OPTIMAL CONSECUTIVE-k-OUT-OF-n: G CYCLE FOR n ≤ 2k + 1∗
DING-ZHU DU†, FRANK K. HWANG‡, XIAOHUA JIA§,AND HUNG Q. NGO¶
Abstract. A cyclic consecutive-k-out-of-n: G system consists of n components lying on a cycle.
Those components are exchangeable but may have different working probabilities. The system works if and only if there are k consecutive components at work. What is the optimal assignment of components for maximizing the reliability of the system? Does the optimal assignment depend on the working probability values of components? For k ≤ n ≤ 2k + 1, Zuo and Kuo in 1990 proposed a solution independent from the working probability values of components, called the invariant optimal assignment. However, their proof is incomplete, pointed out recently by Jalali et al. [The Optimal Consecutive-k-out-of-n: G Line for n ≤ 2k, manuscript, 1999]. We present a complete proof in this paper.
Key words. invariant optimal assignment, consecutive-k-out-of-n: G cycle AMS subject classifications. 60K10, 90B25
PII. S0895480100375041
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 components all work. Suppose n components with working probabilities p[1] ≤ p[2] ≤ · · · ≤ 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 [10] showed that if k = 2, then the optimal assignment is invariant; i.e., it depends only on the ordering of working probabilities of the components but not their value. They also claimed that for k ≥ 3 and n > 2k + 1, ConC(k, n : G) has no invariant
optimal assignment. For n ≤ 2k + 1, Zuo and Kuo [13] claimed that there exists an invariant optimal assignment
(p[1], p[3], p[5], . . . , p[6], p[4], p[2]).
However, Jalali et al. [9] found that their proof is incomplete. A proof in case n = 2k + 1 was given in [6]. In this paper, we give a complete proof for this invariant optimal assignment with n ≤ 2k + 1.
2. Main result. In this section, we show the following.
Theorem 2.1. For k ≤ n ≤ 2k + 1, there exists an invariant optimal assignment (p[1], p[3], p[5], . . . , p[6], p[4], p[2]).
∗Received by the editors July 11, 2000; accepted for publication (in revised form) March 18, 2002;
published electronically May 8, 2002.
http://www.siam.org/journals/sidma/15-3/37504.html
†Department of Computer Science and Engineering, University of Minnesota, Minneapolis, MN
55455 ([email protected]). This author’s research was supported in part by the National Science Foundation under grant CCR-9530306, by the Institute of Applied Mathematics, Chinese Academy of Sciences, Beijing, People’s Republic of China, and in part by National 973 Information Technology and High-Performance Software Program of China under grant G1998030402.
‡Department of Applied Mathematics, National Chiao-Tung University, Hsinchu, Taiwan 9
§Department of Computer Science, City University of Hong Kong, Kowloon Tong, Hong Kong
¶Department of Computer Science, State University of New York, Buffalo, NY 14260 (hungngo@
cse.buffalo.edu).
305
Let p1, p2, . . . , pn be reliabilities of the n components on the cycle in the
counter-clockwise direction. For simplicity of the proof, we first assume that 0 < p[1]< p[2]< · · · < p[n]< 1.
Our proof is based on the following representation of the reliability of consecutive-k-out-of-n: G cycle for n ≤ 2k + 1.
Lemma 2.2. The reliability of consecutive-k-out-of-n: G cycle for n ≤ 2k + 1 under assignment C can be represented as
R(C) = p1· · · pn+ n i=1 qipi+1· · · pi+k = p1· · · pn+ n i=1 pi· · · pi+k−1− n 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 n ≤ 2k + 1, there exists at most one such i. Therefore,
R(C) = p1· · · pn+ n i=1 qipi+1· · · pi+k. Note that
qipi+1· · · pi+k= pi+1· · · pi+k− pi· · · pi+k.
This implies the second representation.
