Optimal Consecutive-$ $i$ $k$ $/i$ $-out-of-$ $i$ $(2k+1)$ $/i$ $: $ $i$ $G$ $/i$ $ Cycle. Technical Report

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1 Optimal Consecutive-$ $i$ $k$ $/i$ $-out-of-$ $i$ $2k1)$ $/i$ $: $ $i$ $G$ $/i$ $ Cycle Technical Report Department of Computer Science and Engineering University of Minnesota EECS Building 2 Union Street SE Minneapolis, MN USA TR -49 Optimal Consecutive-$ $i$ $k$ $/i$ $-out-of-$ $i$ $2k1)$ $/i$ $: $ $i$ $G$ $/i$ $ Cycle Ding-zhu Du, Frank K. Hwang, unjae Jung, and Hung Q. Ngo October 2, 2

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3 Optimal Consecutiv e-k-out-of-2k 1): G Cycle Ding-Zhu Du z Frank K. Hwang y unjae Jung z Hung Q. Ngo z May 2, 2 Abstract We present a complete proof for the invariant optimal assignment for consecutive-kout-of-2k1): G cycle, which was proposed by Zuo and Kao in 199 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. Institute of Applied Mathematics, Chinese Academy of Sciences, Beijing. y Department of Applied Mathematics, National Chiao-Tung University, Hsinchu, Taiwan. z Department of Computer Science and Engineering, University of Minnesota, Minneapolis, MN 55455, USA. fdzd,yunjae,hngog@cs.umn.edu. Support in part by the National Science Foundation under grant CCR

4 1 Introduction A cyclic consecutive-k-out-of-n: G system con C 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 reliabilities 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 [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 know 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 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, Con C k; n : G) has no invariant optimal assignment. For n 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: G system Con L k; n : G) can be dened in a similar way to the cyclic system Con C k; n : G). Kuo, Zhang, and Zuo [2] presented an invariant optimal assignment for Con L k; n : G) with n 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 [2k1?n], the cycle system Con C k; 2k 1 : G) becomes the line system Con L k; n : G). Thus, our result yields the result of Jalali et al. [1]. The cycle case is in general much more dicult than the line case, and our proof adopts a new approach dierent from previous attempts. 2

5 2 Main Result In this section, we show the following. Theorem 1 Con C k; 2k 1 : G) has invariant optimal assignment p [1] ; p [3] ; p [5] ; ; p [2k1] ; p [2k] ; ; p [6] ; p [4] ; p [2] ): Let p 1 ; p 2 ; ; p 2k1 be reliabilities of the 2k 1 components on the cycle in counterclockwise direction. For simplicity of the proof, we rst assume that < p [1] < p [2] < < p [2k1] < 1: Our proof is based on the following representation of the reliability of consecutive-kout-of-n: G cycle for n 2k 1 Lemma 1 The reliability of consecutive-k-out-of-n: G cycle for n 2k 1 under assignment C can be represented as RC) p 1 p n p 1 p n where q i 1? p i and p ni p i. n n q i p i1 p ik p i p ik?1? n p i p ik 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, RC) p 1 p n Note that n q i p i1 p ik : q i p i1 p ik p i1 p ik? p i p ik : This implies the second representation. 2 3

6 To prove Theorem 1, it suces to show that in any optimal assignment, p i? )p i?1? 1 ) > for 1 < i < j < 2k 1: 1) That is, selecting any component to be labeled p 1, we always have p i? p 2k2?i )p i1? p 2k1?i ) > for i 1; ; k: 2) For simplicity of representation, we denote i 2k 2? i. Note that k 1) k 1 and i ) i. Furthermore, without loss of generality, we assume p 1 > p 1 throughout this proof. Then the condition 2) can be rewritten as p i > p i for i 1; ; k: 3) We will employ an inductive argument to prove these k inequalities in the following ordering: p 1 > p 1 p k > p k p 2 > p 2 p k?1 > p k?1) Let C i be the assignment obtained from C by exchanging components i and i and C i;j;;y;z C i;j;;y ) z. Let C be an optimal assignment. We divide this inductive argument into the following lemmas. Lemma 2 Under assumption that p 1 > p 1, we must have p k > p k and p 2 p k?1 p 2 p k?1). Proof. Consider RC )? RC 1) p 1? p 1 )p 2 p k? p 2 p k )q k1 : Since p 1 > p 1, we have p 2 p k p 2 p k : 4) 4

