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On the optimal allocation of an active redundancy in a two-component series system
Statistics & Probability Letters 63 (2003) 325 – 332
On the optimal allocation of an active redundancy in a two-component series system Jos$e E. Vald$esa;∗ , R$omulo I. Zequeirab a
Facultad de Matem atica y Computaci on, Universidad de La Habana, C. San L azaro s/n, CP 10400 La Habana, Cuba b Laboratoire de Mod elisation et Sˆuret e des Syst#emes, Universit e de Technologie de Troyes, 12 rue Marie Curie, BP 2060, 10010 Troyes Cedex, France Received September 2002; received in revised form March 2003
Abstract Let X1 and X2 be independent random variables representing the lifetimes of the components C1 and C2 , and Y1 and Y2 be random variables, independent of X1 and X2 , representing the lifetimes of the spares R1 and R2 , respectively. Suppose that the components C1 and C2 form a series system. There are two options for using one of the spares R1 or R2 as an active redundancy: to allocate R1 in parallel with C1 or R2 in parallel with C2 . In this paper we compare these options using failure rate ordering. c 2003 Elsevier Science B.V. All rights reserved. Keywords: Reliability; Active redundancy; Stochastic orderings
1. Introduction One way of improving the reliability of a system is the allocation of an active redundancy. The problem of where to allocate active redundancies in a system has been studied using a variety of stochastic orderings (see Boland et al., 1992, 1994; Singh and Misra, 1994; Boland, 1998). An extensive treatment of stochastic orders is given in Shaked and Shanthikumar (1994) (see also Ross, 1983). In this work we will assume that all distribution functions G(t) satisfy G(0) = 0 and G(t) ¡ 1, t ¿ 0. We will also suppose that for each distribution function G(t) there exists a probability density < = 1 − G(t). function g(t). We will use the notation G(t) ∗
Corresponding author. E-mail addresses: [email protected] (J.E. Vald$es), [email protected] (R.I. Zequeira).
c 2003 Elsevier Science B.V. All rights reserved. 0167-7152/03/$ - see front matter doi:10.1016/S0167-7152(03)00098-1
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J.E. Vald es, R.I. Zequeira / Statistics & Probability Letters 63 (2003) 325 – 332
Recall that the failure rate associated to a random variable with distribution function G(t) is < deCned by g(t)= G(t). Let X1 , X2 and Y be independent random variables representing the lifetimes of the components C1 , C2 and of the spare R, respectively. Suppose that the components C1 and C2 form a series system and that we can allocate R in parallel with C1 or C2 . Then comparing the lifetimes V1 = min{max(X1 ; Y ); X2 }
and
V2 = min{X1 ; max(X2 ; Y )};
we can decide which of these two options is better. Boland et al. (1992) showed that X2 ¿st X1 is a necessary and suDcient condition for V1 ¿st V2
(1)
to hold. Here the symbol ¿st denotes the stochastic order in the usual classical sense. Let us denote by r1 (t) and r2 (t) the failure rates corresponding to the systems with lifetimes V1 and V2 , respectively. Singh and Misra (1994) proved that if X1 , X2 and Y have exponential distribution with means 1=1 ; 1=2 and 1=, respectively, then 1 ¿ max(2 ; ) is a suDcient condition for the failure (hazard) rate ordering V1 ¿hr V2
(2)
to hold, that is, for the inequality r1 (t) 6 r2 (t);
t ¿ 0;
to hold. The orderings given by Eqs. (1) and (2) can be interpreted in the following way: it is better to allocate the spare component in parallel with the weakest component. Sometimes it is more natural to suppose that the spare to be allocated in parallel depends on the component with which it is to be allocated. Let us deCne the random variables Y1 and Y2 as the lifetimes of the spares R1 and R2 , respectively. Suppose we have two options for using one spare, R1 or R2 , as a componentwise active redundancy. We can allocate R1 in parallel with C1 or, otherwise, allocate R2 with C2 . To Cnd the best option we must compare the lifetimes U1 and U2 deCned as U1 = min{max(X1 ; Y1 ); X2 }
and
U2 = min{X1 ; max(X2 ; Y2 )}:
That is, U1 (U2 ) is the lifetime of the system formed when the spare R1 (R2 ) is allocated as an active redundancy with C1 (C2 ). Observe that the case in which there is only one spare available, R1 = R2 = R, may be considered as a particular case of this, with Y1 = Y2 = Y . Throughout this paper we will assume that the lifetimes X1 and X2 are independent, and that the lifetimes Y1 and Y2 are independent of X1 and X2 .
