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European Journal of Operational Research 241 (2015) 402–411
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Stochastics and Statistics
Redundancy allocation at component level versus system level ✩ Peng Zhao a,∗, Yiying Zhang b , Long Li b a b
School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou 221116, China School of Mathematics and Statistics, Lanzhou University, Lanzhou 730000, China
a r t i c l e
i n f o
Article history: Received 24 August 2013 Accepted 31 August 2014 Available online 22 September 2014 Keywords: Reliability Active redundancy Standby redundancy Component level System level
a b s t r a c t In this paper, we treat the problem of stochastic comparison of standby [active] redundancy at component level versus system level. In the case of standby redundancy, we present some interesting comparison results of both series systems and parallel systems in the sense of various stochastic orderings for both the matching spares case and non-matching spares case, respectively. In the case of active redundancy, a likelihood ratio ordering result of series systems is presented for the matching spares case; and for the non-matching spares case, a counterexample is provided to show that there does not exist similar result even for the hazard rate ordering. The results established here strengthen and generalize some of those known in the literature. Some numerical examples are also provided to illustrate the theoretical results. © 2014 Elsevier B.V. All rights reserved.
1. Introduction In order to optimize the lifetime of a coherent system, one efficient method is to add redundancy components to the system. This topic has posed many interesting problems to which many researchers have devoted themselves in the past decades; see, for example, Boland, El-Neweihi, and Proschan (1992); Shaked and Shanthikumar (1992); Singh and Misra (1994); Singh and Singh (1997); Mi (1998); Ramirez-Marquez and Coit (2004); Yalaoui, Chu, and Châtelet (2005); da Costa Bueno (2005); Valdés and Zequeira (2006); da Costa Bueno and do Carmo (2007); Hu and Wang (2009); Valdés, Arango, and Zequeira (2010); Brito, Zequeira, and Valdès (2011); Misra, Dhariyal, and Gupta (2009); Misra, Misra, and Dhariyal (2011a,b); Zhao, Chan, and Ng (2012); Zhao, Chan, Li, and Ng (2013a,b); Laniado and Lillo (2014) and the references therein. In general, there are two types of redundancies called active redundancy and standby redundancy commonly used in reliability engineering and system security. The former [active redundancy] is used when replacement of components during the operation of the system is impossible; in this case the redundancies are put in parallel to components of the system which leads to taking the maximum of random lifetimes. The latter [standby redundancy] is used when replacement of components during the operation of the system is possible; in this case the redundancy starts functioning immediately after the ✩ This work was supported by National Natural Science Foundation of China (11422109), Natural Science Foundation of Jiangsu Province (BK20141155) and A Project Funded by the Priority Academic Program Development of Jiangsu Higher Education Institutions. ∗ Corresponding author. Tel.: +86 516 83500332; fax: +86 516 83500387. E-mail address: [email protected], [email protected] (P. Zhao).
http://dx.doi.org/10.1016/j.ejor.2014.08.040 0377-2217/© 2014 Elsevier B.V. All rights reserved.
corresponding original component in the system fails which leads to taking the convolution of random variables. It is known that the performance of a coherent system consisting of independent components can be enhanced by putting spares to each of its components or by creating a duplicate system consisting of spares similar to the original coherent system. For example, there is a series system with n components and we have another n components at hand as spares. In order to enhance the lifetime of the original series system, we can assemble these n spares either on component level (Fig. 1(a)) or system level (Fig. 1(b)). Naturally, it is of great interest and importance to make sure which type is superior to the other. To begin, we suppose the general coherent system φ consists of n components C1 , . . . , Cn having independent lifetimes X1 , . . . , Xn . Denote τ (X) = τ (X1 , . . . , Xn ) the lifetime of the coherent system φ . Suppose now n standby spares R1 , . . . , Rn have independent lifetimes Y1 , . . . , Yn , and (X1 , . . . , Xn ) and (Y1 , . . . , Yn ) are statistically independent. In component redundancy case, we allocate a standby spare Ri to the component Ci , i = 1, . . . , n. Then the resultant coherent system, denoted by TC , has lifetime τ (X + Y) = τ (X1 + Y1 , . . . , Xn + Yn ). In system redundancy case, we duplicate the coherent system φ with components C1 , . . . , Cn by R1 , . . . , Rn and make it available as a standby redundant spare to the coherent system φ . The resultant coherent system, denoted by TS , has lifetime τ (X) + τ (Y). A well-known principle among design engineers states that redundancy at the component level is always better than at the system level (cf. Barlow & Proschan, 1981). This statement, however, is not true in general for the standby redundancy case though it is for the active redundancy case. Boland and El-Neweihi (1995) concluded that, in the sense of the usual stochastic order, redundancy at component level is better than redundancy at system level for series systems, while the reverse is true for parallel
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
X1
X2
Xn
X1
X2
It would be very interesting to examine some stronger stochastic orderings. In the case of matching spares, i.e., Xi =st Yi , i = 1, . . . , n, Boland and El-Neweihi (1995) proved, for series systems, that the component redundancy is superior to the system redundancy in the sense of the hazard rate ordering, i.e.,
Xn
∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥hr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )].
Y1
Y2
Yn
(a) Component level
Y1
Y2
Yn
(b) System level
Fig. 1. Component level versus system level.
systems, i.e.,
∧(X1 + Y1 , . . . , Xn + Yn ) ≥st [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )]
(1)
and
∨(X1 + Y1 , . . . , Xn + Yn ) ≤st [∨(X1 , . . . , Xn )] + [∨(Y1 , . . . , Yn )],
(2)
where the symbols ‘∧’ and ‘∨’ mean min and max, respectively. Boland and El-Neweihi (1995) also showed that, for series systems, if all random lifetimes are exponentially i.i.d., then the usual stochastic order in (1) can be strengthened to the hazard rate order. On the other hand, Meng (1996) proved that (1) holds if and only if the system is series and (2) holds if and only if the system is parallel. What does this suggest about the non-i.i.d. case? Here, we shall present some comparison results in the sense of various stochastic orderings when the original components and spares have independent but nonidentical exponential distributed lifetimes. Specifically, for the standby redundancy of series systems, if either Xi =st Yi (i = 1, . . . , n) or X1 =st . . . =st Xn and Y1 =st . . . =st Yn , it is established that
∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )]. For the standby redundancy of parallel systems, if X1 =st X2 and Y1 =st Y2 , we establish
∨(X1 + Y1 , X2 + Y2 ) ≤lr [∨(X1 , X2 )] + [∨(Y1 , Y2 )], and if Xi =st Yi (i = 1, 2), we prove
∨(X1 + Y1 , X2 + Y2 ) ≤st [∨(X1 , X2 )] + [∨(Y1 , Y2 )]. Consider a series system with two components and two active spares. In the case of matching spares, i.e., Xi =st Yi , i = 1, 2, the redundancy at component level is better in the hazard rate ordering than the redundancy at system level (see Boland & El-Neweihi, 1995). An analogous result was proved for the reversed hazard rate ordering in Gupta and Nanda (2001). Brito, Zequeira, and Valdès (2011) further considered the case of matching spares and gave a sufficient condition under which the likelihood ratio ordering holds. Gupta and Nanda (2001) proved the redundancy at component level is better in the reversed hazard rate ordering than the redundancy at system level in the case of non-matching spares, X1 =st X2 and Y1 =st Y2 . Boland and El-Neweihi (1995) showed with a counterexample that the hazard rate ordering does not always exist between them. In this regard, Brito, Zequeira, and Valdès (2011) proved that, if the reversed hazard rates of X1 and Y1 are proportional, then the hazard rate ordering holds. Obviously, the exponential case is not satisfied with the proportional condition, and we provide a counterexample to depict the hazard rate ordering does not hold for the exponential case in Section 4 (Example 4.1). We next consider the general coherent systems with n components and n active spares. Similar to the standby case, in component redundancy case, the resultant coherent system, denoted by SC , has lifetime τ (X ∨ Y) = τ (X1 ∨ Y1 , . . . , Xn ∨ Yn ). In system redundancy case, the resultant coherent system, denoted by SS , has lifetime τ (X) ∨ τ (Y). For general coherent systems, it is well known (cf. Barlow and Proschan, 1981) that the component redundancy is better than system redundancy in the sense of the usual stochastic ordering, i.e.,
τ (X ∨ Y) ≥st τ (X) ∨ τ (Y).
403
(3)
The above result can be readily extended to series–parallel systems (cf. Corollary 1-1 in Boland and El-Neweihi 1995). When all components and spares are i.i.d., they also proved, for general coherent systems, τ (X ∨ Y) ≥hr τ (X) ∨ τ (Y) under some mild conditions. Though this result cannot be applied to the general k-out-of-n system, but they showed the result is true for the special case of 2-out-of-n system and left the general case for an open problem. Gupta and Nanda (2001) obtained a similar result for the reversed hazard rate ordering. In fact, Singh and Singh (1997) have solved the open problem posed by Boland and El-Neweihi (1995) and established a stronger result in the sense of the likelihood ratio ordering for the general k-out-of-n system with all components and spares having i.i.d. absolutely continuous lifetimes. Also, there remains an open problem that whether this result still holds for the case when we just have the condition Xi =st Yi , i = 1, . . . , n (matching case) and such a result for two-component series system has been shown in Brito, Zequeira, and Valdès (2011) wherein they require that the hazard rates are proportional. On the other hand, Misra, Dhariyal, and Gupta (2009) studied the non-matching case and obtained some new results. In this paper, we present a likelihood ratio order result for the series system with n exponential components under the setup of matching case (Theorem 4.2). Specifically, for the active redundancy of series systems, if Xi =st Yi (i = 1, . . . , n), we establish
∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥lr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )]. This paper is organized as follows. In Section 2, we recall some pertinent notions of stochastic orderings which will be used in the sequel. Section 3 presents some comparison results of series systems and parallel systems in the sense of various stochastic orderings for standby redundancy at component level versus system level. Section 4 carries out stochastic comparisons of series systems between component level and system level in the sense of the likelihood ratio ordering under the setup of matching spares as well as for the nonmatching spares: a counterexample is provided to show that there does not exist similar comparison result even for the hazard rate ordering. Conclusion is finally made in Section 5. The technical details of the proofs of all the main results are presented in Appendix A. 2. Definitions and notations Throughout this paper, the term increasing is used for monotone non-decreasing and decreasing is used for monotone non-increasing. Definition 2.1. For two random variables X and Y with densities fX and fY , and distribution functions FX and FY , respectively, let F X = 1 − FX and F Y = 1 − FY be the corresponding survival functions. Then: (i) X is said to be smaller than Y in the likelihood ratio order (denoted by X ≤lr Y) if fY (x)/fX (x) is increasing in x; (ii) X is said to be smaller than Y in the hazard rate order (denoted by X ≤hr Y) if F Y (x)/F X (x) is increasing in x; (iii) X is said to be smaller than Y in the reversed hazard rate order (denoted by X ≤rh Y) if FY (x)/FX (x) is increasing in x; (iv) X is said to be smaller than Y in the stochastic order (denoted by X ≤st Y) if F Y (x) ≥ F X (x), ∀ x. It is known that the likelihood ratio order implies both the hazard rate order and the reversed hazard rate order which in turn implies the usual stochastic order, but neither the hazard rate order nor the reversed hazard rate order order implies the other; see Shaked and Shanthikumar (2007).
404
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411 300
3000 f (x)/f (x) T
S
250
Likelihood ratio function
Likelihood ratio function
T
C
2500
2000
1500
1000
500
200
150
100
50
fT (x)/fT (x) C
0
0
0.5
1
1.5
2
2.5 x
3
3.5
4
4.5
0
5
Fig. 2. Plot of the likelihood ratio function fTC (x)/fTS (x) when n = 4, λ1 = 3, λ2 = 5, λ3 = 4, and λ4 = 6.
3. Standby redundancy at component level vs. system level In this section, we present some comparison results for series systems and parallel systems having standby redundancy at component level versus system level. 3.1. Series systems In the case of series system having matching spares, i.e., Xi =st Yi , i = 1, 2, . . . , n, we have the following result which states that the redundancy at component level is more efficient than the redundancy at system level in the sense of the likelihood ratio order. Theorem 3.1. Let X1 , . . . , Xn be independent exponential lifetimes with respective parameters λ1 , . . . , λn , and Y1 , . . . , Yn be another independent set of exponential lifetimes with respective parameters λ1 , . . . , λn . Then,
τ (X + Y) = ∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )] = τ (X) + τ (Y). Example 3.2. Set n = 4, λ1 = 3, λ2 = 5, λ3 = 4, and λ4 = 6 in Theorem 3.1. Fig. 2 plots the likelihood ratio function fTC (x)/fTS (x). It can be seen that fTC (x)/fTS (x) is increasing in x ∈ + , which shows the validity of the result in Theorem 3.1. In the case of non-matching spares, we have the following similar result. Theorem 3.3. Let X1 , . . . , Xn be independent exponential random lifetimes with common parameter λ1 , and Y1 , . . . , Yn be independent exponential random lifetimes with common parameter λ2 . Then,
τ (X + Y) = ∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )] = τ (X) + τ (Y). Example 3.4. Set n = 5, λ1 = 4 and λ2 = 3 in Theorem 3.3. By the symmetry, we know that it is same as the case λ1 = 3 and λ2 = 4. Fig. 3 plots the ratio of the two density functions. It can be seen fTC (x)/fTS (x) is increasing in x ∈ + , which is in accordance with the theoretical result in Theorem 3.3. 3.2. Parallel systems We next discuss the case of parallel systems. In this case, due to the extreme complexity of distribution theory, we only deal with the special case when n = 2 and it can be found that the comparison result is reversed compared to the case of series systems. Now, let us first go into the non-matching spares.
