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Some stochastic comparisons in series systems with active redundancy
Statistics and Probability Letters 80 (2010) 945–949
Contents lists available at ScienceDirect
Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
Some stochastic comparisons in series systems with active redundancy José E. Valdés a,∗ , Gerardo Arango b , Romulo I. Zequeira c , Gerandy Brito a a
Facultad de Matemática y Computación, Universidad de La Habana, San Lázaro y L, 10400, La Habana, Cuba
b
Departamento de Ciencias Básicas, Universidad EAFIT, Carrera 49 N 7 Sur - 50, Medellín, Colombia
c
Technology and Operations Management Laboratory, College of Management of Technology, EPFL, Lausanne, Switzerland
article
info
Article history: Received 23 January 2009 Received in revised form 9 February 2010 Accepted 10 February 2010 Available online 17 February 2010
abstract We compare the lifetimes of series systems with different allocations of active redundancy using a variety of stochastic comparisons. It is assumed that only one spare can be allocated to each component of the system. © 2010 Elsevier B.V. All rights reserved.
Keywords: Reliability Series systems Active redundancy Stochastic comparisons
1. Introduction One way to increase the reliability of a system is the use of redundant components. In general, there are three kinds of redundancies, i.e. active (hot), passive (cold) and warm redundancy. Active redundancy means that the redundant component undergoes the regular stress level, that is, the same stress level as when it is the principal component. In a system with active redundancy the principal component and the redundant one form a parallel system. In the state of passive redundancy the redundant component has zero failure rate and then it cannot fail while it remains in this state. Warm redundancy is an intermediate case. The component in a warm state operates under a milder stress level than when it is the principal component and at the failure of the principal component it will immediately operate under the regular stress level. Recently Cha et al. (2008) proposed a general standby system that includes the cases of cold, hot and warm standby. In this paper we consider series systems with active redundancies. We study the problem of where to allocate the redundancies in order to maximize, in different senses of stochastic comparison, the lifetime of the systems. The problem of where to allocate redundancies in a system to obtain optimal configurations has been studied using a variety of stochastic comparisons. There is a great amount of work on the study of redundancy allocation using stochastic orders. Some remarkable works are Boland et al. (1988, 1989, 1994), Meng (1996) and Singh and Singh (1997). Extensive references on stochastic orders are Shaked and Shanthikumar (2007) and Müller and Stoyan (2002). The former of these books summarizes many works on redundancy allocation and contains a large list of references therein. Stochastic comparisons may be also made through what in Boland et al. (2004) is called the stochastic precedence order. Some other works where this type of stochastic comparison is used are Blyth (1972), Li and Hu (2008), Romera et al. (2004) and Singh and Misra (1994).
∗
Corresponding author. E-mail addresses: [email protected] (J.E. Valdés), [email protected] (G. Arango), [email protected] (R.I. Zequeira), [email protected] (G. Brito). 0167-7152/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2010.02.005
946
J.E. Valdés et al. / Statistics and Probability Letters 80 (2010) 945–949
We interpret all random variables as lifetimes of components or systems; therefore we will consider nonnegative random variables. We will assume that all distribution functions B(t ) satisfy B(0) = 0. We use the notation B¯ ≡ 1 − B. The maximum and minimum will be denoted as ∨ and ∧, respectively. The terms increasing and decreasing will be used in the non-strict sense. Let X and Y be random variables with corresponding distribution functions F (t ) and G(t ). If X and Y are absolutely continuous, let us denote by λ(t ) and µ(t ) their respective hazard rate functions and by r (t ) and q(t ) their corresponding reversed hazard rate functions. The following definitions of stochastic comparisons between two random variables will be needed (a/0 will be taken equal to ∞ whenever a > 0). X is said to be smaller than Y in the:
¯ (t ) for all real t. We will write X =st Y if F (t ) = G(t ) for all t. 1. Usual stochastic order (denoted as X ≤st Y ) if F¯ (t ) ≤ G ¯ (t )/F¯ (t ) increases in t ≥ 0. If X and Y are absolutely continuous, then X ≤hr Y 2. Hazard rate order (denoted as X ≤hr Y ) if G is equivalent to λ(t ) ≥ µ(t ) for all t ≥ 0. 3. Reversed hazard rate order (denoted as X ≤rh Y ) if G(t )/F (t ) increases in t > 0. If X and Y are absolutely continuous, then X ≤rh Y is equivalent to r (t ) ≤ q(t ) for all t ≥ 0. Rt Rt ¯ (x)dx for all t ≥ 0. 4. Increasing concave order (denoted as X ≤ic v Y ) if 0 F¯ (x)dx ≤ 0 G 5. Probabilistic relation (denoted as X ≤pr Y ) if P (X > Y ) ≤ P (Y > X ). The relation ≤pr in Boland et al. (2004) is called the stochastic precedence order. This relation is not a partial order since it does not meet the transitive property requirement (see Blyth (1972)). Unlike the stochastic orders that we consider in this paper, which only depend on the marginal distributions of X and Y , the probabilistic relation depends on the joint distribution of these random variables. Therefore, the relation ≤pr may be of special interest when comparing X and Y because it does take into account the possible dependence between these random variables. The relation X ≤st Y in general does not imply X ≤pr Y ; nevertheless if X and Y are independent this implication holds (see Boland et al. (2004)). X ≤hr Y and also X ≤rh Y imply X ≤st Y and this in turn implies X ≤ic v Y . Further details about these stochastic orders may be found in Shaked and Shanthikumar (2007). Throughout the paper we implicitly assume that the lifetimes X1 , X2 , . . . , Xn , Y1 , Y2 , . . . , Yn are independent. We denote the distribution function of the lifetime Xi (Yi ) by Fi (t ) (Gi (t )). If Xi (Yi ) is absolutely continuous then we will denote its probability density function by fi (t ) (gi (t )), i = 1, 2, . . . , n. In Section 2 of this paper we analyze the allocation of one spare in a series system. In Section 3 we study the allocation of more than one spare. 