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327
In this paper we examine the comparison of U1 and U2 in the sense of the failure rate ordering.
2. Optimal allocation of the spare Let us denote the distribution functions of the random variables X1 , X2 , Y1 and Y2 , respectively, by F1 (t), F2 (t), G1 (t) and G2 (t), their corresponding probability densities by f1 (t); f2 (t), g1 (t) and g2 (t), and the respective failure rates by 1 (t), 2 (t), 1 (t) and 2 (t). Now we will Cnd out conditions for the order U1 ¿hr U2 to be satisCed. Let us redeCne r1 (t) and r2 (t) as the failure rates corresponding to the systems with lifetimes U1 and U2 , respectively. The following lemmas will be useful in the proof of Proposition 1. Lemma 1. r1 (0) = 2 (0) and r2 (0) = 1 (0). Proof. Observe that the failure rates r1 (t) and r2 (t) can be expressed as r1 (t) =
f1 (t)G1 (t) + g1 (t)F1 (t) + 2 (t); 1 − F1 (t)G1 (t)
r2 (t) =
f2 (t)G2 (t) + g2 (t)F2 (t) + 1 (t); 1 − F2 (t)G2 (t)
from which obviously follows the statement of the lemma. Lemma 2. sup
06x61; 0¡y61
[xy − x − (1 − y)] 6 0:
Proof. Let us denote h(x) = xy − x − (1 − y): If y = 1 then h(x) = 0. Let now 0 ¡ y ¡ 1. Observe that h(0) = h(1) = y − 1 ¡ 0 and h (x) = yxy−1 − 1;
h (x) = (y − 1)yxy−2 :
On the other hand, h (x) ¡ 0 for 0 ¡ x ¡ 1. Hence from the condition h (x) = 0 we obtain the value x0 = y1=(1−y) where the function h(x) is maximized. But h(x0 ) = yy=(1−y) − y1=(1−y) − (1 − y) = (1 − y)(yy=(1−y) − 1) ¡ 0; hence the lemma is proved. Let us suppose that 1 (t) ¿ 0 for all t ¿ 0. Let us denote (t) = 2 (t)=1 (t), t ¿ 0.
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J.E. Vald es, R.I. Zequeira / Statistics & Probability Letters 63 (2003) 325 – 332
Proposition 1. Let 2 (t) ¿ 0, t ¿ 0. Suppose that 1 (t) ¿ max(2 (t); 1 (t)), 2 (t) ¿ 1 (t), t ¿ 0, and that (t) is non-increasing. Then r1 (t) 6 r2 (t);
t ¿ 0;
that is, U1 ¿hr U2 . Proof. For t = 0 the result of the proposition is obtained from Lemma 1. Let t ¿ 0. Observe that r1 (t) and r2 (t) can be rewritten as r1 (t) =
1 (t)G1 (t)F< 1 (t) + 1 (t)F1 (t)G< 1 (t) + 2 (t); 1 − F1 (t)G1 (t)
r2 (t) =
2 (t)G2 (t)F< 2 (t) + 2 (t)F2 (t)G< 2 (t) + 1 (t): 1 − F2 (t)G2 (t)
After some algebraic transformations it is not hard to see that the inequality r1 (t) 6 r2 (t) is equivalent to the inequality G< 1 (t) 1 − F2 (t)G2 (t) (1 (t) − 1 (t)F1 (t)) ¿ 2 (t) − 2 (t)F2 (t): G< 2 (t) 1 − F1 (t)G1 (t)
(3)
Observe that inequality G< 1 (t) 1 − F2 (t)G2 (t) ¿1 G< 2 (t) 1 − F1 (t)G1 (t)
(4)
is equivalent to the inequality G2 (t)G< 1 (t)F< 2 (t) ¿ G1 (t)G< 2 (t)F< 1 (t): But G2 (t) ¿ G1 (t) since 2 (t) ¿ 1 (t) and F1 (t) ¿ F2 (t) since 1 (t) ¿ 2 (t), t ¿ 0. Hence inequality (4) holds. On the other hand, 1 (t) ¿ 1 (t) implies 1 (t) − 1 (t)F1 (t) ¿ 0. Therefore (3) holds if the inequality 1 (t) − 1 (t)F1 (t) ¿ 2 (t) − 2 (t)F2 (t)
(5)
is satisCed. Note that inequality (5) can be written as 1 (t) 2 (t) F1 (t) − F2 (t) 6 1 − (t): 1 (t) 1 (t)
(6)
Hence, since 2 (t) ¿ 1 (t), for (6) to hold it is suDcient that 1 (t) < (F 2 (t) − F< 1 (t)) 6 1 − (t): 1 (t)
(7)
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Thus, to prove (3) it is suDcient to show that inequality F< 2 (t) − F< 1 (t) 6 1 − (t);
t ¿0
(8)
holds. Observe now that as (t) is non-increasing we obtain t t < F 2 (t) = exp − (x)1 (x) d x 6 exp −(t) 1 (x) d x = (F< 1 (t))(t) : 0