7 Note that RC )? RC k) p k? p k )[q 1 p 1 q 1 p k1 )p 2 p k?1? p 1 q 1 q 1 p k1 )p 2 p k?1) ] 5) and q 1 p 1 q 1 p k1 )? p 1 q 1 q 1 p k1 ) p 1? p 1 )1? p k1 ) > : We rst claim that p k > p k. In fact, if p 2 p k?1 p 2 p k?1), then it follows from 5) that p k > p k ; If p 2 p k?1 < p 2 p k?1), then it follows from 4) that p k > p k. Now, we show that p 2 p k?1 p 2 p k?1). Note that RC )? RC1;k) [p 1 p k? p 1 p k ) p k? p k )p k1? p 1 p 1 p k? p k )]p 2 p k?1? p 2 p k?1) )q k1 [p 1 p k q 1? p 1 p k q 1 )q k1 1? p 1 p 1 )p k? p k )p k1 ]p 2 p k?1? p 2 p k?1) ): Moreover, p 1 q 1?p 1 q 1 p 1?p 1 > and p k > p k. Therefore, p 2 p k?1 p 2 p k?1). 2 To show inductive step, we rst prove an equality. Lemma 3 Suppose m < k2. Then 2m1 k?1 m?1 Proof. Note that m i 1? p?mij? k )p i1 q i1) ) i ) q 1 p 1 p k q k1 m p?mij ) )[1? k j1 )p 1 q 1) ) ) )[1? ] k j2 k j1 ]: p 1 p 1 p k1 5

8 p 1 p 1 p k1 q k p 1 p 1 p k1 p k p 1 p 1 p k1 q k q 2 p 1 p 1 p k1 p k p 2 p 1 p 1 p k1 p k m?1 m m i i ) ) )p i1 q i1) ) )p 1 q 1) ) ) k j1 ) k j2 k j1 and p 1 p k? p 1 p 1 p k? p 1 p k1 q 1 p 1 p k q k1? p 1 p 1 p k1 : Thus, 2m1 k?1 m i2 p?mij? q?mi?1 k?1 k p?mij p?mij ) 2m1 im2 k?1 q?mik p?mij p?m1 p?mk p 1 p k? p p k? p 1 p k1 m?1 m m i m?1 m )p i1 q i1) ) ) i 1? )p 1 q 1) ) )p i1 q i1) ) i ) ) k j2 ) p 1 p k? p 1 p 1 p k? p 1 p k1 )[1? k j1 )p 1 q 1) ) ] k j2 6

9 m q 1 p 1 p k q k1 ) )[1? k j1 ]: 2 Lemma 4 Suppose m < k2. If p i > p i for i 1; ; m; k? m 1; ; k 6) and 7) then p m1 > p m1) and : Proof. We rst show p m1 > p m1). For contradiction, suppose p m1 < p m1). It follows from 7) that By Lemma 3, > : 8) RC )? RC m1) 2m1 k?1 2m1? p?mij? [ p m1) p m1 k?1 k m?1 p m1? p m1) )f?p i1) q i1 ) m 1? i m k?1 p?mij p?mij)? p?mij? )[ i ) m k k p?mij ) p m1 p m1) )[p i1 q i1) ) ) )][1? k m j1 ) p?mij) ) k?1 ] )p 1 q 1) ) 7 p?mij? ) k p?mij) )]