2. Allocation of one spare Consider a series system with n components. Suppose that there are two spares R1 and R2 to be allocated as active redundancies to the components C1 and C2 , respectively, but only one of the spares can be allocated. Let X1 , X2 , . . . , Xn be the lifetimes of components. Let X1 and X2 be the lifetimes of components C1 and C2 and Y1 and Y2 be the lifetimes of spares R1 and R2 , respectively. Let Z = ∧(X3 , X4 , . . . , Xn ) and denote by H (t ) the distribution function of Z . Then U1 = ∧ [∨(X1 , Y1 ), X2 , Z ]
and
U2 = ∧ [X1 , ∨(X2 , Y2 ), Z ]
represent the lifetimes of the two possible configurations of the system. Proposition 1. Suppose that either of the following conditions holds: (a) X1 ≤ic v X2 and Y1 ≥st Y2 , (b) X1 ≤ic v X2 , Y2 ≤st X2 and X1 ≤st Y1 . Then U1 ≥ic v U2 . Proof. Note that
¯ (t ) P (U1 > t ) = F¯2 (t ) − F1 (t )G1 (t ) + G1 (t )F1 (t )F2 (t ) H
and
¯ (t ). P (U2 > t ) = F¯1 (t ) − F2 (t )G2 (t ) + G2 (t )F1 (t )F2 (t ) H
Thus,
¯ 1 (t ) − F2 (t )F¯1 (t )G¯ 2 (t )]H¯ (t ). P (U1 > t ) − P (U2 > t ) = [F1 (t )F¯2 (t )G Suppose that condition (a) holds. Using that Y1 ≥st Y2 we obtain
¯ 2 (t )H¯ (t ). P (U1 > t ) − P (U2 > t ) ≥ (F¯2 (t ) − F¯1 (t ))G
J.E. Valdés et al. / Statistics and Probability Letters 80 (2010) 945–949
947
Because X1 ≤ic v X2 , then from part (b) of Lemma 7.1, p. 120, in Barlow and Proschan (1981) we deduce t
Z
(F¯2 (x) − F¯1 (x))G¯ 2 (x)H¯ (x)dx ≥ 0 0
and the first part of the proposition is proved. Suppose now that condition (b) holds. Then using Y2 ≤st X2 and X1 ≤st Y1 we get
¯ 2 (t )F¯1 (t )H¯ (t ). P (U1 > t ) − P (U2 > t ) ≥ (F¯2 (t ) − F¯1 (t ))G As X1 ≤ic v X2 , then the second part of the proposition follows using part (b) of Lemma 7.1 in Barlow and Proschan (1981) again. It is worth mentioning that if Y1 =st Y2 , then Theorem 1 in Li and Hu (2008) follows from part (a) of Proposition 1. Under condition (a), Proposition 1 means that in order to maximize the lifetime of the series system in the sense of the increasing concave ordering, the strongest spare must be allocated with the weakest component of the series system. No ordering between Y1 and Y2 is required in condition (b) of Proposition 1. Nevertheless, if the condition Y1 ≤ic v Y2 is added (i.e. X1 ≤st Y1 ≤ic v Y2 ≤st X2 ) then the lifetime of the system is maximized when the weakest spare is allocated with the weakest component. If in particular X1 ≤st Y1 ≤st Y2 ≤st X2 , then U1 ≥st U2 as was noted in Valdés and Zequeira (2003). ¯ 1 (x)F¯2 (x) ≥ G¯ 2 (x)F¯1 (x) then U1 ≥pr U2 . Therefore the probabilistic Romera et al. (2004) proved that if X1 ≤st X2 and G relation U1 ≥pr U2 holds if X1 ≤st Y1 ≤st Y2 ≤st X2 . Note that in this case also the weakest spare must be allocated with the weakest component. The following example shows that the condition X1 ≤st X2 is not necessary for U1 ≥pr U2 to hold. Example 1. Consider a two-component series system composed of the components C1 and C2 and the spares R1 and R2 . In this case U1 = ∧[∨(X1 , Y1 ), X2 ] and U2 = ∧[X1 , ∨(X2 , Y2 )]. Let X1 , X2 , Y1 and Y2 be exponentially distributed with means 1/λ1 , 1/λ2 , 1/µ1 and 1/µ2 , respectively. Then P (U1 > U2 ) − P (U2 > U1 ) =
λ1 (λ1 + λ2 + µ2 ) − λ2 (λ1 + λ2 + µ1 ) . (λ1 + λ2 + µ1 )(λ1 + λ2 + µ2 )
Observe that given values of λ1 and λ2 such that λ1 < λ2 (X1 ≥st X2 ) and a value of µ1 , we can select a value of µ2 such that µ2 > µ1 (Y1 ≥st Y2 ) and P (U1 > U2 ) − P (U2 > U1 ) > 0. 3. Allocation of more than one spare We will assume in this section that the lifetimes X1 , X2 , Y1 and Y2 are absolutely continuous. As in the preceding section, we consider a series system with n components and we suppose that there are two spares R1 and R2 to be allocated as active redundancies to the components C1 and C2 , respectively. Let X1 , X2 , . . . , Xn be the lifetimes of the components of the system. Let X1 and X2 be the lifetimes of the components C1 and C2 and Y1 and Y2 be the lifetimes of the spares R1 and R2 , respectively. We consider now that there are two options: allocate R1 with C1 and R2 with C2 , or otherwise, allocate R1 with C2 and R2 with C1 . As in Section 2 we define Z = ∧(X3 , X4 , . . . , Xn ) and denote by H (t ) the distribution function of this random variable. Then V1 = ∧ [∨(X1 , Y1 ), ∨(X2 , Y2 ), Z ]
and
V2 = ∧ [∨(X1 , Y2 ), ∨(X2 , Y1 ), Z ]
represent the lifetimes of the two possible configurations of the system. The following proposition is proved in Romera et al. (2004). Proposition 2. If X1 ≤hr X2 and Y1 ≥hr Y2 , then V1 ≥pr V2 . The hazard rate orderings of Proposition 2 can be substituted by the reversed hazard rate ordering. We will prove this fact in Proposition 3. In the proof of this proposition we use a result taken from Shaked and Shanthikumar (2007), p. 37, which we state next as Lemma 1. Lemma 1. Let X and Y be random variables with distribution functions F (t ) and G(t ), and probability density functions f (t ) and g (t ), respectively. Then X ≤rh Y if and only if f (y) F ( x)
≤
g (y) G(x)
,
for all x ≤ y.
