Optimal Consecutive-k-out-of-(2k
+ 1): G Cycle
?DING-ZHU DU1,3, FRANK K. HWANG2, YUNJAE JUNG3and HUNG Q. NGO3
1Institute of Applied Mathematics, Chinese Academy of Sciences, Beijing, China;2Department of
Applied Mathematics, National Chiao-Tung University, Hsinchu, Taiwan;3Department of Computer Science and Engineering, University of Minnesota, Minneapolis, MN 55455, USA (e-mail: [email protected].)
(Received May 20, 2000)
Abstract. We present a complete proof for the invariant optimal assignment for
consecutive-k-out-of-(2k+1): G cycle, which was proposed by Zuo and Kao in 1990 with an incomplete proof, pointed out recently by Jalali, Hawkes, Cui and Hwang.
Key words: Invariant optimal assignment, Consecutive-k-out-of-n: G cycle
1. Introduction
A cyclic consecutive-k-out-of-n: G system conC(k, n : G) is a cycle of n(> k)
components such that the system works if and only if some k consecutive compon-ents all work. Suppose n componcompon-ents with reliabilities p[1] 6 p[2] 6 · · · 6 p[n]
are all exchangeable. How can they be assigned to the n positions on the cycle to maximize the reliability of the system? Kuo, Zhang and Zuo [2] showed that if
k= 2, then the optimal assignment is invariant, i.e., it depends only on the ordering
of reliabilities of the components, but not their value. Invariant optimal assignment is very important in practice. In fact, in the real world, one usually knows the ranking of reliabilities of components, but not their exact values. For example, the ages of the components are known and one cannot compute the exact value of reliability from the age of each component. However, one may rank reliabilities of components according to their age by the rule that the older the less reliable. Kuo, Zhang and Zuo [2] also showed that for k> 3 and n > 2k + 1, ConC(k, n: G) has no invariant optimal assignment. For n 6 2k + 1, Zuo and Kuo [3] claimed that there exists an invariant optimal assignment
(p[1], p[3], p[5], . . . , p[6], p[4], p[2], ).
However, Jalali, Hawkes, Cui and Hwang [1] found that their proof is incomplete. In this paper, we give a complete proof for this invariant optimal assignment in case n= 2k + 1.
A similar situation occurred in a line system. A linear consecutive-k-out-of-n:
Gsystem ConL(k, n : G) can be defined in a similar way to the cyclic system ConC(k, n : G). Kuo, Zhang and Zuo [2] presented an invariant optimal assign-ment for ConL(k, n: G) with n 6 2k. However, the proof is incomplete too. This
was also pointed out by Jalali et al. [1]. In addition, they gave a complete proof for the line system. It is worth pointing out that by setting p[1]= · · · = p[2k+1−n]= 0, the cycle system ConC(k,2k+ 1 : G) becomes the line system ConL(k, n : G). Thus, our result yields the result of Jalali et al. [1]. The cycle case is in gen-eral much more difficult than the line case, and our proof adopts a new approach different from previous attempts.
2. Main Result
In this section, we show the following.
THEOREM 1. ConC(k,2k+ 1 : G) has invariant optimal assignment
(p[1], p[3], p[5], . . . , p[2k+1], p[2k], . . . , p[6], p[4], p[2]).
Let p1, p2, . . . , p2k+1be reliabilities of the 2k+ 1 components on the cycle in counter-clockwise direction. For simplicity of the proof, we first assume that
0 < p[1] < p[2]<· · · < p[2k+1]<1.
Our proof is based on the following representation of the reliability of consecu-tive-k-out-of-n: G cycle for n6 2k + 1.
LEMMA 1. The reliability of consecutive-k-out-of-n: Gcycle for n6 2k+1 under
assignment C can be represented as R(C)= p1. . . pn+ n X i=1 qipi+1. . . pi+k = p1. . . pn+ n X i=1 pi. . . pi+k−1− n X i=1 pi. . . pi+k where qi = 1 − pi and pn+i = pi.
Proof. The system works if and only if all components work or for some i, the ith component fails and the (i+ 1)st component, ..., the (i + k)th component all work. Since n6 2k + 1, there exists at most one such i. Therefore,
R(C)= p1. . . pn+ n X i=1 qipi+1. . . pi+k. Note that qipi+1. . . pi+k = pi+1. . . pi+k− pi. . . pi+k.
This implies the second representation. 2 To prove Theorem 1, it suffices to show that in any optimal assignment,
(pi − pj)(pi−1− pj+1) >0 for 1 < i < j < 2k+ 1. (1) That is, selecting any component to be labeled p1, we always have
(pi − p2k+2−i)(pi+1− p2k+1−i) >0 for i = 1, . . . , k. (2) For simplicity of representation, we denote i0 = 2k + 2 − i. Note that (k + 1)0 =
k+ 1 and (i0)0 = i. Furthermore, without loss of generality, we assume p1 > p10 throughout this proof. Then the condition (2) can be rewritten as
pi > pi0 for i = 1, . . . , k. (3)
We will employ an inductive argument to prove these k inequalities in the following ordering: p1> p10 pk > pk0 p2> p20 pk−1 > p(k−1)0 . . .