0
1
2
3 x
4
S
5
6
Fig. 3. Plot of the likelihood ratio function fTC (x)/fTS (x) when n = 5, λ1 = 4 and λ2 = 3.
Theorem 3.5. Let X1 , X2 be independent exponential random lifetimes with common parameter λ1 and Y1 , Y2 be independent exponential random lifetimes with common parameter λ2 . Then,
τ (X + Y) = ∨(X1 + Y1 , X2 + Y2 ) ≤lr [∨(X1 , X2 )] + [∨(Y1 , Y2 )] = τ (X) + τ (Y). Example 3.6. Under the setup in Theorem 3.5, we consider the likelihood ratio fTS (x)/fTC (x) under the condition λ1 ≥ λ2 . Based on this, we suppose that the parameters in (a), (b), (c) and (d) below satisfy the cases λ1 = λ2 , λ2 < λ1 < 2λ2 , λ1 > 2λ2 and λ1 = 2λ2 , respectively. It can be seen from Fig. 4 that the likelihood ratio functions for all these four cases are increasing, which coincides with the theoretic result in Theorem 3.5. (a) (b) (c) (d)
λ1 = 0.6, λ2 = 0.6; λ1 = 3, λ2 = 2; λ1 = 5, λ2 = 0.7; λ1 = 8, λ2 = 4.
For the matching spares case, we only have the following comparison result in the sense of the usual stochastic order. Theorem 3.7. Let X1 , X2 be independent exponential lifetimes with respective parameters λ1 , λ2 , and Y1 , Y2 be another independent exponential set with respective parameters λ1 , λ2 . Then,
τ (X + Y) = ∨(X1 + Y1 , X2 + Y2 ) ≤st [∨(X1 , X2 )] + [∨(Y1 , Y2 )] = τ (X) + τ (Y). Example 3.8. Set λ1 = 5 and λ2 = 3 in Theorem 3.7. Fig. 5 plots the survival functions of TC and TS . It can be seen that the survival function of TS is larger than that of TC for x ∈ + , which shows the validity of the result in Theorem 3.7. One may wonder whether the result in Theorem 3.7 can be strengthened to the hazard rate ordering. The following Example 3.9 gives a negative answer to this problem. Example 3.9. Denote by hTC (x)) = fTC (x)/F TC (x) and hTS (x) = fTS (x)/F TS (x) the hazard rate functions of TC and TS , respectively. Fig. 6 plots the shape of hTC (x) and hTS (x). It can be observed that we cannot compare these two hazard rate functions in this regard. Consequently, the likelihood ratio order does not hold as well. 4. Active redundancy at component level vs. system level In this section, we carry out stochastic comparisons of series systems at component level and system level for the matching spares
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
405
1.5
1
1.4 0.9 1.3 0.8 Likelihood ratio function
Likelihood ratio function
1.2
0.7
0.6
1.1
1
0.9 0.5 0.8 f (x)/f (x)
f (x)/f (x)
0.4
T
T
S
0
5
10
15
20
25
30
35
T
0.7
C
40
0
1
2
3
4
x
x
(a)
(b)
1.15
T
S
5
C
6
7
1.4
1.1
1.3
1.05 1.2
Likelihood ratio function
Likelihood ratio function
1
0.95
0.9
0.85
1.1
1
0.9
0.8 0.8 0.75 f (x)/f (x) T
0.7
0.65
0
1
2
3
4
5
f (x)/f (x)
0.7
T
S
T
C
6
S
7
0
0.5
1
1.5
x
x
(c)
(d)
2
2.5
T C
3
Fig. 4. Plot of likelihood ratio functions fTS (x)/fTC (x) when λ1 ≥ λ2 .
case in the sense of the likelihood ratio order. First we provide a counterexample to show the comparison result does not hold even for the hazard rate order in the non-matching spares case. Example 4.1. Set λ1 = 3.5 and λ2 = 2. The survival functions of Q1 = ∧(X1 ∨ Y1 , X2 ∨ Y2 ) and Q2 = ∨(X1 ∧ X2 , Y1 ∧ Y2 ) can be written as
F Q1 (x) = [1 − (1 − e−3.5x )(1 − e−2x )]2 and
F Q2 (x) = 1 − [(1 − e−2×3.5x )][(1 − e−2×2x )]. But, the ratio is not monotone as can be seen in Fig. 7.
Now, we are ready to present our main result in the case of matching spares. Theorem 4.2. Let X1 , . . . , Xn be independent exponential lifetimes with respective parameters λ1 , . . . , λn , and Y1 , . . . , Yn be another independent exponential set with respective parameters λ1 , . . . , λn . Then,
τ (X ∨ Y) = ∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥lr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )] = τ (X) ∨ τ (Y). (4) Example 4.3. Set n = 4, λ1 = 2, λ2 = 0.8, λ3 = 1.6 and λ4 = 2.5 in Theorem 4.2. Fig. 8 plots the ratio of the two density functions. One
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P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
8
1 7
Survival function of TC
0.8
Survival function of TS
6 Likelihood ratio function
0.9
0.7 0.6 0.5 0.4
4 3 2
0.3
f (x)/f (x) S
1
0.2 0.1 0
5
0
0
0.5
1
1.5 x
2
2.5
c
0
0.5
1
3
1.5
2 x
2.5
3
S
s
3.5
4
Fig. 8. Plot of likelihood ratio function fSC (x)/fSS (x) when λ1 = 2, λ2 = 0.8, λ3 = 1.6 and λ4 = 2.5.
Fig. 5. Plot of F TC (x) and F TS (x) when λ1 = 5 and λ2 = 3.
can observe that fSC (x)/fSS (x) is increasing in x ∈ + , which fits the theoretical result in Theorem 4.2. 4
5. Concluding remarks 3.5 3 2.5 2 1.5 1
hT (x)
0.5
hT (x)
C
0
S
0
0.5
1
1.5 x
2
2.5
3
Fig. 6. Plot of hTC (x) and hTS (x) when λ1 = 7 and λ2 = 4.
Let us suppose the general coherent system φ consists of n components C1 , . . . , Cn having independent lifetimes X1 , . . . , Xn . Denote τ (X) = τ (X1 , . . . , Xn ) the lifetime of the coherent system φ . Suppose now n standby[active] spares R1 , . . . , Rn have independent lifetimes Y1 , . . . , Yn , and (X1 , . . . , Xn ) and (Y1 , . . . , Yn ) are statistically independent. In component redundancy case, we allocate a standby[active] spare Ri to the component Ci , i = 1, . . . , n. Then the resultant coherent system, denoted by TC [SC ], has lifetime τ (X + Y) = τ (X1 + Y1 , . . . , Xn + Yn )[τ (X ∨ Y) = τ (X1 ∨ Y1 , . . . , Xn ∨ Yn )]. In system redundancy case, we duplicate the coherent system φ with components C1 , . . . , Cn by R1 , . . . , Rn and make it available as a standby[active] redundant spare to the coherent system φ . The resultant coherent system, denoted by TS [SS ], has lifetime τ (X) + τ (Y)[τ (X) ∨ τ (Y)]. Let X1 , . . . , Xn be independent exponential lifetimes with respective parameters λ1 , . . . , λn , and Y1 , . . . , Yn be another independent exponential set with respective parameters λ1 , . . . , λn . In this article, for the standby redundancy of series systems, it is established that
τ (X + Y) = ∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )] = τ (X) + τ (Y)
1.45
and for the standby redundancy of parallel systems, we establish 1.4
τ (X + Y) = ∨(X1 + Y1 , X2 + Y2 ) ≤st [∨(X1 , X2 )] + [∨(Y1 , Y2 )] = τ (X) + τ (Y).
1.35
For the active redundancy of series systems, we have
1.3
τ (X ∨ Y) = ∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥lr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )] = τ (X) ∨ τ (Y).
1.25 1.2
Let W1 , . . . , Wn be independent exponential lifetimes with common parameter λ1 , and V1 , . . . , Vn be another set of independent exponential lifetimes with common parameter λ2 . For the standby redundancy of series systems, it is established that
1.15 1.1
τ (W + V) = ∧(W1 + V1 , . . . , Wn + Vn ) ≥lr [∧(W1 , . . . , Wn )] + [∧(V1 , . . . , Vn )] = τ (W) + τ (V)
1.05 1
0
0.5
1
1.5 x
2
Fig. 7. Plot of the ratio function F Q1 (x)/F Q2 (x).
2.5
3
and for the standby redundancy of parallel systems, we establish
τ (W + V) = ∨(W1 + V1 , W2 + V2 ) ≤lr [∨(W1 , W2 )] + [∨(V1 , V2 )] = τ (W) + τ (V).
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
For the active redundancy of series systems, we can conclude that the comparison result does not hold even for the hazard rate order by a counterexample.
We now have to show the ratio
Acknowledgements
is increasing in x. Denote by
The authors thank the editor and two anonymous referees for the insight comments and suggestions, which resulted in an improvement in the presentation of this manuscript.
ξ (x) =
In this appendix, we provide detailed proofs for all of the theorems in this paper. Proof of Theorem 3.1
i=1
fTC (x) =
fXi +Yi (x)
n
n
λ2i xe−λi x
(λj x + 1)e−λj x
j =i
= xe−x
n
λ2i
n
(λj x + 1),
where = ni=1 λi . On the other hand, observe that TS has a gamma distribution with shape parameter 2 and scale parameter with its density function
fTS (x) = 2 xe−x . Now it can be readily verified that the ratio
n
i=1
λ2i
n
j =i
(λj x + 1)
n
and thus its density function is given by
nλ1 λ2 [λ e−λ2 x − λ2 e−λ1 x ]n−1 [e−λ2 x − e−λ1 x ]. (λ1 − λ2 )n 1
On the other hand, the survival function of TS is
F TS (x) =
λ1 − λ2
e
−nλ2 x
and
β = (x1 − x2 )[en(x1 −x2 ) − nex1 −x2 + n − 1]. It is known from the assumption that x1 > x2 > 0, which implies α ≥ 0. On the other hand, observe that
β (y) = neny − ney = ney [e(n−1)y − 1] ≥ 0.
λ1 λ2 e−λ2 x − e−λ1 x λ1 − λ2 λ1 − λ2
λ1
α = nx2 (ex1 −x2 − 1)[e(n−1)(x1 −x2 ) − 1]
Let y = x1 − x2 and we then have β(y) = eny − ney + n − 1. It can be verified that
Without loss of generality, we suppose that λ1 ≥ λ2 . For the case λ1 = λ2 , the result can be derived from Theorem 3.1 and we only need to prove the case when λ1 > λ2 . The survival function of TC can be written as
fTC (x) =
Let x1 = λ1 x and x2 = λ2 x, it is equivalent to proving that ϕ(x1 , x2 ) ≥ 0, where
sgn
Proof of Theorem 3.3
F TC (x) =
sgn
β = en(x1 −x2 ) − nex1 −x2 + n − 1.
2
is increasing in x.
ε(x) = nλ2 [en(λ1 −λ2 )x − e(n−1)(λ1 −λ2 )x ] − nλ1 [e(λ1 −λ2 )x − 1] + (λ1 − λ2 )[en(λ1 −λ2 )x − 1].
where
j =i
i=1
ξ (x) = nλ2 (λ1 − λ2 ){e−[λ1 +(n+1)λ2 ]x − e−[2λ1 +nλ2 ]x } + nλ1 (λ1 − λ2 ){e−[λ2 +(n+1)λ1 ]x − e−[2λ2 +nλ1 ]x } + (λ1 − λ2 )2 {e−[λ1 +(n+1)λ2 ]x − e−[λ2 +(n+1)λ1 ]x } = ε(x).
ϕ(x1 , x2 ) = nx2 [en(x1 −x2 ) − e(n−1)(x1 −x2 ) − (ex1 −x2 − 1)] + (x1 − x2 )[en(x1 −x2 ) − 1] − n(x1 − x2 )(ex1 −x2 − 1) = nx2 (ex1 −x2 − 1)[e(n−1)(x1 −x2 ) − 1] + (x1 − x2 )[en(x1 −x2 ) − nex1 −x2 + n − 1] = α + β , say,
F Xj +Yj (x)
n
i=1
fTC (x) = fTS (x)
and taking the derivative of ξ (x) with respect to x yields that
Observe that
j =i
i=1
=
e−nλ2 x − e−nλ1 x
ϕ(x1 , x2 ) = nx2 [en(x1 −x2 ) − e(n−1)(x1 −x2 )] − nx1 (ex1 −x2 − 1) + (x1 − x2 )[en(x1 −x2 ) − 1].
