Page 58 - A First Course In Stochastic Models

RENEWAL-REWARD PROCESSES 49

a proof. The result that the time until the first occurrence of a rare event in a
regenerative process is approximately exponentially distributed is very useful. It
gives not only quantitative insight, but it also implies that the computation of the
mean of the first-passage time suffices to get the whole distribution.
In the next example we obtain the above result by elementary arguments.

Example 2.2.4 A reliability problem with periodic inspections
High reliability of an electronic system is often achieved by employing redundant
components and having periodic inspections. Let us consider a reliability system
with two identical units, where one unit is in full operation and the other unit is in
warm standby. The operating unit has a constant failure rate of λ 0 and the unit in
standby has a constant failure rate of λ 1 , where 0 ≤ λ 1 < λ 0 . Upon failure of the
operating unit, the standby unit is put into full operation provided the standby is not
in the failure state. Failed units are replaced only at the scheduled times T, 2T, . . .
when the system is inspected. The time to replace any failed unit is negligible.
A system failure occurs if both units are down. It is assumed that (λ 0 + λ 1 )T is
sufficiently small so that a system failure is a rare event. In designing highly reliable
systems a key measure of system performance is the probability distribution of the
time until the first system failure.
To find the distribution of the time until the first system failure, we first compute
the probability q defined by

q = P {system failure occurs between two inspections}.

To do so, observe that a constant failure rate λ for the lifetime of a unit implies that
the lifetime has an exponential distribution with mean 1/λ. Using the fact that the
minimum of two independent exponentials with respective means 1/λ 0 and 1/λ 1
is exponentially distributed with mean 1/(λ 0 + λ 1 ), we find by conditioning on the
epoch of the first failure of a unit that

T −λ 0 (T −x) −(λ 0 +λ 1 )x
q = 1 − e (λ 0 + λ 1 ) e dx
0
(λ 0 + λ 1 ) −λ 0 T λ 0 −(λ 0 +λ 1 )T
= 1 − e + e .
λ 1 λ 1
Assuming that both units are in good condition at epoch 0, let

U = the time until the first system failure.

Since the process describing the state of the two units regenerates itself at each
inspection, it follows that
n
P {U > nT } = (1 − q) , n = 0, 1, . . . .