Proposition 3. If X1 ≤rh X2 and Y1 ≥rh Y2 , then V1 ≥pr V2 . Proof. Romera et al. (2004) show that the inequality V1 > V2 is satisfied if and only if one of the two inequalities ∨(X1 , Y2 , Z ) < ∧(X2 , Y1 ) or ∨(X2 , Y1 , Z ) < ∧(X1 , Y2 ) holds. An analogous equivalence may be obtained for the inequality
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V1 < V2 . Then to prove the proposition it is sufficient to show that
∆ = P (∧(X2 , Y1 ) > ∨(X1 , Y2 , Z )) + P (∧(X1 , Y2 ) > ∨(X2 , Y1 , Z )) − P (∧(X2 , Y2 ) > ∨(X1 , Y1 , Z )) − P (∧(X1 , Y1 ) > ∨(X2 , Y2 , Z )) ≥ 0. But
∆=
∞
Z
∞
Z
F1 (∧(x, y))G2 (∧(x, y))H (∧(x, y))dG1 (x)dF2 (y) 0
0
∞
Z
∞
Z
F2 (∧(x, y))G1 (∧(x, y))H (∧(x, y))dG2 (x)dF1 (y)
+ Z0 ∞ Z0 ∞ − Z0 ∞ Z0 ∞ − 0
F1 (∧(x, y))G1 (∧(x, y))H (∧(x, y))dG2 (x)dF2 (y) F2 (∧(x, y))G2 (∧(x, y))H (∧(x, y))dG1 (x)dF1 (y).
0
Note that ∆ ≥ 0 if the inequality F1 (u)G2 (u)g1 (x)f2 (y) + F2 (u)G1 (u)g2 (x)f1 (y) − F1 (u)G1 (u)g2 (x)f2 (y) − F2 (u)G2 (u)g1 (x)f1 (y) ≥ 0 holds for all x, y ≥ 0, where u = ∧(x, y). But the last inequality can be written as [g1 (x)G2 (u) − g2 (x)G1 (u)] [f2 (y)F1 (u) − f1 (y)F2 (u)] ≥ 0. Then using Lemma 1 and the assumptions, the proof is completed.
Example 2. Consider a two-component series system with the components C1 and C2 and spares R1 and R2 . In this case V1 = ∧[∨(X1 , Y1 ), ∨(X2 , Y2 )] and V2 = ∧[∨(X1 , Y2 ), ∨(X2 , Y1 )]. If X1 , X2 , Y1 , Y2 are exponentially distributed with means 1/λ1 , 1/λ2 , 1/µ1 and 1/µ2 , respectively, then P (V1 > V2 ) =
λ 2 µ1 λ 1 µ2 + . (λ1 + λ2 + µ1 )(λ2 + µ1 + µ2 ) (λ1 + λ2 + µ2 )(λ1 + µ1 + µ2 )
An analogous formula can be obtained for P (V1 < V2 ). After some transformations we obtain that P (V1 > V2 ) − P (V1 < V2 ) ≥ 0 if and only if
(λ1 − λ2 )(µ2 − µ1 )(λ1 + λ2 + µ1 + µ2 )2 ≥ 0. Therefore, if λ1 ≥ λ2 (X1 ≤hr X2 and X1 ≤rh X2 ) and µ2 ≥ µ1 (Y1 ≥hr Y2 and Y1 ≥rh Y2 ) then V1 ≥pr V2 . This result agrees with Propositions 2 and 3. Moreover, if V1 ≥pr V2 then X1 ≤hr X2 (X1 ≤rh X2 ) and Y1 ≥hr Y2 (Y1 ≥rh Y2 ), or X1 ≥hr X2 (X1 ≥rh X2 ) and Y1 ≤hr Y2 (Y1 ≤rh Y2 ). Proposition 4. Suppose that Yi =st Xi , i = 1, 2. If X1 ≤rh X2 or X2 ≤rh X1 , then V1 ≤pr V2 . Proof. We will use the same notation as in the proof of Proposition 3. A sufficient condition for ∆ ≤ 0 to hold is that the inequality [g1 (x)G2 (u) − g2 (x)G1 (u)] [f2 (y)F1 (u) − f1 (y)F2 (u)] ≤ 0 is satisfied for all x, y ≥ 0. This inequality can be written as
− [f1 (x)F2 (u) − f2 (x)F1 (u)] [f1 (y)F2 (u) − f2 (y)F1 (u)] ≤ 0 and then using Lemma 1 the proposition is proved.
Remark 1. The relation V1 ≤pr V2 also holds if in Proposition 4 the order ≤rh is substituted by the order ≤hr . Remark 2. Observe that the condition X1 ≤rh X2 (X2 ≤rh X1 ) of Proposition 4 holds when the hazard rate functions or the reversed hazard rate functions of X1 and X2 are proportional. The results of Propositions 2–4 mean in practice that, in order to maximize the lifetime of the series system in the sense of the probabilistic relation, the stronger spare must be allocated with the weakest component of the system. Suppose now that there are n spares R1 , . . . , Rn that may be allocated as active redundancies to the components C1 , . . . , Cn . Let X1 , X2 , . . . , Xn and Y1 , Y2 , . . . , Yn be the corresponding lifetimes of the components and the spares, respectively. Consider a series system constituted by the components C1 , . . . , Cn . We will assume that only one spare can be allocated to each component of the systems. It is also assumed that only r, 1 ≤ r < n, of the spares can be allocated. The following result generalizes part (a) of Proposition 1.
