A note on warm standby system

Accepted Manuscript A note on warm standby system Nil Kamal Hazra, Asok K. Nanda PII: DOI: Reference:

S0167-7152(15)00236-9 http://dx.doi.org/10.1016/j.spl.2015.07.004 STAPRO 7343

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Statistics and Probability Letters

Received date: 6 March 2015 Revised date: 1 July 2015 Accepted date: 1 July 2015 Please cite this article as: Hazra, N.K., Nanda, A.K., A note on warm standby system. Statistics and Probability Letters (2015), http://dx.doi.org/10.1016/j.spl.2015.07.004 This is a PDF file of an unedited manuscript that has been accepted for publication. As a service to our customers we are providing this early version of the manuscript. The manuscript will undergo copyediting, typesetting, and review of the resulting proof before it is published in its final form. Please note that during the production process errors may be discovered which could affect the content, and all legal disclaimers that apply to the journal pertain.

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A Note on Warm Standby System Nil Kamal Hazra and Asok K. Nanda ∗ Department of Mathematics and Statistics IISER Kolkata, India Submitted: March, 2015 Revision-I: June, 2015 Revision-II: July, 2015

Abstract We investigate, under what circumstances, a warm standby system has more number of surviving standby components, in terms of different stochastic orders, than another similar system at the time of kth failure of the active component of the respective systems.

Key Words and Phrases: Order statistics, permanent of a matrix, RR2 function, stochastic orders

1

Introduction

The failure of a system could happen at any time, and we can enhance the lifetime of the system by incorporating standby (or redundant) components. Standby components are mostly of three types − hot (or active) standby, warm standby and cold standby. Here we study the system with warm standby components, called warm standby system. For this system, a redundant component undergoes two operational environments. Initially, it functions in a milder environment (in which a redundant component has non-zero failure rate which is less than its actual failure rate), thereafter it switches over to the usual environment (in which the system is running) after the original component fails. It might happen that the redundant component fails before switching over to the usual ∗

e-mail: [email protected]; [email protected], corresponding author.

1

environment. Warm standby is used in a case where the switching time from the failed component to the standby component cannot be allowed. One practical example of such a system is shadowless lamp used in case of surgery. In this case the censoring and switching device is not allowed to take any time to start operating the standby component. In this case, the standby component is kept in a low-charging state so that the failure is neither zero (as in case of cold standby) nor same as the failure rate of the system in case of hot standby, but a state where the failure rate is positive and small so that once the component fails, the standby starts working immediately. Depending on the importance of the situation, one may look for quite a few standby components. If such systems are available from different producers, as a customer, we must look for the one whose standby components will not fail before the failure of the original component or the one which has less number of failures of the standby components prior to failure of the original components. In the present competitive market scenario, producers would like to build a system where such failures of standby components will be the least possible (along with other criteria). Warm standby system is well studied in the literature by different researchers, namely, Cha et al. [7], Li et al. [11], Eryilmaz [8], Hazra and Nanda [9] and the references therein. The rest of the paper is organized as follows. In Section 2, we discuss the useful definitions. In Section 3, we consider two warm standby systems where each system is formed by n active components and m warm standby components. We assume that one system has stochastically stronger active components than those of the other system. Then we show that the total number of surviving warm standby components at the time of kth failure of the active component of a warm standby system dominates that of another warm standby system with respect to different stochastic orders, viz. usual stochastic order, hazard rate order, reversed hazard rate order and likelihood ratio order. In Section 4, we study a similar kind of comparison result with respect to the usual stochastic order. Here we assume that two warm standby systems have different sets of warm standby components, and one set of warm standby components is superior to that of the other set with respect to the usual stochastic order. In Section 5, we carry out some simulation to verify the proposed results. Conclusion is given in Section 6. Throughout the paper, increasing and decreasing properties are not used in strict sense. For any differentiable function k(·), we write k ′ (t) to denote the first derivative of def. k(t) with respect to t. By a = b we mean that a is defined as b.

