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    A unit for a telecommunications system is to be designed so as to have a minimum availability of 0.9994. The MTTR (mean time to repair) for the unit is estimated to be 24 hours. The unit is to consist of (n) effectively identical elements in parallel, from a reliability point of view. If the hazard rate of each of the elements is 5x(10 to the power -5) per hour, estimate the minimum value of (n) which is needed if the unit is to meet the availability specification assuming that the elements are repaired only when the whole system fails.

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    Solution Preview

    MTTR = Mean Time To Repair = 24 hours
    MTBF = Mean Time To Failure
    A = Availability = 0.9994
    FR = Failure (Hazard) Rate of one element = 5*10^-5/Hr
    If there are n elements in the unit, then
    TFR = Total Failure Rate = FR^n (all the ...

    Solution Summary

    MTTR = Mean Time To Repair = 24 hours
    MTBF = Mean Time To Failure
    A = Availability = 0.9994
    FR = Failure (Hazard) Rate of one element = 5*10^-5/Hr
    If there are n elements in the unit, then
    TFR = Total Failure Rate = FR^n (all the individual FRs are multiplied together since the elements are in Parallel)
    MTBF is the inverse of the Total failure Rate
    MTBF = 1 / TFR

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