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Infinite Parallel Redundancy
Posted:
Jul 12, 1996 6:30 PM


A common strategy for increasing the mean time between failure (MTBF) of a system is to add redundant paths in parallel. If each path has an exponential failure rate of q per hour, then its MTBF is 1/q hours, and the MTBF of two parallel paths is 3/(2q). In other words, the redundant path increases the MTBF by 50%.
It isn't uncommon for critical systems to have 3 or more parallel redundant channels. In general the overall system with n channels can be modelled as a simple Markov chain as illustrated below:
_______ _______ _______ _______  n   n1   n2   0   paths  nq  paths  (n1)q  paths (n2)q 1q  paths  working>working>working ... >working _______ _______ _______ _______
The mean time from the "n paths working" state to the "0 paths working" state is just the sum of the mean times from each state to the next. This immediately gives the well known result that the MTBF of a system with n parallel redundant paths (without repairs) is proportional to the partial sum of the harmonic series 1 / 1 1 1 1 MTBF(n) =  ( 1 +  +  +  + ... +  ) q \ 2 3 4 n /
Since the harmonic series diverges, this implies a system consisting of an infinite number of parallel paths, each of which has a mean time to failure of 1 hour, would NEVER fail.
Now consider the variance of the MTBF for a highly parallel system. The variance of an exponential distribution with mean 1/q is simply (1/q)^2. Since each transition in the above Markov chain is exponential with means 1/q, 1/2q, ..., 1/nq, and since the variance of the sum of two (or more) random variables is the sum of their variances, it follows that the variance of the MTBF for an npath system is given by
1 / 1 1 1 1 VARIANCE(n) =  ( 1 +  +  +  + ... +  ) q^2 \ 2^2 3^2 4^2 n^2 /
Interestingly, although the MTBF itself goes to infinity as more parallel paths are added, the variance on the MTBF converges on the finite value pi^2 VARIANCE(inf) =  6 q^2
I wonder if any aspect of nature can be modelled as an infinitely parallel system?
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