2

2

Preliminaries

For an absolutely continuous random variable Z, we denote the probability density function by fZ (·), the cumulative distribution function by FZ (·), and the survival or reliability function by F¯Z (·). For a collection {X1 , X2 , . . . , Xn } of random variables, the corresponding order statistics are denoted as X1,n ≤ X2,n ≤ · · · ≤ Xn,n . In order to compare the lifetimes of two systems stochastic orders are very useful tool. In the literature many different types of stochastic orders have been developed. The following well known definitions may be obtained in Shaked and Shanthikumar [14]. Definition 2.1 Let X and Y be two absolutely continuous random variables with respective supports (lX , uX ) and (lY , uY ), where uX and uY may be positive infinity, and lX and lY may be negative infinity. Then, X is said to be smaller than Y in (a) likelihood ratio (lr) order, denoted as X ≤lr Y , if fY (t) is increasing in t ∈ (lX , uX ) ∪ (lY , uY ); fX (t) (b) hazard rate (hr) order, denoted as X ≤hr Y , if

F¯Y (t) is increasing in t ∈ (−∞, max(uX , uY )); F¯X (t)

(c) reversed hazard rate (rhr) order, denoted as X ≤rhr Y , if FY (t) is increasing in t ∈ (min(lX , lY ), ∞); FX (t) (d) usual stochastic (st) order, denoted as X ≤st Y , if F¯X (t) ≤ F¯Y (t) for all t ∈ (−∞, ∞).

2

In the following diagram we present a chain of implications of the stochastic orders (cf. Shaked and Shanthikumar [14]): X ≤hr Y ↑ ց X ≤lr Y → X ≤st Y. ↓ ր X ≤rhr Y For the sake of completeness, below we give the definition of an RR2 function, which has been borrowed from Karlin [10]. 3

Definition 2.2 Let X and Y be two linearly ordered sets. Then, a real-valued function κ(·, ·) defined on X × Y, is said to be reverse regular of order 2 (written as RR2 ) if κ(x1 , y1 )κ(x2 , y2 ) ≤ κ(x1 , y2 )κ(x2 , y1 ), for all x1 < x2 and y1 < y2 .

2

We assume throughout the discussion that there is no repair for the original or the standby components.

3

Comparison Based on Single Set of Standby Components

Consider a warm standby system formed by n active components having lifetimes X = (X1 , X2 , . . . , Xn ) and m warm standby components having lifetimes Y = (Y1 , Y2 , . . . , Ym ). We denote Nk (X, Y) as the total number of surviving warm standby components at the time when the kth active component of the system fails. Below we compare two warm standby systems where the active components of one system dominate those of the other system with respect to the usual stochastic order. We show that the total number of surviving warm standby components at the time of kth failure of the active component of a warm standby system (formed by stochastically stronger active components) is less than that of another warm standby system (formed by stochastically weaker active components) with respect to the usual stochastic order. Theorem 3.1 Let X = (X1 , X2 , . . . , Xn ) and X∗ = (X1∗ , X2∗, . . . , Xn∗ ) be the lifetimes of the two groups of active components, and Y = (Y1 , Y2 , . . . , Ym ) be those of a group of standby components. Further assume that all Xi , Xi∗ and Yi are independent. If Xi ≤st Xi∗ for all i = 1, 2, . . . , n, then Nk (X∗ , Y) ≤st Nk (X, Y). Proof: Note that F¯Nk (X,Y) (r) = P (Nk (X, Y) > r) Z∞ = P (Nk (X, Y) > r|Xk,n = t)dFXk,n (t) 0

=

Z∞ 0

P

m X j=1

4

I[Yj >t]

!

!

> r dFXk,n (t)

=

Z∞

F¯Ym−r,m (t)dFXk,n (t)

(3.1)

=

Z∞

FXk,n (t)dFYm−r,m (t),

(3.2)

0

0

where I[a>b] = 1 if a > b and, = 0 otherwise. Similarly, F¯Nk (X∗ ,Y) (r) =

Z∞

∗ (t)dFY FXk,n m−r,m (t).

(3.3)

0

Since, Xi ≤st Xi∗ for all i = 1, 2, . . . , n, by Corollary 3.2 of Belzunce et al. [5], we have ∗ ∗ (t) for all t ∈ (0, ∞). Hence Nk (X , Y) ≤st Nk (X, Y). FXk,n (t) ≥ FXk,n 2 ∗ The following counterexample shows that the condition Xi ≤st Xi given in the above theorem cannot be relaxed. Counterexample 3.1 Let X = (X1 , X2 , X3 ) and X∗ = (X1∗ , X2∗ , X3∗ ) be the lifetimes of the two groups of active components with hazard rates (4, 5, 6) and (7, 8, 2), respectively. Further, let Y = (Y1 , Y2, Y3 ) be those of a group of standby components with hazard rates (1, 3, 4). Assume that all Xi , Xi∗ and Yi are independent. Clearly, X1 ≥st X1∗ , X2 ≥st X2∗ and X3 ≤st X3∗ . Thus, Xi ≤st Xi∗ does not hold for all i = 1, 2, 3. Now, F¯N2 (X,Y) (1) =

Z∞

e−4t + e−7t + e−5t − 2e−8t

0

= 0.737





9e−9t + 11e−11t + 10e−10t − 30e−15t dt

and F¯N2 (X∗ ,Y) (1) =

Z∞

e−4t + e−7t + e−5t − 2e−8t

0

= 0.753.