0
Then to prove (8) it is suDcient to verify that (F< 1 (t))(t) − F< 1 (t) − (1 − (t)) 6 0;
t ¿ 0:
< ¡ 1 and 0 ¡ (t) 6 1 for t ¿ 0. Thus the But this inequality follows from Lemma 2, since 0 ¡ F(t) proposition is proved. Remark 1. From (7) it follows that if 2 (t) ¿ 1 (t) and 1 (t) ¿ 2 (t)+1 (t), t ¿ 0, then U1 ¿hr U2 . The suDcient conditions of Proposition 1 require the spare to be allocated as a redundancy be stronger (in the failure rate order) than the other spares, and also stronger than the weakest component of the series system. However, with respect to the components of the system, together with the failure rate ordering of these components, it is required that the function (t) be non-increasing. The following examples show that this condition is satisCed in some cases of practical interest. Example 1. Let 1 (t) = at + b and 2 (t) = ct + d, t ¿ 0, with a ¿ c ¿ 0, b ¿ d ¿ 0. Evidently, 2 (t) 6 1 (t). If furthermore cb − ad 6 0, then (t) is non-increasing. Example 2. One special case in which the function (t) is non-increasing is when the failure rates 1 (t) and 2 (t) are proportional, that is, 2 (t) = c1 (t) for all t ¿ 0, where c is a constant. This is the case, for example, of two Weibull distributions with the same shape parameter. If 0 ¡ c 6 1, then 2 (t) 6 1 (t). The proof of the following corollary is straightforward and is omitted. Corollary 1. Let Y1 =st Y2 =st Y , and Y =st X1 or Y =st X2 . Suppose 1 (t) ¿ 0 and 2 (t) = c1 (t), t ¿ 0, where c ¿ 0. Then r1 (t) 6 r2 (t)
if ; and only if
1 (t) ¿ 2 (t);
t ¿ 0;
that is, U1 ¿hr U2 if, and only if X1 6hr X2 . The non-increasing condition of the function (t) in Proposition 1 can be substituted as is shown in the following proposition. Proposition 2. Let 2 (t) ¿ 0, t ¿ 0. Suppose that 1 (t) ¿ max(2 (t); 1 (t)), 2 (t) ¿ 1 (t), t ¿ 0, and that (t) 6 1 − mm=(1−m) + m1=(1−m) ;
t ¿ 0;
(9)
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J.E. Vald es, R.I. Zequeira / Statistics & Probability Letters 63 (2003) 325 – 332
where m = inf t¿0 (t) and m ¿ 0. Then r1 (t) 6 r2 (t);
t ¿ 0:
Proof. The proof is similar to the proof of Proposition 1. In fact, it is the same proof until inequality (8): F< 2 (t) − F< 1 (t) 6 1 − (t);
t ¿ 0:
Since F< 2 (t) 6 (F< 1 (t))m ; to prove (8) it is suDcient to verify that (F< 1 (t))m − F< 1 (t) − (1 − (t)) 6 0;
t ¿ 0:
But denoting x = F< 1 (t) and operating as in the proof of Lemma 2 we obtain that this last inequality is satisCed under condition (9). Then the proposition is proved. The function (m) = 1 − mm=(1−m) + m1=(1−m) − m is represented in Fig. 1. The function (m) attains its maximum at m = 0:31, with (0:31) = 0:28. For m = 0:31 we obtain that if supt¿0 (t) 6 0:59, then r1 (t) 6 r2 (t), t ¿ 0. The following example gives a practical case where Proposition 1 is not applicable but Proposition 2 is applicable. Example 3. Let 1 (t) and 2 (t), t ¿ 0, be deCned as in Example 1. If furthermore cb − ad ¿ 0, then (t) is strictly increasing. In this case m = d=b and supt¿0 (t) = c=a. Let m = 0:3, then (m) = 0:28. Suppose that c=a 6 0:58 and that 1 (t) and 2 (t) are such that 1 (t) 6 min(1 (t); 2 (t)), t ¿ 0. Thus, Proposition 1 is not applicable ((t) is strictly increasing) but by Proposition 2 it is shown that r1 (t) 6 r2 (t).