J.E. Valdés et al. / Statistics and Probability Letters 80 (2010) 945–949
949
Proposition 5. Suppose that X1 ≤ic v X2 ≤ic v · · · ≤ic v Xn and Y1 ≥st Y2 ≥st · · · ≥st Yn . Then
" r ^ ^
∨(Xi , Yi ),
i=1
n ^ i=r +1
# Xi ≥ic v
" r ^ ^
# ∨(Xsi , Ysi ),
i=1
^
Xj ,
j≤n,j6=si ,i=1,...,r
where r < n and si , i = 1, . . . , r, are not equal integers such that si ≤ n. Proof. We will construct the proof for the case n = 3, r = 2. The general proof can be obtained using mathematical induction, where the inductive step is proved using basically the same argument. From Proposition 1 we have
∧[X1 , ∨(X2 , Y2 ), ∨(X3 , Y3 )] ≤ic v ∧[∨(X1 , Y1 ), X2 , ∨(X3 , Y3 )] ≤ic v ∧[∨(X1 , Y1 ), ∨(X2 , Y2 ), X3 ]. Boland et al. (1988) proved an analogous result when in Proposition 5 the order ≤ic v is replaced by the order ≤st . Moreover, they proved a similar statement for k-out-of-n systems using the order ≤st . The example below shows that for coherent systems an analogous result is in general not true even for very simple systems. Example 3. Consider a three-component system. The components C1 and C2 form a series subsystem which in turn is in parallel with the component C3 . The spares R1 , R2 and R3 must be allocated one to one as active redundancies with the components of the system. Assume that Y1 =st X3 , Y2 =st X2 and Y3 =st X1 and that X1 ≤st X2 ≤st X3 (then Y1 ≥st Y2 ≥st Y3 ). We will consider two system with redundancies. In the first system the spare Ri is allocated with the component Ci , i = 1, 2, 3. In the second system the spares are allocated as follows: R2 with C1 , R3 with C2 and R1 with C3 . The lifetimes for the first and second systems are U123 = ∨ [∧(X1 ∨ Y1 , X2 ∨ Y2 ), X3 ∨ Y3 ] and U231 = ∨ [∧(X1 ∨ Y2 , X2 ∨ Y3 ), X3 ∨ Y1 ] , respectively. Thus, P (U123 > t ) = [1 − F1 (t )F3 (t )][1 − F22 (t )] + [1 − F1 (t )F3 (t )] − [1 − F1 (t )F3 (t )]2 [1 − F22 (t )]
(1)
P (U231 > t ) = [1 − F1 (t )F2 (t )]2 + [1 − F32 (t )] − [1 − F1 (t )F2 (t )]2 [1 − F32 (t )].
(2)
and
Now let us take for some t = t0 , F1 (t0 ) = 0.6, F2 (t0 ) = 0.5 and F3 (t0 ) = 0.4. Substituting these values for t = t0 in (1) and (2) we obtain P (U123 > t0 ) = 0.8968 < P (U231 > t0 ) = 0.9184. Acknowledgements The work of G. Arango was supported by Universidad EAFIT, Colombia, under project n-45-000031 of 2008. The work of J.E. Valdés was partially supported by the same university and project. The authors are very grateful to the referees for their valuable comments which helped to improve the presentation of this paper. References Barlow, R.E., Proschan, F., 1981. Statistical Theory of Reliability and Life Testing. McArdle Press, Inc. Blyth, C.R., 1972. Some probability paradoxes in choice from among random alternatives. Journal of the American Statistical Association 67, 366–373. Boland, P.J., El-Neweihi, E., Proshan, F., 1988. Active redundancy allocation in coherent systems. Probability in the Engineering and Informational Sciences 2, 343–353. Boland, P.J., El-Neweihi, E., Proshan, F., 1994. Applications of the hazard rate orderings in reliability and orders statistics. Journal of Applied Probability 31, 180–192. Boland, P.J., Proshan, F., Tong, Y.L., 1989. Optimal arrangement of components via pairwise rearrangements. Naval Research Logistics 36, 807–815. Boland, P.J., Singh, H., Cukic, B., 2004. The stochastic precedence ordering with applications in sampling an testing. Journal of Applied Probability 41, 73–82. Cha, J.H., Mi, J., Yun, W.Y., 2008. Modelling a general standby system and evaluation of its performance. Applied Stochastic Models in Business and Industry 24, 159–169. Li, X., Hu, X., 2008. Some new stochastic comparisons for redundancy allocations in series and parallel systems. Statistics and Probability Letters 78 (18), 3388–3394. Meng, F.C., 1996. More on optimal allocation of components in coherent systems. Journal of Applied Probability 33, 548–556. Müller, A., Stoyan, D., 2002. Comparison Methods for Stochastic Models and Risks. Wiley. Romera, R., Valdés, J.E., Zequeira, R.I., 2004. Active-redundancy allocation in systems. IEEE Transactions on Reliability 53, 313–318. Shaked, M., Shanthikumar, J.G., 2007. Stochastic Orders. Springer, New York. Singh, H., Misra, N., 1994. On redundancy allocation in systems. Journal of Applied Probability 31, 1004–1014. Singh, H., Singh, R.S., 1997. Optimal allocation of resources to nodes of series systems with respect to failure-rate ordering. Naval Research Logistics 44, 147–152. Valdés, J.E., Zequeira, R.I., 2003. On the optimal allocation of an active redundancy in a two-component series system. Statistics and Probability Letters 63, 235–332.