15e−15t + 10e−10t + 9e−9t − 34e−17t

Thus, F¯N2 (X∗ ,Y) (1) ≥ F¯N2 (X,Y) (1), and hence N2 (X∗ , Y) st N2 (X, Y).


2

In the following theorem we extend the above result to the hazard rate order. But before that we state the following lemma which may be obtained in Cap´eraa˙ [6]. This lemma gives an equivalent condition of the hazard rate order.

5

Lemma 3.1 Let α1 (·) and β1 (·) be two nonnegative real-valued functions such that β1 (·) and α1 (·)/β1(·) are increasing. Further, let U1 and U2 be two continuous nonnegative random variables. Then R∞ R∞ α1 (t)dFU2 (t) α1 (t)dFU1 (t) 0 0 ≤ R∞ R∞ β1 (t)dFU1 (t) β1 (t)dFU2 (t) 0

0

if, and only if, U1 ≤hr U2 .

2

The following theorem shows that, if the components of a system dominate those of another system (in the sense of reversed hazard rate order), then the first system will have less number of surviving warm standby components than the latter (in the sense of hazard rate order) at the time of kth failure of the active component of the respective systems. Theorem 3.2 Let X = (X1 , X2 , . . . , Xn ) and X∗ = (X1∗ , X2∗, . . . , Xn∗ ) be the lifetimes of the two groups of active components, and Y = (Y1 , Y2 , . . . , Ym ) be those of a group of standby components. Further assume that all Xi , Xi∗ and Yi are independent. If Xi ≤rhr Xj∗ for all i, j = 1, 2, . . . , n, then Nk (X∗ , Y) ≤hr Nk (X, Y). Proof: Note, from (3.2) and (3.3), that Nk (X∗ , Y) ≤hr Nk (X, Y) if, and only if, R∞

0 R∞ 0

FXk,n (t)dFYm−r,m (t) is increasing in r. ∗ (t)dFY FXk,n m−r,m (t)

This is equivalent to the fact that, for r ≤ s, R∞ R∞ FXk,n (t)dFYm−r,m (t) FXk,n (t)dFYm−s,m (t) 0 0 ≤ R∞ , R∞ ∗ (t)dFY ∗ (t) (t)dF (t) FXk,n F Ym−s,m Xk,n m−r,m 0

0

or equivalently,

R∞

0 R∞ 0

α1 (t)dFU1 (t) β1 (t)dFU1 (t)

≤

R∞

0 R∞

α1 (t)dFU2 (t) ,

(3.4)

β1 (t)dFU2 (t)

0

∗ (t), β1 (t) = FX (t), FU1 (t) = FYm−s,m (t), and FU2 (t) = FYm−r,m (t). where α1 (t) = FXk,n k,n Now, by Theorem 1.B.26 of Shaked and Shanthikumar [14] we have

U1 ≤hr U2 . 6

(3.5)

Further, since Xi ≤rhr Xj∗ for all i, j, by Theorem 1.B.61 of Shaked and Shanthikumar [14] ∗ we have Xk,n ≤rhr Xk,n , which is equivalent to the fact that α1 (t) is increasing in t. β1 (t)

(3.6)

Thus, on using (3.5) and (3.6), Lemma 3.1 gives (3.4), and hence Nk (X∗ , Y) ≤hr Nk (X, Y).2 Remark 3.1 Counterexample 3.1 can be used to show that the condition Xi ≤rhr Xj∗ for all i, j, given in Theorem 3.2 cannot be removed. 2 The following lemma may be obtained in Shaked and Shanthikumar [14], which will be used in proving the next theorem. This lemma gives an equivalent condition of the reversed hazard rate order. Lemma 3.2 Let α2 (·) and β2 (·) be two nonnegative real-valued functions such that β2 (·) and α2 (·)/β2 (·) are decreasing. Further, let W1 and W2 be two continuous nonnegative random variables. Then R∞

0 R∞

α2 (t)dFW1 (t) β2 (t)dFW1 (t)

≥

0

R∞

0 R∞

α2 (t)dFW2 (t) β2 (t)dFW2 (t)

0

if, and only if, W1 ≤rhr W2 .