Fig. 1. Graph of the bound (m) for (t) − m in Proposition 2.
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331
Observe that if we assume that Y1 =st X1 and Y2 =st X2 , the conditions of Propositions 1 and 2 are satisCed only in the trivial case Y1 =st X1 =st X2 =st Y2 . Observe also that in these propositions the condition 1 (t) ¿ 2 (t), t ¿ 0, is assumed. Proposition 3. Let Y1 =st X1 , Y2 =st X2 and X1 =st X2 . Suppose that 1 (t) ¿ 2 (t), t ¿ 0, and 1 (0) ¿ 2 (0). Then there is no failure rate ordering between U1 and U2 . Proof. Let r1 (t) 6 r2 (t), t ¿ 0. It is readily seen from (3) that, for t ¿ 0, this inequality can be written as f2 (t) 1 + F2 (t) ¿ 1 + F1 (t) f1 (t) or, equivalently, f1 (t) + f1 (t)F2 (t) ¿ f2 (t) + f2 (t)F1 (t):
(10)
On the other hand, the inequality 1 (t) ¿ 2 (t) is equivalent to f1 (t) − f1 (t)F2 (t) ¿ f2 (t) − f2 (t)F1 (t):
(11)
Now summing (10) and (11) we obtain the contradictory inequality f1 (t) ¿ f2 (t), t ¿ 0. Let now r1 (t) ¿ r2 (t), t ¿ 0. Then from Lemma 1 we have 1 (0) 6 2 (0). But this contradicts the assumption that 1 (0) ¿ 2 (0). Thus the proof is completed. In the results presented until now it is better regarding the failure rate ordering to allocate the stronger spare in parallel with the weaker component. Nevertheless, as we show in the next proposition, considering the usual stochastic order, in some cases allocating the weaker spare in parallel with the weaker component, can be the best option. Proposition 4. Suppose either of the following conditions (a) or (b) holds: (a) X2 ¿st X1 and Y1 ¿st Y2 ; (b) X2 ¿st X1 ; X2 ¿st Y2 and
Y1 ¿st X1 .
Then U1 ¿st U2 : Proof. It is easily seen that the relation U1 ¿st U2 is equivalent to the inequality F1 (t)F< 2 (t)G< 1 (t) ¿ F2 (t)F< 1 (t)G< 2 (t); t ¿ 0: But this inequality is satisCed under assumption (a) or (b). The result of Proposition 4 may be interpreted as follows: under condition (a) the stronger spare must be allocated with the weakest component of the series system; however, under condition (b), if in particular, X2 ¿st Y2 ¿st Y1 ¿st X1 , the weakest spare must be allocated with the weakest component. Remark 2. From Proposition 4 we obtain that when Y1 =st Y2 , or Y1 =st X1 and Y2 =st X2 , the stochastic order X2 ¿st X1 is a necessary and suDcient condition for the order U1 ¿st U2 to hold.
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Acknowledgements The authors are grateful to the referee for his valuable comments. References Boland, P.J., 1998. A reliability comparison of basic systems using hazard rate functions. Appl. Stochastic Models Data Anal. 13, 377–384. Boland, P.J., El-Neweihi, E., Proschan, F., 1992. Stochastic order for redundancy allocations in series and parallel systems. Adv. Appl. Probab. 24, 161–171. Boland, P.J., El-Neweihi, E., Proschan, F., 1994. Applications of the hazard rate ordering in reliability and order statistics. J. Appl. Probab. 31, 180–192. Ross, S.M., 1983. Stochastic Processes. Wiley, New York. Shaked, M., Shanthikumar, J.G., 1994. Stochastic Orders and their Applications. Academic Press, New York. Singh, H., Misra, N., 1994. On redundancy allocation in systems. J. Appl. Probab. 31, 1004–1014.
