Page 406 - A First Course In Stochastic Models

THE GI/G/c QUEUE 401

This expression can be further simplified. To show this, we use that
π j+1
= η, j ≥ c − 1 (9.7.8)
π j
for some constant 0 < η < 1. The proof of this result is a replica of the proof of
the corresponding result for the GI/M/1 queue; see (3.5.15). Hence
j−c+1
π j = η π c−1 , j ≥ c − 1. (9.7.9)

As a by-product of (9.7.6) and (9.7.7) we have
p j = η j−c p c , j ≥ c. (9.7.10)

Substituting (9.7.9) into (9.7.8) yields

η −cµ(1−η)x
1 − W q (x) = π c−1 e , x ≥ 0. (9.7.11)
1 − η
The constant η is the unique solution of the equation
∞

η = e −cµ(1−η)t a(t) dt (9.7.12)
0
on the interval (0,1). To see this, note that {π j } is the equilibrium distribution of the
embedded Markov chain describing the number of customers present just before
an arrival epoch. Substituting (9.7.9) into the balance equations

∞ k+1−j
∞
−cµt (cµt)
π j = π k e a(t) dt, j ≥ c
0 (k + 1 − j)!
k=j−1
easily yields the result (9.7.12).
By the relations (9.7.6), (9.7.9) and (9.7.10), the probability distributions {p j }
and {π j } are completely determined once we have computed π 0 , . . . , π c−1 or
p 0 , . . . , p c . These c unknowns can be rather easily computed for the special cases
of deterministic, Coxian-2 and Erlangian interarrival times. If one is only inter-
ested in the waiting-time probabilities (9.7.11), these computations can be avoided.
An explicit expression for the delay probability ηπ c−1 /(1 − η) is given in Tak´ acs
(1962). For the case of c = 1 (GI/M/1 queue), ηπ c−1 /(1 − η) = η.

Deterministic arrivals
Suppose there is a constant time D between two consecutive arrivals. Define the
embedded Markov chain {X n } by

X n = the number of customers present just before the nth arrival.