This representation is a key point to show the main theorem. It explains why invariant optimal assignment exists for n ≤ 2k + 1 but does not exist for n > 2k + 1. For n = k, k + 1, by Lemma 2.2, R(C) has the same value for all assignment C, and hence Theorem 2.1 is trivially true. Next, we assume k + 2 ≤ n ≤ 2k + 1.
To prove Theorem 2.1, it suffices to show that in any optimal assignment, (pi− pj)(pi−1− pj+1) > 0 for 1 < i < j < n.
(2.1)
In fact, the optimal assigment described in Theorem 2.1 can be determined uniquely by condition (2.1) (see [9]). Selecting any component to be labeled p1, we always have
from condition (2.1) that
(pi− pn−i+1)(pi+1− pn−i) > 0 for i = 1, . . . , h,
(2.2)
where h = n/2. For simplicity of representation, we denote i = n − i + 1. When
n is odd, (n+1
2 )= n+12 . Furthermore, without loss of generality, we assume p1> p1
throughout this proof. Then the condition (2.2) can be rewritten as pi> pi for i = 1, . . . , h.
Let I = {i | 1 < i ≤ h, pi < pi}. Let CI be the assignment obtained from C by
exchanging components i and i for all i ∈ I. To prove (2.1), it suffices to show that
for any assignment C, if I = ∅,
R(C) < R(CI).
Denote I= {i| i ∈ I} and (yi1· · · yid)I = 1≤j≤d,ij∈I∪I yij− 1≤j≤d,ij∈I∪I yi j 1≤j≤d,ij∈I∪I yi j− 1≤j≤d,ij∈I∪I yij , where yi= pi or qi. It is easy to verify that
yi1· · · yid 1≤j≤d,ij∈I∪I yi j yij + yi 1· · · yid 1≤j≤d,ij∈I∪I yij yi j −(yi1· · · yid+ yi1· · · yid) = (yi1· · · yid)I.
Denote Qk(C) =ni=1pi· · · pi+k−1. Then
R(C) = p1· · · pn+ Qk(C) − Qk+1(C).
Let a = k/2 and s =
h if n is even and k is odd, h + 1 otherwise.
Then we have the following.
Lemma 2.3. Qk(CI) − Qk(C) =si=1(p−a+i· · · p−a+i+k−1)I.
Proof. Consider four cases.
Case 1. n and k both are even. In this case, s = h + 1 = 1 + n/2 and a = k/2. Note that n i=h+2 p−a+i· · · p−a+i+k−1 = h i=2 p−a+(2h+2−i)· · · p−a+(2h+2−i)+k−1 =h i=2 pn−(−a+i+k−1)+1· · · pn−(−a+i)+1 = h i=2 p(−a+i+k−1)· · · p(−a+i). Thus, Qk(C) = p−a+1· · · pa+ h i=2
(p−a+i· · · p−a+i+k−1+ p(−a+i)· · · p(−a+i+k−1))
+ p−a+h+1· · · p−a+h+k = a j=1 (pjpj) + h i=2
(p−a+i· · · p−a+i+k−1+ p(−a+i)· · · p(−a+i+k−1))
+
a
j=1
(ph−a+jp(h−a+j)).
So, Qk(CI) − Qk(C) = h i=2 (p−a+i· · · p−a+i+k−1)I. However,
(p−a+1· · · p−a+k)I = (p−a+h+1· · · p−a+h+k)I = 0.
Therefore, Qk(CI) − Qk(C) = h+1 i=1 (p−a+i· · · p−a+i+k−1)I.
Case 2. n is even and k is odd. In this case, s = h = n/2 and a = (k − 1)/2. Note that
Qk(C) = h
i=1
(p−a+i· · · p−a+i+k−1+ p(−a+i)· · · p(−a+i+k−1)).
Therefore, Qk(CI) − Qk(C) = h i=1 (p−a+i· · · p−a+i+k−1)I.