On the optimal allocation of an active redundancy in a two-component series system Jos$e E. Vald$esa;∗ , R$omulo I. Zequeirab a
Facultad de Matem atica y Computaci on, Universidad de La Habana, C. San L azaro s/n, CP 10400 La Habana, Cuba b Laboratoire de Mod elisation et Sˆuret e des Syst#emes, Universit e de Technologie de Troyes, 12 rue Marie Curie, BP 2060, 10010 Troyes Cedex, France Received September 2002; received in revised form March 2003
Abstract Let X1 and X2 be independent random variables representing the lifetimes of the components C1 and C2 , and Y1 and Y2 be random variables, independent of X1 and X2 , representing the lifetimes of the spares R1 and R2 , respectively. Suppose that the components C1 and C2 form a series system. There are two options for using one of the spares R1 or R2 as an active redundancy: to allocate R1 in parallel with C1 or R2 in parallel with C2 . In this paper we compare these options using failure rate ordering. c 2003 Elsevier Science B.V. All rights reserved. Keywords: Reliability; Active redundancy; Stochastic orderings
1. Introduction One way of improving the reliability of a system is the allocation of an active redundancy. The problem of where to allocate active redundancies in a system has been studied using a variety of stochastic orderings (see Boland et al., 1992, 1994; Singh and Misra, 1994; Boland, 1998). An extensive treatment of stochastic orders is given in Shaked and Shanthikumar (1994) (see also Ross, 1983). In this work we will assume that all distribution functions G(t) satisfy G(0) = 0 and G(t) ¡ 1, t ¿ 0. We will also suppose that for each distribution function G(t) there exists a probability density < = 1 − G(t). function g(t). We will use the notation G(t) ∗
Corresponding author. E-mail addresses: [email protected] (J.E. Vald$es), [email protected] (R.I. Zequeira).
c 2003 Elsevier Science B.V. All rights reserved. 0167-7152/03/$ - see front matter doi:10.1016/S0167-7152(03)00098-1
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J.E. Vald es, R.I. Zequeira / Statistics & Probability Letters 63 (2003) 325 – 332
Recall that the failure rate associated to a random variable with distribution function G(t) is < deCned by g(t)= G(t). Let X1 , X2 and Y be independent random variables representing the lifetimes of the components C1 , C2 and of the spare R, respectively. Suppose that the components C1 and C2 form a series system and that we can allocate R in parallel with C1 or C2 . Then comparing the lifetimes V1 = min{max(X1 ; Y ); X2 }
and
V2 = min{X1 ; max(X2 ; Y )};
we can decide which of these two options is better. Boland et al. (1992) showed that X2 ¿st X1 is a necessary and suDcient condition for V1 ¿st V2
(1)
to hold. Here the symbol ¿st denotes the stochastic order in the usual classical sense. Let us denote by r1 (t) and r2 (t) the failure rates corresponding to the systems with lifetimes V1 and V2 , respectively. Singh and Misra (1994) proved that if X1 , X2 and Y have exponential distribution with means 1=1 ; 1=2 and 1=, respectively, then 1 ¿ max(2 ; ) is a suDcient condition for the failure (hazard) rate ordering V1 ¿hr V2
(2)
to hold, that is, for the inequality r1 (t) 6 r2 (t);
t ¿ 0;
to hold. The orderings given by Eqs. (1) and (2) can be interpreted in the following way: it is better to allocate the spare component in parallel with the weakest component. Sometimes it is more natural to suppose that the spare to be allocated in parallel depends on the component with which it is to be allocated. Let us deCne the random variables Y1 and Y2 as the lifetimes of the spares R1 and R2 , respectively. Suppose we have two options for using one spare, R1 or R2 , as a componentwise active redundancy. We can allocate R1 in parallel with C1 or, otherwise, allocate R2 with C2 . To Cnd the best option we must compare the lifetimes U1 and U2 deCned as U1 = min{max(X1 ; Y1 ); X2 }
and
U2 = min{X1 ; max(X2 ; Y2 )}:
That is, U1 (U2 ) is the lifetime of the system formed when the spare R1 (R2 ) is allocated as an active redundancy with C1 (C2 ). Observe that the case in which there is only one spare available, R1 = R2 = R, may be considered as a particular case of this, with Y1 = Y2 = Y . Throughout this paper we will assume that the lifetimes X1 and X2 are independent, and that the lifetimes Y1 and Y2 are independent of X1 and X2 .
