10.7.1.1 __Case (1)__

The steady state availability in Case (1) is from Reference
[25]:

where:

m
= subsystem repair
rate

l = subsystem failure
rate

n = number of subsystems
in series

m/l = “operability ratio” as opposed to l/m
(the utilization
factor)

For example, if n = 4 and A_{s} = 0.90, the
allocation equation becomes

"1">

or
m/l = 38.9

The complexities of allocating failure and repair rates for even
simple examples are apparent. If the subsystems are not identical, the
allocation must be solved using the state matrix approach to compute
availability.