This preview shows page 1. Sign up to view the full content.
Unformatted text preview: B ) is irreducible, show that the new process is reversible and ﬁnd its
steady state probabilities.
Exercise 6.24. Consider a queueing system with two classes of customers. Type A customer arrivals are Poisson with rate ∏A and Type B customer arrivals are Poisson with rate
∏B . The service time for type A customers is exponential with rate µA and that for type
B is exponential with rate µB . Each service time is independent of all other service times
and of all arrival epochs.
a) First assume there are inﬁnitely many identical servers, and each new arrival immediately
enters an idle server and begins service. Let the state of the system be (i, j ) where i and
j are the numbers of type A and B customers respectively in service. Draw a graph of the
state transitions for i ≤ 2, j ≤ 2. Find the steady state PMF, {p(i, j ); i, j ≥ 0}, for the
Markov process. Hint: Note that the type A and type B customers do not interact.
b) Assume for the rest of the exercise that there is some ﬁnite number m of servers. Customers who arrive when all servers are occupied are turned away. Find the steady state
PMF, {p(i, j ); i, j ≥ 0, i + j ≤ m}, in terms of p(0, 0) for this Markov process. Hint:
Combine part (a) with the result of Exercise 6.23.
c) Let Qn be the probability that there are n customers in service at some given time in
steady state. Show that Qn = p(0, 0)ρn /n! for 0 ≤ n ≤ m where ρ = ρA + ρB , ρA = ∏A /µA ,
and ρB = ∏B /µB . Solve for p(0, 0).
Exercise 6.25. a) Generalize Exercise 6.24 to the case in which there are K types of
customers, each with independent Poisson arrivals and each with independent exponential
service times. Let ∏k and µk be the arrival rate and service rate respectively for the kh user
type, 1 ≤ k ≤ K . Let ρk = ∏k /µk and ρ = ρ1 + ρ2 + · · · + ρK . In particular, show, as before,
that the probability of n customers in the system is Qn = p(0, . . . , 0)ρn /n! for 0 ≤ n ≤ m. 6.9. EXERCISES 277 b) View the customers in part (a) as a single type of customer with Poisson arrivals of rate
P
P
∏ = k ∏k and with a service density k (∏k /∏)µk exp(−µk x). Show that the expected
service time is ρ/∏. Note that what you have shown is that, if a service distribution can
be represented as a weighted sum of exponentials, then the distribution of customers in the
system is the same as for the M/M/m,m queue with equal mean service time.
Exercise 6.26. Consider a sampledtime M/D/m/m queueing system. The arrival process
is Bernoulli with probability ∏ << 1 of arrival in each time unit. There are m servers; each
arrival enters a server if a server is not busy and otherwise the arrival is discarded. If an
arrival enters a server, it keeps the server busy for d units of time and then departs; d is
some integer constant and is the same for each server.
Let n, 0 ≤ n ≤ m be the number of customers in service at a given time and let xi be
the number of time units that the ith of those n customers (in order of arrival) has been
in service. Thus the state of the system can be taken as (n, x ) = (n, x1 , x2 , . . . , xn ) where
0 ≤ n ≤ m and 1 ≤ x1 < x2 < . . . < xn ≤ d.
Let A(n, x ) denote the next state if the present state is (n, x ) and a new arrival enters
service. That is,
A(n, x ) = (n + 1, 1, x1 + 1, x2 + 1, . . . , xn + 1)
A(n, x ) = (n, 1, x1 + 1, x2 + 1, . . . , xn−1 + 1) for n < m and xn < d (6.90)
for n ≤ m and xn = d. (6.91) That is, the new customer receives one unit of service by the next state time, and all the
old customers receive one additional unit of service. If the oldest customer has received d
units of service, then it leaves the system by the next state time. Note that it is possible for
a customer with d units of service at the present time to leave the system and be replaced
by an arrival at the present time (i.e., (6.91) with n = m, xn = d). Let B (n, x ) denote the
next state if either no arrival occurs or if a new arrival is discarded.
B (n, x ) = (n, x1 + 1, x2 + 1, . . . , xn + 1) for xn < d B (n, x ) = (n − 1, x1 + 1, x2 + 1, . . . , xn−1 + 1) for xn = d. (6.92)
(6.93) a) Hypothesize that the backward chain for this system is also a sampledtime M/D/m/m
queueing system, but that the state (n, x1 , . . . , xn )(0 ≤ n ≤ m, 1 ≤ x1 < x2 < . . . < xn ≤ d)
has a diﬀerent interpretation: n is the number of customers as before, but xi is now the
remaining service required by customer i. Explain how this hypothesis leads to the following
steady state equations:
∏πn,x
∏πn,x
(1 − ∏)πn,x
(1 − ∏)πn,x = (1 − ∏)πA(n,x )
= ∏πA(n,x ) = ∏πB (n,x )
= (1 − ∏)πB (n,x ) ; n < m, xn < d
; n ≤ m, xn = d ; n ≤ m, xn = d ; n ≤ m, xn < d. (6.94)
(6.95)
(6.96)
(6.97) b) Using this hypothesis, ﬁnd πn,x in terms of π0 , where π0 is the probability of an empty
system. Hint: Use (6.96) and (6.97); your answer should depend on n, but not x . 278 CHAPTER 6. MARKOV PROCESSES WITH C...
View
Full
Document
This note was uploaded on 09/27/2010 for the course EE 229 taught by Professor R.srikant during the Spring '09 term at University of Illinois, Urbana Champaign.
 Spring '09
 R.Srikant

Click to edit the document details