Jounal of Mathematics and Statistics 5 (): 0-4, 009 ISSN 549-3644 009 Science Publications Failue Pobability of -within-consecutive-(, )-out-of-(n, m): F System fo Special Values of m E.M.E.. Sayed Depatment of Mathematics, Faculty of Science, Ain Shams Univesity, Caio, Egypt Abstact: Poblem statements: In this eseach, the eseache aimed to discuses the system failue pobability of the model -within-consecutive (, ) out of (n, m) system fo special values of m. Appoach: The basic idea fo evaluating the failue pobability was the usage of the numbe of configuation of k (k =, 3, 4) paallel columns each contained n components in a -matix. Results: The equation fo the linea k-within (, s) out of (n, m) system wee eached. In this study the failue pobability of -within-consecutive (, ) out of (n, m) system fo m =, 3, 4. Conclusion/Recommendations: In geneal, it was difficult to evaluate the failue pobability in the two-dimensional eliability stuctues such as the linea k-within (, s) out of (n, m) system. The eseache established the failue pobability and then the eliability of thee special cases. It was ecommended to genealize the esults fo any values of k,, s and m. Key wods: System eliability, consecutive-k-out-of-n: F system, -dimensional k-withinconsecutive-(, s)-out-of-(m, n): F system INTRODUCTION The consecutive k-out-of-n: F system has been extensively studied in ecent yeas [-3]. The system is specified by n, the numbe of components and k, the numbe of consecutive failed components that lead to system failue. Genealizations of the consecutive-kout-of-n: F system has been epoted in a consideable numbe of papes [4]. One of the genealizations is the linea connected-(, s)-out-of-(n, m): F lattice system. It consists of mn components aanged in m ows and n columns. The system fails wheneve thee is at least one ectangle of dimension s which contains all failed components. Bounds, eliability evaluation and invaiant optimal design of the linea connected-(, s)- out-of-(n, m): F lattice system is studied in [4-6]. Zuo, in and Wu [7] popose combined k-out-of-n: F, consecutivek-out-of-n: F and linea connected-(, s)-out-of-(n, m): F lattice system stuctues and povide ecusive fomulas fo the eliability of the combined system stuctues. Koutas [8] uses the Makov chain appoach fo eliability evaluation of Makov chain embeddable systems. He shows that the tool is vey useful to a geat vaiety of well-known one-dimensional eliability stuctues. Howeve, it is vey difficult o impossible to apply the idea to two-dimensional eliability stuctues such as the linea connected-(, s)-out-of-(n, m): F lattice system and the above-mentioned combined system stuctues. A futhe genealization of the linea connected-(, s)-out-of-(n, m): F lattice system is the linea k-within (, s)-out-of-(n, m): F lattice system. It consists of mn components aanged in m ows and n columns. The system fails wheneve thee is at least one cluste of size s, which contains k o moe failed components. It becomes a linea connected-(, s)-out-of-(n, m): F lattice system when k = s. The linea k-within (, s)- out-of-(n, m): F lattice system is applied to Thin Film Tansisto iquid Cystal Display failue model. If XGA (04 768 = total 78643 dot) TFT display system fails if and only if moe than o equal to 0 dot fail in 0 0 dot matix, then the system become to be linea 0-within (0, 0)-out-of (04, 768): F lattice system [9]. Akiba and Yammoto [0] poposed appoximate values of eliability of this system. This study gives a simple, diect combinatoial method fo detemining the system failue pobability of the following models: -within consecutive-(, )-out-of-(n, ): F system -within consecutive-(, )-out-of-(n, 3): F system -within consecutive-(, )-out-of-(n, 4): F system Notation: p(n, k) Pobability of system failue, k =, 3, 4 T k Tables with ows indexed by = 0,, and columns indexed by c = 0,,,, k =, 3, 4 p Pobability that a component functions q -p α(n, k, ) Numbe of configuation of k(k =, 3, 4) paallel columns each contains n components having total failues with at least two failed components in a -matix C k (n, ) Enty in T k k =, 3,4 0
