Optimizing replacement policy for a cold-standby system with waiting repair times

Optimizing replacement policy for a cold-standby system with waiting repair times

Applied Mathematics and Computation 214 (2009) 133–141 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepag...

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Applied Mathematics and Computation 214 (2009) 133–141

Contents lists available at ScienceDirect

Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Optimizing replacement policy for a cold-standby system with waiting repair times Jishen Jia a, Shaomin Wu b,* a b

Department of Scientific Research, Hennan Mechanical and Electrical Engineering College, Xinxiang 453002, PR China School of Applied Sciences, Cranfield University, Bedfordshire MK43 0AL, UK

a r t i c l e

i n f o

Keywords: Geometric process Cold-standby system Long-run cost per unit time Replacement policy Maintenance policy

a b s t r a c t This paper presents the formulas of the expected long-run cost per unit time for a coldstandby system composed of two identical components with perfect switching. When a component fails, a repairman will be called in to bring the component back to a certain working state. The time to repair is composed of two different time periods: waiting time and real repair time. The waiting time starts from the failure of a component to the start of repair, and the real repair time is the time between the start to repair and the completion of the repair. We also assume that the time to repair can either include only real repair time with a probability p, or include both waiting and real repair times with a probability 1  p. Special cases are discussed when both working times and real repair times are assumed to be geometric processes, and the waiting time is assumed to be a renewal process. The expected long-run cost per unit time is derived and a numerical example is given to demonstrate the usefulness of the derived expression. Ó 2009 Elsevier Inc. All rights reserved.

0. Introduction A two-component cold-standby system is composed of a primary component and a backup component, where the backup component is only called upon when the primary component fails. Cold-standby systems are commonly used for non-critical applications. However, cold-standby systems are one of most important structures in the reliability engineering and have been widely applied in reality. An example of such a system is the data backup system in computer networks. Reliability analysis and maintenance policy optimisation for cold-standby systems has attracted attentions from many researchers. Zhang and Wang [11,12] and Zhang et al. [10] derived the expected long-run cost per unit time for a repairable system consisting of two identical components and one repairman when a geometric process depicting working times is assumed. Utkin [5] proposed imprecise reliability models of cold-standby systems when he assumed that arbitrary probability distributions of the component time to failure are possible and they are restricted only by available information in the form of lower and upper probabilities of some events. Coit [1] described a solution methodology to optimal design configurations for non-repairable series–parallel systems with cold-standby redundancy when he assumed non-constant component hazard functions and imperfect switching. Yu et al. [9] considers a framework to optimally design a maintainable previous term cold-stand by next term system, and determine the maintenance policy and the reliability character of the components. Due to various reasons, repair might start immediately after a component fails. In some scenarios, from the failure of a component to the completion of repair, there might be two periods: waiting time and real repair time. The waiting time starts from the failure of the component to the start of repair; and the real repair time is the time between the start to repair

* Corresponding author. E-mail address: shaomin.wu@cranfield.ac.uk (S. Wu). 0096-3003/$ - see front matter Ó 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2009.03.064

