replacement policies have been done by Stadje and Zuckerman [12], Stanley [13], Zhang et al. [14,15], Zhang [16], Lam and Zhang [17], Zhang [18], Wang and Zhang [19] and others under the geometric process repair model. However, the above various research is related to the simple repairable system. For multi-component systems, Zhang and Wu [20] first reported a two-component series repairable system under the geometric process repair model. When the working time of component 1 in the system follows the exponential distribution while that of component 2 and the repair times of both components follow general distributions, they derived some reliability indices of the system. Lam and Zhang [21] provided a more in depth analysis of the system studied by Zhang and Wu [20] under the assumptions that the working and repair times of both components all follow the exponential distributions. Lam [22] investigated a two-component parallel repairable system assuming that all working and repair times follow the exponential distributions and one component after repair is ‘‘as good as new’’ while the other is not. The system ROCOF is determined through the Monte Carlo numerical method. Lam and Zhang [23] studied a similar model, more reliability indices of the system are obtained by using the Laplace-transform technique. Lam [24] reported a maintenance model for two-unit redundant system with one repairman. Under this model, he studied two kinds of replacement policy, respectively based on the number of failure and the working age for two units. The long-run average cost per unit time for each kind of replacement policy is derived. Zhang [25] applied the geometric process repair model to a two-component cold standby repairable system with one repairman. He also assumed that each component after repair is not ‘‘as good as new’’. Under this assumption, by using a geometric process, he studied a replacement policy N based on the number of repairs of component 1. The problem is to determine an optimal replacement policy N* such that the long-run expected reward per unit time is maximized. In practical applications, a two-component cold standby repairable system with one repairman and priority in use is often used. For example, in the operating room of a hospital, a patient on the operating table has to discontinue his/her operation as soon as the power source is cut (i.e. the power station failures). Usually, there is a standby power station (e.g. a storage battery) in the operating room. Thus, the power station (e.g. write as component 1) and the storage battery (e.g. write as component 2) form a cold standby repairable lighting system. Obviously, it is reasonable to assume that the power station has priority in use due to the operating cost of the power station is cheaper than the operating cost of the storage battery, and the storage battery after repair is as good as new due to its used time being smaller than the power station, and the repair of the storage battery is also convenient. Besides a cold standby lighting system in a hospital, some similar examples can be found from Lam [24]. The purpose of this paper is to apply the geometric process repair model to a two-component cold standby repairable system with one repairman. Now we may assume that the component 2 after repair is ‘‘as good as new’’ while component 1 follows the geometric process repair, but component 1 has priority in use.
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Furthermore, we assume that the working time and the repair time of both components are exponentially distributed. Under these assumptions, by using the geometric process and a supplementary variable technique we not only obtain some important reliability indices such as availability, reliability, MTTFF, ROCOF, and the idle probability of the repairman, but also determine the replacement policy based on the working age T of component 1. The problem is to choose an optimal replacement policy T* such that the long-run average cost per unit time is minimized. The explicit expression for the long-run average cost per unit time of the system is evaluated, and the corresponding optimal replacement policy T* can be found analytically or numerically. Finally, two numerical examples are respectively given for the system reliability R(t) and the system replacement policy T. For further reference, we first state the definitions of stochastic order and a geometric process. Definition 1. Given two random variables n and g, n is said to be stochastically larger than g or g is stochastically smaller than n, if P ðn > aÞ P P ðg > aÞ;
for all real a;
denoted by n P stg or g 6 stn (see e.g. Ross [26]). Furthermore, we say that a stochastic process {Xn, n = 1, 2,. . .} is stochastically decreasing if Xn P st Xn+1 and stochastically increasing if Xn 6 st Xn+1 for all n = 1, 2,. . . Definition 2. Assume that {nn, n = 1, 2,. . .} is a sequence of independent non-negative random variables. If the distribution function of nn is Fn(t) = F(an1t), n = 1, 2, . . . , and if a is a positive constant, then {nn, n = 1, 2,. . .} is called a geometric process, and a is the ratio of the geometric process (see e.g. Lam [8] and Zhang [11] for further details). Obviously, if a > 1, then {nn, n = 1, 2,. . .} is stochastically decreasing. If 0 < a < 1, then {nn, n = 1, 2,. . .} is stochastically increasing. If a = 1, then the geometric process is the sequence of interarrival times of a renewal process. 2. Model We study a two-component cold standby repairable system with one repairman and priority in use by making the following assumptions: Assumption 1. Initially, the two components are both new, and component 1 is in working state while component 2 is in cold standby state. Assumption 2. Assume that component 2 after repair is as good as new while component 1 follows a geometric process repair. The repair rule is ‘‘first-in-first-out’’. If a component fails during the repair of the other, it must wait for repair and the system is down. Component 1 has priority in use. Assumption 3. Assume that the time interval between the completion of the (n 1)th repair and the completion of the nth repair of component i is called the nth cycle of component i, i = 1, 2; n = 1, 2,. . . ðiÞ Let X ðiÞ n and Y n be, respectively, the working time and the repair time of component i in the nth cycle, i = 1, ðiÞ 2; n = 1, 2,. . . We assume that the distributions of X ðiÞ n and Y n ði ¼ 1; 2Þ are, respectively, n1 tÞ ¼ 1 expðan1 k1 tÞ; F ð1Þ n ðtÞ ¼ F ða
Gnð1Þ ðtÞ ¼ Gðbn1 tÞ ¼ 1 expðbn1 l1 tÞ; F ð2Þ ðtÞ ¼ 1 expðk2 tÞ; Gð2Þ ðtÞ ¼ 1 expðl2 tÞ; where t P 0, a P 1, 0 < b 6 1, ki > 0, li > 0; i = 1, 2.
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281
ðiÞ Assumption 4. Assume that X ðiÞ n and Y n ; i ¼ 1; 2; n ¼ 1; 2; . . . are independent.
Obviously, such a model for a two-component cold standby repairable system with one repairman and priority in use based on these assumptions above takes the deterioration into account, it is more realistic than the model in which two components after repair are both ‘‘as good as new’’. 3. System analysis Let {N(t), t P 0} be a stochastic process characterized by the following mutually exclusive events: {N(t) = 0}: {N(t) = 1}: {N(t) = 2}: {N(t) = 3}: {N(t) = 4}:
Component Component Component Component Component
1 1 2 1 2
is is is is is
working, component 2 is working, component 2 is working, component 1 is under repair, component under repair, component
in cold standby at time t. under repair at time t. under repair at time t. 2 is waiting for repair at time t. 1 is waiting for repair at time t.