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327
In this paper we examine the comparison of U1 and U2 in the sense of the failure rate ordering.
2. Optimal allocation of the spare Let us denote the distribution functions of the random variables X1 , X2 , Y1 and Y2 , respectively, by F1 (t), F2 (t), G1 (t) and G2 (t), their corresponding probability densities by f1 (t); f2 (t), g1 (t) and g2 (t), and the respective failure rates by 1 (t), 2 (t), 1 (t) and 2 (t). Now we will Cnd out conditions for the order U1 ¿hr U2 to be satisCed. Let us redeCne r1 (t) and r2 (t) as the failure rates corresponding to the systems with lifetimes U1 and U2 , respectively. The following lemmas will be useful in the proof of Proposition 1. Lemma 1. r1 (0) = 2 (0) and r2 (0) = 1 (0). Proof. Observe that the failure rates r1 (t) and r2 (t) can be expressed as r1 (t) =
f1 (t)G1 (t) + g1 (t)F1 (t) + 2 (t); 1 − F1 (t)G1 (t)
r2 (t) =
f2 (t)G2 (t) + g2 (t)F2 (t) + 1 (t); 1 − F2 (t)G2 (t)
from which obviously follows the statement of the lemma. Lemma 2. sup
06x61; 0¡y61
[xy − x − (1 − y)] 6 0:
Proof. Let us denote h(x) = xy − x − (1 − y): If y = 1 then h(x) = 0. Let now 0 ¡ y ¡ 1. Observe that h(0) = h(1) = y − 1 ¡ 0 and h (x) = yxy−1 − 1;
h (x) = (y − 1)yxy−2 :
On the other hand, h (x) ¡ 0 for 0 ¡ x ¡ 1. Hence from the condition h (x) = 0 we obtain the value x0 = y1=(1−y) where the function h(x) is maximized. But h(x0 ) = yy=(1−y) − y1=(1−y) − (1 − y) = (1 − y)(yy=(1−y) − 1) ¡ 0; hence the lemma is proved. Let us suppose that 1 (t) ¿ 0 for all t ¿ 0. Let us denote (t) = 2 (t)=1 (t), t ¿ 0.
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Proposition 1. Let 2 (t) ¿ 0, t ¿ 0. Suppose that 1 (t) ¿ max(2 (t); 1 (t)), 2 (t) ¿ 1 (t), t ¿ 0, and that (t) is non-increasing. Then r1 (t) 6 r2 (t);
t ¿ 0;
that is, U1 ¿hr U2 . Proof. For t = 0 the result of the proposition is obtained from Lemma 1. Let t ¿ 0. Observe that r1 (t) and r2 (t) can be rewritten as r1 (t) =
1 (t)G1 (t)F< 1 (t) + 1 (t)F1 (t)G< 1 (t) + 2 (t); 1 − F1 (t)G1 (t)
r2 (t) =
2 (t)G2 (t)F< 2 (t) + 2 (t)F2 (t)G< 2 (t) + 1 (t): 1 − F2 (t)G2 (t)
After some algebraic transformations it is not hard to see that the inequality r1 (t) 6 r2 (t) is equivalent to the inequality G< 1 (t) 1 − F2 (t)G2 (t) (1 (t) − 1 (t)F1 (t)) ¿ 2 (t) − 2 (t)F2 (t): G< 2 (t) 1 − F1 (t)G1 (t)
(3)
Observe that inequality G< 1 (t) 1 − F2 (t)G2 (t) ¿1 G< 2 (t) 1 − F1 (t)G1 (t)
(4)
is equivalent to the inequality G2 (t)G< 1 (t)F< 2 (t) ¿ G1 (t)G< 2 (t)F< 1 (t): But G2 (t) ¿ G1 (t) since 2 (t) ¿ 1 (t) and F1 (t) ¿ F2 (t) since 1 (t) ¿ 2 (t), t ¿ 0. Hence inequality (4) holds. On the other hand, 1 (t) ¿ 1 (t) implies 1 (t) − 1 (t)F1 (t) ¿ 0. Therefore (3) holds if the inequality 1 (t) − 1 (t)F1 (t) ¿ 2 (t) − 2 (t)F2 (t)
(5)
is satisCed. Note that inequality (5) can be written as 1 (t) 2 (t) F1 (t) − F2 (t) 6 1 − (t): 1 (t) 1 (t)
(6)
Hence, since 2 (t) ¿ 1 (t), for (6) to hold it is suDcient that 1 (t) < (F 2 (t) − F< 1 (t)) 6 1 − (t): 1 (t)
(7)
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329
Thus, to prove (3) it is suDcient to show that inequality F< 2 (t) − F< 1 (t) 6 1 − (t);
t ¿0
(8)
holds. Observe now that as (t) is non-increasing we obtain t t < F 2 (t) = exp − (x)1 (x) d x 6 exp −(t) 1 (x) d x = (F< 1 (t))(t) : 0