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and the completion of the repair. This is especially true for cold-standby systems as they are not critical enough for a standby repairman be equipped for it. For example, when a component fails to work, its owner will call its contracted maintenance company or return the component to its suppler for repair. The waiting time can be the time spent by repairmen from their working place to the place where the component fails, or the time on delivering the failed component to its supplier. Usually, the waiting time can be seen as a random variable independent of the age of the component, whereas the real repair time can become longer and longer when the component becomes older. On the other hand, the working time of the component can become shorter and shorter due to various reasons such as ageing and deterioration. Such working and real repair time patterns can be depicted by geometric processes as many authors have studied [2–4,7,8,10–13]. The geometric processes introduced by Lam [4] define an alternative to the non-homogeneous Poisson processes: a sequence of random variables {Xk, k = 1, 2, . . .} is a geometric process if the distribution function of Xk is given by F(ak1t) for k = 1, 2, . . . and a is a positive constant. Wang and Pham [6] later refer the geometric process as a quasi-renewal process. Finkelstein [2] develops a very similar model: he defines a general deteriorating renewal process such that Fk+1(t) 6 Fk(t). Wu and Clements-Croome [7] extend the geometric process by replacing its parameter ak1 with a1ak1 + b1bk1, where a > 1 and 0 < b < 1. The geometric process has been applied in reliability analysis and maintenance policy optimisation for various systems by many authors; for example, see Wang and Pham [6], and Wu and Clements-Croome [8]. This paper presents the formulas of the expected long-run cost per unit time for a cold-standby system that consists of two identical components with perfect switching. When a component fails, a repairman will be called in to bring the component back to a certain working state. The time to repair is composed of two different time periods: waiting time and real repair time. The waiting time starts from the component failure to the start to repair, and the real repair time is the time between the start to repair and the completion of the repair. Both the working times and real repair times are assumed to be geometric processes, and the waiting time is assumed to be a renewal process. We also assume that the time to repair can either include only real repair time with a probability p, or include both waiting time and real repair time with a probability 1  p. The expected long-run cost per unit time is derived and a numerical example is given to demonstrate the usefulness of the derived expression. The paper is structured as follows. The coming section introduces geometric processes defined by Lam [4], denotation and assumptions. Section 3 discusses special cases. Section 4 offers numerical examples. Concluding remarks are offered in the last section. 1. Definitions and model assumptions This section first borrows the definition of the geometric process from Lam [4], and then makes assumptions for the paper. 1.1. Definition Definition 1. Assume n, g are the two random variables. For arbitrary real number a, there is

Pðn P aÞ > Pðg P aÞ; then n is called stochastically bigger than g. Similarly, if n stochastically smaller than g. Definition 2 [4]. Assume that {Xn, n = 1, 2, . . .} is a sequence of independent non-negative random variables. If the distribution function of Xn is F(an1t), for some a > 0 and all, n = 1, 2, . . ., then {Xn, n = 1, 2, . . .} is called a geometric process. Obviously, if a > 1, then {Xn, n = 1, 2, . . .} is stochastically decreasing, if a < 1, then {Xn, n = 1, 2, . . .} is stochastically increasing, and if a = 1, {Xn, n = 1, 2, . . .} is a renewal process.

1.2. Assumptions and denotation The following assumptions are assumed to hold in what follows. A. At the beginning, both components are new, component 1 is working and component 2 is under cold standby. B. When both of the two components are in good condition, one is working and the other is under cold standby. When the working component fails, a repairman repairs the failed component immediately with probability p, or repairs it with a waiting time with probability 1  p. As soon as the working component fails, the standby one will start to work. Assume the switching is perfect. After the failed one has been repaired, it is either put in use the other one fails or put in standby if the other is working. If one fails while the other is still under repair, the failed one must wait for repair until the repair for the first failed is completed.

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C. The time interval from the completion of the (n  1)th repair to that of the nth repair of component i is called the nth cycle of component i, where i = 1, 2; n = 1, 2, . . .. Denote the working time and the repair time of component i in the nth ðiÞ cycle (i = 1, 2; n = 1, 2, . . .) as X ðiÞ n and Y n , respectively, the waiting time of component i (i = 1, 2) in the nth cycle as ðiÞ ðiÞ ; n ¼ 1; 2; . . .g, the cumulative distribution functions of X ðiÞ fZ ðiÞ n n , Y n and Z n , as Fn(x) Gn(x), and S(x), respectively. ðiÞ ðiÞ ; Y ; and Z (i = 1, 2, and n = 1, 2, . . .) are statistically independent. D. X ðiÞ n n n E. When a replacement is required, a brand new but identical component will be used to replace, and the replacement time is negligible. F. Denote the repair cost per unit time of two components as Cm, the working reward per unit time as Cw, the replacement cost as Cr.