Then {N(t), t P 0} is a stochastic process with state space X = {0, 1, 2, 3, 4}. The set of working states is W = {0, 1, 2} and the set of failure states is F = {3, 4}. According to the model assumptions, {N(t), t P 0} is not a Markov process. However, it can be extended to a generalized Markov process by introducing a supplementary variable. Let S(t) be the number of the cycle of component 1 at time t. Then {N(t), S(t), t P 0} constitutes a generalized (i.e. two-dimensional) Markov process. The state marginal probabilities of the system at time t are defined by pjk ðtÞ ¼ P fN ðtÞ ¼ j; SðtÞ ¼ kg;
j 2 X;
k ¼ 1; 2; . . .
According to the model assumptions and the supplementary variable technique, we can obtain the following differential equations for the system. By straightforward probability arguments, for example, we have p0k ðt þ DtÞ ¼ pfN ðt þ DtÞ ¼ 0; Sðt þ DtÞ ¼ kg ¼ p0k ðtÞð1 ak1 k1 DtÞ þ p1k ðtÞl2 Dt þ p2 letting Dt tend to zero, we can get d k1 þ a k1 p0k ðtÞ ¼ l2 p1k ðtÞ þ bk2 l1 p2 dt
k1 ðtÞ
k1 ðtÞb
k2
l1 Dt þ oðDtÞ;
ðk P 2Þ:
In the same way, we have d k1 þ a k1 þ l2 p1k ðtÞ ¼ bk2 l1 p3 k1 ðtÞ ðk P 2Þ; dt d þ bk1 l1 þ k2 p2k ðtÞ ¼ ak1 k1 p0k ðtÞ þ l2 p4k ðtÞ ðk P 2Þ; dt d þ bk1 l1 p3k ðtÞ ¼ k2 p2k ðtÞ ðk P 2Þ; dt d þ l2 p4k ðtÞ ¼ ak1 k1 p1k ðtÞ ðk P 2Þ; dt d þ k1 p01 ðtÞ ¼ 0; dt p11 ðtÞ ¼ 0; d þ k2 þ l1 p21 ðtÞ ¼ k1 p01 ðtÞ þ l2 p41 ðtÞ; dt d þ l1 p31 ðtÞ ¼ k2 p21 ðtÞ; dt p41 ðtÞ ¼ 0:
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The initial conditions are p01 ð0Þ ¼ 1;
p0k ð0Þ ¼ 0;
ðk P 2Þ;
pjk ð0Þ ¼ 0;
j ¼ 1; 2; 3; 4;
k ¼ 1; 2; . . .
According to the theory of Markov process, it is easy to see that above equations are actually Kolmogorov forward equations. Based on the property (Lipschitz condition) of differential equation, the solution will exist and is unique. Obviously, a numerical method should be applicable for the solution of the system of differential equations above. Here, we adopt the method of Laplace transform to determine Laplace transform solution of the system of differential equations. In Section 4.6, a numerical example for the system reliability is given to illustrate the numerical method. R1 Denote the Laplace transform of Pjk(t) by pjk ðsÞ ¼ 0 est pjk ðtÞdt. The Laplace transforms of the above differential equations are respectively given by ðs þ ak1 k1 Þp0k ðsÞ ¼ l2 p1k ðsÞ þ bk2 l1 p2 ðs þ ðs þ ðs þ ðs þ
k1 ðsÞ ðk P 2Þ; k2 a k1 þ l2 Þp1k ðsÞ ¼ b l1 p3 k1 ðsÞ ðk P 2Þ; bk1 l1 þ k2 Þp2k ðsÞ ¼ ak1 k1 p0k ðsÞ þ l2 p4k ðsÞ ðk P 2Þ; bk1 l1 Þp3k ðsÞ ¼ k2 p2k ðsÞ ðk P 2Þ; l2 Þp4k ðsÞ ¼ ak1 k1 p1k ðsÞ ðk P 2Þ; k1 Þp01 ðsÞ ¼ 1;
ð1Þ
k1
ð2Þ ð3Þ ð4Þ ð5Þ
ðs þ p11 ðsÞ ¼ 0;
ð6Þ ð7Þ
ðs þ k2 þ l1 Þp21 ðsÞ ¼ k1 p01 ðsÞ; ðs þ l1 Þp31 ðsÞ ¼ k2 p21 ðsÞ;
ð8Þ ð9Þ
p41 ¼ 0:
ð10Þ
According to Eqs. (4) and (2), respectively, we obtain s þ bk2 l1 p3 k1 ðsÞ ðk P 2Þ; k2 s þ ak1 k1 þ l2 p1k ðsÞ ðk P 2Þ: p3 k1 ðsÞ ¼ bk2 l1
p2
k1 ðsÞ
¼
Therefore, p2
k1 ðsÞ
¼
ðs þ ak1 k1 þ l2 Þðs þ bk2 l1 Þ p1k ðsÞ bk2 l1 k2
ðk P 2Þ:
ð11Þ
Substituting Eq. (11) into Eq. (1) entails that ðs þ ak1 k1 Þp0k ðsÞ ¼
k2 l2 þ ðs þ ak1 k1 þ l2 Þðs þ bk2 l1 Þ p1k ðsÞ ðk P 2Þ: k2
ð12Þ
Hence p0
k1 ðsÞ
¼
k2 l2 þ ðs þ ak2 k1 þ l2 Þðs þ bk3 l1 Þ p1 k2 ðs þ ak2 k1 Þ
k1 ðsÞ
ðk P 3Þ:
ð13Þ
Substitution of Eqs. (5) into Eq. (3) yields ðs þ bk1 l1 þ k2 Þp2k ðsÞ ¼ ak1 k1 p0k ðsÞ þ
ak1 k1 l2 p ðsÞ s þ l2 1k
ðk P 2Þ:
Or, ðs þ bk2 l1 þ k2 Þp2
k1 ðsÞ
¼ ak2 k1 p0
k1 ðsÞ
þ
ak2 k1 l2 p s þ l2 1
k1 ðsÞ
ðk P 3Þ:
ð14Þ
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283
Substitution of Eqs. (11) and (13) into Eq. (14) yields k2
p1k ðsÞ ¼
k1 l1 ðabÞ f ðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ p1 ðs þ l2 Þðs þ ak2 k1 Þðs þ bk2 l1 Þðs þ bk2 l1 þ k2 Þðs þ ak1 k1 þ l2 Þ
k1 ðsÞ
¼ Ak p12 ðsÞ
ðk P 3Þ; ð15Þ
where f ðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ ¼ ðs þ l2 Þ½k2 l2 þ ðs þ bk3 l1 Þðs þ ak2 k1 þ l2 Þ þ k2 l2 ðs þ ak2 k1 Þ; k Y k1 l1 ðabÞm2 f ðs; a; b; k1 ; k2 ; l1 ; l2 ; mÞ Ak ¼ : m2 k Þðs þ bm2 l Þðs þ bm2 l þ k Þðs þ am1 k þ l Þ 1 2 1 1 1 2 m¼3 ðs þ l2 Þðs þ a According to Eq. (15), Eqs. (5) and (12) become, respectively, ak1 k1 Ak p12 ðsÞ ðk P 3Þ; p4k ðsÞ ¼ s þ l2 p0k ðsÞ ¼