Contents lists available at ScienceDirect
Statistics and Probability Letters journal homepage: www.elsevier.com/locate/stapro
Some stochastic comparisons in series systems with active redundancy José E. Valdés a,∗ , Gerardo Arango b , Romulo I. Zequeira c , Gerandy Brito a a
Facultad de Matemática y Computación, Universidad de La Habana, San Lázaro y L, 10400, La Habana, Cuba
b
Departamento de Ciencias Básicas, Universidad EAFIT, Carrera 49 N 7 Sur - 50, Medellín, Colombia
c
Technology and Operations Management Laboratory, College of Management of Technology, EPFL, Lausanne, Switzerland
article
info
Article history: Received 23 January 2009 Received in revised form 9 February 2010 Accepted 10 February 2010 Available online 17 February 2010
abstract We compare the lifetimes of series systems with different allocations of active redundancy using a variety of stochastic comparisons. It is assumed that only one spare can be allocated to each component of the system. © 2010 Elsevier B.V. All rights reserved.
Keywords: Reliability Series systems Active redundancy Stochastic comparisons
1. Introduction One way to increase the reliability of a system is the use of redundant components. In general, there are three kinds of redundancies, i.e. active (hot), passive (cold) and warm redundancy. Active redundancy means that the redundant component undergoes the regular stress level, that is, the same stress level as when it is the principal component. In a system with active redundancy the principal component and the redundant one form a parallel system. In the state of passive redundancy the redundant component has zero failure rate and then it cannot fail while it remains in this state. Warm redundancy is an intermediate case. The component in a warm state operates under a milder stress level than when it is the principal component and at the failure of the principal component it will immediately operate under the regular stress level. Recently Cha et al. (2008) proposed a general standby system that includes the cases of cold, hot and warm standby. In this paper we consider series systems with active redundancies. We study the problem of where to allocate the redundancies in order to maximize, in different senses of stochastic comparison, the lifetime of the systems. The problem of where to allocate redundancies in a system to obtain optimal configurations has been studied using a variety of stochastic comparisons. There is a great amount of work on the study of redundancy allocation using stochastic orders. Some remarkable works are Boland et al. (1988, 1989, 1994), Meng (1996) and Singh and Singh (1997). Extensive references on stochastic orders are Shaked and Shanthikumar (2007) and Müller and Stoyan (2002). The former of these books summarizes many works on redundancy allocation and contains a large list of references therein. Stochastic comparisons may be also made through what in Boland et al. (2004) is called the stochastic precedence order. Some other works where this type of stochastic comparison is used are Blyth (1972), Li and Hu (2008), Romera et al. (2004) and Singh and Misra (1994).
∗
Corresponding author. E-mail addresses: [email protected] (J.E. Valdés), [email protected] (G. Arango), [email protected] (R.I. Zequeira), [email protected] (G. Brito). 0167-7152/$ – see front matter © 2010 Elsevier B.V. All rights reserved. doi:10.1016/j.spl.2010.02.005
946
J.E. Valdés et al. / Statistics and Probability Letters 80 (2010) 945–949
We interpret all random variables as lifetimes of components or systems; therefore we will consider nonnegative random variables. We will assume that all distribution functions B(t ) satisfy B(0) = 0. We use the notation B¯ ≡ 1 − B. The maximum and minimum will be denoted as ∨ and ∧, respectively. The terms increasing and decreasing will be used in the non-strict sense. Let X and Y be random variables with corresponding distribution functions F (t ) and G(t ). If X and Y are absolutely continuous, let us denote by λ(t ) and µ(t ) their respective hazard rate functions and by r (t ) and q(t ) their corresponding reversed hazard rate functions. The following definitions of stochastic comparisons between two random variables will be needed (a/0 will be taken equal to ∞ whenever a > 0). X is said to be smaller than Y in the:
¯ (t ) for all real t. We will write X =st Y if F (t ) = G(t ) for all t. 1. Usual stochastic order (denoted as X ≤st Y ) if F¯ (t ) ≤ G ¯ (t )/F¯ (t ) increases in t ≥ 0. If X and Y are absolutely continuous, then X ≤hr Y 2. Hazard rate order (denoted as X ≤hr Y ) if G is equivalent to λ(t ) ≥ µ(t ) for all t ≥ 0. 3. Reversed hazard rate order (denoted as X ≤rh Y ) if G(t )/F (t ) increases in t > 0. If X and Y are absolutely continuous, then X ≤rh Y is equivalent to r (t ) ≤ q(t ) for all t ≥ 0. Rt Rt ¯ (x)dx for all t ≥ 0. 4. Increasing concave order (denoted as X ≤ic v Y ) if 0 F¯ (x)dx ≤ 0 G 5. Probabilistic relation (denoted as X ≤pr Y ) if P (X > Y ) ≤ P (Y > X ). The relation ≤pr in Boland et al. (2004) is called the stochastic precedence order. This relation is not a partial order since it does not meet the transitive property requirement (see Blyth (1972)). Unlike the stochastic orders that we consider in this paper, which only depend on the marginal distributions of X and Y , the probabilistic relation depends on the joint distribution of these random variables. Therefore, the relation ≤pr may be of special interest when comparing X and Y because it does take into account the possible dependence between these random variables. The relation X ≤st Y in general does not imply X ≤pr Y ; nevertheless if X and Y are independent this implication holds (see Boland et al. (2004)). X ≤hr Y and also X ≤rh Y imply X ≤st Y and this in turn implies X ≤ic v Y . Further details about these stochastic orders may be found in Shaked and Shanthikumar (2007). Throughout the paper we implicitly assume that the lifetimes X1 , X2 , . . . , Xn , Y1 , Y2 , . . . , Yn are independent. We denote the distribution function of the lifetime Xi (Yi ) by Fi (t ) (Gi (t )). If Xi (Yi ) is absolutely continuous then we will denote its probability density function by fi (t ) (gi (t )), i = 1, 2, . . . , n. In Section 2 of this paper we analyze the allocation of one spare in a series system. In Section 3 we study the allocation of more than one spare. 