0
Then to prove (8) it is suDcient to verify that (F< 1 (t))(t) − F< 1 (t) − (1 − (t)) 6 0;
t ¿ 0:
< ¡ 1 and 0 ¡ (t) 6 1 for t ¿ 0. Thus the But this inequality follows from Lemma 2, since 0 ¡ F(t) proposition is proved. Remark 1. From (7) it follows that if 2 (t) ¿ 1 (t) and 1 (t) ¿ 2 (t)+1 (t), t ¿ 0, then U1 ¿hr U2 . The suDcient conditions of Proposition 1 require the spare to be allocated as a redundancy be stronger (in the failure rate order) than the other spares, and also stronger than the weakest component of the series system. However, with respect to the components of the system, together with the failure rate ordering of these components, it is required that the function (t) be non-increasing. The following examples show that this condition is satisCed in some cases of practical interest. Example 1. Let 1 (t) = at + b and 2 (t) = ct + d, t ¿ 0, with a ¿ c ¿ 0, b ¿ d ¿ 0. Evidently, 2 (t) 6 1 (t). If furthermore cb − ad 6 0, then (t) is non-increasing. Example 2. One special case in which the function (t) is non-increasing is when the failure rates 1 (t) and 2 (t) are proportional, that is, 2 (t) = c1 (t) for all t ¿ 0, where c is a constant. This is the case, for example, of two Weibull distributions with the same shape parameter. If 0 ¡ c 6 1, then 2 (t) 6 1 (t). The proof of the following corollary is straightforward and is omitted. Corollary 1. Let Y1 =st Y2 =st Y , and Y =st X1 or Y =st X2 . Suppose 1 (t) ¿ 0 and 2 (t) = c1 (t), t ¿ 0, where c ¿ 0. Then r1 (t) 6 r2 (t)
if ; and only if
1 (t) ¿ 2 (t);
t ¿ 0;
that is, U1 ¿hr U2 if, and only if X1 6hr X2 . The non-increasing condition of the function (t) in Proposition 1 can be substituted as is shown in the following proposition. Proposition 2. Let 2 (t) ¿ 0, t ¿ 0. Suppose that 1 (t) ¿ max(2 (t); 1 (t)), 2 (t) ¿ 1 (t), t ¿ 0, and that (t) 6 1 − mm=(1−m) + m1=(1−m) ;
t ¿ 0;
(9)
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where m = inf t¿0 (t) and m ¿ 0. Then r1 (t) 6 r2 (t);
t ¿ 0:
Proof. The proof is similar to the proof of Proposition 1. In fact, it is the same proof until inequality (8): F< 2 (t) − F< 1 (t) 6 1 − (t);
t ¿ 0:
Since F< 2 (t) 6 (F< 1 (t))m ; to prove (8) it is suDcient to verify that (F< 1 (t))m − F< 1 (t) − (1 − (t)) 6 0;
t ¿ 0:
But denoting x = F< 1 (t) and operating as in the proof of Lemma 2 we obtain that this last inequality is satisCed under condition (9). Then the proposition is proved. The function (m) = 1 − mm=(1−m) + m1=(1−m) − m is represented in Fig. 1. The function (m) attains its maximum at m = 0:31, with (0:31) = 0:28. For m = 0:31 we obtain that if supt¿0 (t) 6 0:59, then r1 (t) 6 r2 (t), t ¿ 0. The following example gives a practical case where Proposition 1 is not applicable but Proposition 2 is applicable. Example 3. Let 1 (t) and 2 (t), t ¿ 0, be deCned as in Example 1. If furthermore cb − ad ¿ 0, then (t) is strictly increasing. In this case m = d=b and supt¿0 (t) = c=a. Let m = 0:3, then (m) = 0:28. Suppose that c=a 6 0:58 and that 1 (t) and 2 (t) are such that 1 (t) 6 min(1 (t); 2 (t)), t ¿ 0. Thus, Proposition 1 is not applicable ((t) is strictly increasing) but by Proposition 2 it is shown that r1 (t) 6 r2 (t).
