By P.R. Kumar, Pravin Varaiya

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**Additional info for Stochastic Systems: Estimation, Identification, and Adaptive Control**

**Example text**

Tn > tn ) the survival funcIf we denote H(t tion of H and C¯ the n-dimensional joint survival function corresponding to C, then we get for the lifetime distribution of a series system ¯ . . , t) = 1 − C(F ¯ 1 (t), . . , Fn (t)). F S (t) = 1 − H(t, In the special case n = 2 we have GΦ,C = t1 + t2 − C(t1 , t2 ) yielding F S (t) = F1 (t) + F2 (t) − C(F1 (t), F2 (t)). (iii) If the n component lifetimes are independent, then the copula C is the product copula (t1 , . . , tn ) = t1 · · · · · tn .

N. 1 Complex Systems pij = P (Xi = xij ), hj = P (Φ(X) ≥ Φj ), a = EΦ(X)/ΦM = j 33 Φj P (Φ(X) = Φj )/ΦM . We call hj the reliability of the system at system level j. For the ﬂow network example above, a represents the expected throughput (ﬂow) relatively to the maximum throughput (ﬂow) level. The problem is to compute hj for one or more values of j, and a, based on the probabilities pij . We assume that the random variables Xi are independent. 22. 21). 01. 941. For the above example it is easy to calculate the system reliability directly by using elementary probability rules.

11. (Reliability of a k-out-of-n structure). The reliability of a k-out-of-n structure of independent components, which all have the same reliability p, equals n h= i=k n i pi (1 − p)n−i . n This formula holds since i=1 Xi has a binomial distribution with parameters n and p under the given assumptions. The case that the component reliabilities are not equal is treated later. Next we look at an arbitrary series–parallel structure. By using the calculation formulae for a series structure and a parallel structure it is relatively straightforward to calculate the reliability of combinations of series and parallel structures, provided that each component is included in just one such structure.