MATERIAS AND METHODS States of the system: We denote a functioning and a o,. The typical failed component espectively by ( ) states of the system fo n =, k =, 3, 4 will be shown in Fig.. The states of the system ae denoted by i, whee the fist digit denotes the numbe of failed components in the fist ow and the second digit denotes the numbe of failed components in the second ow Now, we denote the states in Fig. a by 0,0,and espectively and denote the states in Fig. b by, and espectively, fo example the state means, the fist ow contains one failed component and two failed components in the second ow. Assumptions: Each component and the system, is eithe woking o failed The failues of the components ae mutually s- independent Theoem: et k k =,3, 4, be the numbe of configuation of n k -matix having total failues and no two failed components occu in each ( ) -matix. Then: k ( ) = ( + ) n, C n, k Poof: Fist, k =, fo = 0,, we have by definition and by using T : Then we have: ( ) = ( ) + ( ) n, n, n, The conclusion can be poved by induction: the hypothesis assets that: (n-, ) = C (n-, ) and (n-, -) = C (n-, -) Howeve, this implies: ( ) = ( ) + ( ) n, C n, C n, and since this is exactly the elation satisfied by the elements of T, this is equivalent to C (n-+, ): Fo k = 3 3 3n =, = 0, If 3n, the equied numbes which end in 0 ae enumeated by 3 (n-, ). Those end in 0, ae enumeated by 3 (n-, -). Those end in 0, ae enumeated by 3 3 (n-, -). Those end in 0, 0, and ae enumeated by 3 ( n i, i) and complete the pove in the same way of the case k = : n = Now, if n, the equied numbes which end in 0 ae enumeated by (n-, ). Those end in 0, ae enumeated by (n-, -). o o o o o o o o o o o o o Fig. a: Functioning states o o o o o o o o o o Fig. b: Failed states Fo k = 4 4 4n =, = 0, If 4n, the equied numbes which end in 0 ae enumeated by 4 (n-, ). Those which end in 0 ae enumeated by 3 4 (n-, -). Those end in 0, ae enumeated by 4 4 (n-, -). Those end in 0, ae enumeated by 4 4 (n-, -). Those which end by 0 ae enumeated by 6 4 (n-3, -). Those end in 0, ae enumeated by (4+6) 4 (n-4, -3). Thus the equied numbe which end in 0, 0, 0, and i 4 whee ae enumeated by a ( n i, i) ai = ai + ai with a = 4,a = 6 and complete the pove in the same way of the case k =.
RESUTS AND DISCUSION With α ( n,k, ) as defined in the notation, numbe of the states which cause system failue, the pobability of system failue is: kn n () = ( ) = α( ) p n,k n,k, p q To enumeate the states which cause system failue, we can evaluate, at fist, the states at which the system is woking and then we subtact this fom kn,k =,3,4. A diect method fo obtaining α ( n,k, ) is by constucting table T k k =,3,4 as follows: Case : When k =, we constuct T. Given n, fom ow 0 though ow (n-) and column 0 though column (n-) of Table T as follows: Enties of ow = 0 ae: A one followed by zeos Any enty in ow > 0 is the sum of the enty ust above it and twice the enty immediate left neighbos, i.e.: ( ) ( ) ( ) C n +, = C n, + C n, () The coefficients in p(n,) ae: n C ( n, ),,3, n + + = K α ( n,, ) n n +, > (3) Theefoe, p(n, ) is complete detemined by (-3). Example : Suppose that n = 7; to find p(7,), we constuct Table T following the pevious steps, then using (3), we get Table : The elements in T n = 7 C 0 3 4 5 6 0 0 0 0 0 0 0 0 0 0 0 0 4 4 0 0 0 0 3 6 8 0 0 0 4 8 4 3 6 0 0 5 0 40 80 80 3 0 6 60 60 40 9 64 4 α ( 7,,) = C ( 6,) = 9 60 = 3 4 α ( 7,,3) = C ( 5,3) = 364 80 = 84 4 α ( 7,, 4) = C ( 4, 4) = 00 6 = 987 4 4 4 α ( 7,,5) = = 00 α ( 7,,6) = = 3003 5 6 4 4 α ( 7,,7) = = 343 α ( 7,,8) = = 3003 7 8 4 4 α ( 7,,9) = = 00 α ( 7,,0) = = 00 0 4 4 α ( 7,,) = = 364 α ( 7,,) = = 9 4 4 α ( 7,,3) = = 4 α ( 7,,4) = = 3 4 4 P 7, 7,, p q = n = 3p q + 84p q + + 4p q + q 3 3 4 (4) Case : When k = 3 we constuct T 3. Given n, fom ow 0 though ow (n-) and column 0 though column (n) of Table T 3 as follows: Enties of ow = 0 ae consecutive uns of one and zeo Enties of any ow >0 ae the sum of the fou tems: The enty peceding its left neighbo The enty ust above it Tiple of the enty immediate left neighbo to the enty of ) Twice of the enties immediate left neighbo to the enty of 3) i.e.: ( ) ( ) C n +, = C N J +, J + 3 3 ( ) + ( ) + ( ) C n, 3C n, C n, i 3 3 3 (5)