Fig. 1. Graph of the bound (m) for (t) − m in Proposition 2.
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Observe that if we assume that Y1 =st X1 and Y2 =st X2 , the conditions of Propositions 1 and 2 are satisCed only in the trivial case Y1 =st X1 =st X2 =st Y2 . Observe also that in these propositions the condition 1 (t) ¿ 2 (t), t ¿ 0, is assumed. Proposition 3. Let Y1 =st X1 , Y2 =st X2 and X1 =st X2 . Suppose that 1 (t) ¿ 2 (t), t ¿ 0, and 1 (0) ¿ 2 (0). Then there is no failure rate ordering between U1 and U2 . Proof. Let r1 (t) 6 r2 (t), t ¿ 0. It is readily seen from (3) that, for t ¿ 0, this inequality can be written as f2 (t) 1 + F2 (t) ¿ 1 + F1 (t) f1 (t) or, equivalently, f1 (t) + f1 (t)F2 (t) ¿ f2 (t) + f2 (t)F1 (t):
(10)
On the other hand, the inequality 1 (t) ¿ 2 (t) is equivalent to f1 (t) − f1 (t)F2 (t) ¿ f2 (t) − f2 (t)F1 (t):
(11)
Now summing (10) and (11) we obtain the contradictory inequality f1 (t) ¿ f2 (t), t ¿ 0. Let now r1 (t) ¿ r2 (t), t ¿ 0. Then from Lemma 1 we have 1 (0) 6 2 (0). But this contradicts the assumption that 1 (0) ¿ 2 (0). Thus the proof is completed. In the results presented until now it is better regarding the failure rate ordering to allocate the stronger spare in parallel with the weaker component. Nevertheless, as we show in the next proposition, considering the usual stochastic order, in some cases allocating the weaker spare in parallel with the weaker component, can be the best option. Proposition 4. Suppose either of the following conditions (a) or (b) holds: (a) X2 ¿st X1 and Y1 ¿st Y2 ; (b) X2 ¿st X1 ; X2 ¿st Y2 and
Y1 ¿st X1 .
Then U1 ¿st U2 : Proof. It is easily seen that the relation U1 ¿st U2 is equivalent to the inequality F1 (t)F< 2 (t)G< 1 (t) ¿ F2 (t)F< 1 (t)G< 2 (t); t ¿ 0: But this inequality is satisCed under assumption (a) or (b). The result of Proposition 4 may be interpreted as follows: under condition (a) the stronger spare must be allocated with the weakest component of the series system; however, under condition (b), if in particular, X2 ¿st Y2 ¿st Y1 ¿st X1 , the weakest spare must be allocated with the weakest component. Remark 2. From Proposition 4 we obtain that when Y1 =st Y2 , or Y1 =st X1 and Y2 =st X2 , the stochastic order X2 ¿st X1 is a necessary and suDcient condition for the order U1 ¿st U2 to hold.
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Acknowledgements The authors are grateful to the referee for his valuable comments. References Boland, P.J., 1998. A reliability comparison of basic systems using hazard rate functions. Appl. Stochastic Models Data Anal. 13, 377–384. Boland, P.J., El-Neweihi, E., Proschan, F., 1992. Stochastic order for redundancy allocations in series and parallel systems. Adv. Appl. Probab. 24, 161–171. Boland, P.J., El-Neweihi, E., Proschan, F., 1994. Applications of the hazard rate ordering in reliability and order statistics. J. Appl. Probab. 31, 180–192. Ross, S.M., 1983. Stochastic Processes. Wiley, New York. Shaked, M., Shanthikumar, J.G., 1994. Stochastic Orders and their Applications. Academic Press, New York. Singh, H., Misra, N., 1994. On redundancy allocation in systems. J. Appl. Probab. 31, 1004–1014.