10 ? m ) )p 1) q 1 )] k j2 q 1 p 1 p m p m2 p k? q 1 p 1 p m p m2) p k )q k1 < ; m )? )[1? k j1 ]g a contradiction. The last inequality holds because it follows from 6) and 8) that every term in f g is positive. Now, we show By Lemma 3, we have : RC )? RC m2;;m?k) m?1 f? i?p i1) q i1 )? m m1 1? ) )[p i1 q i1) ) i ) m1 )[ j1 ) m1 m1 j1 ) ) j1 )][1? j1 )p 1) q 1 )] k ) j1 ] )p 1 q 1) ) k j2 q 1 p 1 p m1 p 1 p k? q 1 p 1 p m1) p 1) p k )q k1 m )p m1? p m1) )[1? k j1 ]g: It follows from 6) and p m1 > p m1) that every term in f g is positive. Therefore, : 8

11 2 Lemma 5 Suppose m < k2. Then 2m2 m k?1 i m 1? 2m2 p?mij? )p i1 q i1) ) i q 1 p 1 p k q k1 1? m ) ) k ) p?m?1ij )[1? k j1 )p 1 q 1) ) k 1 : ] k j2 Proof. It is similar to the proof of Lemma 3. 2 Lemma 6 Suppose m < k2. If p i > p i for i 1; ; m 1; k? m 1; ; k 9) and 1) then p > p ) 11) and?1?1 : 12) Proof. We rst show 11). For contradiction, suppose p > p ).?1 >?1 : 13) 9

12 By Lemma 5, RC )? RC ) p? p ) )f m i?1?p i1) q i1 ) ) < ; m 1??1? i )?1 )[ j1?1 )[p i1 q i1) ) ) j1 ) )][1? j1 )p 1) q 1 )] k j1 j1 ] )p 1 q 1) ) k j2 q 1 p 1 p?1 p 1 p k? q 1 p 1 p?1) p 1) p k )q k1 1? a contradiction. m?1?1 )? ) Now, we prove 12). Consider RC )? RC m2;;?1)?1?1? ) f m i?p i1) q i1 )? m 1? m1 )[p i1 q i1) ) i ) m1 )[ j ) m1 j ) m1 k 1 ) j )][1? j )p 1) q 1 )] g ) k jm?i1 ] )p 1 q 1) ) k j2 q 1 p 1 p m1 p p k? q 1 p 1 p m1) p ) p k )q k1 1? m )p m1 p? p m1) p ) ) 1 k 1 ) g:

13 It follows from 9) and 11) that every term in f g is positive. Thus, 12) holds. 2 By Lemmas 2, 4, and 6, we know that 3) holds. Therefore, Theorem 1 is proved for < p [1] < < p [2k1] < 1. Finally, we deal with the case that some equality signs hold in p [1] p [2] p [2k1] 1. If p [1] p [2] p [2k1], then Theorem 1 is trivially true. If there exists i, 1 i < 2k 1, such that p [i] < p [i1], then for suciently small " >, we have < p [1] " < < p [i] i" < p [i1]? 2k 1? i)" < < p [2k1]? " < 1: For them, we already proved the optimality of assignment C in Theorem 1, that is, for any assignment C, RC ) RC). Now, we can complete our proof of Theorem 1 by setting "!. 3 Discussion Zuo and Kuo [3] proved only that RC ) RCi ) for any i. RC ) RC) for any assignment C. Thus, their proof is incomplete. This cannot imply that Note that Con C k; n : G) for n 2k cannot be induced from Con C k; 2k 1 : G). Moreover, our inductive argument does not work in cycle system Con C k; n : G) for n 2k. Thus, some additional techniques are required to solve the general case. References [1] A. Jalali, A.G. Hawkes, L. Cui, and F.K. Hwang, The optimal consecutive-k-out-of-n: G line for n 2k, to appear in Naval Res. Logist.. [2] W. Kuo, W. Zhang, and M. Zuo, A consecutive-k-out-of-n: G system: The mirror image of a consecutive-k-out-of-n: F system, IEEE Trans. Rel., ), pp

14 [3] M. Zuo and W. Kuo, Design and performance analysis of consecutive-k-out-of-n structure, Naval Res. Logist., ), pp