Let Ci be the assignment obtained from C by exchanging components i and i0 and Ci,j,... ,y,z = (Ci,j, . . . , y)z. Let C∗be an optimal assignment. We divide this inductive argument into the following lemmas.
LEMMA 2. Under assumption that p1 > p10, we must have pk > pk0 and
p2. . . pk−1 > p20. . . p(k−1)0. Proof. Consider 06 R(C∗)− R(C1∗)= (p1− p10)(p2. . . pk− p20. . . pk0)qk+1. Since p1> p10, we have p2. . . pk > p20. . . pk0. (4) Note that 06 R(C∗)− R(Ck∗) = (pk−pk0)[(q10p1+q1pk+1)p2. . . pk−1−(p10q1+q10pk+1)p20. . . pk−1)0] (5) and (q10p1+ q1pk+1)− (p10q1+ q10pk+1)= (p1− p10)(1− pk+1) >0.
We first claim that pk > pk0.In fact, if p2. . . pk−1 > p20. . . p(k−1)0, then it follows from (5) that pk > pk0; If p2. . . pk−1< p20. . . p(k−1)0, then it follows from (4) that
pk > pk0.
Now, we show that p2. . . pk−1 > p20. . . p(k−1)0. Note that
06R(C∗)− R(C1,k∗ )
=[(p1pk− p10pk0)+ (pk− pk0)pk+1− p10p1(pk− pk0)](p2. . . pk−1− p20. . . p(k−1)0)qk+1
=[(p1pkq10− p10pk0q1)qk+1+ (1 − p1p10)(pk− pk0)p2. . . pk−1− p20. . . p(k−1)0).
Moreover, p1q10− p10q1 = p1− p10 > 0 and pk > pk0. Therefore, p2. . . pk−1 >
p20. . . p(k−1)0. 2
To show inductive step, we first prove an equality. LEMMA 3. Suppose m < k/2. Then
2mX+1 i=1 kY−1 j=0 p−m+i+j − k Y j=0 p−m+i+j = mX−1 i= 1 Yi j=1 pkpj0 (pi+1q(i+1)0 Yk−i j=i+2 pj 1 − Yk j=k−i+1 pjpj0 pk+1 + m X i=1 1 −Yi j=1 pjpj0 kY−i j=i+1 pj (pk−i+1q(k−i+1)0) Yk j=k−i+2 pjpj0 pk+1 + q10p1. . . pkqk+1+ Ym j=1 pjpj0 kY−m j=m+1 pj 1 − Yk j=k−m+1 pjpj0 pk+1 .
Proof. Note that p10p1. . . pk+1 = p10p1. . . pk+1qk0+ p10p1. . . pk+1pk0 = p10p1. . . pk+1qk0+ q20p10p1. . . pk+1pk0 + p20p10p1. . . pk+1pk0 = · · · = m−1 X i=1 Yi j=1 pjpj0 Yk−i j=i+1 pj (pk−i+1q(k−i+1)0) Yk j=k−i+2 pjpj0 pk+1 + m X i=1 Yi j=1 pjpj0 (pi+1q(i+1)0) kY−1 j=i+2 pj Yk j=k−1+1 pjpj0 pk+1 + Ym j=1 pjpj0 kY−m j=m+1 pj Yk j=k−m+1 pjpj0 pk+1
and p1. . . pk− p10p1. . . pk− p1. . . pk+1 = q10p1. . . pkqk+1− p10p1. . . pk+1. Thus, 2mX+1 i=1 kY−1 j=0 p−m+i+j− k Y j=0 p−m+i+j = m X i=2 q−m+i−1 kY−1 j=0 p−m+i+j+ 2mX+1 i=m+2 q−m+i+k kY−1 j=0 p−m+i+j + p−m+1. . . p−m+k+ p1. . . pk− p0. . . pk− p1. . . pk+1 = mX−1 i=1 Xi j=1 pjpj0 (pi+1q(i+1)0) kY−i j=i+2 pj + m X i=1 kY−i j=i+1 pj (pk−i+1q(k−i+1)0) Yk j=ik−i+2 pjpj0 pk+1 + Ym j=1 pjpj0 kY−m j=m+1 pj + p1. . . pk− p10p1. . . pk− p1. . . pk+1 = mX−1 i=1 Yi j=1 pjpj0 (pi+1q(i+1)0) kY−i j=i+2 pj 1 − Yk j=k−i+1 pjpj0 pk+1 + m X i=1 1 −Yi j=1 pjpj0 kY−i j=i+1 pj (pk−i+1q(k−i+1)0) Yk j=k−i+2 pjp0j pk+1 + q10p1. . . pkqk+1+ Ym j=1 pjpj0 kY−m j=m+1 pj 1 − Yk j=k−m+1 pjpj0 pk+1 . 2 LEMMA 4. Suppose m < k/2. If pi > pi0 for i = 1, . . . , m, k − m + 1, . . . , k (6) and kY−m j=m+1 pj > kY−m j=m+1 pj0 (7) then pm+1 > p(m+1)0
and kY−m j=m+2 pj > kY−m j=m+2 pj0.