F Xi +Yi (x)
with its density function n
[λ1 e−λ2 x − λ2 e−λ1 x ]n−1 [e−λ2 x − e−λ1 x ]
Note that
It is easy to see that Xi + Yi (i = 1, . . . , n) has a gamma distribution with shape parameter 2 and scale parameter λi . Denote by fXi +Yi and F Xi +Yi the corresponding density and survival functions, respectively. Then, we can write the survival function of TC as n
fTC (x) [λ1 e−λ2 x − λ2 e−λ1 x ]n−1 [e−λ2 x − e−λ1 x ] ∝ fTS (x) e−nλ2 x − e−nλ1 x
sgn
Appendix A
F TC (x) =
δ(x) =
407
−
λ2
λ1 − λ2
with its density function as
nλ1 λ2 −nλ2 x fTS (x) = [e − e−nλ1 x ]. λ1 − λ2
e
−nλ1 x
from which and the fact β(0) = 1 − n + n − 1 = 0 we know β(y) ≥ 0. Therefore, the theorem follows. Proof of Theorem 3.5 Without loss of generality, we only need to show the result for the case λ1 ≥ λ2 due to the symmetry. The proof will be proceeded by discussing the following two cases. Case 1: λ1 = λ2 Let λ1 = λ2 = λ. Then, the density function of TC can be written as
fTC (x) = 2λ[λxe−λx − λx(λx + 1)e−2λx ]. Also, it is easy to derive the density function of TS as
fTS (x) = 4λ2
0
x
e−λx [1 − e−λ(x−t)][1 − e−λt ]dt
= 4λ[(λx + 2)e−2λx + (λx − 2)e−λx ].
408
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
We need to show that
(x) =
f (x) (x) = TS fTC (x) ∝
∝
λx + 2 + (λx − 2)eλx λxeλx − λx(λx + 1)
is an increasing function in y ∈ + . Taking the derivative of W (y) with respect to y, we have sgn
W (y) = [1 + ey + (y − 2)ey ] × [yey − y(y + 1)] − [ey + yey − 2y − 1] × [y + 2 + (y − 2)ey ] = 2e2y − (y3 − y2 + 4y + 4)ey + y2 + 4y + 2
ω1 (y).
Since ω1 (0) = 0, we just need to prove that ω1 (y) ≥ 0. Notice that
ω1 (y) = 4e2y − (3y2 − 2y + 4)ey − (y3 − y2 + 4y + 4)ey + 2y + 4 = 4e2y − (y3 + 2y2 + 2y + 8)ey + 2y + 4 = ω2 (y), and ω2 (0) = 0. Now, it is sufficient to show that ω2 (y) ≥ 0. We can see that
ω2 (y) = 8e2y − (y3 + 5y2 + 6y + 10)ey + 2 = ω3 (y). Again, it is easy to calculate that ω3 (0) = 0 and sgn
ω3 (y) = 16ey − y3 − 8y2 − 16y − 16 = ω4 (y). Similarly, we have ω4 (0) = 0 and
ω4 (y) = 16ey − 3y2 − 16y − 16 = ω5 (y). Based on the fact ω5 (0) = 0, it is enough to show that ω5 (y) ≥ 0. Observe that
ω5 (y) = 16ey − 6y − 16 = ω6 (y), and then ω6 (0) = 0 and
ω6 (y) = 16ey − 6 ≥ 16 − 6 = 10 > 0. Thus, we have the following relationship
ω6 (y) ≥ 0 =⇒ ω5 (y) ≥ 0 =⇒ ω4 (y) ≥ 0 =⇒ ω3 (y) ≥ 0 =⇒ ω2 (y) ≥ 0 =⇒ ω1 (y) ≥ 0, which yields that ω1 (y) ≥ 0 and then (x) is an increasing function in x ∈ + . Case 2: λ1 > λ2 Subcase 2.1: λ1 > λ2 and λ1 = 2λ2 For this case, we have
fTC (x) =
2λ1 λ2 [λ (1 − e−λ2 x ) − λ2 (1 − e−λ1 x )](e−λ2 x − e−λ1 x ) (λ1 − λ2 )2 1
and
fTS (x) = 4λ1 λ2 = 4λ1 λ2 −
x 0
e−λ1 x e(λ1 −λ2 )y [1 − e−λ1 (x−y)][1 − e−λ2 y ]dy 1
λ1 − λ2
[e−λ2 x − e−λ1 x ] −
1
[e−2λ2 x − e−λ1 x ]
[e−2λ2 x − e−2λ1 x ] .
λ1 − 2λ2
1 1 [e−λ2 x − e−2λ1 x ] + 2λ1 −λ2 2(λ1 − λ2 )
We need to show that
λ1 λ2 λ1 e−λ1 x + e−λ2 x − e−2λ1 x − e−2λ2 x λ1 − 2 λ 2 2 λ1 − λ2 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) −λ1 x −λ2 x −2λ1 x −2λ2 x −(λ1 +λ2 )x −(λ1 − λ2 )e + (λ1 − λ2 )e − λ2 e − λ1 e + (λ1 + λ2 )e
is increasing in x ∈ + . Since
y + 2 + (y − 2)ey yey − y(y + 1)
=
S
C λ2
is increasing in x ∈ + . Denote y = λx, we then need to prove that
W (y) =
fT (x) fT (x)
λ1 λ2 −λ1 x λ1 λ2 −λ2 x e − e λ1 − 2λ2 2λ1 − λ2 λ1 λ2 −2λ1 x λ1 λ2 −2λ2 x + e + e 2λ1 − λ2 λ1 − 2λ2 × [−(λ1 − λ2 )e−λ1 x + (λ1 − λ2 )e−λ2 x − λ2 e−2λ1 x − λ1 e−2λ2 x + (λ1 + λ2 )e−(λ1 +λ2 )x ] − [λ1 (λ1 − λ2 )e−λ1 x − λ2 (λ1 − λ2 )e−λ2 x + 2λ1 λ2 e−2λ1 x + 2λ1 λ2 e−2λ2 x − (λ1 + λ2 )2 e−(λ1 +λ2 )x ] λ2 λ1 × e−λ1 x + e−λ2 x λ1 − 2λ2 2λ1 − λ2 λ2 λ1 − e−2λ1 x − e−2λ2 x 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) λ2 λ21 − λ22 −(3λ1 +λ2 )x λ1 λ21 − λ22 −(λ1 +3λ2 )x = e − e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) 2 λ1 λ2 λ1 − λ22 + e−(2λ1 +2λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) λ21 λ2 (λ1 + λ2 ) − e−3λ1 x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ22 (λ1 + λ2 ) + e−3λ2 x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ2 (λ1 + λ2 )(λ1 − 4λ2 ) −(2λ1 +λ2 )x − e 2(λ1 − 2λ2 ) λ1 (λ1 + λ2 )(4λ1 − λ2 ) −(λ1 +2λ2 )x + e 2(2λ1 − λ2 ) (λ1 + λ2 )(λ1 − λ2 )3 −(λ1 +λ2 )x − e (2λ1 − λ2 )(λ1 − 2λ2 ) 2 2 λ1 λ21 − λ22 −2λ2 x sgn λ2 λ1 − λ2 = e−2λ1 x − e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) 2 λ1 λ2 λ1 − λ22 + e−(λ1 +λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) λ21 λ2 (λ1 + λ2 ) − e−(2λ1 −λ2 )x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ22 (λ1 + λ2 ) + e−(2λ2 −λ1 )x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ2 (λ1 + λ2 )(λ1 − 4λ2 ) −λ1 x − e 2(λ1 − 2λ2 ) λ1 (λ1 + λ2 )(4λ1 − λ2 ) −λ2 x + e 2(2λ1 − λ2 ) (λ1 + λ2 )(λ1 − λ2 )3 − (2λ1 − λ2 )(λ1 − 2λ2 ) = ϕ1 (x), sgn
(x) = −
and it suffices to show that ϕ1 (x) ≥ 0 for any x ∈ + . Observe that
ϕ1 (0) =
λ2 λ21 − λ22 λ1 λ21 − λ22 λ1 λ2 λ21 − λ22 − + 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) (2λ1 − λ2 )(λ1 − 2λ2 ) λ21 λ2 (λ1 + λ2 ) λ1 λ22 (λ1 + λ2 ) − + 2(2λ1 − λ2 )(λ1 − 2λ2 ) 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ2 (λ1 + λ2 )(λ1 − 4λ2 ) λ1 (λ1 + λ2 )(4λ1 − λ2 ) − + 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) (λ1 + λ2 )(λ1 − λ2 )3 − (2λ1 − λ2 )(λ1 − 2λ2 )
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
2 λ21 − λ22 (λ1 − λ2 )2 λ1 λ2 λ21 − λ22 − (2λ1 − λ2 )(λ1 − 2λ2 ) 2(2λ1 − λ2 )(λ1 − 2λ2 ) 2 λ1 − λ22 4λ21 − 7λ1 λ2 + 4λ22 + 2(2λ1 − λ2 )(λ1 − 2λ2 ) = 0, =−
and now it suffices to show that ϕ1 (x) ≥ 0 for x ∈ + . Taking the derivative of ϕ1 (x) with respect to x yields
λ1 λ2 λ21 − λ22 −2λ1 x λ1 λ2 λ21 − λ22 −2λ2 x e + e 2λ1 − λ2 λ1 − 2λ2 2 λ1 λ2 (λ1 + λ2 ) λ1 − λ22 −(λ1 +λ2 )x − e (2λ1 − λ2 )(λ1 − 2λ2 ) λ2 λ2 (λ1 + λ2 ) −(2λ1 −λ2 )x λ1 λ22 (λ1 + λ2 ) −(2λ2 −λ1 )x + 1 e + e 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 λ2 (λ1 + λ2 )(λ1 − 4λ2 ) −λ1 x + e 2(λ1 − 2λ2 ) λ1 λ2 (λ1 + λ2 )(4λ1 − λ2 ) −λ2 x − e 2(2λ1 − λ2 ) λ1 − λ2 −2λ1 x λ1 − λ2 −2λ2 x sgn = − e + e 2λ1 − λ2 λ1 − 2λ2 λ21 − λ22 − e−(λ1 +λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ2 + e−(2λ1 −λ2 )x + e−(2λ2 −λ1 )x 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) 4λ1 − λ2 −λ2 x λ1 − 4λ2 −λ1 x + e − e 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 − λ2 −(2λ1 −λ2 )x λ1 − λ2 −λ2 x sgn = − e + e 2λ1 − λ2 λ1 − 2λ2 2 2 λ1 − λ2 λ1 − e−λ1 x + e−2(λ1 −λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) 2(λ1 − 2λ2 ) λ2 λ1 − 4λ2 −(λ1 −λ2 )x + e(λ1 −λ2 )x + e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) 4λ1 − λ2 − 2(2λ1 − λ2 ) = ϕ2 (x).
ϕ1 (x) = −
Notice that
λ21 − λ22 λ1 − λ2 λ1 − λ2 + − 2λ1 − λ2 λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ2 + + 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 − 4λ2 4λ1 − λ2 + − 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 λ2 = + 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) 4λ1 − λ2 λ1 − 4λ2 + − 2(λ1 − 2λ2 ) 2(2λ1 − λ2 )
ϕ2 (0) = −
= 0, it is enough to show that ϕ2 (x) ≥ 0 in x ∈ (0, +∞). Taking the derivative of ϕ2 (x) with respect to x, we have
λ2 (λ1 − λ2 ) −λ2 x e λ1 − 2λ2 2 λ1 λ1 − λ22 λ1 (λ1 − λ2 ) −2(λ1 −λ2 )x + e−λ1 x − e (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 − 2λ2 λ2 (λ1 − λ2 ) (λ1 −λ2 )x (λ1 − 4λ2 )(λ1 − λ2 ) −(λ1 −λ2 )x + e − e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) λ2 λ1 (λ1 + λ2 ) sgn −(2λ1 −λ2 )x = e − e−λ2 x + e−λ1 x λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 )
ϕ2 (x) = (λ1 − λ2 )e−(2λ1 −λ2 )x −
409
λ1 λ2 e−2(λ1 −λ2 )x + e(λ1 −λ2 )x λ1 − 2λ2 2(2λ1 − λ2 ) λ1 − 4λ2 −(λ1 −λ2 )x − e 2(λ1 − 2λ2 ) λ2 λ1 (λ1 + λ2 ) sgn −λ1 x = e − e(λ1 −2λ2 )x + e−λ2 x λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ2 − e−(λ1 −λ2 )x + e2(λ1 −λ2 )x λ1 − 2λ2 2(2λ1 − λ2 ) λ1 − 4λ2 − 2(λ1 − 2λ2 ) = ϕ3 (x). −
It is easy to check that
λ2 λ1 (λ1 + λ2 ) λ1 + − λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 − 2λ2 λ2 λ1 − 4λ2 + − 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) λ1 (λ1 + λ2 ) λ1 + 2λ2 λ2 = − + (2λ1 − λ2 )(λ1 − 2λ2 ) 2(λ1 − 2λ2 ) 2(2λ1 − λ2 )
ϕ3 (0) = 1 −
= 0. Thus, the desired result can be obtained if we could show that ϕ3 (x) ≥ 0. Observe further that
λ1 λ2 (λ1 + λ2 ) e−λ2 x (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 (λ1 − λ2 ) −(λ1 −λ2 )x λ2 (λ1 − λ2 ) 2(λ1 −λ2 )x + e + e λ1 − 2λ2 2λ1 − λ2 λ1 (λ1 − λ2 ) −(λ1 −2λ2 )x sgn = −λ1 e−(λ1 −λ2 )x − λ2 e(λ1 −λ2 )x + e λ1 − 2λ2 λ2 (λ1 − λ2 ) (2λ1 −λ2 )x λ1 λ2 (λ1 + λ2 ) + e − 2λ1 − λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) = ϕ4 (x).