Case 3. n and k both are odd. In this case, s = h+1 = (n+1)/2 and a = (k−1)/2. Note that Qk(C) = h i=1
(p−a+i· · · p−a+i+k−1+ p(−a+i)· · · p(−a+i+k−1)) + p−a+h+1· · · p−a+h+k
=h
i=1
(p−a+i· · · p−a+i+k−1+ p(−a+i)· · · p(−a+i+k−1))
+ p−a+h−1p(−a+h−1)· · · phphph+1. Thus, Qk(CI) − Qk(C) = h i=1 (p−a+i· · · p−a+i+k−1)I. However, (p−a+h+1· · · p−a+h+k)I = 0. Therefore, Qk(CI) − Qk(C) = h+1 i=1 (p−a+i· · · p−a+i+k−1)I.
Case 4. n is odd and k is even. In this case, s = h + 1 = (n + 1)/2 and a = k/2. Note that Qk(C) = p−a+1· · · p−a+k+ h+1 i=2
(p−a+i· · · p−a+i+k−1+ p(−a+i)· · · p(−a+i+k−1))
=a
j=1
(pjpj) +
h+1 i=2
(p−a+i· · · p−a+i+k−1+ p(−a+i)· · · p(−a+i+k−1)).
Thus Qk(CI) − Qk(C) = h+1 i=2 (p−a+i· · · p−a+i+k−1)I. However, (p−a+1· · · p−a+k)I = 0. Therefore, Qk(CI) − Qk(C) = h+1 i=1 (p−a+i· · · p−a+i+k−1)I. Define t =
h if n is even and k + 1 is odd, h + 1 otherwise
and b = (k + 1)/2. We have a useful representation of R(CI) − R(C) as follows.
Lemma 2.4. R(CI) − R(C) = t i=2 (p−b+i· · · p−b+i+k−1)I− t i=1 (p−b+i· · · p−b+i+k)I.
Proof. By Lemma 2.3, we have R(CI) − R(C) = s i=1 (p−a+i· · · p−a+i+k−1)I − t i=1 (p−b+i· · · p−b+i+k)I.
Note that if k is even and n is odd, then a = b, s = t, and (p−a+1· · · p−a+k)I = (p−k/2+1· · · pk/2)I = 0;
if k is odd and n is odd, then a = b − 1, s = t, and
(p−a+s· · · p−a+s+k−1)I = (p−(n−k)/2· · · p−(n+k)/2−1)I = 0;
if k is even and n is even, then a = b, s = t + 1, and
(p−a+1· · · p−a+k)I = (p−a+s· · · p−a+s+k−1)I = 0;
if k is odd and n is even, then a = b − 1 and s = t − 1. Therefore, we always have s i=1 (p−a+i· · · p−a+i+k−1)I = t i=2 (p−a+i· · · p−a+i+k−1)I.
Note that (p−b+i· · · p−b+i+k−1)I ≥ 0 for 2 ≤ i ≤ t and (p−b+i· · · p−b+i+k)I ≥ 0
for 1 ≤ i ≤ t. Therefore, to prove R(CI) < R(C), we need to compare (p−b+i· · ·
p−b+i+k−1)I with (p−b+i· · · p−b+i+k)I.
Lemma 2.5. Suppose I = {i | 1 < i ≤ h, pi< pi}. Then, for i = 1, . . . , b,
(q−b+ip−b+i+1· · · p−b+i+k)I = (p−b+i+1· · · p−b+i+k)I−(p−b+ip−b+i+1· · · p−b+i+k)I ≥ 0,
and the strict inequality sign holds if and only if
{j | b − i + 1 ≤ j ≤ min(−b + i + k, n + b − i − k), j ∈ I} = ∅, {j | b − i + 1 ≤ j ≤ min(−b + i + k, n + b − i − k), j ∈ I} = ∅. Proof. First, assume −b + i ∈ I ∪ I. Then we have
(q−b+ip−b+i+1· · · p−b+i+k)I
= q−b+i −b+i+1≤j≤−b+i+k,j∈I∪I pj− q(−b+i) −b+i+1≤j≤−b+i+k,j∈I∪I pj · −b+i+1≤j≤−b+i+k,j∈I∪I pj− −b+i+1≤j≤−b+i+k,j∈I∪I pj = −b+i+1≤j≤−b+i+k,j∈I∪I pj− −b+i+1≤j≤−b+i+k,j∈I∪I pj · −b+i+1≤j≤−b+i+k,j∈I∪I pj− −b+i+1≤j≤−b+i+k,j∈I∪I pj − −b+i≤j≤−b+i+k,j∈I∪I pj− −b+i≤j≤−b+i+k,j∈I∪I pj · −b+i+1≤j≤−b+i+k,j∈I∪I pj− −b+i+1≤j≤−b+i+k,j∈I∪I pj = (p−b+i+1· · · p−b+i+k)I− (pi· · · pi+k)I.