Proof. We first show pm+1 > p(m+1)0. For contradiction, suppose pm+1 <
p(m+1)0. It follows from (7) that kY−m j=m+2 pj > kY−m j=m+2 pj0. (8) By Lemma 3, 06R(C∗)− R(C∗m+1) = 2mX+1 i=1 kY−1 j=0 p−m+i+j− k Y j=0 p−m+i+j+ kY−1 j=0 p(−m+i+j)0− k Y j=0 p(−m+i+j)0 − 2mX+1 i=1 p(m+1)0 pm+1 kX−1 j=0 p−m+i+j− k Y j=0 p−m+i+j + pm+1 p(m+1)0 kY−1 j=0 p−m+i+j− k Y j=0 p(−m+i+j)0 =(pm+1− p(m+1)0) mX−1 i=1 Yi j=1 pjpj0 (pi+1q(i+1)0 Ym j=i+2 pj kY−i j=m+2 pj −(p(i+1)0qi+1) Ym j=i+2 pj0 kY−i j=m+2 pj0 1 − Yk j=k−i+1 pjpj0 pk+1 + m X i=1 1 −Yi j=1 pipj0 Ym j=i+1 pj kY−i j=m+2 pj (pk−i+1q(k−i+1)0) − Ym j=i+1 pj0 kY−i j=m+2 pj0 (p(k−i+1)0qk−i+1) Yk j=k−i+2 pjpj0 pk+1 + (q10p1. . . pmpm+2. . . pk− q1p10. . . pm0p(m+2)0. . . pk0)qk+1 + Ym j=1 pjpj0 kY−m j=m+2 pj− kY−m j=m+2 pj0 1 − Yk j=k−m+1 pjpj0 pk+1 <0,
a contradiction. The last inequality holds because it follows from (6) and (8) that every term in{· · · } is positive.
Now, we show kY−m j=m+2 pj > kY−m j=m+2 pj0.
By Lemma 3, we have 06R(C∗)− R(C∗m+2,... ,m−k) = kY−m j=m+2 pj− kY−m j=m+2 pj0 · mX−1 i=1 Yi j=1 pjpj0 (pi+1q(i+1)0 mY+1 j=i+2 pj kY−i j=k−m+1 pj − (p(i+1)0qi+1) mY+1 j=i+2 pj0 kY−i j=k−m+1 pj0 1 − Yk j=k−i+1 pjpj0 pk+1 + m X i=1 1 −Yi j=1 pjpj0 mY+1 j=i+1 pj kY−i j=k−m+1 pj (pk−i+1q(k−i+1)0) − mY+1 j=i+1 pj0 kY−i j=k−m+1 pj0 (p(k−i+1)0qk−i+1) Yk j=k−i+2 pjpj0 pk+1 + (q10p1. . . pm+1pk−m+1· pk− q1p10. . . p(m+1)0p(k−m+1)0. . . pk0)qk+1 + Ym j=1 pjpj0 (pm+1− p(m+1)0 1 − Yk j=k−m+1 pjpj0 pk+1 .
It follows from (6) and pm+1 > p(m+1)0 that every term in{· · · } is positive. There-fore, kY−m j=m+2 pj > kY−m j=m+2 pj0. 2
LEMMA 5. Suppose m < k/2. Then
2mX+2 i=1 kY−1 j=0 p−m+i+j− 2mX+2 i=1 k Y j=0 p−m−1+i+j = m X i=1 Yi j=1 pjpj0 (pi+1q(i+1)0) kY−i j=i+2 pj 1 − Yk j=k−i+1 pjpj0 pk+1 + m X i=1 1 −Yi j=1 pjpj0 kY−i j=i+1 pj (pk−i+1q(k−i+1)0) Yk j=k−i+2 pjpj0 pk+1 + q10p1. . . pkqk+1 + 1 −Ym j=1 pjpj0 kY−m j=m+1 pj Yk k−m+1 pjpj0 pk+1.