ϕ3 (x) = −λ1 e−λ1 x − λ2 e(λ1 −2λ2 )x −
It is sufficient to show that ϕ4 (0) ≥ 0 and ϕ4 (x) ≥ 0 for x ∈ + . On one hand,
λ1 (λ1 − λ2 ) λ2 (λ1 − λ2 ) + λ1 − 2λ2 2λ1 − λ2 λ1 λ2 (λ1 + λ2 ) − (2λ1 − λ2 )(λ1 − 2λ2 ) 2(λ1 − λ2 )2 (λ1 + λ2 ) λ1 λ2 (λ1 + λ2 ) = − − (λ1 + λ2 ) (2λ1 − λ2 )(λ1 − 2λ2 ) (2λ1 − λ2 )(λ1 − 2λ2 )
ϕ4 (0) = − λ1 − λ2 +
= 0, and on the other hand,
ϕ4 (x) = λ1 (λ1 − λ2 )e−(λ1 −λ2 )x − λ2 (λ1 − λ2 )e(λ1 −λ2 )x − λ1 (λ1 − λ2 )e−(λ1 −2λ2 )x + λ2 (λ1 − λ2 )e(2λ1 −λ2 )x sgn = λ1 − λ2 e2(λ1 −λ2 )x − λ1 eλ2 x + λ2 e(3λ1 −2λ2 )x = ϕ5 (x). Since ϕ5 (0) = 0, it needs to prove that ϕ5 (x) ≥ 0. Taking the derivative of ϕ5 (x) with respect to x, we have
ϕ5 (x) = −2(λ1 − λ2 )e2(λ1 −λ2 )x − λ1 eλ2 x + (3λ1 − 2λ2 )e(3λ1 −2λ2 )x sgn
= −2(λ1 − λ2 )e(2λ1 −3λ2 )x + (3λ1 − 2λ2 )e3(λ1 −λ2 )x − λ1
sgn
=
ϕ6 (x).
Note that ϕ6 (0) = 0, and that
ϕ6 (x) = −2(λ1 − λ2 )(2λ1 − 3λ2 )e(2λ1 −3λ2 )x + 3(λ1 − λ2 )(3λ1 − 2λ2 )e3(λ1 −λ2 )x sgn = −2(2λ1 − 3λ2 )e(2λ1 −3λ2 )x + 3(3λ1 − 2λ2 )e3(λ1 −λ2 )x
410
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
= 3(3λ1 − 2λ2 )eλ1 x − 2(2λ1 − 3λ2 )
sgn
From the above two density functions, we have
F TC (x) =
≥ 3(3λ1 − 2λ2 ) − 2(2λ1 − 3λ2 ) = 5λ1 > 0.
and
Thus, we now can conclude by the following relationship
ϕ6 (x) ≥ 0 =⇒ ϕ5 (x) ≥ 0 =⇒ ϕ4 (x) ≥ 0 =⇒ ϕ3 (x) ≥ 0 =⇒ ϕ2 (x) ≥ 0 =⇒ ϕ1 (x) ≥ 0, Subcase 2.2: λ1 = 2λ2
fTC (x) = 4λ[−e
+ 3e
− 3e
−2λx
+e
−λx
]
λ1 x + 1 −
λ1 λ2
Denote λ2 = λ and we then have λ1 = 2λ. One can compute that −3λx
2λ21 e−λ1 x λ2 (λ1 − λ2 )
2λ22 e−λ2 x + λ2 x + 1 + λ1 (λ1 − λ2 )
2 λ21 + λ22 e−(λ1 +λ2 )x . + (λ1 + λ2 )x + 1 +
F TS (x) =
that ϕ1 (x) ≥ 0. Hence, it holds that (x) is increasing in x ∈ + .
−4λx
(λ1 x + 1)e−λ1 x + (λ2 x + 1)e−λ2 x − (λ1 x + 1)(λ2 x + 1)e−(λ1 +λ2 )x
Thus,
F TS (x) − F TC (x) =
and
fTS (x) =
4 λ[−e−4λx − (6λx + 3)e−2λx + 4e−λx ]. 3
We need to show that
fTS (x) fTC (x)
(x) =
= 2λ32 eλ1 x − 2λ31 eλ2 x + λ21 λ22 (λ1 − λ2 )x2 + 2λ1 λ2 λ21 − λ22 x + 2 λ31 − λ32
sgn
−e−4λx − (6λx + 3)e−2λx + 4e−λx
∝
−e−4λx + 3e−3λx − 3e−2λx + e−λx
= 1+
=
3(e3λx − 2λxe2λx − eλx ) e3λx − 3e2λx + 3eλx − 1
is increasing in x ∈ + for λ > 0. Let y = λx and then it is equivalent to proving that
ψ(y) =
e
− 3e
2y
y
+ 3e − 1
is increasing in y ∈ + . Taking the derivative of ψ(y) with respect to y, we get sgn
ψ (y) = [3e3y − 2(2y + 1)e2y − ey ] × (e3y − 3e2y + 3ey − 1) − (3e3y − 6e2y + 3ey ) × (e3y − 2ye2y − ey ) sgn
= (2y − 5)e4y + 14e3y − (6y + 12)e2y + (4y + 2)ey + 1
ψ1 (y).
=
fTC (x) = x{λ21 e−λ1 x + λ22 e−λ2 x 2 −(λ1 +λ2 )x 2 ]e
− [λ1 λ2 (λ1 + λ2 )x + λ + λ
0
=
}
=
Since ξ2 (0) = 0, we now need to show that ξ2 (x) ≥ 0. One can see that
−λ1 (x−z)
λ21 xe−λ1 x −
λ2 eλ1 x − λ1 eλ2 x + λ1 − λ2 = ξ3 (x).
sgn
=
ξ3 (x) = λ1 λ2 eλ1 x − λ1 λ2 eλ2 x = eλ1 x − eλ2 x ≥ 0,
+ λ2 e
−λ2 (x−z)
− (λ1 + λ2 )e
−(λ1 +λ2 )(x−z)
2λ e−λ1 x + λ22 xe−λ2 x λ2 (λ1 − λ2 ) 3 1
2λ32 + e−λ2 x + (λ1 + λ2 )2 xe−(λ1 +λ2 )x λ1 (λ1 − λ2 ) 2(λ1 + λ2 ) λ21 + λ22 −(λ1 +λ2 )x + e .
λ1 λ2
it follows that ξ1 (x) ≥ 0, i.e., F TS (x) ≥ F TC (x). Hence, the theorem follows.
Denote
SC = ∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ),
[λ1 e−λ1 z + λ2 e−λ2 z − (λ1 + λ2 )e−(λ1 +λ2 )z ]
×[λ1 e
λ22 eλ1 x − λ21 eλ2 x + λ1 λ2 (λ1 − λ2 )x + λ21 − λ22 = ξ2 (x).
sgn
Proof of Theorem 4.2 2 1
x
ξ1 (x) = 2λ1 λ32 eλ1 x − 2λ31 λ2 eλ2 x + 2λ21 λ22 (λ1 − λ2 )x + 2λ1 λ2 λ21 − λ22
sgn
Without loss of generality, we suppose that λ1 ≥ λ2 . For the case λ1 = λ2 , the result is trivially true from Theorem 3.5. In the following, we need to treat the case when λ1 > λ2 . In this case, we can write
ξ1 (0) = 2λ32 − 2λ31 + 2 λ31 − λ32 = 0,
Since ξ3 (0) = 0 and
Proof of Theorem 3.7
fTS (x) =
So, F TS (x) − F TC (x) ≥ 0 is equivalent to ξ1 (x) ≥ 0 for any positive real number x. Since
ξ2 (x) = λ1 λ22 eλ1 x − λ21 λ2 eλ2 x + λ1 λ2 (λ1 − λ2 )
Upon using the method similar to that of Subcase 2.1, the desired result can be reached for this case and we omit the proof details for brevity.
and
ξ1 (x), say.
it suffices to show that ξ1 (x) ≥ 0. Taking the derivative of ξ1 (x) with respect to x, we have
e3y − 2ye2y − ey 3y
(λ1 x + 1)(λ2 x + 1)e−(λ1 +λ2 )x 2λ21 2λ22 − e−λ1 x + e−λ2 x λ2 (λ1 − λ2 ) λ1 (λ1 − λ2 )
2 λ21 + λ22 e−(λ1 +λ2 )x + (λ1 + λ2 )x + 1 + λ1 λ2
SS = [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )]. ]dz
The survival function of SC can be written as
F SC (x) =
n
[1 − (1 − e−λi x )2 ],
i=1
and thus its density function is given by
fSC (x) = 2
n i=1
λi e−λi x (1 − e−λi x ) F S (x). 1 − (1 − e−λi x )2 C
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
On the other hand, the survival function of SS can be written as
F SS (x) = 1 − (1 − e
),
n
i=1 λi , and its density function is given by
where =
2e−x (1 − e−x ) fSS (x) = 2e−x (1 − e−x ) = F SS (x). 1 − (1 − e−x )2 Now we need to show that the ratio
n
λi e−λi x (1 − e−λi x ) i=1 1 − (1 − e−λi x )2 F SC (x) e−x (1 − e−x ) F SS (x) 1 − (1 − e−x )2
f (x) (x) = SC ∝ fSS (x)
is increasing in x. From Theorem 3.1 in Boland and El-Neweihi (1995), the ratio F SC (x)/F SS (x) is increasing in x and so it is enough if we could show the function
n
λi e−λi x (1 − e−λi x ) n λi (1 − e−λi x ) i=1 1 − (1 − e−λi x )2 i=1 2 − e−λi x = e−x (1 − e−x ) 1 − e−x 1 − (1 − e−x )2 2 − e−x
g(x) =
is increasing in x. The derivative of g(x) is
g (x) =
1
1 − e−x 2 − e−x
−
n i=1
sgn
=
−
i=1 sgn
=
−
(2 − e−λi x )2
n i=1
λ2i e−λi x
(2 − e−λi x )2 e−x
1 − e−x 2 − e−x
(2 − e−x )2
(1 − e−x )
λi (1 − e−λi x ) e−x 2 − e−λi x (2 − e−x ) λ2i e−λi x
n i=1
2 − e−λi x
(2 − e−λi x )2
n
i=1
λi (1 − e−λi x )
λ2i e−λi x
n i=1
2
n
(1 − e−x )(2 − e−x ) e−x
λi (1 − e−λi x ) 2 − e−λi x
= h(x).
It is well known that, for an exponential random variable with hazard rate λ, its reversed hazard rate is decreasing in λ. Based on this fact, we have the following inequality:
λi e−λi x 1−
e−λi x
≥
for any i. Now it follows that
h(x) ≥
−x 2
e−x
1 − e−x
for any i, which in turn implies that
(1 − e−x )(2 − e−x ) (1 − e−λi x )(2 − e−λi x ) ≥ e−x λi e−λi x
411
n i=1
−
λ2i e−λi x
(2 −
n i=1
(1 − e−λi x )(2 − e−λi x ) ) λi e−λi x
e−λi x 2
λi (1 − e−λi x ) 2 − e−λi x
= 0.
We finish the entire proof. References Barlow, R., & Proschan, R. (1981). Statistical theory of reliability and life testing, Silver Spring, MD: Madison. Boland, P. J., & El-Neweihi, E. (1995). Component redundancy versus system redundancy in the hazard rate ordering. IEEE Transactions on Reliability 44, 614–619. Boland, P. J., El-Neweihi, E., & Proschan, F. (1992). Stochastic order for redundancy allocation in series and parallel systems. Advances in Applied Probability 24, 161– 171. Brito, G., Zequeira, R. I., & Valdès, J. E. (2011). On the hazard rate and reversed hazard rate orderings in two-component series systems with active redundancies. Statistics & Probability Letters 81, 201–206. da Costa Bueno, V. (2005). Minimal standby redundancy allocation in a k-out-of-n: F system of dependent components. European Journal of Operational Research 165, 786–793. da Costa Bueno, V., & do Carmo, I. M. (2007). Active redundancy allocation for a k-outof-n: F system of dependent components. European Journal of Operational Research 176, 1041–1051. Gupta, R. D., & Nanda, A. K. (2001). Some results on reversed hazard rate orderings. Communications in Statistics – Theory and Methods 30, 2447–2457. Hu, T., & Wang, T. (2009). Optimal allocation of active redundancies of r-out-of-n systems. Journal of Statistical and Planning Inference 139, 3733–3737. Laniado, H., & Lillo, R. E. (2014). Allocation policies of redundancies in two-parallelseries and two-series-parallel systems. IEEE Transactions on Reliability 63, 223–229. Meng, F. C. (1996). More on optimal allocation of components in coherent systems. Journal of Applied Probability 33, 548–556. Mi, J. (1998). Bolstering components for maximizing system lifetime. Naval Research Logistics 45, 497–509. Misra, N., Dhariyal, I. D., & Gupta, N. (2009). Optimal allocation of active spares in series systems and comparison of component and system redundancies. Journal of Applied Probability 46, 19–34. Misra, N., Misra, A. K., & Dhariyal, I. D. (2011a). Active reduandancy allocations in series systems. Probability in the Engineering and Informational Sciences 25, 219–235. Misra, N., Misra, A. K., & Dhariyal, I. D. (2011b). Standby redundancy allocations in series and parallel systems. Journal of Applied Probability 48, 43–45. Ramirez-Marquez, J. E., & Coit, D. W. (2004). A heuristic for solving the redundancy allocation problem for multi-state series-parallel systems. Reliability Engineering & System Safety 83, 341–349. Shaked, M., & Shanthikumar, J. G. (1992). Optimal allocation of resources to nodes of parallel and series systems. Advances in Applied Probability 24, 894–914. Shaked, M., & Shanthikumar, J. G. (2007). Stochastic orders. New York: Springer-Verlag. Singh, H., & Misra, H. (1994). On redundancy allocation in systems. Journal of Applied Probability 31, 1004–1014. Singh, H., & Singh, R. S. (1997). On allocation of spares at component level versus system level. Journal of Applied Probability 34, 283–287. Valdés, J. E., Arango, G., & Zequeira, R. I. (2010). Some stochastic comparisons in series systems with active redundancy. Statistics & Probability Letters 80, 945–949. Valdés, J. E., & Zequeira, R. I. (2006). On the optimal allocation of two active redundancies in a two-component series systems. Operations Research Letters 34, 49–52. Yalaoui, A., Chu, C., & Châtelet, E. (2005). Reliability allocation problem in a seriesparallel system. Reliability Engineering & System Safety 90, 55–61. Zhao, P., Chan, P., & Ng, H. K. T. (2012). Optimal allocation of redundancies in series systems. European Journal of Operational Research 220, 673–683. Zhao, P., Chan, P. S., Li, L., & Ng, H. K. T. (2013a). On allocation of redundancies in two-component series systems. Operations Research Letters 41, 690–693. Zhao, P., Chan, P. S., Li, L., & Ng, H. K. T. (2013b). Allocation of two redundancies in two-component series systems. Naval Research Logistics 60, 588–598.
