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Infinite Parallel Redundancy |
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A common strategy for increasing the mean time between failure (MTBF) of a system is to add redundant parallel paths. If each path has an exponential failure rate of λ per hour, then its MTBF is 1/λ hours, and it’s easy to show that the MTBF of two parallel paths is 3/(2λ), i.e., the redundant path increases the MTBF by 50%. Each additional parallel path increases the MTBF still further, although the increase per additional path is less as we add more and more paths. In general the overall system with N channels can be modeled as a simple sequence of states, beginning with the state in which all N components are working, and ending in the state with all of the components failed, as illustrated below: |
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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 |
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Since the harmonic series diverges, it follows that a system consisting of an infinite number of parallel paths, each of which has a mean time to failure of (say) 1 hour, would never fail. (Conversely, since the exponential distribution is non-zero for all time, it is obvious that in an infinite population we will never find all failed, the correspondence furnishes another proof that the harmonic series diverges.) |
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It’s interesting to consider the variance of the MTBF for a highly parallel system. The variance of an exponential distribution with mean 1/λ is simply (1/λ)2. Since each transition in the above sequence of states is exponential with means 1/λ, 1/(2λ), ..., 1/(Nλ), 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 N-path system is given by |
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Interestingly, although the MTBF itself goes to infinity as more parallel paths are added, the variance on the MTBF converges on the finite value |
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