Page 416 - A First Course In Stochastic Models

FINITE-CAPACITY QUEUES 411

∞
(∞)
(1 − ρ) p
j
j=N+c
P rej = . (9.8.4)
∞
(∞)
1 − ρ p
j
j=N+c
Proof The proof of (9.8.3) is based on the theory of regenerative processes.
The process describing the number of customers present is a regenerative stochas-
tic process in both the finite-capacity model and the infinite-capacity model. For
both models, let a cycle be defined as the time elapsed between two consecutive
arrivals that find the system empty. For the finite-capacity model, we define the
random variables
T = the length of one cycle,
T j = the amount of time that j customers are present during one cycle.

The corresponding quantities for the infinite-capacity model are denoted by T (∞)
(∞)
and T . By the theory of regenerative processes,
j
(∞)
E(T j ) (∞) E(T j )
p j = and p j = (∞) , j = 0, 1, . . . , N + c. (9.8.5)
E(T ) E(T )
The crucial observation is that the random variable T j has the same distribution
(∞)
as T for any 0 ≤ j ≤ N + c − 1 both in the M/M/c/c + N queue and
j
in the M/G/1/N + 1 queue. This result can be roughly explained as follows.
Suppose that at epoch 0 a cycle starts and let the processes {L(t)} and {L (∞) (t)}
describe the number of customers present in the finite-capacity system and in the
infinite-capacity system. During the first cycle the behaviour of the process {L(t)} is
identical to that of the process {L (∞) (t)} as long as the processes have not reached
the level N + c. Once the level N + c has been reached, the process {L (∞) (t)}
may temporarily make an excursion above the level N + c. However, after having
reached the level N + c, both the process {L(t)} and the process {L (∞) (t)} will
return to the level N + c − 1. This return to the level N + c − 1 occurs at a
service completion epoch. At a service completion epoch the elapsed service times
of the other services in progress are not relevant. In the M/G/1/N + 1 queue
the reason is simply that no other services are in progress at a service completion
epoch and in the M/M/c/c + N queue the explanation lies in the memoryless
property of the exponential service-time distribution. Also, it should be noted that
at a service completion epoch the elapsed time since the last arrival is not relevant
since the arrival process is a Poisson process. Thus we can conclude that after a
downcrossing to the level N + c − 1 the behaviour of the process {L (∞) (t)} is
again probabilistically the same as the behaviour of the process {L(t)} as long as
the number of customers present stays below the level N + c. These arguments
(∞)
make it plausible that the distribution of T j is the same as that of T for any
j