2. Expected cost under replacement policy N Fig. 1 shows a typical scenario, given that the above-mentioned assumptions hold. In what follows, we consider a replacement policy N, where a replacement is carried out if the number of failures reaches N for component 1. Denote the time between the (n  1)th replacement and the nth replacement of the system as Tn. Obviously, {T1, T2, . . .} forms a renewal process. Let C(N) be the expected long-run cost per unit time of the system under the policy N. Because {T1, T2, . . .} is a renewal process, the interval time between two consecutive replacements is a renewal cycle. Then, according to renewal reward theorem, we can know that the long-run average cost per unit time is given by

CðNÞ ¼

Expected cost incurred in a cycle : Expected length of a cycle

ð1Þ

Let W be the length of a renewal cycle of the system, then ð1Þ

W ¼ X1 þ

N1 X ð1Þ ð1Þ ð2Þ ð1Þ ð1Þ ð2Þ ð1Þ ½maxfZ i þ Y i ; X i gIfAi g þ maxfY i ; X i gIfBi g i¼1

N2 X ð2Þ ð2Þ ð1Þ ð2Þ ð2Þ ð1Þ ð2Þ ð1Þ ½maxfZ i þ Y i ; X iþ1 gIfAi g þ maxfY i ; X iþ1 gIfBi g þ X N : þ i¼1

The expected length of a renewal cycle is ð1Þ

ð1Þ

EðWÞ ¼ E½X 1  þ E½X N  þ

N1 X

ð1Þ

ð1Þ

ð2Þ

ð1Þ

ð1Þ

ð2Þ

ð1Þ

E½maxfZ i þ Y i ; X i gIfAi g þ maxfY i ; X i gIfBi g

i¼1

þ

N2 X

ð2Þ E½maxfZ i

ð2Þ

ð1Þ

ð2Þ

ð2Þ

ð1Þ

ð2Þ

þ Y i ; X iþ1 gIfAi g þ maxfY i ; X iþ1 gIfBi g:

ð2Þ

i¼1

Let C be the cost of a renewal cycle of the system under the policy N, then

C ¼ Cr þ Cm  Cx

( N1 X

" N X

N2 X

i¼1 ð1Þ Xi

i¼1 ð1Þ

ð1Þ

Yi þ

ð2Þ

)

i¼1

þ

N1 X

ð2Þ

ð2Þ

ð1Þ

ð2Þ

ð2Þ

ð2Þ

ð1Þ

ð2Þ

Y i þ ½Y N1 IfAg þ ðX N  Z N1 ÞIfAgIðAN1 Þ þ ½Y N1 IfBg þ X N IfBgIðBN1 Þ

ð2Þ Xi

# ð3Þ

;

i¼1 ð2Þ

ð1Þ

ð2Þ

ð2Þ

ð1Þ

ð2Þ

ð1Þ

ð2Þ

where A ¼ fX N  Z N1 > Y N1 g, A ¼ fX N  Z N1 < Y N1 g, B ¼ fX N  Y N1 > 0g, and B ¼ fX N  Y N1 < 0g.

Fig. 1. A possible progressive figure of the system.

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If X and Y are two independent non-negative random variables and their cumulative distribution functions are F(x) and G(x), respectively, we have following three lemmas. Denote E(C) as the expected value of C. By substituting the numerator and denominator of Eq. (1) with E(C) and E(W), respectively, we have

CðNÞ ¼

EðCÞ : EðWÞ

ð4Þ

Then the optimal replacement number can be obtained by minimising the value of C(N) in Eq. (4). Lemma 1

Z

EðmaxfX; YgÞ ¼ EX þ

1

Z0 1

¼ EY þ

FðxÞ½1  GðxÞdx

ð5Þ

GðxÞ½1  FðxÞdx:

ð6Þ

0

The proof of Lemma 1 is given in Appendix. Lemma 2

E½IfY > XgX þ E½If0 < Y < XgY ¼

Z

1

½1  FðxÞ½1  GðxÞdx:

ð7Þ

0

The proof of Lemma 2 is given in Appendix. Similarly, we have Lemma 3

E½ðY  XÞIfY  X > 0g ¼

Z

1

½1  GðxÞFðxÞdx:

ð8Þ

0

3. Special cases and discussion ðiÞ n1 t) and G(bn1t), respectively, where a > 1, 0 < b < 1. fY ðiÞ Denote the distributions of X ðiÞ n and Y n as F(a n ; n ¼ 1; 2; . . .g constitutes an increasing geometric process, whereas fX ðiÞ n ; n ¼ 1; 2; . . .g constitutes a decreasing geometric process. Then we have the following theorem.  n1   n1    n1 Theorem 1. Assume F n ðtÞ ¼ Fðan1 tÞ ¼ 1  exp  a k t , Gn ðtÞ ¼ Gðb tÞ ¼ 1  exp  b l t , and SðtÞ ¼ 1  exp  ct ,

(t P 0), respectively. Then the expected length of a renewal cycle is given by

EðWÞ ¼ k þ þ2

k þ ð2N  3Þð1  pÞc aN1 N2 X i¼1

l b

i1

þ

l b

N2

þ ð1  pÞk

N2

þ

pk2 b aN2 ðb

l

N2

k þ aN2 lÞ

þ

N2 X

! 1 l2  aN2 ðcaN2 þ kÞ ðaN2 l þ bN2 kÞð2bN2 ck þ lk þ aN2 lcÞ

2

ð1  pÞk2

i¼1

1

2 i1

ðai l þ b

i1

kÞð2b

ck þ lk þ ai lcÞ i1

þp



 1 1 þ ai ðcai þ kÞ ai1 ðcai1 þ kÞ

þ

!

1 ðai1 l þ b !!

i1

i1

kÞð2b

ck þ lk þ ai1 lcÞ

kð1 þ aÞ klðai l þ ai1 l þ 2b kÞ  i1 i1 ai ðb k þ ai lÞðb k þ ai1 lÞ

ð9Þ

and the expected cost is a cycle is

EðCÞ ¼ C m 2

N 2 X

l i1

i¼1

b

þ Cr  Cx 2

þ

l b

N2

þ ð1  pÞ

N1 X k k þ N1 i1 a a i¼1

!

2l N2

b



aN1 lc2 N2

ðaN1 c þ kÞðb



k2 l

!

c  lÞ ðaN1 c þ kÞðbN2 k þ aN1 lÞ

þ

!

kl N2

b

k þ aN1 l

p

ð10Þ

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J. Jia, S. Wu / Applied Mathematics and Computation 214 (2009) 133–141

and the expected long-run cost per unit time is given by

CðNÞ ¼

EðCÞ : EðWÞ

ð11Þ

If one sets a = 1 and b = 1, the above results E(W) and E(C) will be the situations where the components can be repaired as good as new.

4. Numerical examples 4.1. Parameter set 1 If we set a = 1.8, b = 0.98, k = 100, l = 10, c = 5, Cx = 500, Cm = 20, Cr = 5000, and p = 0.8, then the optimum number for a replacement will be N = 6, and the corresponding expected long-run cost per unit time is 433.41. The expected long-run cost per unit time is shown in Table 1, which corresponds to Fig. 1. 4.2. Parameter set 2 If we set a = 1.1, b = 0.98, k = 100, l = 1, c = 0.2, Cx = 500, Cm = 20, Cr = 5000, and p = 0.8, then the optimum number for a replacement will be N = 35, and the corresponding expected long-run cost per unit time is 491.85. The expected long-run cost per unit time is shown in Table 2, which corresponds to Fig. 2. Compare Figs. 2 and 3, we can find that the optimum replacement time becomes longer in the second situation. In both situations, we can easily find an optimum replacement time point. However, due to the complexity of Eq. (11), we are not able to prove that there exists a unique optimal value N.

