Dependability for Systems with a Partitioned State Space: by Attila Csenki

By Attila Csenki

Probabilistic versions of technical structures are studied right here whose finite nation area is partitioned into or extra subsets. The platforms thought of are such that every of these subsets of the country area will correspond to a undeniable functionality point of the process. The crudest process differentiates among 'working' and 'failed' procedure states purely. one other, extra subtle, method will differentiate among some of the degrees of redundancy supplied through the procedure. The dependability features tested listed below are random variables linked to the nation space's partitioned constitution; a few general ones are as follows • The series of the lengths of the system's operating classes; • The sequences of the days spent via the approach on the a variety of functionality degrees; • The cumulative time spent by way of the method within the set of operating states throughout the first m operating classes; • the entire cumulative 'up' time of the process until eventually ultimate breakdown; • The variety of fix occasions in the course of a fmite time period; • The variety of fix occasions till ultimate procedure breakdown; • Any mixture of the above. those dependability features might be mentioned in the Markov and semi-Markov frameworks.

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56) MAl = KAI,oo = lim KAI,n. n~oo where n-I KAI,n = I{ Zo E Al } + L I{ Zi-I E A2, Zi E Al }. n ::::: 1. • Zn-I }. • by Z during the first n 'time' instances. 45). 46). n was the finite-horizon version of NSAJ,oo, Similarly to LAJ,n. KAI,n is also finite even if Z is not absorbing. , it need not be absorbing or irreducible. 14. The probability mass function o/KAI,I is given by l a A 2T Pr{ KAJ,n=k} = 1 if k = O. oa AI T 1 if k = 1, if k ::::: 2. For n ::::: 2, the probability mass function 0/ KA I,n is given by if k = 0, if k E {I, ...

4) since NAI,l = 0 <=> 26 NAj,j =,.. =NAj,rn =O. 3), we may assume that m' =m - 1 ;::: 1 since NAj,rn'+1 =0 ~ NAj,rn'+1 = ... =NAJ,rn =O. It is thus nl, ... , nrn-I ;::: 1 and nrn =O. 2) the following Pr{ NAj,1 = nl, ... , NAj,rn-1 =nrn-I, NA),rn =nrn } = Dm=1 To conclude the proof, it suffices to note that all the other events hitherto not considered carry zero probability due to the following two equivalences: NA),i = 0 for all i;::: m + 1 ~ NA),rn+1 = 0, andNAj,l, ... , NA),rn+1 ;::: 1 ~ NA),rn+1 ;::: 1.

9); this will not be pursued here, however. 6, will now be rederived. 11) for m ~ 1.

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