ð16Þ
k2 l2 þ ðs þ bk2 l1 Þðs þ ak1 k1 þ l2 Þ Ak p12 ðsÞ ðk P 3Þ: k2 ðs þ ak1 k1 Þ
ð17Þ
Substituting Eqs. (16) and (17) into Eq. (3) yields p2k ðsÞ ¼ gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞAk p12 ðsÞ ðk P 3Þ;
ð18Þ
where gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ ¼
ak1 k1 f ðs; a; b; k1 ; k2 ; l1 ; l2 ; k þ 1Þ : k2 ðs þ l2 Þðs þ ak1 k1 Þðs þ bk1 l1 þ k2 Þ
Finally, using Eqs. (2), (4) and (6)–(10) for k = 2, we have p3k ðsÞ ¼
k2 gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ Ak p12 ðsÞ s þ bk1 l1
p01 ðsÞ ¼
1 ; s þ k1
ðk P 3Þ;
p11 ðsÞ ¼ 0; p41 ðsÞ ¼ 0; p21 ðsÞ ¼
k1 ; ðs þ k1 Þðs þ k2 þ l1 Þ
p31 ðsÞ ¼
k1 k2 ; ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þ
p12 ðsÞ ¼
k1 k2 l1 : ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ
With these explicit expressions obtained from the differential equations and the associated Laplace transforms, we are ready to derive some important reliability indices of the repairable system. 4. Reliability indices 4.1. System availability By the definition, the system availability at time t is given by AðtÞ ¼ P ðthe system is working at time tÞ ¼ P ðN ðtÞ 2 W Þ ¼
1 X k¼1
½p0k ðtÞ þ p1k ðtÞ þ p2k ðtÞ:
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Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
The Laplace-transform of A(t) is A ðsÞ ¼
1 X
½p0k ðsÞ þ p1k ðsÞ þ p2k ðsÞ
k¼1
¼ ½p01 ðsÞ þ p11 ðsÞ þ p21 ðsÞ þ p02 ðsÞ þ p12 ðsÞ þ p22 ðsÞ þ
1 X ½p0k ðsÞ þ p1k ðsÞ þ p2k ðsÞ; k¼3
where, according to (1) for k = 2, (3) for k = 2 and (5) for k = 2 yields p02 ðsÞ ¼
k1 l1 ½k2 l2 þ ðs þ l1 Þðs þ ak1 þ l2 Þ ; ðs þ k1 Þðs þ l1 Þðs þ ak1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ
p22 ðsÞ ¼
ak21 l1 hðs; a; b; k1 ; k2 ; l1 ; l2 Þ ðs þ k1 Þðs þ k2 þ l1 Þðs þ k2 þ bl1 Þðs þ l1 Þðs þ l2 Þðs þ ak1 þ l2 Þ
and where hðs; a; b; k1 ; k2 ; l1 ; l2 Þ ¼ k2 l2 ½ðs þ l2 Þ þ ðs þ ak1 Þðs þ k2 þ bl1 Þ þ ðs þ l1 Þðs þ l2 Þðs þ ak1 þ l2 Þ: Thus, A ðsÞ ¼
1 k1 l1 1þ 1þ s þ k1 s þ k 2 þ l1 ðs þ l1 Þðs þ ak1 þ l2 Þ k2 l2 þ ðs þ l1 Þðs þ ak1 þ l2 Þ ak1 hðs; a; b; k1 ; k2 ; l1 ; l2 Þ k2 þ þ s þ ak1 ðs þ l2 Þðs þ k2 þ bl1 Þ k1 k2 l1 ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ 1 X k2 l2 þ ðs þ bk2 l1 Þðs þ ak1 k1 þ l2 Þ þ gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞ Ak : 1þ k2 ðs þ ak1 k1 Þ k¼3
þ
In reliability theory, the system availability is one of the important indices. However, in engineering application, the steady state availability or the limiting availability of the system is more important. Now, using a Tauberian theorem, when a, b 5 1 the steady state availability or the limiting availability of the system is given by A ¼ lim AðtÞ ¼ lim sA ðsÞ ¼ 0; t!þ1
s!0
which is consistent with a physical interpretation. In fact, since component 1 after repair is not as good as new, it implies that the availability tends to 0 as t ! + 1. 4.2. System reliability In order to determine the reliability R(t) of the system, we consider the two-dimensional continuous-time e ðtÞ; SðtÞ; t P 0g. The difference between the Markov processes {N(t), S(t), homogeneous Markov process f N e t P 0} and f N ðtÞ; SðtÞ; t P 0g is that the set of failure states of {N(t), S(t), t P 0} is now the set of absorbing e ðtÞ; SðtÞ; t P 0g. Now, let states of f N e ðtÞ ¼ j; SðtÞ ¼ kg; qjk ðtÞ ¼ P f N
j 2 X;
k ¼ 1; 2; . . . ;
then RðtÞ ¼ P fthe working time of the system > tg ¼
1 X k¼1
½q0k ðtÞ þ q1k ðtÞ þ q2k ðtÞ:
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
285
By a similar argument as in Section 3, we obtain the following differential equations: d k1 þ a k1 q0k ðtÞ ¼ l2 q1k ðtÞ þ bk2 l1 q2 k1 ðtÞ ðk P 2Þ; dt d k1 þ a k1 þ l2 q1k ðtÞ ¼ 0 ðk P 2Þ; dt d k1 þ k2 þ b l1 q2k ðtÞ ¼ ak1 k1 q0k ðtÞ ðk P 2Þ; dt d þ k1 q01 ðtÞ ¼ 0; dt q11 ðtÞ ¼ 0; d þ k2 þ l1 q21 ðtÞ ¼ k1 q01 ðtÞ: dt The initial conditions are q01 ð0Þ ¼ 1;
q0k ð0Þ ¼ 0; k ¼ 2; 3; . . . ;
qjk ð0Þ ¼ 0;
j ¼ 1; 2; k ¼ 1; 2; . . .