Contents lists available at ScienceDirect
European Journal of Operational Research journal homepage: www.elsevier.com/locate/ejor
Stochastics and Statistics
Redundancy allocation at component level versus system level ✩ Peng Zhao a,∗, Yiying Zhang b , Long Li b a b
School of Mathematics and Statistics, Jiangsu Normal University, Xuzhou 221116, China School of Mathematics and Statistics, Lanzhou University, Lanzhou 730000, China
a r t i c l e
i n f o
Article history: Received 24 August 2013 Accepted 31 August 2014 Available online 22 September 2014 Keywords: Reliability Active redundancy Standby redundancy Component level System level
a b s t r a c t In this paper, we treat the problem of stochastic comparison of standby [active] redundancy at component level versus system level. In the case of standby redundancy, we present some interesting comparison results of both series systems and parallel systems in the sense of various stochastic orderings for both the matching spares case and non-matching spares case, respectively. In the case of active redundancy, a likelihood ratio ordering result of series systems is presented for the matching spares case; and for the non-matching spares case, a counterexample is provided to show that there does not exist similar result even for the hazard rate ordering. The results established here strengthen and generalize some of those known in the literature. Some numerical examples are also provided to illustrate the theoretical results. © 2014 Elsevier B.V. All rights reserved.
1. Introduction In order to optimize the lifetime of a coherent system, one efficient method is to add redundancy components to the system. This topic has posed many interesting problems to which many researchers have devoted themselves in the past decades; see, for example, Boland, El-Neweihi, and Proschan (1992); Shaked and Shanthikumar (1992); Singh and Misra (1994); Singh and Singh (1997); Mi (1998); Ramirez-Marquez and Coit (2004); Yalaoui, Chu, and Châtelet (2005); da Costa Bueno (2005); Valdés and Zequeira (2006); da Costa Bueno and do Carmo (2007); Hu and Wang (2009); Valdés, Arango, and Zequeira (2010); Brito, Zequeira, and Valdès (2011); Misra, Dhariyal, and Gupta (2009); Misra, Misra, and Dhariyal (2011a,b); Zhao, Chan, and Ng (2012); Zhao, Chan, Li, and Ng (2013a,b); Laniado and Lillo (2014) and the references therein. In general, there are two types of redundancies called active redundancy and standby redundancy commonly used in reliability engineering and system security. The former [active redundancy] is used when replacement of components during the operation of the system is impossible; in this case the redundancies are put in parallel to components of the system which leads to taking the maximum of random lifetimes. The latter [standby redundancy] is used when replacement of components during the operation of the system is possible; in this case the redundancy starts functioning immediately after the ✩ This work was supported by National Natural Science Foundation of China (11422109), Natural Science Foundation of Jiangsu Province (BK20141155) and A Project Funded by the Priority Academic Program Development of Jiangsu Higher Education Institutions. ∗ Corresponding author. Tel.: +86 516 83500332; fax: +86 516 83500387. E-mail address: [email protected], [email protected] (P. Zhao).
http://dx.doi.org/10.1016/j.ejor.2014.08.040 0377-2217/© 2014 Elsevier B.V. All rights reserved.
corresponding original component in the system fails which leads to taking the convolution of random variables. It is known that the performance of a coherent system consisting of independent components can be enhanced by putting spares to each of its components or by creating a duplicate system consisting of spares similar to the original coherent system. For example, there is a series system with n components and we have another n components at hand as spares. In order to enhance the lifetime of the original series system, we can assemble these n spares either on component level (Fig. 1(a)) or system level (Fig. 1(b)). Naturally, it is of great interest and importance to make sure which type is superior to the other. To begin, we suppose the general coherent system φ consists of n components C1 , . . . , Cn having independent lifetimes X1 , . . . , Xn . Denote τ (X) = τ (X1 , . . . , Xn ) the lifetime of the coherent system φ . Suppose now n standby spares R1 , . . . , Rn have independent lifetimes Y1 , . . . , Yn , and (X1 , . . . , Xn ) and (Y1 , . . . , Yn ) are statistically independent. In component redundancy case, we allocate a standby spare Ri to the component Ci , i = 1, . . . , n. Then the resultant coherent system, denoted by TC , has lifetime τ (X + Y) = τ (X1 + Y1 , . . . , Xn + Yn ). In system redundancy case, we duplicate the coherent system φ with components C1 , . . . , Cn by R1 , . . . , Rn and make it available as a standby redundant spare to the coherent system φ . The resultant coherent system, denoted by TS , has lifetime τ (X) + τ (Y). A well-known principle among design engineers states that redundancy at the component level is always better than at the system level (cf. Barlow & Proschan, 1981). This statement, however, is not true in general for the standby redundancy case though it is for the active redundancy case. Boland and El-Neweihi (1995) concluded that, in the sense of the usual stochastic order, redundancy at component level is better than redundancy at system level for series systems, while the reverse is true for parallel
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
X1
X2
Xn
X1
X2
It would be very interesting to examine some stronger stochastic orderings. In the case of matching spares, i.e., Xi =st Yi , i = 1, . . . , n, Boland and El-Neweihi (1995) proved, for series systems, that the component redundancy is superior to the system redundancy in the sense of the hazard rate ordering, i.e.,
Xn
∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥hr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )].
Y1
Y2
Yn
(a) Component level
Y1
Y2
Yn
(b) System level
Fig. 1. Component level versus system level.
systems, i.e.,
∧(X1 + Y1 , . . . , Xn + Yn ) ≥st [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )]
(1)
and
∨(X1 + Y1 , . . . , Xn + Yn ) ≤st [∨(X1 , . . . , Xn )] + [∨(Y1 , . . . , Yn )],
(2)
where the symbols ‘∧’ and ‘∨’ mean min and max, respectively. Boland and El-Neweihi (1995) also showed that, for series systems, if all random lifetimes are exponentially i.i.d., then the usual stochastic order in (1) can be strengthened to the hazard rate order. On the other hand, Meng (1996) proved that (1) holds if and only if the system is series and (2) holds if and only if the system is parallel. What does this suggest about the non-i.i.d. case? Here, we shall present some comparison results in the sense of various stochastic orderings when the original components and spares have independent but nonidentical exponential distributed lifetimes. Specifically, for the standby redundancy of series systems, if either Xi =st Yi (i = 1, . . . , n) or X1 =st . . . =st Xn and Y1 =st . . . =st Yn , it is established that
∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )]. For the standby redundancy of parallel systems, if X1 =st X2 and Y1 =st Y2 , we establish
∨(X1 + Y1 , X2 + Y2 ) ≤lr [∨(X1 , X2 )] + [∨(Y1 , Y2 )], and if Xi =st Yi (i = 1, 2), we prove
∨(X1 + Y1 , X2 + Y2 ) ≤st [∨(X1 , X2 )] + [∨(Y1 , Y2 )]. Consider a series system with two components and two active spares. In the case of matching spares, i.e., Xi =st Yi , i = 1, 2, the redundancy at component level is better in the hazard rate ordering than the redundancy at system level (see Boland & El-Neweihi, 1995). An analogous result was proved for the reversed hazard rate ordering in Gupta and Nanda (2001). Brito, Zequeira, and Valdès (2011) further considered the case of matching spares and gave a sufficient condition under which the likelihood ratio ordering holds. Gupta and Nanda (2001) proved the redundancy at component level is better in the reversed hazard rate ordering than the redundancy at system level in the case of non-matching spares, X1 =st X2 and Y1 =st Y2 . Boland and El-Neweihi (1995) showed with a counterexample that the hazard rate ordering does not always exist between them. In this regard, Brito, Zequeira, and Valdès (2011) proved that, if the reversed hazard rates of X1 and Y1 are proportional, then the hazard rate ordering holds. Obviously, the exponential case is not satisfied with the proportional condition, and we provide a counterexample to depict the hazard rate ordering does not hold for the exponential case in Section 4 (Example 4.1). We next consider the general coherent systems with n components and n active spares. Similar to the standby case, in component redundancy case, the resultant coherent system, denoted by SC , has lifetime τ (X ∨ Y) = τ (X1 ∨ Y1 , . . . , Xn ∨ Yn ). In system redundancy case, the resultant coherent system, denoted by SS , has lifetime τ (X) ∨ τ (Y). For general coherent systems, it is well known (cf. Barlow and Proschan, 1981) that the component redundancy is better than system redundancy in the sense of the usual stochastic ordering, i.e.,
τ (X ∨ Y) ≥st τ (X) ∨ τ (Y).
403
(3)
The above result can be readily extended to series–parallel systems (cf. Corollary 1-1 in Boland and El-Neweihi 1995). When all components and spares are i.i.d., they also proved, for general coherent systems, τ (X ∨ Y) ≥hr τ (X) ∨ τ (Y) under some mild conditions. Though this result cannot be applied to the general k-out-of-n system, but they showed the result is true for the special case of 2-out-of-n system and left the general case for an open problem. Gupta and Nanda (2001) obtained a similar result for the reversed hazard rate ordering. In fact, Singh and Singh (1997) have solved the open problem posed by Boland and El-Neweihi (1995) and established a stronger result in the sense of the likelihood ratio ordering for the general k-out-of-n system with all components and spares having i.i.d. absolutely continuous lifetimes. Also, there remains an open problem that whether this result still holds for the case when we just have the condition Xi =st Yi , i = 1, . . . , n (matching case) and such a result for two-component series system has been shown in Brito, Zequeira, and Valdès (2011) wherein they require that the hazard rates are proportional. On the other hand, Misra, Dhariyal, and Gupta (2009) studied the non-matching case and obtained some new results. In this paper, we present a likelihood ratio order result for the series system with n exponential components under the setup of matching case (Theorem 4.2). Specifically, for the active redundancy of series systems, if Xi =st Yi (i = 1, . . . , n), we establish
∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥lr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )]. This paper is organized as follows. In Section 2, we recall some pertinent notions of stochastic orderings which will be used in the sequel. Section 3 presents some comparison results of series systems and parallel systems in the sense of various stochastic orderings for standby redundancy at component level versus system level. Section 4 carries out stochastic comparisons of series systems between component level and system level in the sense of the likelihood ratio ordering under the setup of matching spares as well as for the nonmatching spares: a counterexample is provided to show that there does not exist similar comparison result even for the hazard rate ordering. Conclusion is finally made in Section 5. The technical details of the proofs of all the main results are presented in Appendix A. 2. Definitions and notations Throughout this paper, the term increasing is used for monotone non-decreasing and decreasing is used for monotone non-increasing. Definition 2.1. For two random variables X and Y with densities fX and fY , and distribution functions FX and FY , respectively, let F X = 1 − FX and F Y = 1 − FY be the corresponding survival functions. Then: (i) X is said to be smaller than Y in the likelihood ratio order (denoted by X ≤lr Y) if fY (x)/fX (x) is increasing in x; (ii) X is said to be smaller than Y in the hazard rate order (denoted by X ≤hr Y) if F Y (x)/F X (x) is increasing in x; (iii) X is said to be smaller than Y in the reversed hazard rate order (denoted by X ≤rh Y) if FY (x)/FX (x) is increasing in x; (iv) X is said to be smaller than Y in the stochastic order (denoted by X ≤st Y) if F Y (x) ≥ F X (x), ∀ x. It is known that the likelihood ratio order implies both the hazard rate order and the reversed hazard rate order which in turn implies the usual stochastic order, but neither the hazard rate order nor the reversed hazard rate order order implies the other; see Shaked and Shanthikumar (2007).
404
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411 300
3000 f (x)/f (x) T
S
250
Likelihood ratio function
Likelihood ratio function
T
C
2500
2000
1500
1000
500
200
150
100
50
fT (x)/fT (x) C
0
0
0.5
1
1.5
2
2.5 x
3
3.5
4
4.5
0
5
Fig. 2. Plot of the likelihood ratio function fTC (x)/fTS (x) when n = 4, λ1 = 3, λ2 = 5, λ3 = 4, and λ4 = 6.