2

The next theorem extends the result discussed in Theorem 3.1 to the reversed hazard rate order. This shows that Nk (X∗ , Y) ≤rhr Nk (X, Y) holds under the condition that Xi ≤hr Xj∗ for all i, j = 1, 2, . . . , n. Theorem 3.3 Let X = (X1 , X2 , . . . , Xn ) and X∗ = (X1∗ , X2∗, . . . , Xn∗ ) be the lifetimes of the two groups of active components, and Y = (Y1 , Y2 , . . . , Ym ) be those of a group of standby components. Further assume that all Xi , Xi∗ and Yi are independent. If Xi ≤hr Xj∗ for all i, j = 1, 2, . . . , n, then Nk (X∗ , Y) ≤rhr Nk (X, Y). Proof: From (3.2) and (3.3) we have FNk (X,Y) (r) = 1 −

Z∞

FXk,n (t)dFYm−r,m (t)

0

=

Z∞

F¯Xk,n (t)dFYm−r,m (t),

0

7

and FNk (X∗ ,Y) (r) =

Z∞

∗ (t)dFY F¯Xk,n m−r,m (t).

0

Hence, Nk (X∗ , Y) ≤rhr Nk (X, Y) if, and only if, R∞

0 R∞ 0

F¯Xk,n (t)dFYm−r,m (t) is increasing in r. ∗ (t)dFY F¯Xk,n m−r,m (t)

This is equivalent to the fact that, for r ≤ s, R∞

0 R∞ 0

or equivalently,

F¯Xk,n (t)dFYm−r,m (t) ∗ (t)dFY F¯Xk,n m−r,m (t)

R∞

0 R∞ 0

≤

0 R∞ 0

α2 (t)dFW1 (t) β2 (t)dFW1 (t)

R∞

≥

R∞

0 R∞

F¯Xk,n (t)dFYm−s,m (t) , ∗ (t)dFY F¯Xk,n m−s,m (t)

α2 (t)dFW2 (t) ,

(3.7)

β2 (t)dFW2 (t)

0

∗ (t), FW (t) = FY where α2 (t) = F¯Xk,n (t), β2 (t) = F¯Xk,n 1 m−s,m (t), and FW2 (t) = FYm−r,m (t). Now, by Theorem 1.B.56 of Shaked and Shanthikumar [14] we have

W1 ≤rhr W2 .

(3.8)

Further, since Xi ≤hr Xj∗ for all i, j, by Theorem 1.B.36 of Shaked and Shanthikumar [14] ∗ we have Xk,n ≤hr Xk,n , which is equivalent to the fact that α2 (t) is decreasing in t. β2 (t)

(3.9)

Thus, on using (3.8) and (3.9), Lemma 3.2 gives (3.7), and hence Nk (X∗ , Y) ≤rhr Nk (X, Y). 2 Remark 3.2 The condition Xi ≤hr Xj∗ for all i, j, given in Theorem 3.3 cannot be removed as Counterexample 3.1 shows. 2

8

In order to extend the above discussed results to the likelihood ratio order we shall take help of the concept of permanent of a matrix, which will be used to prove a few lemmas that are required to establish the desired result. Let A = ((ai,j )) be an n × n matrix. Then the permanent of A is defined as Per A =

n XY S

where

P

aj,ij ,

j=1

denotes the sum over all n! permutations (i1 , i2 , . . . , in ) of (1, 2, . . . , n). If

S

a1 , a2 , . . . , an are column vectors of A, then the permanent of A can be written as   Per A =  a1 , a2 , . . . , an  , |{z} |{z} |{z} r1

r2

rn

where the matrix is formed by r1 copies of a1 , r2 copies of a2 and so on. For more discussion on permanent of a matrix we may refer the reader to Minc [12], Bapat [2], Bapat and Beg [3], Bapat and Kochar [4], and Balakrishnan [1]. The following lemma is borrowed from Bapat [2], which is used to prove Lemma 3.4. Lemma 3.3 Let z, b, c, d be nonnegative vectors of order m. Then, for 1 ≤ r ≤ m, " #" # " #" # r |{z} z , |{z} b , |{z} c r−1

m−r

1

z , |{z} b , |{z} d ≥ (r − 1) |{z} z , |{z} b |{z} r−1

m−r

1

r

m−r

z , |{z} b , |{z} c , |{z} d . |{z} r−2

m−r

1

1

The next lemma, which will be used in proving the upcoming theorem, gives the RR2 property of a particular type of function. ! P Q Q FYj (t) , for r = 0, 1, . . . , m, and F¯Yi (t) Lemma 3.4 Let D(r, t) = {A: |A|=r}