2. Allocation of one spare Consider a series system with n components. Suppose that there are two spares R1 and R2 to be allocated as active redundancies to the components C1 and C2 , respectively, but only one of the spares can be allocated. Let X1 , X2 , . . . , Xn be the lifetimes of components. Let X1 and X2 be the lifetimes of components C1 and C2 and Y1 and Y2 be the lifetimes of spares R1 and R2 , respectively. Let Z = ∧(X3 , X4 , . . . , Xn ) and denote by H (t ) the distribution function of Z . Then U1 = ∧ [∨(X1 , Y1 ), X2 , Z ]
and
U2 = ∧ [X1 , ∨(X2 , Y2 ), Z ]
represent the lifetimes of the two possible configurations of the system. Proposition 1. Suppose that either of the following conditions holds: (a) X1 ≤ic v X2 and Y1 ≥st Y2 , (b) X1 ≤ic v X2 , Y2 ≤st X2 and X1 ≤st Y1 . Then U1 ≥ic v U2 . Proof. Note that
¯ (t ) P (U1 > t ) = F¯2 (t ) − F1 (t )G1 (t ) + G1 (t )F1 (t )F2 (t ) H
and
¯ (t ). P (U2 > t ) = F¯1 (t ) − F2 (t )G2 (t ) + G2 (t )F1 (t )F2 (t ) H
Thus,
¯ 1 (t ) − F2 (t )F¯1 (t )G¯ 2 (t )]H¯ (t ). P (U1 > t ) − P (U2 > t ) = [F1 (t )F¯2 (t )G Suppose that condition (a) holds. Using that Y1 ≥st Y2 we obtain
¯ 2 (t )H¯ (t ). P (U1 > t ) − P (U2 > t ) ≥ (F¯2 (t ) − F¯1 (t ))G
J.E. Valdés et al. / Statistics and Probability Letters 80 (2010) 945–949
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Because X1 ≤ic v X2 , then from part (b) of Lemma 7.1, p. 120, in Barlow and Proschan (1981) we deduce t
Z
(F¯2 (x) − F¯1 (x))G¯ 2 (x)H¯ (x)dx ≥ 0 0
and the first part of the proposition is proved. Suppose now that condition (b) holds. Then using Y2 ≤st X2 and X1 ≤st Y1 we get
¯ 2 (t )F¯1 (t )H¯ (t ). P (U1 > t ) − P (U2 > t ) ≥ (F¯2 (t ) − F¯1 (t ))G As X1 ≤ic v X2 , then the second part of the proposition follows using part (b) of Lemma 7.1 in Barlow and Proschan (1981) again. It is worth mentioning that if Y1 =st Y2 , then Theorem 1 in Li and Hu (2008) follows from part (a) of Proposition 1. Under condition (a), Proposition 1 means that in order to maximize the lifetime of the series system in the sense of the increasing concave ordering, the strongest spare must be allocated with the weakest component of the series system. No ordering between Y1 and Y2 is required in condition (b) of Proposition 1. Nevertheless, if the condition Y1 ≤ic v Y2 is added (i.e. X1 ≤st Y1 ≤ic v Y2 ≤st X2 ) then the lifetime of the system is maximized when the weakest spare is allocated with the weakest component. If in particular X1 ≤st Y1 ≤st Y2 ≤st X2 , then U1 ≥st U2 as was noted in Valdés and Zequeira (2003). ¯ 1 (x)F¯2 (x) ≥ G¯ 2 (x)F¯1 (x) then U1 ≥pr U2 . Therefore the probabilistic Romera et al. (2004) proved that if X1 ≤st X2 and G relation U1 ≥pr U2 holds if X1 ≤st Y1 ≤st Y2 ≤st X2 . Note that in this case also the weakest spare must be allocated with the weakest component. The following example shows that the condition X1 ≤st X2 is not necessary for U1 ≥pr U2 to hold. Example 1. Consider a two-component series system composed of the components C1 and C2 and the spares R1 and R2 . In this case U1 = ∧[∨(X1 , Y1 ), X2 ] and U2 = ∧[X1 , ∨(X2 , Y2 )]. Let X1 , X2 , Y1 and Y2 be exponentially distributed with means 1/λ1 , 1/λ2 , 1/µ1 and 1/µ2 , respectively. Then P (U1 > U2 ) − P (U2 > U1 ) =
λ1 (λ1 + λ2 + µ2 ) − λ2 (λ1 + λ2 + µ1 ) . (λ1 + λ2 + µ1 )(λ1 + λ2 + µ2 )
Observe that given values of λ1 and λ2 such that λ1 < λ2 (X1 ≥st X2 ) and a value of µ1 , we can select a value of µ2 such that µ2 > µ1 (Y1 ≥st Y2 ) and P (U1 > U2 ) − P (U2 > U1 ) > 0. 3. Allocation of more than one spare We will assume in this section that the lifetimes X1 , X2 , Y1 and Y2 are absolutely continuous. As in the preceding section, we consider a series system with n components and we suppose that there are two spares R1 and R2 to be allocated as active redundancies to the components C1 and C2 , respectively. Let X1 , X2 , . . . , Xn be the lifetimes of the components of the system. Let X1 and X2 be the lifetimes of the components C1 and C2 and Y1 and Y2 be the lifetimes of the spares R1 and R2 , respectively. We consider now that there are two options: allocate R1 with C1 and R2 with C2 , or otherwise, allocate R1 with C2 and R2 with C1 . As in Section 2 we define Z = ∧(X3 , X4 , . . . , Xn ) and denote by H (t ) the distribution function of this random variable. Then V1 = ∧ [∨(X1 , Y1 ), ∨(X2 , Y2 ), Z ]
and
V2 = ∧ [∨(X1 , Y2 ), ∨(X2 , Y1 ), Z ]
represent the lifetimes of the two possible configurations of the system. The following proposition is proved in Romera et al. (2004). Proposition 2. If X1 ≤hr X2 and Y1 ≥hr Y2 , then V1 ≥pr V2 . The hazard rate orderings of Proposition 2 can be substituted by the reversed hazard rate ordering. We will prove this fact in Proposition 3. In the proof of this proposition we use a result taken from Shaked and Shanthikumar (2007), p. 37, which we state next as Lemma 1. Lemma 1. Let X and Y be random variables with distribution functions F (t ) and G(t ), and probability density functions f (t ) and g (t ), respectively. Then X ≤rh Y if and only if f (y) F ( x)
≤
g (y) G(x)
,
for all x ≤ y.