If −b + i + k < n + b − i − k + 1, then −b + i + k ≤ h, and hence we have (q−b+ip−b+i+1· · · p−b+i+k)I
= b−i j=1 pjpj b−i+2≤j≤−b+i+k,j∈I pj− b−i+1≤j≤−b+i+k,j∈I pj pb−i+1q(b−i+1) b−i+1≤j≤−b+i+k,j∈I pj− p(b−i+1)qb−i+1 b−i+1≤j≤−b+i+k,j∈I pj ≥ 0
since −b + i ∈ I ∪ I implies that
pb−i+1q(b−i+1)− p(b−i+1)qb−i+1= pb−i+1− p(b−i+1) > 0.
Moreover, it is easy to verify that
(q−b+ip−b+i+1· · · p−b+i+k)I > 0
if and only if
{j | b − i ≤ j ≤ −b + i + k, j ∈ I} = ∅, {j | b − i ≤ j ≤ −b + i + k, j ∈ I} = ∅.
If −b + i + k ≥ n + b − i − k + 1, then n + b − i − k ≤ h, and hence (q−b+ip−b+i+1· · · p−b+i+k)I
= b−i j=1 pjpj b−i+2≤j≤n+b−i−k,j∈I pj− b−i+1≤j≤n+b−i−k,j∈I pj pb−i+1q(b−i+1) b−i+1≤j≤n+b−i−k,j∈I pj− p(b−i+1)qb−i+1 b−i+1≤j≤n+b−i−k,j∈I pj · n+b−i−k+1≤j≤h pjpj γ ≥ 0, where γ = 1 if n is even, ph+1 if n is odd. Moreover,
(q−b+ip−b+i+1· · · p−b+i+k)I > 0
if and only if
{j | b − i + 1 ≤ j ≤ n + b − i − k, j ∈ I} = ∅, {j | b − i + 1 ≤ j ≤ n + b − i − k, j ∈ I} = ∅.
Finally, we note that a similar argument works in the case that −b + i ∈ I ∪ I.
Similarly, we can show the following.
Lemma 2.6. Suppose I = {i | 1 ≤ i ≤ h, pi< pi}. Then, for i = b, . . . , t,
(p−b+i· · · p−b+i+k−1q−b+i+k)I = (p−b+i· · · p−b+i+k−1)I− (p−b+ip−b+i+1· · · p−b+i+k)I
≥ 0.
Lemma 2.7. Suppose I = {i | 1 ≤ i ≤ h, pi< pi}. Then
(q0p1· · · pkqk+1)I
= (p1· · · pk)I − (p0· · · pk)I− (p1· · · pk+1)I+ (p0· · · pk+1)I
≥ 0,
and the strict inequality sign holds if and only if
{j | 1 ≤ j ≤ n − k, j ∈ I} = ∅, {j | 1 ≤ j ≤ n − k, j ∈ I} = ∅. By Lemmas 2.4–2.7, we have R(CI) − R(C) = b−1 i=1
(q−b+ip−b+i+1· · · p−b+i+k)I+ t
i=b+2
(p−b+i· · · p−b+i+k−1q−b+i+k)I
+ (q0p1· · · pkqk+1)I − k+1 j=0 pj I . Let d = (n − k)/2 − 1. Note that
k+1 j=0 pj I = q−1 k+1 j=0 pj I + k+1 j=−1 pj I = · · · = d i=1 q−i k+i j=−i+1 pj I + n−k−2−d i=1 qi+k+1 i+k j=−i pj I + n−d−1 j=−d pj I and n−d−1 j=−d pj I = 0. Thus, we have R(CI) − R(C) (2.3) = b−d−1 i=1 q−b+i −b+i+k j=−b+i+1 pj I + d i=1 q−i −i+k j=−i+1 pj I − q−i i+k j=−i+1 pj I
+ t i=b+n−k−d q−b+i+k −b+i+k−1 j=−b+i pj I + n−k−2−d i=1 qi+k+1 i+k j=i+1 pj I − qi+k+1 i+k j=−i pj I + (q0p1· · · pkqk+1)I.