Taking Laplace-transforms on both sides of the above differential equations and using the initial conditions, we obtain the following for k P 2 q0k ðsÞ ¼
bk2 l1 q s þ ak1 k1 2
k1 ðsÞ; k2
¼
k1 l1 ðabÞ q0 ðs þ ak1 k1 Þðs þ bk2 l1 þ k2 Þ
k1 ðsÞ;
¼ Bk q01 ðsÞ; ak1 k1 q0k ðsÞ s þ bk1 l1 þ k2 ak1 k1 Bk q01 ðsÞ; ¼ s þ bk1 l1 þ k2
q2k ðsÞ ¼
where Bk ¼
k Y m¼2
ðk1 l1 ÞðabÞm2 ; ðs þ am1 k1 Þðs þ bm2 l1 þ k2 Þ
q01 ðsÞ ¼
1 : s þ k1
Clearly q1k ðsÞ ¼ 0
ðk P 1Þ; k1 : q21 ðsÞ ¼ ðs þ k1 Þðs þ k2 þ l1 Þ
Therefore, the Laplace transform of R(t) is given by R ðsÞ ¼
1 1 X X ½q0k ðsÞ þ q2k ðsÞ ¼ ½q01 ðsÞ þ q21 ðsÞ þ ½q0k ðsÞ þ q2k ðsÞ k¼1
1 ¼ s þ k1
"
k1 1þ s þ k 2 þ l1
k¼2
1 X 1þ þ k¼2
# ak1 k1 Bk : s þ bk1 l1 þ k2
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4.3. System MTTFF The system MTTFF is one of the most important reliability indices for a repairable system. In particular, it will produce the calamity result as soon as the system fails. Based on the system reliability and a property of the Laplace transform, the system MTTFF is given by X Y Z þ1 1 k 1 1 1 ak1 bm2 l1 MTTFF ¼ RðtÞdt ¼ lim R ðsÞ ¼ þ þ k1 : þ s!0 k 1 k 2 þ l1 k1 b l1 þ k2 m¼2 aðbm2 l1 þ k2 Þ 0 k¼2 Let Ck ¼
Y k 1 ak1 bm2 l1 þ k1 ; k1 b l1 þ k2 m¼2 aðbm2 l1 þ k2 Þ
P1 then k¼2 C k is a positive series under k1, k2, l1, a and b are the positive P1 constants. And when a P 1, 0 < b 6 1, according to the ratio test of the positive series, it is easy to see k¼2 C k < 1. Thus, the result of the system MTTFF is finite. Obviously, if any component in the system is always no failure, the system MTTFF is infinite. 4.4. System ROCOF Let Nf(t) be the number of failures of a system that have occurred by time t, then the mean number of failures of the system in (0, t] is given by M f ðtÞ ¼ E½N f ðtÞ: Its derivative mf(t) is called the rate of occurrence of failure (ROCOF) or the failure frequency of the system at time t. Hence Z t mf ðxÞdx: M f ðtÞ ¼ 0
Obviously, if mf(t) is increasing, the mean number of failures that have occurred in (s, s + t] (where s is positive constant) is larger than the mean number in (0, t], and hence the system is deteriorating. On the other hand, if mf(t) is decreasing, the system is improving. Therefore, the ROCOF is one of the important indices in reliability theory, in particular in the study of a deteriorating or improving system. According to Lam [22], we have mf ðtÞ ¼
1 X X
pjk ðtÞajlk ;
j2W ;l2F k¼1
where ajlk is transition rate from state j to l in kth cycle. In view of the model assumptions, we have a03k = a04k = a13k = a24k = 0, a14k = ak1k1, a23k = k2. Therefore, mf ðtÞ ¼
1 X
½p0k ðtÞða03k þ a04k Þ þ p1k ðtÞða13k þ a14k Þ þ p2k ðtÞða23k þ a24k Þ ¼
k¼1
1 X
½ak1 k1 p1k ðtÞ þ k2 p2k ðtÞ:
k¼1
The Laplace transform of mf(t) is given by mf ðsÞ ¼
1 1 X X ½ak1 k1 p1k ðsÞ þ k2 p2k ðsÞ ¼ ½k1 p11 ðsÞ þ k2 p21 ðsÞ þ ½ak1 k1 p1k ðsÞ þ k2 p2k ðsÞ k¼1
k¼2
1 X k1 k2 k1 k2 l1 ½ak1 k1 þ gðs; a; b; k1 ; k2 ; l1 ; l2 ; kÞk2 þ Ak : ¼ ðs þ k1 Þðs þ k2 þ l1 Þ k¼2 ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þ
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287
4.5. The idle time probability of the repairman Clearly, the repairman will be idle if and only if one component is functioning while the other component is in cold standby. Thus, the idle time probability of the repairman at time t is given by 1 X p0k ðtÞ: IðtÞ ¼ P fN ðtÞ ¼ 0g ¼ k¼1
The Laplace-transform of I(t) is I ðsÞ ¼
1 X
p0k ðsÞ ¼ ½p01 ðsÞ þ p02 ðsÞ þ
k¼1
1 X
p0k ðsÞ
k¼3
1 k1 l1 k2 l2 1þ 1þ ¼ s þ k1 ðs þ ak1 Þðs þ k2 þ l1 Þ ðs þ l1 Þðs þ ak1 þ l2 Þ 1 k2 X k1 l1 ½k2 l2 þ ðs þ b l1 Þðs þ ak1 k1 þ l2 Þ Ak : þ ðs þ k1 Þðs þ l1 Þðs þ k2 þ l1 Þðs þ ak1 þ l2 Þðs þ ak1 k1 Þ k¼3 Using a Tauberian theorem, we can obtain the idle probability of the repairman in the steady- state as follows: I ¼ lim IðtÞ ¼ lim sI ðsÞ ¼ 0; t!þ1
s!0
i.e. the steady-state idle time probability of the repairman tends to 0. In fact, this conclusion is also in agreement with our intuition. Because component 1 after repair is not as good as new and its consecutive repair times after failure are stochastically increasing, it will be unrepairable finally. This implies that the idle time probability of the repairman will tends to zero as t ! + 1. In particular, if a = b = 1, ki = k, li = l, i = 1, 2 the system reduces to a two-identical-component cold standby repairable system with one repairman and without priority rule. This is the classical case which has been studied by e.g., Barlow and Proschan [1] and Ascher and Feingold [2]. 4.6. A numerical example for the system reliability R(t) Based on these results above, we can find that it is very difficult to obtain the transient results of the reliability indices for the system proposed in this paper. We can only obtain the steady-state results of some reliability indices of the system. Moreover, the method by using the inverse Laplace transform to get the transient results of the reliability indices is also very tedious. Thus, in engineering applications, a numerical method based on the Runge–Kutta method is often adopted. Here, an approximative solution of R(t) is given to illustrate the numerical method. Let a = 1.1, b = 0.9, k1 = 0.01, k2 = 0.1, l1 = 0.1, l2 = 0.5. According to the differential equations and the initial conditions about qjk(t) in Section 4.2, when k = 1, 2, 3; t = 10, 20, 30, 40, 50, we can obtain these approximative solutions of qjk(t), j = 0, 2 (see Table 1).