Table : The elements in T 3 n = 4 C 0 3 4 0 0 0 3 4 8 9 6 6 34 65 3 9 37 05 50 The coefficients in p(n,3) ae: 3n n + C3 ( n +, ), =,3, K α ( n,3, ) 3n n +, > (6) Example : Suppose that n = 4; to find p(4, 3), we constuct Table T 3 following the pevious steps, then using (6), we get: α ( 4,3, ) = C3 ( 3, ) = 66 37 = 9 α ( 4,3,3) = C3 (,3) = 0 34 = 86 Table 3: The elements in T 4 n = 4 C 0 3 4 0 0 3 0 9 4 4 79 8 37 30 44 3 78 37 486 Given n, fom ow 0 though ow n and column 0 though column n of table T 4 as follows: Enties of ow = 0 ae consecutive uns of 3 and zeo fo = 0,, Enties of any ow > 0 ae the sum of thee tems: Tiple of the enty peceding its left neighbo The enty ust above it. a i multiple of enty immediate left neighbos to the enty ) whee a i = a i- +a i- with a = 4, a = 6 i.e.: ( + ) = ( + ) C n, 3C n, 4 4 ( ) ( ) + C n, + a C n, i 4 i 4 whee, a i = a i- +a i- with a = 4, a = 6. The coefficient in p(n, 4) ae: (8) α ( 4,3, 4) = C3 (, 4) = 495 9 = 486 4 α ( 4,3,5) = = 79 α ( 4,3,6) = = 94 5 6 4n n + C4 ( n +, ), =,3, K α ( n,4, ) 4n n +, > (9) α ( 4,3,7) = = 79 α ( 4,3,8) = = 495 7 8 α ( 4,3,9) = = 0 α ( 4,3,0) = = 66 0 α ( 4,3,) = = α ( 4,3,) = = Example 3: Suppose that n = 4; to find P(4,4), we constuct Table 3 T 4 following the pevious steps, then using (9), we get: 6 α ( 4, 4, ) = C4 ( 3, ) = 0 78 = 4 6 α ( 4, 4,3) = C4 (,3) = 560 0 = 440 6 α ( 4, 4, 4) = C4 (, 4) = 80 47 = 773 4 P 4,3 4,3, p q = 3n = 9 p q + 8p q + + p q + q 0 9 3 Case 3: When k = 4 we constuct T 4. (7) 3 6 6 α ( 4, 4,5) = = 4368 α ( 4, 4,6) = = 8008 5 6 6 6 α ( 4, 4,7) = = 440 α ( 4, 4,8) = = 870 7 8
6 6 α ( 4, 4,9) = = 440 α ( 4,4,0) = = 8008 0 6 6 α ( 4, 4,) = = 4368 α ( 4,4,) = = 80 6 6 α ( 4, 4,3) = = 560 α ( 4,4,4) = = 0 3 4 6 6 α ( 4, 4,5) = = 6 α ( 4, 4,6) = = 5 6 6 P 4,4 4,4, p q = 4n = 4 p q + 440p q + + 6p q + q 4 3 3 5 6 CONCUSION (0) In this study, we study the failue pobability of the model -Within consecutive-(, )-out-of-(n, m): F system, fo m =, 3, 4. We make exact simple fomula of the failue pobability in these cases. We constuct Table -3 fom which we get α(n,, k) when m =, 3, 4. ACKNOWEDGEMENT I m pleased to acknowledge many helpful comments made by the editos and efeees. REFERENCES. Zhang, Y.. and Y. am, 998. Reliability of consecutive-k-out-of-n: G epaiable system. Int. J. Syst. Sci., 9: 375-379. DOI: 0.080/0007798089963. Kossow, A. and W. Peuss, 989. Reliability of consecutive-k-out-of-n: F systems with nonidentical component eliabilities. IEEE Tans. Reliabil., 38: 9-33. DOI: 0.09/4.3 3. Papastavidis, S.G., 989 ifetime distibution of cicula consecutive-k-out-of-n: F systems with exchangeable lifetimes. IEEE Tans. Reliabil., 38: 460-46. DOI: 0.09/4.46464 4. Cui,. and M. Xie, 005. On a genealized k-outof-n system and its eliability. Int. J. Syst. Sci., 36: 67-74. DOI: 0.080/00077050006470 5. Salvia, A.A. and W.C. ashe, 990. Twodimensional consecutive-k-out-of-n: F models. IEEE Tans. Reliabil., 39: 38-385. DOI: 0.09/4.0303 6. Koutas, M.V., G.K. Papadopoulos and S.G. Papastavidis, 993. Reliability and design of -dimensional consecutive-k-out-of-n: F systems. IEEE Tans. Reliabil., 4: 658-66. DOI: 0.09/4.7360 7. Zuo, M.J., D. in and Y. Wu, 000. Reliability evaluation of combined k-out-of-n: F, consecutivek-out-of-n: F and linea connected-(, s)-out-of-(m, n): F system stuctues. IEEE Tans. Reliabil., R-49, pp.99-04. DOI: 0.09/4.85554 8. Koutas, M.V., 996. On a makov chain appoach fo the study of eliability stuctues. J. Applied Pobabil., 33: 357-367. http://cat.inist.f/?amodele=affichen&cpsidt=37387 9. Akiba and Yamamoto, 005. Evaluating methods fo the eliability of a lage -dimensional ectangula k-within-consecutive-(, s)-out-of-(m, n): F system. Naval Res. ogist., 5: 43-5. DOI: 0.00/nav.0067 0. Papastavidis, S.G. and M.V. Koutas, 993. Bounds fo eliability of consecutive k-within-mout-of-n: F systems. IEEE Tans. Reliabil., 4: 56-60. DOI: 0.09/4.088 4