This representation suggests that we show Lemmas 2.8 and 2.9. Lemma 2.8. For i = 1, . . . , d, q−i −i+k j=−i+1 pj I ≥ q−i i+k j=−i+1 pj I , and the inequality holds strictly if and only if
q−i −i+k j=−i+1 pj I > 0.
Proof. First, consider the case that −i ∈ I and n is even. Denote
A = −i+1≤j≤−i+k,j∈I∪I pj− −i+1≤j≤−i+k,j∈I∪I pj, B = −i+1≤j≤i+k,j∈I∪I pj− −i+1≤j≤i+k,j∈I∪I pj, A= q (−i) −i+1≤j≤−i+k,j∈I∪I pj− q−i −i+1≤j≤−i+k,j∈I∪I pj, B= q (−i) −i+1≤j≤i+k,j∈I∪I pj− q−i −i+1≤j≤−i+k,j∈I∪I pj. Then q−i −i+k j=−i+1 pj I − q−i i+k j=−i+1 pj I = AA− BB.
If −i + k < n + i − k + 1, then −i + k ≤ h. Hence A − B = 1≤j≤i,j∈I pjpj i+1≤j≤n−i−k,j∈I pj α(1 − βδ) − i+1≤j≤n−i−k,j∈I pj β(1 − αδ) ,
where α = n−i−k+1≤j≤−i+k,j∈I pj, β = n−i−k+1≤j≤−i+k,j∈I pj, δ = −i+k+1≤j≤h,j∈I pjpj, and α(1 − βδ) − β(1 − αδ) = α − β ≥ 0. Thus, A ≥ B.
If −i + k ≥ n + i − k + 1, then n + i − k ≤ h. Hence A − B = 1≤j≤i,j∈I pjpj n+i−k+1≤j≤h,j∈I pjpj i+1≤j≤n−i−k,j∈I pj α(1 − β) − i+1≤j≤n−i−k,j∈I pj β(1 − α) , where α = n−i−k+1≤j≤n+i−k,j∈I pj, β = n−i−k+1≤j≤n+i−k,j∈I pj, and α(1 − β) − β(1 − α) = α − β ≥ 0. Thus, A ≥ B.
Similarly, we have A ≥ B. Therefore, AA ≥ BB. By similar arguments, we
can prove the inequalities in other cases.
Now it is easy to verify that AA> BB if and only if
{j | i + 1 ≤ j ≤ min(−i + k, n + i − k), h ∈ I} = ∅, {j | i + 1 ≤ j ≤ min(−i + k, n + i − k), h ∈ I} = ∅ if and only if, or, by Lemma 2.5,
q−i −i+k j=−i+1 pj I > 0. Similarly, we can prove the following.
Lemma 2.9. For i = 1, . . . , n − k − 2 − d, qi+k+1 i+k j=i+1 pj I ≥ qi+k+1 i+k j=−i pj I .
By Lemmas 2.5–2.9, all terms in the right-hand side of (2.3) are nonnegative. Next, we show that if I = ∅, then at least one term in (2.3) is positive.