Table 1 Some results of qjk(t) q01(t) q21(t) q02(t) q22(t) q03(t) q23(t)
t = 10
t = 20
t = 30
t = 40
t = 50
0.8187 0.0281 0.1159 0.0037 0.0061 1.9176 · 105
0.6703 0.0231 0.2074 0.0077 0.0287 2.0088 · 104
0.5488 0.0189 0.2601 0.0101 0.0588 6.5363 · 104
0.4493 0.0155 0.2854 0.0113 0.0893 0.0014
0.3679 0.0127 0.2919 0.0117 0.1162 0.0023
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Thus, we can further obtain these approximative solutions of R(t), i.e. RðtÞ ¼
1 3 X X ½q0k ðtÞ þ q2k ðtÞ ½q0k ðtÞ þ q2k ðtÞ ¼ q01 ðtÞ þ q21 ðtÞ þ q02 ðtÞ þ q22 ðtÞ þ q03 ðtÞ þ q23 ðtÞ: k¼1
k¼1
When t = 10, 20, 30, 40, 50, the results of R(t) are, respectively, Rð10Þ 0:9725;
Rð20Þ 0:9374;
Rð30Þ 0:8974;
Rð40Þ 0:8522;
Rð50Þ 0:8027:
Clearly, we can see that the value of the system reliability R(t) is decreasing when t increases. For a deteriorating system, the observation is consistent with our intuition. 5. Replacement model under policy T 5.1. Long-run average cost under policy T In this section, we consider a repair-replacement policy based on the working age T of component 1 which is other than the policy N in Zhang [25]. The system will be replaced by a new one when the working age of component 1 reaches T. Our objective is to determine an optimal replacement policy T* such that the long-run average cost per unit time of the system is minimized. To do this, based on Section 2 we add the following assumptions. Assumption 5. Assume that a component in the system is replaced sometime by a new and identical one, and the replacement time is negligible. Assumption 6. Assume that a component in the system cannot produce the working reward during cold standby, and no cost is incurred during waiting for repair. Assumption 7. Assume that the repair cost rate of component i is cðiÞ r ; i ¼ 1; 2 while the working reward rate of two components is same cw. And the replacement cost of the system is C. Let s1 be the first replacement time of the system, and sn(n P 2) be the time between the (n 1)th replacement and the nth replacement of the system under policy T. Obviously, {s1, s2,. . .} forms a renewal process, and the interarrival time between two consecutive replacements is called a renewal cycle. Let C(T) be the long-run average cost per unit time of the system under policy T. Thus, according to renewal reward theorem (see, for example Ross [26]), we have CðT Þ ¼
the expected cost incurred in a renewal cycle : the expected length of a renewal cycle
ð19Þ
Because component 1 has priority in use, component 1 has only the working state, the repair state and the waiting for repair state. Thus, when the working age of component 1 reaches T, then component 2 is either under a cold standby state or under a repair state (see Fig. 1). Let W be the length of a renewal cycle of the system under the policy T, then W ¼T þ
K X j¼1
ð1Þ
Yj þ
K X ð2Þ ð1Þ Y j1 X j vn j¼2
ð2Þ
ð1Þ
o vn
Y j1 X j >0
ð1Þ
ð2Þ
o;
Y j1 X j1 >0
where, the second term and the third term are respectively the length of repair time and the length of time of waiting for repair of component 1 before the working age of component 1 reaches T; v is a indicator function such that 1; If event A occurs vA ¼ 0; If event A does not occur:
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
289
Fig. 1. A possible course of the system.
K is the failure number of component 1 before the working age of component 1 reaches T. Obviously, K is a random variable. Hence
EW ¼ T þ E
K X
2
!
þ E4
ð1Þ Yj
j¼1
K X
ð2Þ Y j1
ð1Þ Xj
vn
j¼2
3 ð2Þ
o vn
ð1Þ
Y j1 X j >0
ð1Þ
ð2Þ
o 5:
ð20Þ
Y j1 X j1 >0
To evaluate the long-run average cost per unit time under the policy T, first of all, we determine the probability mass function of K. Since ( P fK P kg ¼ P
k X
) ð1Þ Xj
ð1Þ
¼ H k ðT Þ;
j¼1 ð1Þ
where H k is the distribution of
Pk
ð1Þ j¼1 X j ,
therefore ð1Þ
ð1Þ
P fK ¼ kg ¼ P fK P kg P fK P k þ 1g ¼ H k ðT Þ H kþ1 ðT Þ:
ð21Þ ð2Þ
ð1Þ
According to Assumption 3 and the definition of the convolution, the distribution functions of Y j1 X j ð2Þ ð1Þ and X j Y j are, respectively,
Uj ðuÞ ¼ Gð2Þ ðuÞ 1 F ð1Þ ðaj1 uÞ ;
Wj ðvÞ ¼ F ð2Þ ðvÞ 1 Gð1Þ ðbj1 vÞ : Therefore 2
ð2Þ ð1Þ E4 Y j1 X j vn
3 ð2Þ
ð1Þ
o vn
Y j1 X j >0
ð1Þ
ð2Þ
o 5 ¼ Wj1 ð0Þ
Y j1 X j1 >0
Z
1
u dUj ðuÞ:
ð22Þ
0
By using the property of the conditional expectation and Eq. (21), we have the following results:
E
K X
"
! ð1Þ Xj
¼E E
j¼1
K X
!# ð1Þ X j jK
j¼1
¼
1 k X X k¼1
j¼1
1 aj1 k1
!
¼
1 X k¼1
E
k X
! ð1Þ Xj
P ðK ¼ kÞ
j¼1
1 h i 1 X 1 ð1Þ ð1Þ ð1Þ H ðT Þ < 1: H k ðT Þ H kþ1 ðT Þ ¼ k1 k¼1 ak1 k
ð23Þ
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Similarly K X
E
! ð2Þ Xj
¼
j¼1
1 1 X ð1Þ H ðT Þ; k2 k¼1 k
ð24Þ
!
