Page 57 - A First Course In Stochastic Models

48 RENEWAL-REWARD PROCESSES

Further, using the fact that a given component fails within a time T with probability
1 − e −µT , it follows that

P {the system as a whole fails between two inspections}
m+r
m + r −µT k −µT (m+r−k)
= (1 − e ) e
k
k=r+1
and
E(number of components that fail between two inspections)
= (m + r)(1 − e −µT ).

Hence
−µT
E(total costs in one cycle) = (m + r)I × T + K + (m + r)(1 − e )R.
This gives
the long-run average cost per time unit
1 −µT
= [(m + r)I × T + K + (m + r)(1 − e )R]
T
with probability 1. The optimal values of the parameters r and T are found from
the following minimization problem:
1 −µT
Minimize [(m + r)I × T + K + (m + r)(1 − e )R]
T
m+r
m + r −µT k −µT (m+r−k)
subject to (1 − e ) e ≤ α.
k
k=r+1
Using the Lagrange method this problem can be numerically solved.

Rare events ∗
In many applied probability problems one has to study rare events. For example,
a rare event could be a system failure in reliability applications or buffer overflow
in finite-buffer telecommunication problems. Under general conditions it holds that
the time until the first occurrence of a rare event is approximately exponentially
distributed. Loosely formulated, the following result holds. Let {X(t)} be a regen-
erative process having a set B of (bad) states such that the probability q that
the process visits the set B during a given cycle is very small. Denote by the
random variable U the time until the process visits the set B for the first time.
Assuming that the cycle length has a finite and positive mean E(T ), it holds that
P {U > t} ≈ e −tq/E(T ) for t ≥ 0; see Keilson (1979) or Solovyez (1971) for

∗ This section may be skipped at first reading.