j∈Ac

i∈A c

0 < t < ∞, where A ⊆ {1, 2, . . . , m}, A is the complement of A and |A| is the cardinality of A. Then D(r, t) is RR2 in (r, t). Proof: To prove that D(r, t) is RR2 , it suffices to show that, for 0 ≤ r ≤ m − 1, and for 0 < t < s < ∞, D(r, t)D(r + 1, s) ≤ D(r, s)D(r + 1, t), or equivalently, D(r, t) is increasing in t ∈ (0, ∞). D(r + 1, t)

9

This is equivalent to the fact that  αr (t)

def.

=



¯ , F(t) F(t) |{z} |{z}  r m−r  is increasing in t ∈ (0, ∞), ¯ , F(t)  F(t) |{z} |{z} r+1

m−r−1

¯ are the column vectors (FY1 (t), FY2 (t), . . . , FYm (t))′ and (F¯Y1 (t), F¯Y2 (t), where F(t) and F(t) . . . , F¯Ym (t))′ , respectively. Because the permanent is a multilinear function of its columns, αr (t) can be differentiated by taking the derivative of one column at a time, keeping the rest fixed, and then adding up the permanent of the resulting matrices. Thus, we have αr′ (t) = ∆r1 (t) + ∆r2 (t), where













¯ , F(t) F(t) ¯ , F(t) , f(t)  − r F(t) ¯ , F(t), f(t)  , ¯ , F(t)  F(t) ∆r1 (t) = (r + 1) F(t) |{z} |{z} |{z} |{z} |{z} |{z} |{z} |{z} |{z} |{z} r

m−r

r

m−r−1

r+1

1

m−r−1





r−1

1



¯ , F(t) , f(t)  ∆r2 (t) = (m − r) ¯ F(t), F(t)  F(t) |{z} |{z} |{z} |{z} |{z} r+1 m−r−1 r m−r−1 1  



¯ , F(t) , f(t)  , −(m − r − 1) ¯ F(t), F(t) F(t) |{z} |{z} |{z} |{z} |{z} r

and

m−r

m−r

r+1

m−r−2

1

f(t) = (fY1 (t), fY2 (t), . . . , fYm (t))′ . ¯ By taking z = F(t), b = F(t) = c and d = f(t), we have, from Lemma 3.3, ∆r1 (t) ≥ 0. ¯ Further, by taking z = F(t), b = F(t) = d and c = f(t), we have, from Lemma 3.3, ∆r2 (t) ≥ 0. Hence αr (t) is increasing in t. 2 The following lemma will be used in the next theorem. The idea of the proof is due to Karlin [10]. Lemma 3.5 Let κ(x, y) > 0, defined on X × Y be RR2 , where X and Y are subsets of real line. Assume that a function f (·, ·) defined on X × Y is such that (i) for each x, f (x, y) changes sign at most once, and if the change of sign does occur, it is from positive to negative, as y traverses Y; 10

(ii) for each y, f (x, y) is increasing in x; R (iii) ω(x) = κ(x, y)f (x, y)dµ(y) exists absolutely and defines a continuous function of Y

x, where µ is a sigma-finite measure.

Then ω(x) changes sign at most once, and if the change of sign does occur, it is from negative to positive. Proof: Let x0 be a point where ω(x) changes its sign, as x traverses X . Then to prove the result, it suffices to show that ω(x) ≥ 0 for all x > x0 . Since ω(x0 ) = 0, corresponding to x0 , there exists a point y0 such that f (x0 , y) ≤ 0 for all y > y0 and f (x0 , y) ≥ 0 for all y < y0 . The existence of such a y0 is guaranteed because of the assumption that ω(x0 ) = 0 and (i) above. Write ω(x) ω(x) ω(x0 ) = − κ(x, y0 ) κ(x, y0 ) κ(x0 , y0)  Z Z  κ(x, y) κ(x0 , y) κ(x, y) f (x0 , y)dµ(y) + [f (x, y) − f (x0 , y)] − dµ(y). = κ(x, y0 ) κ(x0 , y0) κ(x, y0) Y