Table 1 The expected long-run cost per unit time versus replacement times for parameter set 1. Times

Cost rate

Times

Cost rate

Times

Cost rate

Times

Cost rate

Times

Cost rate

1 2 3 4 5 6 7 8 9 10 11 12 13 14

88.74 107.92 279.47 389.09 428.42 433.41 424.96 411.2 395.37 378.99 362.83 347.25 332.41 318.37

15 16 17 18 19 20 21 22 23 24 25 26 27 28

305.11 292.59 280.78 269.62 259.06 249.07 239.6 230.61 222.07 213.94 206.21 198.84 191.8 185.08

29 30 31 32 33 34 35 36 37 38 39 40 41 42

178.66 172.52 166.64 161 155.6 150.42 145.44 140.66 136.06 131.63 127.38 123.27 119.32 115.51

43 44 45 46 47 48 49 50 51 52 53 54 55 56

111.84 108.29 104.87 101.56 98.37 95.28 92.29 89.4 86.6 83.89 81.27 78.73 76.27 73.88

57 58 59 60 61 62 63 64 65 66 67 68 69 70

71.56 69.32 67.14 65.02 62.97 60.97 59.04 57.15 55.33 53.55 51.82 50.14 48.5 46.91

Table 2 The expected long-run cost per unit time versus replacement times for parameter set 2. Times

Cost rate

Times

Cost rate

Times

Cost rate

Times

Cost rate

Times

Cost rate

1 2 3 4 5 6 7 8 9 10 11 12 13 14

17.37 18.96 37.01 58.53 83.54 111.78 142.75 175.68 209.62 243.55 276.47 307.51 336.03 361.63

15 16 17 18 19 20 21 22 23 24 25 26 27 28

384.14 403.59 420.14 434.04 445.6 455.12 462.91 469.24 474.35 478.46 481.74 484.34 486.39 487.99

29 30 31 32 33 34 35 36 37 38 39 40 41 42

489.22 490.15 490.84 491.32 491.63 491.8 491.85 491.79 491.65 491.43 491.13 490.77 490.36 489.88

43 44 45 46 47 48 49 50 51 52 53 54 55 56

489.36 488.78 488.15 487.48 486.76 485.99 485.18 484.33 483.43 482.5 481.52 480.5 479.44 478.35

57 58 59 60 61 62 63 64 65 66 67 68 69 70

477.22 476.05 474.85 473.61 472.34 471.03 469.7 468.33 466.93 465.5 464.05 462.56 461.05 459.5

Note: times in Tables 1 and 2 stands for replacement times; cost rate stands for the expected long-run cost per unit time.

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J. Jia, S. Wu / Applied Mathematics and Computation 214 (2009) 133–141

Fig. 2. The change of C(N) over N for parameter set 1.

Fig. 3. The change of C(N) over N for parameter set 2.

5. Conclusions Cold-standby systems are a category of important reliability structure in engineering. Searching an optimal replacement time point for such systems is of interest and important. This paper derived the expected long-run cost per unit time for a cold-standby system when time to repair is composed of two time periods: waiting time and real repair time. We also considered a special scenario where the working times and real repair times are geometric processes. Numerical examples are given to demonstrate the usefulness of the derived expression. Appendix. Proof of Lemma 1 Proof. Because X, Y are two independent random variables, therefore

EðmaxfX; YgÞ ¼

Z Z Z

maxfx; yg  f ðxÞgðyÞdxdy ¼ D

Z

Z

yf ðxÞgðyÞdxdy þ x6y

y

yf ðxÞgðyÞdxdy þ

0

1

x

Z

Z Z

xf ðxÞgðyÞdxdy x>y

Z 1 xf ðxÞgðyÞdydx ¼ ygðyÞFðyÞdy þ xGðxÞf ðxÞdx 0 0 0 0 0 0 Z 1 Z 1 Z 1 Z 1 ¼ xFðxÞd½1  GðxÞ þ xGðxÞf ðxÞdx ¼ ½xf ðxÞ½1  GðxÞ þ xGðxÞf ðxÞdx þ FðxÞ½1  GðxÞdx 0 0 0 0 Z 1 ¼ EX þ FðxÞ½1  GðxÞdx ¼