Note that 1 ∈ I. Since I = ∅, there exists a positive integer r such that 1 ≤ r < h, r ∈ I, and r + 1 ∈ I. If r + 1 ≤ n − k, then
r ∈ {j | 1 ≤ j ≤ n − k, j ∈ I}, r + 1 ∈ {j | 1 ≤ j ≤ n − k, j ∈ I}, and hence
q0p1· · · pkqk+1> 0.
If r + 1 > n − k, then choose i = b + n − k − (r + 1) < b, and we have b − i + 1 ≤ r , r + 1 ≤ min(−b + i + k, n + b − i − k). Hence, by Lemma 2.5,
qipi+1· · · pi+k> 0.
Finally, we deal with the case that some equality signs hold in 0 ≤ p[1] ≤ p[2]≤ · · · ≤ p[n] ≤ 1. If p[1] = p[2] = · · · = p[n], then Theorem 2.1 is trivially true. If p[i]< p[i+1], then, for sufficiently small ε > 0, we have
0 < p[1]+ ε < · · · < p[i]+ iε < p[i+1]− (n − i)ε < · · · < p[n]− ε < 1.
For them, we already proved the optimality of assignment C∗in Theorem 2.1; that is,
for any assignment C, R(C∗) ≥ R(C). Now we can complete our proof of Theorem
2.1 by setting ε → 0.
3. Discussion. An invariant optimal assignment is a nice thing to have in
prac-tice and also an interesting mathematical problem to solve. The existence of an invariant optimal assignment has been widely studied for the consecutive-k-out-of-n: F systems and G systems, where a F system works if and only if there do not exist k consecutive components that all fail. Usually, the nonexistence of invariant optimal assignments was demonstrated [12, 7, 13]. There are only four nontrivial cases that invariant optimal assignments may exist. The first is an invariant optimal assignment for the consecutive-2-out-of-n: F line conjectured by Derman, Lieberman, and Ross [2] and independently proved by Du and Hwang [3] and Malon [11]. In fact, the former proved the harder cycle version which is the second case of existence. Note that the cycle version implies the line version since by setting p[n] = 1 (p[1] = 0 in
the G system), the line problem is reduced to the cycle problem. The third case is an invariant optimal assignment for the consecutive-k-out-of-n: G line for n ≤ 2k conjectured by Kuo, Zhang, and Zuo [10], and proved by Jalali et al. [9]. The fourth case is its cycle version, the current case. Note that again the cycle version implies the line version but is much harder. In the line version, one needs only to prove the case
n = 2k, and the n < 2k case can be reduced to the n = 2k case. No similar reduction is possible for the cycle case. One may wonder whether a simpler proof exists by considering other pairings. In the current paper, we break the term qipi+1· · · pi+k
into two parts, pi+1· · · pi+k and −pi· · · pi+k. Use the pairing of p−a+i· · · p−a+i+k−1
with p(−a+i)· · · p(−a+i+k−1) for the first part and a similar one for the second part;
then compare the C assignment with the CI assignment. However, since the
com-parison of one part is positive and the other is negative, we have to further compare their sizes, thus complicating the proof. Can we not break the term qipi+1· · · pi+k
and find a pairing to work? One such possibility is also to consider the clockwise representation of R(C), namely, R(C) = p1· · · pn+ni=1qipi−1· · · pi−k. We then
pair each term qipi+1· · · pi+kfrom the counterclockwise representation with the term
qip(i+1)· · · p(i+k) from the clockwise representation. The CI assignment is better
than the C assignment in all cases except when qi < qi and i does not belong to
i, . . . , i + k. The determination of invariant optimal assignments on lines and cycles is an application of the broader problem of finding an optimal permutation, linear or cyclic, under a certain objective function. This type of problem has been considered before [1, 8] when the arguments of the objective function are |xi− xi+1| for all i.
In the optimal assignment problem, the arguments are products like qipi+1· · · pi+k,
which seems to raise a new type of optimal permutation problem. In this paper we give a solution to one such problem and hope the approach may work for other similar problems [1, 8, 4, 5].
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