3. Standby redundancy at component level vs. system level In this section, we present some comparison results for series systems and parallel systems having standby redundancy at component level versus system level. 3.1. Series systems In the case of series system having matching spares, i.e., Xi =st Yi , i = 1, 2, . . . , n, we have the following result which states that the redundancy at component level is more efficient than the redundancy at system level in the sense of the likelihood ratio order. Theorem 3.1. Let X1 , . . . , Xn be independent exponential lifetimes with respective parameters λ1 , . . . , λn , and Y1 , . . . , Yn be another independent set of exponential lifetimes with respective parameters λ1 , . . . , λn . Then,
τ (X + Y) = ∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )] = τ (X) + τ (Y). Example 3.2. Set n = 4, λ1 = 3, λ2 = 5, λ3 = 4, and λ4 = 6 in Theorem 3.1. Fig. 2 plots the likelihood ratio function fTC (x)/fTS (x). It can be seen that fTC (x)/fTS (x) is increasing in x ∈ + , which shows the validity of the result in Theorem 3.1. In the case of non-matching spares, we have the following similar result. Theorem 3.3. Let X1 , . . . , Xn be independent exponential random lifetimes with common parameter λ1 , and Y1 , . . . , Yn be independent exponential random lifetimes with common parameter λ2 . Then,
τ (X + Y) = ∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )] = τ (X) + τ (Y). Example 3.4. Set n = 5, λ1 = 4 and λ2 = 3 in Theorem 3.3. By the symmetry, we know that it is same as the case λ1 = 3 and λ2 = 4. Fig. 3 plots the ratio of the two density functions. It can be seen fTC (x)/fTS (x) is increasing in x ∈ + , which is in accordance with the theoretical result in Theorem 3.3. 3.2. Parallel systems We next discuss the case of parallel systems. In this case, due to the extreme complexity of distribution theory, we only deal with the special case when n = 2 and it can be found that the comparison result is reversed compared to the case of series systems. Now, let us first go into the non-matching spares.
0
1
2
3 x
4
S
5
6
Fig. 3. Plot of the likelihood ratio function fTC (x)/fTS (x) when n = 5, λ1 = 4 and λ2 = 3.
Theorem 3.5. Let X1 , X2 be independent exponential random lifetimes with common parameter λ1 and Y1 , Y2 be independent exponential random lifetimes with common parameter λ2 . Then,
τ (X + Y) = ∨(X1 + Y1 , X2 + Y2 ) ≤lr [∨(X1 , X2 )] + [∨(Y1 , Y2 )] = τ (X) + τ (Y). Example 3.6. Under the setup in Theorem 3.5, we consider the likelihood ratio fTS (x)/fTC (x) under the condition λ1 ≥ λ2 . Based on this, we suppose that the parameters in (a), (b), (c) and (d) below satisfy the cases λ1 = λ2 , λ2 < λ1 < 2λ2 , λ1 > 2λ2 and λ1 = 2λ2 , respectively. It can be seen from Fig. 4 that the likelihood ratio functions for all these four cases are increasing, which coincides with the theoretic result in Theorem 3.5. (a) (b) (c) (d)
λ1 = 0.6, λ2 = 0.6; λ1 = 3, λ2 = 2; λ1 = 5, λ2 = 0.7; λ1 = 8, λ2 = 4.
For the matching spares case, we only have the following comparison result in the sense of the usual stochastic order. Theorem 3.7. Let X1 , X2 be independent exponential lifetimes with respective parameters λ1 , λ2 , and Y1 , Y2 be another independent exponential set with respective parameters λ1 , λ2 . Then,
τ (X + Y) = ∨(X1 + Y1 , X2 + Y2 ) ≤st [∨(X1 , X2 )] + [∨(Y1 , Y2 )] = τ (X) + τ (Y). Example 3.8. Set λ1 = 5 and λ2 = 3 in Theorem 3.7. Fig. 5 plots the survival functions of TC and TS . It can be seen that the survival function of TS is larger than that of TC for x ∈ + , which shows the validity of the result in Theorem 3.7. One may wonder whether the result in Theorem 3.7 can be strengthened to the hazard rate ordering. The following Example 3.9 gives a negative answer to this problem. Example 3.9. Denote by hTC (x)) = fTC (x)/F TC (x) and hTS (x) = fTS (x)/F TS (x) the hazard rate functions of TC and TS , respectively. Fig. 6 plots the shape of hTC (x) and hTS (x). It can be observed that we cannot compare these two hazard rate functions in this regard. Consequently, the likelihood ratio order does not hold as well. 4. Active redundancy at component level vs. system level In this section, we carry out stochastic comparisons of series systems at component level and system level for the matching spares
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
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1.5
1
1.4 0.9 1.3 0.8 Likelihood ratio function
Likelihood ratio function
1.2
0.7
0.6
1.1
1
0.9 0.5 0.8 f (x)/f (x)
f (x)/f (x)
0.4
T
T
S
0
5
10
15
20
25
30
35
T
0.7
C
40
0
1
2
3
4
x
x
(a)
(b)
1.15
T
S
5
C
6
7
1.4
1.1
1.3
1.05 1.2
Likelihood ratio function
Likelihood ratio function
1
0.95
0.9
0.85
1.1
1
0.9
0.8 0.8 0.75 f (x)/f (x) T
0.7
0.65
0
1
2
3
4
5
f (x)/f (x)
0.7
T
S
T
C
6
S
7
0
0.5
1
1.5
x
x
(c)
(d)
2
2.5
T C
3
Fig. 4. Plot of likelihood ratio functions fTS (x)/fTC (x) when λ1 ≥ λ2 .
case in the sense of the likelihood ratio order. First we provide a counterexample to show the comparison result does not hold even for the hazard rate order in the non-matching spares case. Example 4.1. Set λ1 = 3.5 and λ2 = 2. The survival functions of Q1 = ∧(X1 ∨ Y1 , X2 ∨ Y2 ) and Q2 = ∨(X1 ∧ X2 , Y1 ∧ Y2 ) can be written as
F Q1 (x) = [1 − (1 − e−3.5x )(1 − e−2x )]2 and
F Q2 (x) = 1 − [(1 − e−2×3.5x )][(1 − e−2×2x )]. But, the ratio is not monotone as can be seen in Fig. 7.
Now, we are ready to present our main result in the case of matching spares. Theorem 4.2. Let X1 , . . . , Xn be independent exponential lifetimes with respective parameters λ1 , . . . , λn , and Y1 , . . . , Yn be another independent exponential set with respective parameters λ1 , . . . , λn . Then,
τ (X ∨ Y) = ∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥lr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )] = τ (X) ∨ τ (Y). (4) Example 4.3. Set n = 4, λ1 = 2, λ2 = 0.8, λ3 = 1.6 and λ4 = 2.5 in Theorem 4.2. Fig. 8 plots the ratio of the two density functions. One
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8
1 7
Survival function of TC
0.8
Survival function of TS
6 Likelihood ratio function
0.9
0.7 0.6 0.5 0.4
4 3 2
0.3
f (x)/f (x) S
1
0.2 0.1 0
5
0
0
0.5
1
1.5 x
2
2.5
c
0
0.5
1
3
1.5
2 x
2.5
3
S
s
3.5
4
Fig. 8. Plot of likelihood ratio function fSC (x)/fSS (x) when λ1 = 2, λ2 = 0.8, λ3 = 1.6 and λ4 = 2.5.
Fig. 5. Plot of F TC (x) and F TS (x) when λ1 = 5 and λ2 = 3.
can observe that fSC (x)/fSS (x) is increasing in x ∈ + , which fits the theoretical result in Theorem 4.2. 4
5. Concluding remarks 3.5 3 2.5 2 1.5 1
hT (x)
0.5
hT (x)
C
0
S
0
0.5
1
1.5 x
2
2.5
3
Fig. 6. Plot of hTC (x) and hTS (x) when λ1 = 7 and λ2 = 4.
Let us suppose the general coherent system φ consists of n components C1 , . . . , Cn having independent lifetimes X1 , . . . , Xn . Denote τ (X) = τ (X1 , . . . , Xn ) the lifetime of the coherent system φ . Suppose now n standby[active] spares R1 , . . . , Rn have independent lifetimes Y1 , . . . , Yn , and (X1 , . . . , Xn ) and (Y1 , . . . , Yn ) are statistically independent. In component redundancy case, we allocate a standby[active] spare Ri to the component Ci , i = 1, . . . , n. Then the resultant coherent system, denoted by TC [SC ], has lifetime τ (X + Y) = τ (X1 + Y1 , . . . , Xn + Yn )[τ (X ∨ Y) = τ (X1 ∨ Y1 , . . . , Xn ∨ Yn )]. In system redundancy case, we duplicate the coherent system φ with components C1 , . . . , Cn by R1 , . . . , Rn and make it available as a standby[active] redundant spare to the coherent system φ . The resultant coherent system, denoted by TS [SS ], has lifetime τ (X) + τ (Y)[τ (X) ∨ τ (Y)]. Let X1 , . . . , Xn be independent exponential lifetimes with respective parameters λ1 , . . . , λn , and Y1 , . . . , Yn be another independent exponential set with respective parameters λ1 , . . . , λn . In this article, for the standby redundancy of series systems, it is established that
τ (X + Y) = ∧(X1 + Y1 , . . . , Xn + Yn ) ≥lr [∧(X1 , . . . , Xn )] + [∧(Y1 , . . . , Yn )] = τ (X) + τ (Y)
1.45
and for the standby redundancy of parallel systems, we establish 1.4
τ (X + Y) = ∨(X1 + Y1 , X2 + Y2 ) ≤st [∨(X1 , X2 )] + [∨(Y1 , Y2 )] = τ (X) + τ (Y).
1.35
For the active redundancy of series systems, we have
1.3
τ (X ∨ Y) = ∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ) ≥lr [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )] = τ (X) ∨ τ (Y).
1.25 1.2
Let W1 , . . . , Wn be independent exponential lifetimes with common parameter λ1 , and V1 , . . . , Vn be another set of independent exponential lifetimes with common parameter λ2 . For the standby redundancy of series systems, it is established that
1.15 1.1
τ (W + V) = ∧(W1 + V1 , . . . , Wn + Vn ) ≥lr [∧(W1 , . . . , Wn )] + [∧(V1 , . . . , Vn )] = τ (W) + τ (V)
1.05 1
0
0.5
1
1.5 x
2
Fig. 7. Plot of the ratio function F Q1 (x)/F Q2 (x).
2.5
3
and for the standby redundancy of parallel systems, we establish
τ (W + V) = ∨(W1 + V1 , W2 + V2 ) ≤lr [∨(W1 , W2 )] + [∨(V1 , V2 )] = τ (W) + τ (V).
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
For the active redundancy of series systems, we can conclude that the comparison result does not hold even for the hazard rate order by a counterexample.
We now have to show the ratio
Acknowledgements
is increasing in x. Denote by
The authors thank the editor and two anonymous referees for the insight comments and suggestions, which resulted in an improvement in the presentation of this manuscript.
ξ (x) =
In this appendix, we provide detailed proofs for all of the theorems in this paper. Proof of Theorem 3.1
i=1
fTC (x) =
fXi +Yi (x)
n
n
λ2i xe−λi x
(λj x + 1)e−λj x
j =i
= xe−x
n
λ2i
n
(λj x + 1),
where = ni=1 λi . On the other hand, observe that TS has a gamma distribution with shape parameter 2 and scale parameter with its density function
fTS (x) = 2 xe−x . Now it can be readily verified that the ratio
n
i=1
λ2i
n
j =i
(λj x + 1)
n
and thus its density function is given by
nλ1 λ2 [λ e−λ2 x − λ2 e−λ1 x ]n−1 [e−λ2 x − e−λ1 x ]. (λ1 − λ2 )n 1
On the other hand, the survival function of TS is
F TS (x) =
λ1 − λ2
e
−nλ2 x
and
β = (x1 − x2 )[en(x1 −x2 ) − nex1 −x2 + n − 1]. It is known from the assumption that x1 > x2 > 0, which implies α ≥ 0. On the other hand, observe that
β (y) = neny − ney = ney [e(n−1)y − 1] ≥ 0.
λ1 λ2 e−λ2 x − e−λ1 x λ1 − λ2 λ1 − λ2
λ1
α = nx2 (ex1 −x2 − 1)[e(n−1)(x1 −x2 ) − 1]
Let y = x1 − x2 and we then have β(y) = eny − ney + n − 1. It can be verified that
Without loss of generality, we suppose that λ1 ≥ λ2 . For the case λ1 = λ2 , the result can be derived from Theorem 3.1 and we only need to prove the case when λ1 > λ2 . The survival function of TC can be written as
fTC (x) =
Let x1 = λ1 x and x2 = λ2 x, it is equivalent to proving that ϕ(x1 , x2 ) ≥ 0, where
sgn
Proof of Theorem 3.3
F TC (x) =
sgn
β = en(x1 −x2 ) − nex1 −x2 + n − 1.
2
is increasing in x.
ε(x) = nλ2 [en(λ1 −λ2 )x − e(n−1)(λ1 −λ2 )x ] − nλ1 [e(λ1 −λ2 )x − 1] + (λ1 − λ2 )[en(λ1 −λ2 )x − 1].
where
j =i
i=1
ξ (x) = nλ2 (λ1 − λ2 ){e−[λ1 +(n+1)λ2 ]x − e−[2λ1 +nλ2 ]x } + nλ1 (λ1 − λ2 ){e−[λ2 +(n+1)λ1 ]x − e−[2λ2 +nλ1 ]x } + (λ1 − λ2 )2 {e−[λ1 +(n+1)λ2 ]x − e−[λ2 +(n+1)λ1 ]x } = ε(x).