Proposition 3. If X1 ≤rh X2 and Y1 ≥rh Y2 , then V1 ≥pr V2 . Proof. Romera et al. (2004) show that the inequality V1 > V2 is satisfied if and only if one of the two inequalities ∨(X1 , Y2 , Z ) < ∧(X2 , Y1 ) or ∨(X2 , Y1 , Z ) < ∧(X1 , Y2 ) holds. An analogous equivalence may be obtained for the inequality
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V1 < V2 . Then to prove the proposition it is sufficient to show that
∆ = P (∧(X2 , Y1 ) > ∨(X1 , Y2 , Z )) + P (∧(X1 , Y2 ) > ∨(X2 , Y1 , Z )) − P (∧(X2 , Y2 ) > ∨(X1 , Y1 , Z )) − P (∧(X1 , Y1 ) > ∨(X2 , Y2 , Z )) ≥ 0. But
∆=
∞
Z
∞
Z
F1 (∧(x, y))G2 (∧(x, y))H (∧(x, y))dG1 (x)dF2 (y) 0
0
∞
Z
∞
Z
F2 (∧(x, y))G1 (∧(x, y))H (∧(x, y))dG2 (x)dF1 (y)
+ Z0 ∞ Z0 ∞ − Z0 ∞ Z0 ∞ − 0
F1 (∧(x, y))G1 (∧(x, y))H (∧(x, y))dG2 (x)dF2 (y) F2 (∧(x, y))G2 (∧(x, y))H (∧(x, y))dG1 (x)dF1 (y).
0
Note that ∆ ≥ 0 if the inequality F1 (u)G2 (u)g1 (x)f2 (y) + F2 (u)G1 (u)g2 (x)f1 (y) − F1 (u)G1 (u)g2 (x)f2 (y) − F2 (u)G2 (u)g1 (x)f1 (y) ≥ 0 holds for all x, y ≥ 0, where u = ∧(x, y). But the last inequality can be written as [g1 (x)G2 (u) − g2 (x)G1 (u)] [f2 (y)F1 (u) − f1 (y)F2 (u)] ≥ 0. Then using Lemma 1 and the assumptions, the proof is completed.
Example 2. Consider a two-component series system with the components C1 and C2 and spares R1 and R2 . In this case V1 = ∧[∨(X1 , Y1 ), ∨(X2 , Y2 )] and V2 = ∧[∨(X1 , Y2 ), ∨(X2 , Y1 )]. If X1 , X2 , Y1 , Y2 are exponentially distributed with means 1/λ1 , 1/λ2 , 1/µ1 and 1/µ2 , respectively, then P (V1 > V2 ) =
λ 2 µ1 λ 1 µ2 + . (λ1 + λ2 + µ1 )(λ2 + µ1 + µ2 ) (λ1 + λ2 + µ2 )(λ1 + µ1 + µ2 )
An analogous formula can be obtained for P (V1 < V2 ). After some transformations we obtain that P (V1 > V2 ) − P (V1 < V2 ) ≥ 0 if and only if
(λ1 − λ2 )(µ2 − µ1 )(λ1 + λ2 + µ1 + µ2 )2 ≥ 0. Therefore, if λ1 ≥ λ2 (X1 ≤hr X2 and X1 ≤rh X2 ) and µ2 ≥ µ1 (Y1 ≥hr Y2 and Y1 ≥rh Y2 ) then V1 ≥pr V2 . This result agrees with Propositions 2 and 3. Moreover, if V1 ≥pr V2 then X1 ≤hr X2 (X1 ≤rh X2 ) and Y1 ≥hr Y2 (Y1 ≥rh Y2 ), or X1 ≥hr X2 (X1 ≥rh X2 ) and Y1 ≤hr Y2 (Y1 ≤rh Y2 ). Proposition 4. Suppose that Yi =st Xi , i = 1, 2. If X1 ≤rh X2 or X2 ≤rh X1 , then V1 ≤pr V2 . Proof. We will use the same notation as in the proof of Proposition 3. A sufficient condition for ∆ ≤ 0 to hold is that the inequality [g1 (x)G2 (u) − g2 (x)G1 (u)] [f2 (y)F1 (u) − f1 (y)F2 (u)] ≤ 0 is satisfied for all x, y ≥ 0. This inequality can be written as
− [f1 (x)F2 (u) − f2 (x)F1 (u)] [f1 (y)F2 (u) − f2 (y)F1 (u)] ≤ 0 and then using Lemma 1 the proposition is proved.
Remark 1. The relation V1 ≤pr V2 also holds if in Proposition 4 the order ≤rh is substituted by the order ≤hr . Remark 2. Observe that the condition X1 ≤rh X2 (X2 ≤rh X1 ) of Proposition 4 holds when the hazard rate functions or the reversed hazard rate functions of X1 and X2 are proportional. The results of Propositions 2–4 mean in practice that, in order to maximize the lifetime of the series system in the sense of the probabilistic relation, the stronger spare must be allocated with the weakest component of the system. Suppose now that there are n spares R1 , . . . , Rn that may be allocated as active redundancies to the components C1 , . . . , Cn . Let X1 , X2 , . . . , Xn and Y1 , Y2 , . . . , Yn be the corresponding lifetimes of the components and the spares, respectively. Consider a series system constituted by the components C1 , . . . , Cn . We will assume that only one spare can be allocated to each component of the systems. It is also assumed that only r, 1 ≤ r < n, of the spares can be allocated. The following result generalizes part (a) of Proposition 1.
