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398    CHAPTER 1  Reliability




                         where regular maintenance is carried out, as a good approximation, it is acceptable to
                         take the output of an AND gate as the product of the input event failure probability, pro-
                         vided the MTTRs are very much shorter than the mean time between failures (MTBFs).



                         RELIABILITY BLOCK DIAGRAMS
                         If a system is broken down into key components or elements, the failure of which
                         would have some effect on the system’s availability, a block diagram may be drawn
                         such that for the continuing operation of the system, a continuous string of elements
                         must exist from one side of the diagram to the other. An example of such a block
                         diagram is shown in Fig. 8.1.7. A reliability block diagram produced in this way
                         provides a clear indication of any critical system components and often can also be
                         numerically evaluated. In order to carry out this numerical evaluation, it is necessary
                         to assign a value for failure rate and MTTR for each system element. An availability
                         may then be calculated for each element as follows:
                                                                  μ
                                                 Availability (A) =
                                                                λ + μ
                            where μ is the reciprocal of the MTTR (in hours) and λ is the failure rate (failures
                              6
                         per 10  h). The unavailability is given by
                                                       A = 1 − A

                            and the unavailability of a system of n components in series is then
                                                  A s = A 1 + A 2 +⋯ + A n
                            If the unavailabilities are calculated for each component, they may be added
                         together where the configuration consists of series components and where the com-
                         ponent MTBFs are very much larger than their MTTRs.
                            The MTBF can be obtained by inverting the failure rate and multiplying by 10
                                                                                           6
                         as follows:
                                                   1          6
                                              (  failures  ) = 10 /λ h/failure
                                               λ    6  h
                                                  10
                            The unavailability of a system of n identical components in parallel, assuming
                         only one component is required to maintain the system, is
                                                    A s = A 1 A 2 ⋯ A n
                            Where more than one parallel component is required to maintain the system, the
                         following can be shown to be the combined availability:
                                                     n
                                                    ∑   n   j     n − 1
                                                A s =    j  A (1 − A)
                                                    j= m
                            where n is the total number of components and m is the number of components
                         required to maintain the system.
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