LEMMA 6. Suppose m < k/2. If pi > pi0 for i = 1, . . . , m + 1, k − m + 1, . . . , k (9) and kY−m j=m+2 pj > kY−m j=m+2 pj0 (10) then pk−m> p(k−m)0 (11) and k−m−1Y j=m+2 pj > k−m−1Y j=m+2 pj0. (12)
Proof. We first show (11). For contradiction, suppose pk−m< p(k−m)0. By (10), k−m−1Y j=m+2 pJ > k−m−1Y j=m+2 pj0. (13) By Lemma 5, 06R(C∗)− R(C∗k−m) = (pk−m− p(k−m)0) m X i=1 Yi j=1 pjpj0 (pi+1q(i+1)0) k−m−1Y j=i+2 pj kY−i j=k−m+1 pj −(p(i+1)0qi+1 k−m−1Y j=i+2 pj0 kY−i j=k−m+1 p+ j0 1 − Yk j=k−i+1 pjpj0 pk+1 + m X i=1 1 −Yi j=1 pjpj0 k−m−1Y j=i+1 pj kY−i j=k−m+1 pj (pk−i+1q(k−i+1)0) − k−m−1Y j=i+1 pj0 kY−i j=k−m+1 pj0 (p(k−i+1)0qk−i+1) Yk j=k−i+2 pjpj0 pk+1 + (q10p1. . . pk−m−1pk−m+1. . . pk− q1p10. . . p(k−m−1)0p(k−m+1)0. . . pk0)qk+1 + 1 −Ym j=1 pjpj0 k−m−1Y j=m+1 pj− k−m−1Y j=m+1 pj0 Yk k−m+1 pjpj0 pk+1 <0, a contradiction.
Now, we prove (12). Consider 06R(C∗)− R(C∗m+2,... ,k−m−1) = k−m−1Y j=m+2 pj − k−m−1Y j=m+2 pj0 · m X i=1 Yi j=1 pjpj0 (pi+1q(i+1)0 mY+1 j=i+2 pj Yk−1 j=k−m pj −(p(i+1)0qi+1) mY+1 j=i+2 pj0 Yk−i j=k−m pj0 1− Yk j=m−i+1 pjpj0 pk+1 + m X i=1 1 −Yi j=1 pjpj0 mY+1 j=i+1 pj kY−1 j=k−m pj (pk−i+1q(k−i+1)0) − mY+1 j=i+1 pj0 Yk−i j=k−m pj0 (p(k−i+1)0qk−i+1) Yk j=k−i+2 pjpj0 pk+1 + (q10p1. . . pm+1pk−m. . . pk− q1p10. . . p(m+1)0p(k−m)0. . . pk0)qk+1 + 1 −Ym j=1 pjpj0 (pm+1pk−m− p(m+1)0p(k−m)0) k Y k−m+1 pjpj0 ! pk+1 .
It follows from (9) and (11) that every term in{. . . } is positive. Thus, (12) holds. 2 By Lemmas 2, 4 and 6, we know that (3) holds. Therefore, Theorem 1 is proved for 0 < p[1]<· · · < p[2k+1]<1.
Finally, we deal with the case that some equality signs hold in 06 p[1] 6 p[2] 6
· · · 6 p[2k+1]6 1. If p[1] = p[2] = · · · = p[2k+1], then Theorem 1 is trivially true.
If there exists i, 16 i < 2k + 1, such that p[i] < p[i+1], then for sufficiently small
ε >0, we have
0 < p[1]+ ε < · · · < p[i]+ iε < p[i+1]
− (2k + 1 − i)ε < · · · < p[2k+1]− ε < 1.
For them, we already proved the optimality of assignment C∗ in Theorem 1, that is, for any assignment C, R(C∗) > R(C). Now, we can complete our proof of
Theorem 1 by setting ε→ 0. 3. Discussion
Zuo and Kuo [3] proved only that R(C∗) > R(Ci∗)for any i. This cannot imply that R(C∗)> R(C) for any assignment C. Thus, their proof is incomplete.
Note that ConC(k, n: G) for n 6 2j cannot be induced from ConC(k,2k+ 1 :
G). Moreover, our inductive argument does not work in cycle system ConC(k, n : G)for n6 2k. Thus, some additional techniques are required to solve the general case.
References
1. Jalali, A., Hawkes, A.G., Cui, L. and Hwang, F.K. The optimal consecutive-k-out-of-n: G line for n6 2k, to appear in Naval Res. Logist.
2. Kuo, W., Zhang, W. and Zuo, M. (1990), A consecutive-k-out-of-n: G system: The mirror image of a consecutive-k-out-of-n: F system, IEEE Trans. Rel. 39: 244–253.
3. Zuo, M. and Kuo, W. (1990), Design and performance analysis of consecutive-k-out-of-n: structure, Naval Res. Logist. 37: (1990), 203–230.