Y

Consider the following two cases. Case I: Let x > x0 and y > y0 . Then the first integral is positive because κ(x, y) is RR2 and f (x0 , y) ≤ 0 for all y > y0 . Further, the second integral is positive because of (ii). Thus ω(x) ≥ 0. Case II: Let x > x0 and y < y0 . Then the first integral is positive because κ(x, y) is RR2 and f (x0 , y) ≥ 0 for all y < y0 . Further, the second integral is positive because of (ii). Thus ω(x) ≥ 0. Hence the result is proved. 2 Below we give a lemma without proof, which will be used in proving the upcoming theorem. This lemma gives an equivalent condition to show that f (x)/g(x) is increasing/decreasing in x. Lemma 3.6 Let f (·) and g(·) be two nonnegative real-valued functions defined on (0, ∞). Then f (x)/g(x) is increasing (resp. decreasing) in x, if, and only if, for any real number c, the difference f (x) − cg(x) changes sign at most once, and if the change of sign does occur, it is from negative (resp. positive) to positive (resp. negative), as x traverses from 0 to ∞. 2 In the next theorem we investigate, under what circumstances, a warm standby system (formed by n active components and m warm standby components) will have more number of surviving warm standby components than another similar system (in the sense of likelihood ratio order) at the time of kth failure of the active component of the respective systems. 11

Theorem 3.4 Let X = (X1 , X2 , . . . , Xn ) and X∗ = (X1∗ , X2∗ , . . . , Xn∗ ) be the lifetimes of the two groups of active components, and Y = (Y1 , Y2, . . . , Ym ) be those of a group of standby components. Further assume that all Xi , Xi∗ and Yi are independent. If Xi ≤lr Xj∗ for all i, j = 1, 2, . . . , n, then Nk (X∗ , Y) ≤lr Nk (X, Y). Proof: Note that, for r = 0, 1, . . . , m, Z∞ P (Nk (X, Y) = r|Xk,n = t)dFXk,n (t) P (Nk (X, Y) = r) = 0

=

=

Z∞

0 Z∞

P

m X

!

I[Yj >t] = r dFXk,n (t)

j=1

D(r, t)dFXk,n (t),

0

where D(r, t) is as defined in Lemma 3.4. Similarly, Z∞ ∗ ∗ (t). P (Nk (X , Y) = r) = D(r, t)dFXk,n 0

Let v be any real number. Consider the relation Z∞   ∗ ∗ (t) dt. P (Nk (X, Y) = r) − vP (Nk (X , Y) = r) = D(r, t) fXk,n (t) − vfXk,n 0

Xi ≤lr Xj∗

Since for all i, j = 1, 2, . . . , n, by Theorem 1.C.33 of Shaked and Shanthiku∗ , which gives that mar [14], we have Xk,n ≤lr Xk,n fXk,n (t) is decreasing in t. ∗ (t) fXk,n ∗ (t) changes sign at most once, Thus, on using Lemma 3.6 we have that fXk,n (t) − vfXk,n and if the change of sign does occur, it is from positive to negative, as t goes from 0 to ∞. Further, by Lemma 3.4 we have that

D(r, t) is RR2 in (r, t). Therefore, on using Lemma 3.5 we have that P (Nk (X, Y) = r) − vP (Nk (X∗ , Y) = r) changes sign at most once, and if the change of sign does occur, it is from negative to positive, as r goes from 0 to m. Thus, on using Lemma 3.6 we get that P (Nk (X, Y) = r)/P (Nk (X∗ , Y) = r) is increasing in r, and hence Nk (X∗ , Y) ≤lr Nk (X, Y). 2 Remark 3.3 Counterexample 3.1 can be used to show that the condition Xi ≤lr Xj∗ for all i, j, given in Theorem 3.4 cannot be removed. 12

4

Comparison Based on Two Sets of Standby Components

In the previous section we have considered same set of warm standby components for two different sets of original components to form two different systems and studied their properties. However, one may be interested to know what kind of properties two systems may have if one set of original components and two sets of warm standby components are used to form two different systems. In order to study the problem as discussed above, we consider the following theorem where we show that the total number of surviving warm standby components at the time of kth failure of the active component of a system is less than that of another system with respect to the usual stochastic order, provided the standby components of one batch is smaller than those of the other batch with respect to the usual stochastic order. Theorem 4.1 Let X = (X1 , X2 , . . . , Xn ) be the lifetimes of a group of active components, and Y = (Y1 , Y2 , . . . , Ym ) and Y ∗ = (Y1∗ , Y2∗ , . . . , Yn∗ ) be those of two groups of standby components. Further assume that all Xi , Yi and Yi∗ are independent. If Yi ≤st Yi∗ for all i = 1, 2, . . . , n, then Nk (X, Y) ≤st Nk (X, Y ∗). Proof: From (3.1) we have F¯Nk (X,Y) (r) =

Z∞

F¯Ym−r,m (t)dFXk,n (t)

Z∞

∗ F¯Ym−r,m (t)dFXk,n (t).