ϕ(x1 , x2 ) = nx2 [en(x1 −x2 ) − e(n−1)(x1 −x2 ) − (ex1 −x2 − 1)] + (x1 − x2 )[en(x1 −x2 ) − 1] − n(x1 − x2 )(ex1 −x2 − 1) = nx2 (ex1 −x2 − 1)[e(n−1)(x1 −x2 ) − 1] + (x1 − x2 )[en(x1 −x2 ) − nex1 −x2 + n − 1] = α + β , say,
F Xj +Yj (x)
n
i=1
fTC (x) = fTS (x)
and taking the derivative of ξ (x) with respect to x yields that
Observe that
j =i
i=1
=
e−nλ2 x − e−nλ1 x
ϕ(x1 , x2 ) = nx2 [en(x1 −x2 ) − e(n−1)(x1 −x2 )] − nx1 (ex1 −x2 − 1) + (x1 − x2 )[en(x1 −x2 ) − 1].
F Xi +Yi (x)
with its density function n
[λ1 e−λ2 x − λ2 e−λ1 x ]n−1 [e−λ2 x − e−λ1 x ]
Note that
It is easy to see that Xi + Yi (i = 1, . . . , n) has a gamma distribution with shape parameter 2 and scale parameter λi . Denote by fXi +Yi and F Xi +Yi the corresponding density and survival functions, respectively. Then, we can write the survival function of TC as n
fTC (x) [λ1 e−λ2 x − λ2 e−λ1 x ]n−1 [e−λ2 x − e−λ1 x ] ∝ fTS (x) e−nλ2 x − e−nλ1 x
sgn
Appendix A
F TC (x) =
δ(x) =
407
−
λ2
λ1 − λ2
with its density function as
nλ1 λ2 −nλ2 x fTS (x) = [e − e−nλ1 x ]. λ1 − λ2
e
−nλ1 x
from which and the fact β(0) = 1 − n + n − 1 = 0 we know β(y) ≥ 0. Therefore, the theorem follows. Proof of Theorem 3.5 Without loss of generality, we only need to show the result for the case λ1 ≥ λ2 due to the symmetry. The proof will be proceeded by discussing the following two cases. Case 1: λ1 = λ2 Let λ1 = λ2 = λ. Then, the density function of TC can be written as
fTC (x) = 2λ[λxe−λx − λx(λx + 1)e−2λx ]. Also, it is easy to derive the density function of TS as
fTS (x) = 4λ2
0
x
e−λx [1 − e−λ(x−t)][1 − e−λt ]dt
= 4λ[(λx + 2)e−2λx + (λx − 2)e−λx ].
408
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
We need to show that
(x) =
f (x) (x) = TS fTC (x) ∝
∝
λx + 2 + (λx − 2)eλx λxeλx − λx(λx + 1)
is an increasing function in y ∈ + . Taking the derivative of W (y) with respect to y, we have sgn
W (y) = [1 + ey + (y − 2)ey ] × [yey − y(y + 1)] − [ey + yey − 2y − 1] × [y + 2 + (y − 2)ey ] = 2e2y − (y3 − y2 + 4y + 4)ey + y2 + 4y + 2
ω1 (y).
Since ω1 (0) = 0, we just need to prove that ω1 (y) ≥ 0. Notice that
ω1 (y) = 4e2y − (3y2 − 2y + 4)ey − (y3 − y2 + 4y + 4)ey + 2y + 4 = 4e2y − (y3 + 2y2 + 2y + 8)ey + 2y + 4 = ω2 (y), and ω2 (0) = 0. Now, it is sufficient to show that ω2 (y) ≥ 0. We can see that
ω2 (y) = 8e2y − (y3 + 5y2 + 6y + 10)ey + 2 = ω3 (y). Again, it is easy to calculate that ω3 (0) = 0 and sgn
ω3 (y) = 16ey − y3 − 8y2 − 16y − 16 = ω4 (y). Similarly, we have ω4 (0) = 0 and
ω4 (y) = 16ey − 3y2 − 16y − 16 = ω5 (y). Based on the fact ω5 (0) = 0, it is enough to show that ω5 (y) ≥ 0. Observe that
ω5 (y) = 16ey − 6y − 16 = ω6 (y), and then ω6 (0) = 0 and
ω6 (y) = 16ey − 6 ≥ 16 − 6 = 10 > 0. Thus, we have the following relationship
ω6 (y) ≥ 0 =⇒ ω5 (y) ≥ 0 =⇒ ω4 (y) ≥ 0 =⇒ ω3 (y) ≥ 0 =⇒ ω2 (y) ≥ 0 =⇒ ω1 (y) ≥ 0, which yields that ω1 (y) ≥ 0 and then (x) is an increasing function in x ∈ + . Case 2: λ1 > λ2 Subcase 2.1: λ1 > λ2 and λ1 = 2λ2 For this case, we have
fTC (x) =
2λ1 λ2 [λ (1 − e−λ2 x ) − λ2 (1 − e−λ1 x )](e−λ2 x − e−λ1 x ) (λ1 − λ2 )2 1
and
fTS (x) = 4λ1 λ2 = 4λ1 λ2 −
x 0
e−λ1 x e(λ1 −λ2 )y [1 − e−λ1 (x−y)][1 − e−λ2 y ]dy 1
λ1 − λ2
[e−λ2 x − e−λ1 x ] −
1
[e−2λ2 x − e−λ1 x ]
[e−2λ2 x − e−2λ1 x ] .
λ1 − 2λ2
1 1 [e−λ2 x − e−2λ1 x ] + 2λ1 −λ2 2(λ1 − λ2 )
We need to show that
λ1 λ2 λ1 e−λ1 x + e−λ2 x − e−2λ1 x − e−2λ2 x λ1 − 2 λ 2 2 λ1 − λ2 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) −λ1 x −λ2 x −2λ1 x −2λ2 x −(λ1 +λ2 )x −(λ1 − λ2 )e + (λ1 − λ2 )e − λ2 e − λ1 e + (λ1 + λ2 )e
is increasing in x ∈ + . Since
y + 2 + (y − 2)ey yey − y(y + 1)
=
S
C λ2
is increasing in x ∈ + . Denote y = λx, we then need to prove that
W (y) =
fT (x) fT (x)
λ1 λ2 −λ1 x λ1 λ2 −λ2 x e − e λ1 − 2λ2 2λ1 − λ2 λ1 λ2 −2λ1 x λ1 λ2 −2λ2 x + e + e 2λ1 − λ2 λ1 − 2λ2 × [−(λ1 − λ2 )e−λ1 x + (λ1 − λ2 )e−λ2 x − λ2 e−2λ1 x − λ1 e−2λ2 x + (λ1 + λ2 )e−(λ1 +λ2 )x ] − [λ1 (λ1 − λ2 )e−λ1 x − λ2 (λ1 − λ2 )e−λ2 x + 2λ1 λ2 e−2λ1 x + 2λ1 λ2 e−2λ2 x − (λ1 + λ2 )2 e−(λ1 +λ2 )x ] λ2 λ1 × e−λ1 x + e−λ2 x λ1 − 2λ2 2λ1 − λ2 λ2 λ1 − e−2λ1 x − e−2λ2 x 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) λ2 λ21 − λ22 −(3λ1 +λ2 )x λ1 λ21 − λ22 −(λ1 +3λ2 )x = e − e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) 2 λ1 λ2 λ1 − λ22 + e−(2λ1 +2λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) λ21 λ2 (λ1 + λ2 ) − e−3λ1 x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ22 (λ1 + λ2 ) + e−3λ2 x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ2 (λ1 + λ2 )(λ1 − 4λ2 ) −(2λ1 +λ2 )x − e 2(λ1 − 2λ2 ) λ1 (λ1 + λ2 )(4λ1 − λ2 ) −(λ1 +2λ2 )x + e 2(2λ1 − λ2 ) (λ1 + λ2 )(λ1 − λ2 )3 −(λ1 +λ2 )x − e (2λ1 − λ2 )(λ1 − 2λ2 ) 2 2 λ1 λ21 − λ22 −2λ2 x sgn λ2 λ1 − λ2 = e−2λ1 x − e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) 2 λ1 λ2 λ1 − λ22 + e−(λ1 +λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) λ21 λ2 (λ1 + λ2 ) − e−(2λ1 −λ2 )x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ22 (λ1 + λ2 ) + e−(2λ2 −λ1 )x 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ2 (λ1 + λ2 )(λ1 − 4λ2 ) −λ1 x − e 2(λ1 − 2λ2 ) λ1 (λ1 + λ2 )(4λ1 − λ2 ) −λ2 x + e 2(2λ1 − λ2 ) (λ1 + λ2 )(λ1 − λ2 )3 − (2λ1 − λ2 )(λ1 − 2λ2 ) = ϕ1 (x), sgn
(x) = −
and it suffices to show that ϕ1 (x) ≥ 0 for any x ∈ + . Observe that
ϕ1 (0) =
λ2 λ21 − λ22 λ1 λ21 − λ22 λ1 λ2 λ21 − λ22 − + 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) (2λ1 − λ2 )(λ1 − 2λ2 ) λ21 λ2 (λ1 + λ2 ) λ1 λ22 (λ1 + λ2 ) − + 2(2λ1 − λ2 )(λ1 − 2λ2 ) 2(2λ1 − λ2 )(λ1 − 2λ2 ) λ2 (λ1 + λ2 )(λ1 − 4λ2 ) λ1 (λ1 + λ2 )(4λ1 − λ2 ) − + 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) (λ1 + λ2 )(λ1 − λ2 )3 − (2λ1 − λ2 )(λ1 − 2λ2 )
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
2 λ21 − λ22 (λ1 − λ2 )2 λ1 λ2 λ21 − λ22 − (2λ1 − λ2 )(λ1 − 2λ2 ) 2(2λ1 − λ2 )(λ1 − 2λ2 ) 2 λ1 − λ22 4λ21 − 7λ1 λ2 + 4λ22 + 2(2λ1 − λ2 )(λ1 − 2λ2 ) = 0, =−
and now it suffices to show that ϕ1 (x) ≥ 0 for x ∈ + . Taking the derivative of ϕ1 (x) with respect to x yields
λ1 λ2 λ21 − λ22 −2λ1 x λ1 λ2 λ21 − λ22 −2λ2 x e + e 2λ1 − λ2 λ1 − 2λ2 2 λ1 λ2 (λ1 + λ2 ) λ1 − λ22 −(λ1 +λ2 )x − e (2λ1 − λ2 )(λ1 − 2λ2 ) λ2 λ2 (λ1 + λ2 ) −(2λ1 −λ2 )x λ1 λ22 (λ1 + λ2 ) −(2λ2 −λ1 )x + 1 e + e 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 λ2 (λ1 + λ2 )(λ1 − 4λ2 ) −λ1 x + e 2(λ1 − 2λ2 ) λ1 λ2 (λ1 + λ2 )(4λ1 − λ2 ) −λ2 x − e 2(2λ1 − λ2 ) λ1 − λ2 −2λ1 x λ1 − λ2 −2λ2 x sgn = − e + e 2λ1 − λ2 λ1 − 2λ2 λ21 − λ22 − e−(λ1 +λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ2 + e−(2λ1 −λ2 )x + e−(2λ2 −λ1 )x 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) 4λ1 − λ2 −λ2 x λ1 − 4λ2 −λ1 x + e − e 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 − λ2 −(2λ1 −λ2 )x λ1 − λ2 −λ2 x sgn = − e + e 2λ1 − λ2 λ1 − 2λ2 2 2 λ1 − λ2 λ1 − e−λ1 x + e−2(λ1 −λ2 )x (2λ1 − λ2 )(λ1 − 2λ2 ) 2(λ1 − 2λ2 ) λ2 λ1 − 4λ2 −(λ1 −λ2 )x + e(λ1 −λ2 )x + e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) 4λ1 − λ2 − 2(2λ1 − λ2 ) = ϕ2 (x).