1 1 X 1 ð1Þ H k ðT Þ; k1 l b 1 j¼1 k¼1 " # Z K 1 X X ð2Þ ð1Þ ð1Þ E X j Y j vfX ð2Þ Y ð1Þ >0g ¼ F j ðT Þ K X
E
ð1Þ Yj
j
j¼1
2 K X ð2Þ E4 Y j vn j¼1
E
T
ð26Þ
1 X ð1Þ o5 ¼ 1 Wk1 ð0ÞH k ðT Þ; l2 k¼1
Wk1 ð0Þ
Z
ð27Þ
1
3 ð2Þ
o vn
ð1Þ
Y j1 X j >0
ð1Þ
ð2Þ
Y j1 X j1 >0
o5
ð1Þ u dUk ðuÞ H k ðT Þ;
ð28Þ
0
k¼2
"
v dWj ðvÞ;
0
j¼1
ð1Þ ð2Þ Y j1 X j1 >0
j¼2 1 X
j
1
3
2 K X ð2Þ ð1Þ E4 Y j1 X j vn
¼
ð25Þ
¼
K X
#
! ð1Þ
vfY ð1Þ X ð2Þ >0g ¼
Xj
K
j¼1
K
1 X
ð1Þ
ð1Þ
Wk ð0Þ H k ðT Þ H kþ1 ðT Þ
T
k¼1
k X j¼1
! 1 : k1 aj1
ð29Þ
Substituting (25) and (28) into (20), we have Z 1 1 1 X 1 X 1 ð1Þ ð1Þ H ðT Þ þ Wk1 ð0Þ u dUk ðuÞ H k ðT Þ: EW ¼ T þ l1 k¼1 bk1 k 0 k¼2
ð30Þ
The total work time of the system in a renewal cycle is U ¼T þ
K X
K K n o X X ð2Þ ð1Þ ð2Þ ð2Þ ð1Þ min X j ; Y j Xj X j Y j v ¼Tþ
j¼1
j¼1
ð2Þ
ð1Þ
:
X j Y j >0
j¼1
According to the expressions (24) and (26), we have EU ¼ T þ
Z 1 1 1 X 1 X ð1Þ ð1Þ H k ðT Þ H k ðT Þ v dWk ðvÞ: k2 k¼1 0 k¼1
ð31Þ
The total repair time of the system in a renewal cycle is V ¼ V 1 þ V 2; where V1 and V2 denotes, respectively, the repair times of the components 1 and 2 in a renewal cycle, i.e. V1 ¼
K X
ð1Þ
Yj ;
j¼1
V2 ¼
K 1 X j¼1
ð2Þ Y j vfY ð1Þ X ð2Þ >0g j
j
þ
T
K X j¼1
! ð1Þ Xj
v
ð1Þ
ð2Þ
:
Y K X K >0
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291
According to the expressions (25), (27) and (29), we have EV ¼ EV 1 þ EV 2 1 1 1 X 1 1 X ð1Þ ð1Þ ¼ H k ðT Þ þ Wk ð0ÞH k ðT Þ k1 l1 k¼1 b l2 k¼1 þ
1 X
Wk ð0Þ
ð1Þ H k ðT Þ
ð1Þ H kþ1 ðT Þ
T
k¼1
k X j¼1
! 1 : k1 aj1
ð32Þ
Therefore, substituting the expressions (30)–(32) into the expression (19), then the long-run average cost per unit time of the system under policy T is given by ð2Þ cð1Þ r EV 1 þ cr EV 2 þ C cw EU EW h P i ð1Þ 1 crð1Þ l11 1 H ðT Þ k1 k k¼1 b ¼
R 1 ð1Þ P1 P1 1 ð1Þ 1 T þ l1 k¼1 bk1 H k ðT Þ þ k¼2 Wk1 ð0Þ 0 u dUk ðuÞ H k ðT Þ h P i P1 Pk ð1Þ ð1Þ ð1Þ 1 1 1 cð2Þ W ð0ÞH ðT Þ þ W ð0Þ H ðT Þ H ðT Þ T k k j1 k k kþ1 r k¼1 k¼1 j¼1 k1 a l2 þ
R 1 ð1Þ P1 1 ð1Þ P1 1 T þ l1 k¼1 bk1 H k ðT Þ þ k¼2 Wk1 ð0Þ 0 u dUk ðuÞ H k ðT Þ h i R1 P1 ð1Þ P ð1Þ cw T þ k12 1 H ðT Þ H ðT Þ u dW ðvÞ C k k¼1 k k¼1 k 0 :
R P P 1 ð1Þ ð1Þ 1 1 1 T þ l11 k¼1 bk1 H k ðT Þ þ k¼2 Wk1 ð0Þ 0 u dUk ðuÞ H k ðT Þ
CðT Þ ¼
ð33Þ
Then the following result will show a sufficient condition that the optimal replacement policy T* can be found by minimizing C(T). To do this, we need a lemma as follows. P ð1Þ a 1 Lemma 1. If t P a1 F 1 ðbÞ; a > 1; 0 < b < 1, then 1 where F1 is the inverse function of the k¼1 bk H k ðtÞ ¼ P1, ð1Þ ð1Þ ð1Þ k distribution function of X 1 , and H k is the distribution function of j¼1 X j . (Its proof can be found from Lam [8].) a TheoremP1. Let d0 ¼ a1 F 1 ðbÞ, where a > 1, 0 < b < l, F1 is the inverse function of the distribution function of ð1Þ ð1Þ 1 X 1 . If k¼1 H k ðT Þ < 1 when T P d0, then the optimal replacement policy T* must be found by minimizing C(T) in (0, d0). ð2Þ
ð1Þ
ð2Þ
ð1Þ
Proof. According to the distribution functions of Y j1 X j and X j Y j , we have Z Z Z þ1 ð2Þ ð1Þ
ðy xÞdGðyÞdF ðak1 xÞ u dUk ðuÞ ¼ E Y k1 X k v ð2Þ ð1Þ ¼ 0
¼
Z
þ1
Z
0
Y k1 X k >0
D1
þ1
ðy xÞdGðyÞ dF ðak1 xÞ 6
Z
þ1
Z
0
x
þ1 0
1 ð2Þ y dGðyÞ dF ðak1 xÞ ¼ EY k1 ¼ ; l2
where D1 = {(x,y)|x > 0,y > 0 and y x > 0}. Based on Lemma 1 and the condition in Theorem 1, when T P d0 we can get Z 1 1 1 1 X X 1 ð1Þ 1 X ð1Þ ð1Þ Wk1 ð0Þ u dUk ðuÞ H k ðT Þ ¼ Wk1 ð0Þ H k ðT Þ 6 H k ðT Þ < 1: l l 0 2 2 k¼2 k¼2 k¼2 Similarly, Z 1
v dWk ðvÞ ¼ E
0
¼
ð2Þ ð1Þ X k Y k v
Z þ1 Z 0
þ1
ð2Þ
ð1Þ
X k Y k >0
Z Z
¼ ðx yÞdF ðxÞdGðbk1 yÞ
ðx yÞdF ðxÞ dGðb y
ð34Þ
D2 k1
Z
þ1
Z
yÞ 6 0
0
þ1
ð2Þ
x dF ðxÞ dGðbk1 yÞ ¼ EX k ¼
1 ; k2
292
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
where D2 = {(x,y)|x > 0,y > 0 and x y > 0}. Thus 1 Z þ1 1 X 1 X ð1Þ ð1Þ v dWk ðvÞ H k ðT Þ 6 H k ðT Þ < 1 k 2 0 k¼1 k¼1 and 1 X
k X ð1Þ ð1Þ Wk ð0Þ H k ðT Þ H kþ1 ðT Þ T
k¼1
j¼1
1 k1 aj1
! 6
ð35Þ
1 X k¼1
ð1Þ H k ðT Þ
ð1Þ H kþ1 ðT Þ
T
k X j¼1
1 k1 aj1
!