1

Z

Z Z

1

139

J. Jia, S. Wu / Applied Mathematics and Computation 214 (2009) 133–141

where f ðxÞ ¼ dFðxÞ and gðyÞ ¼ dGðyÞ . dx dy

h

Proof of Lemma 2

Proof. As X and Y are two independent non-negative random variables,

E½IfY > XgX ¼

Z Z

xf ðxÞgðyÞdxdy ¼

Z

x
¼

Z

1

Z

0

x

x½1  GðxÞd½1  FðxÞ ¼

E½If0 < Y < XgY ¼

Z Z

yf ðxÞgðyÞdxdy ¼

xf ðxÞ½1  GðxÞdx

1

½1  GðxÞ  xgðxÞ½1  FðxÞdx; 0

Z

0
1 0

Z

1 0

 Z xf ðxÞgðyÞdy dx ¼

1

1

ygðyÞ½1  FðyÞdy ¼

Z

0

1

xgðxÞ½1  FðxÞdx

0

and

E½IfY > XgX þ E½If0 < Y < XgY ¼

Z

1

½1  FðxÞ½1  GðxÞdx:



0

Proof of Theorem Proof. According to the above theorems and formulas (2) and (3), we have ð1Þ

ð1Þ

EðWÞ ¼ E½X 1  þ E½X N  þ

N 1 X

ð1Þ

ð1Þ

ð2Þ

ð1Þ

ð1Þ

ð2Þ

ð1Þ

E½maxfZ i þ Y i ; X i gIfAi g þ maxfY i ; X i gIfBi g

i¼1

þ

N2 X

ð2Þ

ð2Þ

ð1Þ

ð2Þ

ð2Þ

ð1Þ

ð2Þ

E½maxfZ i þ Y i ; X iþ1 gIfAi g þ maxfY i ; X iþ1 gIfBi g

i¼1 ð1Þ

ð1Þ

¼ E½X 1  þ E½X N  þ

N1 X ð1Þ ð1Þ ð2Þ ð1Þ ð2Þ fE½maxfZ i þ Y i ; X i gð1  pÞ þ E½maxfY i ; X i gpg i¼1

þ

N2 X

ð2Þ

ð2Þ

ð1Þ

ð2Þ

ð1Þ

fE½maxfZ i þ Y i ; X iþ1 gð1  pÞg þ E½maxfY i ; X iþ1 gpg

i¼1

( ¼

ð1Þ E½X 1 

þ

ð1Þ E½X N 

þ

N 1 X

ð1Þ E½maxfZ i

þ

ð1Þ ð2Þ Y i ; X i g

N2 X

þ

i¼1

( þ

N1 X

ð1Þ ð2Þ E½maxfY i ; X i g

( ð1Þ

þ

N2 X

) ð2Þ ð1Þ E½maxfY i ; X iþ1 g

ð1Þ

N1 X

ð1Þ

ð1Þ

E½Z i þ Y i  þ

i¼1 N1 X

ð1Þ

E½Y i  þ

i¼1

¼kþ

þ

k aN1

k aN1

N2 Z X i¼1

p

N1 Z X i¼1

ð2Þ

E½Y i  þ

N1 Z X i¼1

þ ð2N  3Þð1  pÞc þ 2

1

Hi ðtÞ½1  F i ðtÞdt þ 0

Gi ðtÞ½1  F i ðtÞdt þ

0

i¼1

N2 Z X i¼1

l b

i1

þ

ð2Þ

ð2Þ

E½Z i þ Y i  þ

i¼1

1

N2 X

N2 X

l N2

b

þ

N1 Z X i¼1

N2 Z X i¼1

)

1

Hi ðtÞ½1  F iþ1 ðtÞdt ð1  pÞ

0

)