ϕ1 (x) = −
Notice that
λ21 − λ22 λ1 − λ2 λ1 − λ2 + − 2λ1 − λ2 λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ2 + + 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 − 4λ2 4λ1 − λ2 + − 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) λ1 λ2 = + 2(λ1 − 2λ2 ) 2(2λ1 − λ2 ) 4λ1 − λ2 λ1 − 4λ2 + − 2(λ1 − 2λ2 ) 2(2λ1 − λ2 )
ϕ2 (0) = −
= 0, it is enough to show that ϕ2 (x) ≥ 0 in x ∈ (0, +∞). Taking the derivative of ϕ2 (x) with respect to x, we have
λ2 (λ1 − λ2 ) −λ2 x e λ1 − 2λ2 2 λ1 λ1 − λ22 λ1 (λ1 − λ2 ) −2(λ1 −λ2 )x + e−λ1 x − e (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 − 2λ2 λ2 (λ1 − λ2 ) (λ1 −λ2 )x (λ1 − 4λ2 )(λ1 − λ2 ) −(λ1 −λ2 )x + e − e 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) λ2 λ1 (λ1 + λ2 ) sgn −(2λ1 −λ2 )x = e − e−λ2 x + e−λ1 x λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 )
ϕ2 (x) = (λ1 − λ2 )e−(2λ1 −λ2 )x −
409
λ1 λ2 e−2(λ1 −λ2 )x + e(λ1 −λ2 )x λ1 − 2λ2 2(2λ1 − λ2 ) λ1 − 4λ2 −(λ1 −λ2 )x − e 2(λ1 − 2λ2 ) λ2 λ1 (λ1 + λ2 ) sgn −λ1 x = e − e(λ1 −2λ2 )x + e−λ2 x λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 λ2 − e−(λ1 −λ2 )x + e2(λ1 −λ2 )x λ1 − 2λ2 2(2λ1 − λ2 ) λ1 − 4λ2 − 2(λ1 − 2λ2 ) = ϕ3 (x). −
It is easy to check that
λ2 λ1 (λ1 + λ2 ) λ1 + − λ1 − 2λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 − 2λ2 λ2 λ1 − 4λ2 + − 2(2λ1 − λ2 ) 2(λ1 − 2λ2 ) λ1 (λ1 + λ2 ) λ1 + 2λ2 λ2 = − + (2λ1 − λ2 )(λ1 − 2λ2 ) 2(λ1 − 2λ2 ) 2(2λ1 − λ2 )
ϕ3 (0) = 1 −
= 0. Thus, the desired result can be obtained if we could show that ϕ3 (x) ≥ 0. Observe further that
λ1 λ2 (λ1 + λ2 ) e−λ2 x (2λ1 − λ2 )(λ1 − 2λ2 ) λ1 (λ1 − λ2 ) −(λ1 −λ2 )x λ2 (λ1 − λ2 ) 2(λ1 −λ2 )x + e + e λ1 − 2λ2 2λ1 − λ2 λ1 (λ1 − λ2 ) −(λ1 −2λ2 )x sgn = −λ1 e−(λ1 −λ2 )x − λ2 e(λ1 −λ2 )x + e λ1 − 2λ2 λ2 (λ1 − λ2 ) (2λ1 −λ2 )x λ1 λ2 (λ1 + λ2 ) + e − 2λ1 − λ2 (2λ1 − λ2 )(λ1 − 2λ2 ) = ϕ4 (x).
ϕ3 (x) = −λ1 e−λ1 x − λ2 e(λ1 −2λ2 )x −
It is sufficient to show that ϕ4 (0) ≥ 0 and ϕ4 (x) ≥ 0 for x ∈ + . On one hand,
λ1 (λ1 − λ2 ) λ2 (λ1 − λ2 ) + λ1 − 2λ2 2λ1 − λ2 λ1 λ2 (λ1 + λ2 ) − (2λ1 − λ2 )(λ1 − 2λ2 ) 2(λ1 − λ2 )2 (λ1 + λ2 ) λ1 λ2 (λ1 + λ2 ) = − − (λ1 + λ2 ) (2λ1 − λ2 )(λ1 − 2λ2 ) (2λ1 − λ2 )(λ1 − 2λ2 )
ϕ4 (0) = − λ1 − λ2 +
= 0, and on the other hand,
ϕ4 (x) = λ1 (λ1 − λ2 )e−(λ1 −λ2 )x − λ2 (λ1 − λ2 )e(λ1 −λ2 )x − λ1 (λ1 − λ2 )e−(λ1 −2λ2 )x + λ2 (λ1 − λ2 )e(2λ1 −λ2 )x sgn = λ1 − λ2 e2(λ1 −λ2 )x − λ1 eλ2 x + λ2 e(3λ1 −2λ2 )x = ϕ5 (x). Since ϕ5 (0) = 0, it needs to prove that ϕ5 (x) ≥ 0. Taking the derivative of ϕ5 (x) with respect to x, we have
ϕ5 (x) = −2(λ1 − λ2 )e2(λ1 −λ2 )x − λ1 eλ2 x + (3λ1 − 2λ2 )e(3λ1 −2λ2 )x sgn
= −2(λ1 − λ2 )e(2λ1 −3λ2 )x + (3λ1 − 2λ2 )e3(λ1 −λ2 )x − λ1
sgn
=
ϕ6 (x).
Note that ϕ6 (0) = 0, and that
ϕ6 (x) = −2(λ1 − λ2 )(2λ1 − 3λ2 )e(2λ1 −3λ2 )x + 3(λ1 − λ2 )(3λ1 − 2λ2 )e3(λ1 −λ2 )x sgn = −2(2λ1 − 3λ2 )e(2λ1 −3λ2 )x + 3(3λ1 − 2λ2 )e3(λ1 −λ2 )x
410
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
= 3(3λ1 − 2λ2 )eλ1 x − 2(2λ1 − 3λ2 )
sgn
From the above two density functions, we have
F TC (x) =
≥ 3(3λ1 − 2λ2 ) − 2(2λ1 − 3λ2 ) = 5λ1 > 0.
and
Thus, we now can conclude by the following relationship
ϕ6 (x) ≥ 0 =⇒ ϕ5 (x) ≥ 0 =⇒ ϕ4 (x) ≥ 0 =⇒ ϕ3 (x) ≥ 0 =⇒ ϕ2 (x) ≥ 0 =⇒ ϕ1 (x) ≥ 0, Subcase 2.2: λ1 = 2λ2
fTC (x) = 4λ[−e
+ 3e
− 3e
−2λx
+e
−λx
]
λ1 x + 1 −
λ1 λ2
Denote λ2 = λ and we then have λ1 = 2λ. One can compute that −3λx
2λ21 e−λ1 x λ2 (λ1 − λ2 )
2λ22 e−λ2 x + λ2 x + 1 + λ1 (λ1 − λ2 )
2 λ21 + λ22 e−(λ1 +λ2 )x . + (λ1 + λ2 )x + 1 +
F TS (x) =
that ϕ1 (x) ≥ 0. Hence, it holds that (x) is increasing in x ∈ + .
−4λx
(λ1 x + 1)e−λ1 x + (λ2 x + 1)e−λ2 x − (λ1 x + 1)(λ2 x + 1)e−(λ1 +λ2 )x
Thus,
F TS (x) − F TC (x) =
and
fTS (x) =
4 λ[−e−4λx − (6λx + 3)e−2λx + 4e−λx ]. 3
We need to show that
fTS (x) fTC (x)
(x) =
= 2λ32 eλ1 x − 2λ31 eλ2 x + λ21 λ22 (λ1 − λ2 )x2 + 2λ1 λ2 λ21 − λ22 x + 2 λ31 − λ32
sgn
−e−4λx − (6λx + 3)e−2λx + 4e−λx
∝
−e−4λx + 3e−3λx − 3e−2λx + e−λx
= 1+
=
3(e3λx − 2λxe2λx − eλx ) e3λx − 3e2λx + 3eλx − 1
is increasing in x ∈ + for λ > 0. Let y = λx and then it is equivalent to proving that
ψ(y) =
e
− 3e
2y
y
+ 3e − 1
is increasing in y ∈ + . Taking the derivative of ψ(y) with respect to y, we get sgn
ψ (y) = [3e3y − 2(2y + 1)e2y − ey ] × (e3y − 3e2y + 3ey − 1) − (3e3y − 6e2y + 3ey ) × (e3y − 2ye2y − ey ) sgn
= (2y − 5)e4y + 14e3y − (6y + 12)e2y + (4y + 2)ey + 1
ψ1 (y).
=
fTC (x) = x{λ21 e−λ1 x + λ22 e−λ2 x 2 −(λ1 +λ2 )x 2 ]e
− [λ1 λ2 (λ1 + λ2 )x + λ + λ
0
=
}
=
Since ξ2 (0) = 0, we now need to show that ξ2 (x) ≥ 0. One can see that
−λ1 (x−z)
λ21 xe−λ1 x −
λ2 eλ1 x − λ1 eλ2 x + λ1 − λ2 = ξ3 (x).
sgn
=
ξ3 (x) = λ1 λ2 eλ1 x − λ1 λ2 eλ2 x = eλ1 x − eλ2 x ≥ 0,
+ λ2 e
−λ2 (x−z)
− (λ1 + λ2 )e
−(λ1 +λ2 )(x−z)
2λ e−λ1 x + λ22 xe−λ2 x λ2 (λ1 − λ2 ) 3 1
2λ32 + e−λ2 x + (λ1 + λ2 )2 xe−(λ1 +λ2 )x λ1 (λ1 − λ2 ) 2(λ1 + λ2 ) λ21 + λ22 −(λ1 +λ2 )x + e .
λ1 λ2
it follows that ξ1 (x) ≥ 0, i.e., F TS (x) ≥ F TC (x). Hence, the theorem follows.
Denote
SC = ∧(X1 ∨ Y1 , . . . , Xn ∨ Yn ),
[λ1 e−λ1 z + λ2 e−λ2 z − (λ1 + λ2 )e−(λ1 +λ2 )z ]
×[λ1 e
λ22 eλ1 x − λ21 eλ2 x + λ1 λ2 (λ1 − λ2 )x + λ21 − λ22 = ξ2 (x).
sgn
Proof of Theorem 4.2 2 1
x
ξ1 (x) = 2λ1 λ32 eλ1 x − 2λ31 λ2 eλ2 x + 2λ21 λ22 (λ1 − λ2 )x + 2λ1 λ2 λ21 − λ22
sgn
Without loss of generality, we suppose that λ1 ≥ λ2 . For the case λ1 = λ2 , the result is trivially true from Theorem 3.5. In the following, we need to treat the case when λ1 > λ2 . In this case, we can write
ξ1 (0) = 2λ32 − 2λ31 + 2 λ31 − λ32 = 0,
Since ξ3 (0) = 0 and
Proof of Theorem 3.7
fTS (x) =
So, F TS (x) − F TC (x) ≥ 0 is equivalent to ξ1 (x) ≥ 0 for any positive real number x. Since
ξ2 (x) = λ1 λ22 eλ1 x − λ21 λ2 eλ2 x + λ1 λ2 (λ1 − λ2 )
Upon using the method similar to that of Subcase 2.1, the desired result can be reached for this case and we omit the proof details for brevity.
and
ξ1 (x), say.
it suffices to show that ξ1 (x) ≥ 0. Taking the derivative of ξ1 (x) with respect to x, we have
e3y − 2ye2y − ey 3y
(λ1 x + 1)(λ2 x + 1)e−(λ1 +λ2 )x 2λ21 2λ22 − e−λ1 x + e−λ2 x λ2 (λ1 − λ2 ) λ1 (λ1 − λ2 )
2 λ21 + λ22 e−(λ1 +λ2 )x + (λ1 + λ2 )x + 1 + λ1 λ2
SS = [∧(X1 , . . . , Xn )] ∨ [∧(Y1 , . . . , Yn )]. ]dz
The survival function of SC can be written as
F SC (x) =
n
[1 − (1 − e−λi x )2 ],
i=1
and thus its density function is given by
fSC (x) = 2
n i=1
λi e−λi x (1 − e−λi x ) F S (x). 1 − (1 − e−λi x )2 C
P. Zhao et al. / European Journal of Operational Research 241 (2015) 402–411
On the other hand, the survival function of SS can be written as
F SS (x) = 1 − (1 − e
),
n
i=1 λi , and its density function is given by
where =
2e−x (1 − e−x ) fSS (x) = 2e−x (1 − e−x ) = F SS (x). 1 − (1 − e−x )2 Now we need to show that the ratio
n
λi e−λi x (1 − e−λi x ) i=1 1 − (1 − e−λi x )2 F SC (x) e−x (1 − e−x ) F SS (x) 1 − (1 − e−x )2
f (x) (x) = SC ∝ fSS (x)
is increasing in x. From Theorem 3.1 in Boland and El-Neweihi (1995), the ratio F SC (x)/F SS (x) is increasing in x and so it is enough if we could show the function
n
λi e−λi x (1 − e−λi x ) n λi (1 − e−λi x ) i=1 1 − (1 − e−λi x )2 i=1 2 − e−λi x = e−x (1 − e−x ) 1 − e−x 1 − (1 − e−x )2 2 − e−x
g(x) =
is increasing in x. The derivative of g(x) is
g (x) =
1
1 − e−x 2 − e−x
−
n i=1
sgn
=
−
i=1 sgn
=
−
(2 − e−λi x )2
n i=1
λ2i e−λi x
(2 − e−λi x )2 e−x
1 − e−x 2 − e−x
(2 − e−x )2
(1 − e−x )
λi (1 − e−λi x ) e−x 2 − e−λi x (2 − e−x ) λ2i e−λi x
n i=1
2 − e−λi x
(2 − e−λi x )2
n
i=1
λi (1 − e−λi x )
λ2i e−λi x
n i=1
2
n
(1 − e−x )(2 − e−x ) e−x
λi (1 − e−λi x ) 2 − e−λi x
= h(x).
It is well known that, for an exponential random variable with hazard rate λ, its reversed hazard rate is decreasing in λ. Based on this fact, we have the following inequality:
λi e−λi x 1−
e−λi x
≥
for any i. Now it follows that
h(x) ≥
−x 2
e−x
1 − e−x
for any i, which in turn implies that
(1 − e−x )(2 − e−x ) (1 − e−λi x )(2 − e−λi x ) ≥ e−x λi e−λi x
411
n i=1
−
λ2i e−λi x
(2 −
n i=1
(1 − e−λi x )(2 − e−λi x ) ) λi e−λi x
e−λi x 2
λi (1 − e−λi x ) 2 − e−λi x
= 0.
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