1 1 X X ð1Þ ð1Þ ð1Þ 6 T H k ðT Þ H kþ1 ðT Þ 6 T H k ðT Þ k¼1
k¼1
< 1:
ð36Þ
According to Lemma 1, the condition in Theorem 1 and Eqs. (34)–(36), when T P d0, we have CðT Þ crð1Þ : Because the total lifetime of the cold standby repairable system is limited, the minimum of the long-run average cost per unit time under the policy T exists. Thus, under the condition of theorem, the optimal replacement policy T * can be found by minimizing C(T) in (0, d0), because for T P d0, we have CðT Þ crð1Þ . Clearly, T * can be found analytically or numerically. h 5.2. A numerical example In this section, we provide an example to illustrate the theoretical results. ð1Þ ð2Þ ð1Þ ð2Þ Now, we denote the probability density functions of X j ; X j ; Y j and Y j , respectively, by fj(t), f(t), gj(t) ð2Þ ð1Þ and g(t) for j = 1, 2,. . . According to Assumption 3, then the probability density functions of Y j1 X j and ð2Þ ð1Þ X j Y j are, respectively, 8 j1 < aj1 k1 l2 el2 u ; u P 0; a k1 þl2 /j ðuÞ ¼ gðuÞ fj ðuÞ ¼ j1 k l j1 a : 1 2 ea k1 u ; u < 0; aj1 k1 þl 8 j1 2 2 l1 < b kj1 ek2 v ; v P 0; k2 þb l1 wj ðvÞ ¼ f ðvÞ gj ðvÞ ¼ : bj1 k2 l1 ebj1 l1 v ; v < 0: k þbj1 l 2
Therefore Z 1
u dUk ðuÞ ¼
0
Z
1
Z Z
1
1
u/k ðuÞdu ¼
0
ak1 k1 ; l2 ðak1 k1 þ l2 Þ
ð37Þ
1
bk1 l1 ; ð38Þ k2 ðk2 þ bk1 l1 Þ 0 0 k2 ð2Þ ð1Þ Wk ð0Þ ¼ P X j Y j < 0 ¼ : ð39Þ k2 þ bk1 l1 Pk ð1Þ ð1Þ To determine the distribution H k of j¼1 X j , the following lemma is essential, see e.g., Chiang [27] for the proof. v dWk ðvÞ ¼
vwk ðvÞdv ¼
Lemma 2. Assume that {Xi, i = 1, 2,. . .} is a sequence of independent random variables,P and Xi has the exponential distribution Exp(ki) with ki 5 kj for i 5 j. Then n the probability density function of ni¼1 X i is given by hn ðxÞ ¼ ð1Þ
n1
k1 . . . kn
n X
eki x kj Þ
Pnj¼1 ðki i¼1 j6¼i
for
x P 0;
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
293
and 0 otherwise. In particular, according to Assumption 3, 4 and ki = ai1k, the density function of this paper is given by n1
hnð1Þ ðxÞ ¼ ð1Þ
a
nðn1Þ 2
k
n X
ea
ð1Þ
Z
T
ð1Þ
hk ðxÞdx ¼ ð1Þ
ð1Þ j¼1 X j
in
i1 k
Pnj¼1 ðai1 i¼1 j6¼i
x aj1 Þ
and 0 otherwise. Thus, the distribution function of H k ðT Þ ¼
Pn
k1 kðk1Þ=2
a
0
for Pk
k X i¼1
t P 0;
ð1Þ j¼1 X j
is given by i1
1 ea kT : i1 a Pkj¼1 ðai1 aj1 Þ
ð40Þ
j6¼i
Thus, substituting the expressions (37)–(40) into the expression (33), then ð2Þ cð1Þ r EV 1 þ cr EV 2 þ C cw EU EW h P i ð1Þ 1 1 crð1Þ l11 k¼1 bk1 H k ðT Þ ¼ P1 P ð1Þ ð1Þ ak1 k1 k2 1 T þ l11 1 k¼1 bk1 H k ðT Þ þ k¼2 ðk2 þbk2 l1 Þðak1 k1 þl2 Þl2 H k ðT Þ h P i P1 Pk ð1Þ ð1Þ ð1Þ 1 k2 k2 1 cð2Þ T j¼1 k1 a1j1 k¼1 k2 þbk1 l1 H k ðT Þ þ k¼1 k2 þbk1 l1 H k ðT Þ H kþ1 ðT Þ r l2 þ P P1 ð1Þ ð1Þ ak1 k1 k2 1 T þ l11 1 k¼1 bk1 H k ðT Þ þ k¼2 ðk2 þbk2 l1 Þðak1 k1 þl2 Þl2 H k ðT Þ h i P1 ð1Þ cw T þ k¼1 k þb1k1 l H k ðT Þ C 2 1 ; P1 1 ð1Þ P1 k1 k ð1Þ 1 1 2 T þ l1 k¼1 bk1 H k ðT Þ þ k¼2 ðk þbk2 al Þðakk1 H ðT Þ k k þl Þl
CðT Þ ¼
2
ð1Þ
1
1
2
ð41Þ
2
ð1Þ
where H k ðT Þ and H kþ1 ðT Þ are both determined by the expression (40). There are a = 1.1, b = 0.9, k1 = 0.01, k2 = 0.1, l1 = 0.1, l2 = 0.5 in Section 4.6. Now, we further assume that crð1Þ ¼ 10; cð2Þ r ¼ 8; cw ¼ 9 and C = 2500. Substituting the above values into the expression (41) with the expression (40) and overpassing numerical calculation, we can obtain some results presented in Table 2 and Fig. 2. It is easy to find that C(500) = 1.7933 is the minimum of the long-run average cost per unit time of the system. In other words, the optimal policy is T * = 500, and we should replace the system when the working age of component 1 reaches 500. Table 2 or Fig. 2 indicates that the optimal policy T * is unique. However, we should point out that T is a continuous variate. To produce the plot of C(T) versus T as quickly as possible, we have to select 50 units as the step size of T. Thus, the above results in Table 2 and Fig. 2 are obtained under the condition of 50 units as the step size of T.
Table 2 Some results obtained using expressions (40) and (41) T
C(T)
T
C(T)
T
C(T)
5 · 50 6 · 50 7 · 50 8 · 50 9 · 50 10 · 50 11 · 50
1.2117 0.0624 0.9180 1.4828 1.7924 1.7933 1.3490
12 · 50 13 · 50 14 · 50 15 · 50 16 · 50 17 · 50 18 · 50
0.3394 1.1587 2.8354 4.3618 5.5755 6.4696 7.1046
19 · 50 20 · 50 21 · 50 22 · 50 23 · 50 24 · 50 25 · 50
7.5491 7.8586 8.0732 8.2204 8.3192 8.3825 8.4195
294
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295 5
4
3
C(T)
2
1
0
1
2
250
300
350
400
450
500 T
550
600
650
700
750
Fig. 2. The plot of C(T) against T.