0

and F¯Nk (X,Y∗ ) (r) =

0

Because, Yi ≤st Yi∗ for all i = 1, 2, . . . , n, by Corollary 3.2 of Belzunce et al. [5], we have ∗ F¯Ym−r,m (t) ≤ F¯Ym−r,m (t) for all t ∈ (0, ∞). Hence Nk (X, Y) ≤st Nk (X, Y ∗). 2 ∗ The following counterexample shows that the condition Yi ≤st Yi given in the above theorem cannot be relaxed. Counterexample 4.1 Let X = (X1 , X2 , X3 ) be the lifetimes of a group of active components with hazard rates (4, 5, 6). Further, let Y = (Y1 , Y2 , Y3 ) and Y ∗ = (Y1∗ , Y2∗ , Y3∗ ) be those of two groups of standby components with hazard rates (1, 3, 4) and (5, 2, 6), respectively. Assume that all Xi , Yi and Yi∗ are independent. Clearly, Y1 ≥st Y1∗ , Y2 ≤st Y2∗ 13

and Y3 ≥st Y3∗ . Thus, Yi ≤st Yi∗ does not hold for all i = 1, 2, 3. Now, F¯N2 (X,Y) (1) =

Z∞ 0

e−4t + e−7t + e−5t − 2e−8t

= 0.737





9e−9t + 11e−11t + 10e−10t − 30e−15t dt

and F¯N2 (X,Y∗ ) (1) =

Z∞ 0

e−7t + e−8t + e−11t − 2e−13t

= 0.529.





9e−9t + 11e−11t + 10e−10t − 30e−15t dt

Thus, F¯N2 (X,Y) (1) ≥ F¯N2 (X,Y∗ ) (1), and hence N2 (X, Y) st N2 (X, Y ∗).

5

Simulation

The result given in Theorem 3.1, with k = 2, n = 5 and m = 10, is analyzed through a data set as given below. The other results given in this paper can also be verified in a similar way. From Theorem 3.1 we have F¯Nk (X,Y) (r) =

Z∞

FXk,n (t)dFYm−r,m (t)

0

∗

m 1 X R(j) − j , = m∗ j=1 n∗

(5.1)

where m∗ and n∗ are the sample sizes of the order statistics Ym−r,m and Xk,n , respectively, and R(j) is the rank of the jth observation of Ym−r,m in the combined increasing arrangement of the samples from Ym−r,m and Xk,n (see Nanda and Paul [13]). Let X = (X1 , X2 , X3 , X4 , X5 ) and X∗ = (X1∗ , X2∗ , X3∗ , X4∗ , X5∗ ) be the lifetimes of the two groups of active components having failure rates (2, 3, 4, 5, 6) and (1, 2, 3, 4, 5), respectively. Further, let Y = (Y1 , Y2 , . . . , Y10 ) be lifetimes of a group of standby components having failure rates (8, 9, . . . , 17). Assume that all Xi , Xi∗ and Yi are independent. Clearly, Xi ≤st Xi∗ for all i = 1, 2, . . . , 5. Now we draw samples for X, X∗ and Y, each of size 500, 000, from their respective distributions. We call them x = (xij ), x∗ = (x∗ij ) and y = (ylj ), for i = 1, 2, . . . , 5, l = 1, 2, . . . , 10 and j = 1, 2, . . . , 500, 000, where A = (aij ) represents a matrix in which aij is the entity of the ith row and the jth column. Now we sort each column of the matrices x and x∗ in ascending order of magnitude, whereas ˜ , respectively. ˜ and x the column of y is sorted in descending order; we call them x ˜∗ and y 14

˜ (say, x ˜ 2 ) and that of x Then the second row of x ˜∗ (say, x ˜∗2 ) give the observations corresponding to the second order statistics of x and x∗ , respectively. Further, the sth row ˜ gives observations corresponding to the (m − s + 1)th (= (11 − s)th) order statistic of y ′ ′ ′ ′ ˜ = (˜ ˜ 2, . . . , y ˜ 10 ), where a denotes the transpose of the of y, s = 1, 2, . . . , 10. Let y y1 , y ˜ 2 ), and row vector a. Then, for s = 1, 2, . . . , 10, we form the combined sample z = (˜ ys , x s arrange them in ascending order. Let R(j) denote the rank of the jth largest observation ˜ s in the combined sample z. Then, for r = 0, 1, . . . , 9, F¯Nk (X,Y) (r) corresponding to y r+1 is calculated from (5.1) with R(j) replaced by R(j) . Similarly, we set F¯Nk (X∗ ,Y) (r), and plot both the survival functions on the same graph paper (see Figure 1). In this figure, c represents F¯Nk (X∗ ,Y) (r) and d represents F¯Nk (X,Y) (r). We see from the figure that d dominates c, which verifies Theorem 3.1.

