Page 56 - A First Course In Stochastic Models

RENEWAL-REWARD PROCESSES 47

Hence, by the renewal-reward theorem,
1 1 2
the long-run average cost per time unit = K + hλT
T 2
with probability 1. When K = 0 and h = 1, the system incurs a cost at rate j
whenever there are j messages in the buffer, in which case the average cost per
time unit gives the average number of messages in the buffer. Hence
1
the long-run average number of messages in the buffer = λT.
2
Putting the derivative of the cost function equal to 0, it follows that the long-run
average cost is minimal for

2K
∗
T = .
hλ
Example 2.2.3 A reliability system with redundancies

An electronic system consists of a number of independent and identical compo-
nents hooked up in parallel. The lifetime of each component has an exponential
distribution with mean 1/µ. The system is operative only if m or more components
are operating. The non-failed units remain in operation when the system as a whole
is in a non-operative state. The system availability is increased by periodic main-
tenance and by putting r redundant components into operation in addition to the
minimum number m of components required. Under the periodic maintenance the
system is inspected every T time units, where at inspection the failed components
are repaired. The repair time is negligible and each repaired component is again
as good as new. The periodic inspections provide the only repair opportunities.
The following costs are involved. For each component there is a depreciation cost
of I > 0 per time unit. A fixed cost of K > 0 is made for each inspection and
there is a repair cost of R > 0 for each failed component. How can we choose
the number r of redundant components and the time T between two consecutive
inspections such that the long-run average cost per time unit is minimal subject to
the requirement that the probability of system failure between two inspections is
no more than a prespecified value α?
We first derive the performance measures for given values of the parameters r
and T . The stochastic process describing the number of operating components is
regenerative. Using the lack of memory of the exponential lifetimes of the compo-
nents, it follows that the process regenerates itself after each inspection. Taking a
cycle as the time interval between two inspections, we have

E(length of one cycle) = T.