6. Concluding remarks In this paper, we introduce the model for a two-component cold standby repairable system with one repairman and use priority. Assume that component 1 after repair is not ‘‘as good as new’’ such that the successive working times of component 1 form a decreasing geometric process and the consecutive repair times of component 1 form a increasing geometric process. Obviously, the model in this paper is more realistic than the model in which two components after repair are both ‘‘as good as new’’. In this model, we study not only some important reliability indices of the system but also a replacement policy of the system when the working time distributions and the repair time distributions of the two components are both exponential. Thus, this paper has definite theoretical interest and potential practical application. However, based on our analyses and the examples above, we have the following remarks: (1) Let N(t) be the state of the system at time t. It is clear from model assumptions that {N(t), t P 0} is not a Markov chain. However, it can be extended as a two-dimensional Markov process by introducing a supplementary variable. To obtain the system reliability indices, we need to determine the state marginal probabilities of the system at time t. Accordingly, we can derive the system of differential equations about Pjk(t), j 2 X, k = 1, 2,. . . Finally, the Laplace transform results of reliability indices of the system and the steady-state results of some reliability indices of the system are obtained. (2) We can see that the method by using the inverse Laplace transform to get the transient results of the reliability indices is very tedious, and is also very difficult. But, the Laplace transform results of reliability indices of the system is discommodious for practical application. Thus, in engineering, a numerical method based on the Runge–Kutta method is often adopted. In Section 4.6, an approximative solution of P1 R(t) is given to illustrate the numerical method. Note that the expression of RðtÞ ¼ k¼1 ½q0k ðtÞ þ q2k ðtÞ is an infinite series. However, we can select the values of k and t according to the practical request. (3) In this paper, we consider a replacement policy based on the working age T of component 1. An optimal replacement policy T * for minimizing C(T) is determined. And under some conditions, we can prove that the optimal replacement policy T * must be found by minimizing C(T) in (0, d0). In fact, when a = 1.1, b = 0.9, we can compute
Y.L. Zhang, G.J. Wang / European Journal of Operational Research 183 (2007) 278–295
d0 ¼
295
a F 1 ðbÞ 2532:8436: a1
Obviously, the optimal policy T * = 500 2 (0, d0) = (0, 2532.8463). It is consistent with the theoretical analysis in replacement model. (4) From Table 2 and Fig. 2 in Section 5.2, we should point out that the optimal solution is obtained under certain conditions. Note that the working age T of component 1 is a continuous variable. To find optimal solution of the objective function C(T) as quickly as possible, we have to select the step sizes of T. Thus, the optimal solution in this paper can be obtained under certain conditions, i.e. the step sizes of T is 50 units (see Table 2 or Fig. 2). Clearly, different step sizes of T will obtain different optimal solution. Because the total life of the system is finite, we can find a corresponding optimal solution.
Acknowledgement The authors are grateful for the Editor’s and the referee’s valuable comments and suggestions, which have considerably improved the presentation of the paper. References [1] R.E. Barlow, F. Proschan, Mathematical Theory of Reliability, Wiley, New York, 1965. [2] H. Ascher, H. Feingold, Repairable Systems Reliability, Marcel Dekker, New York, 1984. [3] T. Nakagawa, S. Osaki, Stochastic behaviour of a two-unit priority standby redundant system with repair, Microelectronics Reliability 14 (3) (1975) 309–313. [4] S. Osaki, Reliability analysis of a two-unit standby redundant system with priority, Canadian Journal of Operations Research Society 8 (1970) 60–62. [5] J.A. Buzacott, Availability of priority standby redundant systems, IEEE Transactions of Reliability R-20 (1971) 60–63. [6] R.E. Barlow, L.C. Hunter, Optimum preventive maintenance policy, Operations Research 8 (1960) 90–100. [7] M. Brown, F. Proschan, Imperfect repair, Journal of Applied Probability 20 (1983) 851–859. [8] Y. Lam, Geometric processes and replacement problem, Acta Mathematica Applied Sinica 4 (4) (1988) 366–377. [9] Y. Lam, A note on the optimal replacement problem, Advances in Applied Probability 20 (1988) 479–482. [10] M.S. Finkelstein, A scale model of general repair, Microelectronics Reliability 33 (1993) 41–44. [11] Y.L. Zhang, A bivariate optimal replacement policy for a repairable system, Journal of Applied Probability 31 (1994) 1123–1127. [12] W. Stadje, D. Zuckerman, Optimal strategies for some repair replacement models, Advance in Applied Probability 22 (1990) 641–656. [13] A.D.J. Stanley, On geometric processes and repair replacement problems, Microelectronics Reliability 33 (1993) 489–491. [14] Y.L. Zhang, R.C.M. Yam, M.J. Zuo, Optimal replacement policy for a deteriorating production system with preventive maintenance, International Journal of System Science 32 (10) (2001) 1193–1198. [15] Y.L. Zhang, R.C.M. Yam, M.J. Zuo, Optimal replacement policy for a multistate repairable system, Journal of Operations Research Society 53 (2002) 336–341. [16] Y.L. Zhang, A geometric-process repair-model with good-as-new preventive repair, IEEE Transactions Reliability R-51 (2) (2002) 223–228. [17] Y. Lam, Y.L. Zhang, A geometric-process maintenance model for a deteriorating system under a random environment, IEEE Transactions Reliability R-52 (1) (2003) 83–89. [18] Y.L. Zhang, An optimal replacement policy for a three-state repairable system with a monotone process model, IEEE Transactions Reliability R-53 (2004) 452–457. [19] G.J. Wang, Y.L. Zhang, A shock model with two-type failures and optimal replacement policy, International Journal of System Science 36 (4) (2005) 209–214. [20] Y.L. Zhang, S.M. Wu, The reliability analysis of a two-unit series repairable system without being repaired ‘‘as good as new’’. In: Proceedings of the Fourth Symposium on Reliability Mathematics, Guilin, China, 1992, pp. 464–469. [21] Y. Lam, Y.L. Zhang, Analysis of a two-component series system with a geometric process model, Naval Research Logistics 43 (2003) 491–502. [22] Y. Lam, Calculating the rate of occurrence of failures for continuous-time Markov chains with application to a two-component parallel system, Journal of Operations Research Society 45 (1995) 528–536. [23] Y. Lam, Y.L. Zhang, Analysis of a parallel system with two different units, Acta Mathematica Applied Sinica 12 (1996) 408–417. [24] Y. Lam, A maintenance model for two-unit redundant system, Microelectronics Reliability 37 (3) (1997) 497–504. [25] Y.L. Zhang, An optimal geometric process model for a cold standby repairable system, Reliability Engineering and Systems Safety 63 (1999) 107–110. [26] S.M. Ross, Stochastic Processes, second ed., Wiley, New York, 1996. [27] C.L. Chiang, An Introduction to Stochastic Processes and Their Applications, Robert E. Krieger Publishing Co., New York, 1980.