Page 426 - A First Course In Stochastic Models

EXERCISES 421

the server starts servicing. Denote by p 0j (p 1j ) the time-average probability that j customers
are present and the server is on vacation (available for service). Use the regenerative approach
to verify the recursion scheme:

1 − ρ ∞ −λt (λt) j
p 0j = e {1 − V (t)} dt, j ≥ 0,
E(V ) 0 j!
j j
1 − ρ
p 1j = ν k a j−k + λ (p 0k + p 1k )a j−k , j ≥ 1,
E(V )
k=1 k=1
where a n is given in Theorem 9.2.1 and ν k is the probability of k arrivals during a single
vacation period. (Hint: take as cycle the time elapsed between two consecutive epochs at
which either the server becomes idle or finds an empty system upon return from vacation.)
9.3 Consider an M/G/1 queueing system in which the service time of a customer depends on
the queue size at the moment the customer enters service. The service time has a probability
distribution function B 1 (x) when R or fewer customers are present at the moment the
customer enters service; otherwise, the service time has probability distribution function
B 2 (x). Denote by p 1j (p 2j ) the time-average probability that j customers are in the system
and service according to B 1 (B 2 ) is provided. Use the regenerative approach to verify the
recursion scheme
min(j,R)
(1) (1)
p 1j = λp 0 a + λ p 1k a , j = 1, 2, . . . .
j−1 j−k
k=1
j
(2)

p 2j = λ (p 1k + p 2k )a , j > R,
j−k
k=R+1
(i)
where a n is the same as the constant a n in Theorem 9.2.1 except that B(t) is replaced by
R R
B i (t), i = 1, 2. Also, argue that 1 − p 0 = λ{µ 1 p 1j + µ 2 (1 − p 1j )}, where µ i
j=0 j=0
is the mean of the distribution function B i . (Hint: note that the long-run fraction of service
completions at which j customers are left behind equals the long-run fraction of customers
finding j other customers present upon arrival.)
9.4 Consider the M/G/1 retrial queue from Exercise 2.33 again. Let p 0j (p 1j ) denote the
long-run fraction of time that the server is idle (busy) and j customers are in orbit for
j = 0, 1, . . . .
(a) Use the regenerative aproach to establish the recursions
jνp 0j = λp 1,j−1 , j = 1, 2, . . . ,
j 2
1 λ
λa j
p 1j = p 00 + λa j−k+1 + a j−k p 1,k−1 , j = 1, 2, . . . ,
1 − λa 0 1 − λa 0 kν
k=1
∞ −λt k

where a k = 0 e (λt) (1/k!){1−B(t)} dt with B(t) denoting the probability distribution
function of the service time of a customer. (Hint: let T 0j (T 1j ) denote the amount of time
during one cycle that the server is idle (busy) and j customers are in orbit and let N 0j
denote the number of service completions in one cycle at which j customers are left behind
in orbit. Argue that λE(T 1,j−1 ) = E(N 0j ) for j ≥ 0, λE(T 1,j−1 ) = jνE(T 0j ) for j ≥ 1
j+1
and E(T 1j ) = E(N 0k )A kj for j ≥ 0, where A kj is defined as the expected amount
k=0