Figure 1: Plot of c and d against r ∈ [0, 10] (Simulation).

6

Conclusion

When any component of a system, upon failure, is replaced by a warm standby component is called warm standby system. Clearly, the number of warm standby components available for use at the time of kth component failure (for any fixed number k) is a random variable. In this article we compare such random variables for two warm standby systems. Two separate cases have been studied in this paper − (i) Two systems having a single set of warm standby components, (ii) Single system having two separate warm standby components. The comparison for the first case is done with respect to usual stochastic order, hazard rate order, reversed hazard rate order and likelihood ratio order, whereas for the second case, the comparison is done only for usual stochastic order. In all 15

the above cases under (i), it is observed that the number of warm standby components is smaller corresponding to the system having stronger active components, whereas in case of (ii), this number is smaller when the warm standby components are weaker. As mentioned in the introduction, this kind of comparison will help the decision makers to decide on which kind of standby components to use in the system in order to survive in the present-day market. The simulation given in Section 5 verifies Theorem 3.1. Although the simulation tells that, at a fixed time point, the system constructed out of weaker components will have the larger (in the sense of usual stochastic order) number of surviving standby components, this may not be true in general (see Theorem 3.2). Corresponding to these two theorems (Theorems 3.1 and 3.2) we see that depending on how strong dominance of one system over the other is needed (in practice), design engineers/decision makers decide on structuring the system. To be more specific, if stronger dominance (say, in terms of hazard rate) is required, the system should be structured in such a way that all the components of one system should be dominated by the weakest component of the other, whereas so strong dominance (of components) is not required for weaker dominance (say, in terms of stochastic order) of the system.

Acknowledgements The authors are thankful to the Editor, and anonymous Reviewers for their valuable constructive comments and suggestions which lead to an improved version of the manuscript. The authors are also thankful to Dr. Satyaki Mazumder, IISER Kolkata for some stimulating discussion. Financial support from Council of Scientific and Industrial Research, New Delhi (Grant No. 09/921(0060)2011-EMR-I) is sincerely acknowledged by Nil Kamal Hazra. Asok K. Nanda acknowledges the financial assistance of NBHM, DAE, Government of India (Grant no. NBHM/R.P.34/2014/Fresh/939).

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[4] Bapat, R.B. and Kochar, S.C. (1994). On likelihood ratio ordering of order statistics. Linear Algebra and Its Applications, 199, 281-291. [5] Belzunce, F., Franco, M., Ruiz, J.M. and Ruiz, M.C. (2001). On partial orderings between coherent systems with different structures. Probability in the Engineering and Informational Sciences, 15, 273-293. [6] Cap´eraa, ˙ P. (1988). Tail ordering and asymptotic efficiency of rank tests. The Annals of Statistics, 16, 470-478. [7] Cha, J.H., Mi, J. and Yun, W.Y. (2008). Modelling a general standby system and evaluation of its performance. Applied Stochastic Models in Business and Industry, 24, 159-169. [8] Eryilmaz, S. (2013). Reliability of a k-out-of-n system equipped with a single warm standby component. IEEE Transactions on Reliability, 62, 499-503. [9] Hazra, N.K. and Nanda, A.K. (2014). General standby component allocation in series and parallel systems. arXiv:1401.0132v1. [10] Karlin, S. (1968). Total Positivity. Stanford University Press, Stanford, California. [11] Li, X., Yan, R. and Zuo, M. (2009). Evaluating a warm standby system with components having proportional hazard rates. Operations Research Letters, 37, 56-60. [12] Minc, H. (1978). Permanents, Encyclopedia of Mathematics and its Applications. Addison-Wesley, Reading, Massachusetts. [13] Nanda, A.K. and Paul, P. (2003). Tests for reversed hazard rate function. Calcutta Statistical Association Bulletin, 54, 181-193. [14] Shaked, M. and Shanthikumar, J.G. (2007). Stochastic Orders. Springer, New York.

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