1

Gi ðtÞ½1  F iþ1 ðtÞdt p 0

1

½ð1  pÞHi ðtÞ þ pGi ðtÞ½1  F i ðtÞdt 0

1

½ð1  pÞHi ðtÞ þ pGi ðtÞ½1  F iþ1 ðtÞdt

0

i¼1

þ

N2 X i¼1

N2 Z X

¼kþ

ð1  pÞ

i¼1

¼ E½X 1  þ E½X N  þ

þ

þ

ð2Þ ð1Þ Y i ; X iþ1 g

i¼1

i¼1

(

) ð2Þ E½maxfZ i

0

þ ð2N  3Þð1  pÞc þ 2

N2 X i¼1

1

½ð1  pÞHi ðtÞ þ pGðb

i1

l b

i1

þ

l N2

b

þ

Z

1

½ð1  pÞHN1 ðtÞ þ pGðb

N2

tÞ½1  FðaN2 tÞdt

0

tÞ½2  Fðai tÞ  Fðai1 tÞdt

ð12Þ

140

J. Jia, S. Wu / Applied Mathematics and Computation 214 (2009) 133–141

and

# Z 1 Z 1 N2 N2 N1 þ ð1  pÞ ½1  R ðtÞ½1  Gðb tÞdt þ p ½1  Gðb tÞ½1  Fða tÞdt þ Cr N i1 0 0 i¼1 b i¼1 b " # N N1 X X k k  Cx þ ai1 i¼1 ai1 i¼1 ( ) Z 1 Z 1 N2 X l l N2 N2 N1 ¼ Cm 2 þ N2 þ ð1  pÞ ½1  RN ðtÞ½1  Gðb tÞdt þ p ½1  Gðb tÞ½1  Fða tÞdt þ C r i1 0 0 b i¼1 b " # N 1 X k k ;  Cx 2 þ ai1 aN1 i¼1

EL ¼ C m

" N1 X

l

þ i1

N2 X

l

ðiÞ

ðiÞ

ð1Þ

ð2Þ

where t P 0, Hi(t) and RN(x) represent the cumulative distribution functions of the random variables Z i þ Y i and X N  Z N1 , respectively. Hence we have Hi (x) = S(t)*Gi(t), and RN(x) = FN(t)*[1  S(t)], where ‘‘*” indicates convolution, and

Hi ðtÞ ¼ SðtÞ  Gðb

i1

" #)    Z t( i1 b u tÞ ¼ 1  exp  ðt  uÞ d 1  exp 

l

0

  t ¼ 1  exp  

l

c

(

cbi1 þ l

c

i1

exp 

b

l

"

! t

 exp 

i1

2b

l

þ

1

! #)

c

t

;

 N1    Z t a u 1  exp  ðt  uÞ d exp k c 0       N1  t k a t 1þ ; exp  ¼ exp t  exp k þ aN1 c c k c

RN ðtÞ ¼ FðaN1 tÞ  ½1  SðtÞ ¼

where

Z

1

½ð1  pÞHN1 ðtÞ þ pGðb

0

N2

tÞ½1  FðaN2 tÞdt

"

# N2 1 l2 pk2 b þ ; ¼ ð1  pÞk  N2 N2 N2 N2 N2 a ðca þ kÞ ðaN2 l þ b kÞð2b ck þ lk þ aN2 lcÞ aN2 ðb k þ aN2 lÞ 2

Z

1

½ð1  pÞHi ðtÞ þ pGðb

0

i1

tÞ½2  Fðai tÞ  Fðai1 tÞdt

(

 1 1 1  l2 ½ þ i1 i1 ai ðcai þ kÞ ai1 ðcai1 þ kÞ ðai l þ b kÞð2b ck þ lk þ ai lcÞ ) " # i1 1 kð1 þ aÞ klðai l þ ai1 l þ 2b kÞ  i1  þp ; þ i1 i1 i1 ai ðai1 l þ b kÞð2b ck þ lk þ ai1 lcÞ ðb k þ ai lÞðb k þ ai1 lÞ

¼ ð1  pÞk2

Z

1

N2

½1  RN ðtÞ½1  Gðb

tÞdt

0

¼

Z

1

   t  2  exp

c

0

¼

2l b

N2



( N2 )     N1  k a t b  exp  exp  t dt t  exp k þ aN1 c k c l

aN1 lc2 ðaN1

N2

c þ kÞðb

c  lÞ



k2 l ðaN1

c þ kÞðbN2 k þ aN1 lÞ

and

Z

1

N2

½1  Gðb

0

N1

tÞ½1  Fða

tÞdt ¼

Z 0

1

( N2 )  N1  a t b kl exp  t dt ¼ N2 :  exp  k l b k þ aN1 l



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