J.E. Valdés et al. / Statistics and Probability Letters 80 (2010) 945–949
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Proposition 5. Suppose that X1 ≤ic v X2 ≤ic v · · · ≤ic v Xn and Y1 ≥st Y2 ≥st · · · ≥st Yn . Then
" r ^ ^
∨(Xi , Yi ),
i=1
n ^ i=r +1
# Xi ≥ic v
" r ^ ^
# ∨(Xsi , Ysi ),
i=1
^
Xj ,
j≤n,j6=si ,i=1,...,r
where r < n and si , i = 1, . . . , r, are not equal integers such that si ≤ n. Proof. We will construct the proof for the case n = 3, r = 2. The general proof can be obtained using mathematical induction, where the inductive step is proved using basically the same argument. From Proposition 1 we have
∧[X1 , ∨(X2 , Y2 ), ∨(X3 , Y3 )] ≤ic v ∧[∨(X1 , Y1 ), X2 , ∨(X3 , Y3 )] ≤ic v ∧[∨(X1 , Y1 ), ∨(X2 , Y2 ), X3 ]. Boland et al. (1988) proved an analogous result when in Proposition 5 the order ≤ic v is replaced by the order ≤st . Moreover, they proved a similar statement for k-out-of-n systems using the order ≤st . The example below shows that for coherent systems an analogous result is in general not true even for very simple systems. Example 3. Consider a three-component system. The components C1 and C2 form a series subsystem which in turn is in parallel with the component C3 . The spares R1 , R2 and R3 must be allocated one to one as active redundancies with the components of the system. Assume that Y1 =st X3 , Y2 =st X2 and Y3 =st X1 and that X1 ≤st X2 ≤st X3 (then Y1 ≥st Y2 ≥st Y3 ). We will consider two system with redundancies. In the first system the spare Ri is allocated with the component Ci , i = 1, 2, 3. In the second system the spares are allocated as follows: R2 with C1 , R3 with C2 and R1 with C3 . The lifetimes for the first and second systems are U123 = ∨ [∧(X1 ∨ Y1 , X2 ∨ Y2 ), X3 ∨ Y3 ] and U231 = ∨ [∧(X1 ∨ Y2 , X2 ∨ Y3 ), X3 ∨ Y1 ] , respectively. Thus, P (U123 > t ) = [1 − F1 (t )F3 (t )][1 − F22 (t )] + [1 − F1 (t )F3 (t )] − [1 − F1 (t )F3 (t )]2 [1 − F22 (t )]
(1)
P (U231 > t ) = [1 − F1 (t )F2 (t )]2 + [1 − F32 (t )] − [1 − F1 (t )F2 (t )]2 [1 − F32 (t )].
(2)
and
Now let us take for some t = t0 , F1 (t0 ) = 0.6, F2 (t0 ) = 0.5 and F3 (t0 ) = 0.4. Substituting these values for t = t0 in (1) and (2) we obtain P (U123 > t0 ) = 0.8968 < P (U231 > t0 ) = 0.9184. Acknowledgements The work of G. Arango was supported by Universidad EAFIT, Colombia, under project n-45-000031 of 2008. The work of J.E. Valdés was partially supported by the same university and project. The authors are very grateful to the referees for their valuable comments which helped to improve the presentation of this paper. References Barlow, R.E., Proschan, F., 1981. Statistical Theory of Reliability and Life Testing. McArdle Press, Inc. Blyth, C.R., 1972. Some probability paradoxes in choice from among random alternatives. Journal of the American Statistical Association 67, 366–373. Boland, P.J., El-Neweihi, E., Proshan, F., 1988. Active redundancy allocation in coherent systems. Probability in the Engineering and Informational Sciences 2, 343–353. Boland, P.J., El-Neweihi, E., Proshan, F., 1994. Applications of the hazard rate orderings in reliability and orders statistics. Journal of Applied Probability 31, 180–192. Boland, P.J., Proshan, F., Tong, Y.L., 1989. Optimal arrangement of components via pairwise rearrangements. Naval Research Logistics 36, 807–815. Boland, P.J., Singh, H., Cukic, B., 2004. The stochastic precedence ordering with applications in sampling an testing. Journal of Applied Probability 41, 73–82. Cha, J.H., Mi, J., Yun, W.Y., 2008. Modelling a general standby system and evaluation of its performance. Applied Stochastic Models in Business and Industry 24, 159–169. Li, X., Hu, X., 2008. Some new stochastic comparisons for redundancy allocations in series and parallel systems. Statistics and Probability Letters 78 (18), 3388–3394. Meng, F.C., 1996. More on optimal allocation of components in coherent systems. Journal of Applied Probability 33, 548–556. Müller, A., Stoyan, D., 2002. Comparison Methods for Stochastic Models and Risks. Wiley. Romera, R., Valdés, J.E., Zequeira, R.I., 2004. Active-redundancy allocation in systems. IEEE Transactions on Reliability 53, 313–318. Shaked, M., Shanthikumar, J.G., 2007. Stochastic Orders. Springer, New York. Singh, H., Misra, N., 1994. On redundancy allocation in systems. Journal of Applied Probability 31, 1004–1014. Singh, H., Singh, R.S., 1997. Optimal allocation of resources to nodes of series systems with respect to failure-rate ordering. Naval Research Logistics 44, 147–152. Valdés, J.E., Zequeira, R.I., 2003. On the optimal allocation of an active redundancy in a two-component series system. Statistics and Probability Letters 63, 235–332.