Title: Queuing Networks: Burke
1Queuing Networks Burkes Theorem, Kleinrocks
Approximation, and Jacksons Theorem
2Lecture Overview
- Network of Queues Introduction
- Queues in Tandem
- Product Form Solutions
- Burkes Theorem
- What is reversibility?
- Kleinrocks Approximation
- Quick Jackson Theorem
3Network of Queues Setup
- In many networking scenarios, a customer or
packet must receive service from many servers
before its final task is completed. - Hence, departures from a queue might become
arrivals at another queue. - All that discussion we did in M/G/1 queues
becomes very important for networks of queues! - Consider two queues in tandem
- Departures from first queue become arrivals at
second queue - First queues arrival process is Poisson with
rate and service time is exponential with
rate - Service time at the second queue is exponential
with rate and independent of first servers
service time - How do we model these two queues together?
- Whats the state and state diagram?
4Two Queues in Tandem
m1
m2
Queue2
Queue1
We need to keep track of N1(t) and N2(t) to
describe the state of the system!
5Two Queues in Tandem State Diagram
3
m2
m2
m2
m2
l
l
l
2
m2
m2
m2
m2
l
l
l
1
m2
m2
m2
m2
l
l
l
n20
n10
1
2
3
Our state diagram must keep track of both N1(t)
and N2(t), and the many transitions that are
possible
6Global Balance Equations for 2 Queues, pg 1
- Define (N1(t), N2(t)) to be the state vector for
the two queues in tandem - Notice, now, that we have a Markov Process in
terms of the state vector - Recall the global balance equations for M/M/1
queues - What this means is At steady state, the amount
entering a state is equal to the amount leaving
the state. - We similarly may find the global balance
equations for this two queue system
7Global Balance Equations for 2 Queues, pg 2
- Global Balance Equations Case 1
- Case 2 ngt0, m0
- Case 3 mgt0, n0
- Case 4 ngt0, mgt0
8Steady State PMFs for Two M/M/1 Queues
- We may show that the following joint probability
mass function satisfies the global balance
equations - where ril/mi
- So, how do we get P(N1n)?
- Easy, its just an M/M/1!
- So P(N1n) (1-r1)r1n for n0,1,2
- How do we get P(N2m)?
- Answer Its a marginal. Integrate out the joint
pmf! - Sum over all n to get
9Steady State PMFs for Two Queues, pg 2
- This looks interesting
- means
- Thus, the number of customers at queue 1 and the
number at queue 2 at a particular time are
independent random variables! - The steady state at queue 2 is the same as for an
M/M/1 queue with Poisson arrival rate l and
exponential service time m2 . - Definition A network of queues is said to have a
product-form solution when the joint pmf of the
number of customers at each queue is the product
of the marginal pmfs of the number of customers
at each queue.
10Burkes Theorem
- Burkes Theorem is the fundamental result
describing product form solutions - Burkes Theorem Consider an M/M/1, M/M/m,
M/M/infinity queuing system at steady state with
arrival rate l, then - The departure process is Poisson with rate l
- At each time t, the number of customers in the
system N(t) is independent of the sequence of
departure times prior to t. - What Burkes Theorem implies
- Two queue problem follows from Burkes Thm
(arrivals to queue 2 are Poisson with rate l). - Arrivals to queue 2 prior to time t are
departures from queue 1 prior to time t, thus
Burkes theorem says queue-1s departures
(queue-2s arrivals) are independent of N1(t). - N2(t) is determined by the sequence of arrivals
from queue-1 prior to time t and independent of
service times, then N1(t) and N2(t) are
independent as random variables. - Note N1(t) and N2(t) are not independent as
processes!
11Example Application of Burkes Thm
- Consider the network of queues
- Here Queue 1 is driven by a Poisson process with
rate l1, , and the departures are randomly routed
to queues 2 and 3. - Queue 3 has an additional, independent Poisson
arrival process with rate l2.
m2
1/2
m1
l1
1/2
m3
l2
12Example Application of Burkes Thm, pg 2
- Burkes Theorem says
- N1(t) and N2(t) are independent
- N1(t) and N3(t) are independent
- Recall that the random split of a Poisson process
yields independent Poisson processes - Hence inputs to Queue 2 and Queue 3 are
independent - Input to Queue 2 is Poisson with rate l1/2
- Input to Queue 3 is Poisson with rate l1/2 l2
- Thus
- where r1l1/m1, r2l1/2m2 , r3(l1/2
l2)/m3. All queues are assumed to be stable.
13Reversible Markov Processes
- In order to prove Burkes theorem, we need the
concept of the reversibility of a Markov process. - A stationary Markov process X(t), with a
countable state space (i.e. a Markov chain will
do), is reversible if X(t) and Y(t)X(-t) have
the same joint distribution at arbitrarily chosen
instants t1, t2, , tN. - A necessary and sufficient condition for
reversibility is - where pi and pij are the stationary
probabilities and transition probabilities of
X(t) - This condition can be easily shown for M/M/1
queuesbut we will show it in more general form - In fact, it holds for any birth-death process,
and N(-t) is statistically identical to N(t)
14Reversible Markov Processes, pg 2
- Time-Reversal Theorem Let X(t) tgt0 be a
stationary Markov process with (infinitesimal)
generator Ppij, and with initial distribution
equal to stationary distribution. Then for all
Tgt0, the time-reversed process - is equivalent to a stationary Markov process
with (infinitesimal) generator given by - for all state pairs (i,j)
15Reversible Markov Processes, pg 3
- Proof Let Q(t)qij(t) denote the transition
probabilities of X(t), - We need to show X(T-t) is a Markov process with
transition probabilities - Then we obtain the necessary and sufficient
condition by differentiating this and setting
t0. (Its an infinitesimal generator)
16Reversible Markov Processes, pg 4
- Consider the interval (0,ts and divide it into
(0,t and (t,ts, i.e. set Tts. - The joint probability of the three random
variables - is
- Similarly we have
17Reversible Markov Processes, pg 5
- The conditional probability
- Hence, we have shown that is a Markov
process with generator (after
differentiation).
18B-D Processes are Reversible
- It is now easy to show the following
- Time Reversibility of B-D Processes The
stationary B-D process N(t) with generator P and
steady state probabilities p is a time-reversible
Markov process. Thus, the time-reversal of the
death process is a birth process. - Burkes Theorem follows from this
- Interdeparture times of the forward-time system
are the interarrival times of the time-reversed
system... Hence we have Poisson with rate l
coming out of the system. - Fix a time t, then the departures before time t
from the forward system are arrivals after time t
in the reverse system. - Arrivals in reverse system are Poisson, and thus
arrivals in reverse system after time t do not
depend on N(t) - Consequently, departures after time t in forward
system do not depend on N(t)
19Step back to Network of Queues
- What we derived held true because the first queue
was M/M/1 and implicitly we assumed it had
achieved steady state and independent service
times between queues! - The problem is more complicated when we have more
general networks of queues. - Again, consider two transmission lines in
sequence. - The arrivals to the first are Poisson of rate l,
but all customers (packets) have deterministic
and equal service times, i.e. we have an M/D/1
queue. - Average packet delay for first queue is given by
Pollaczek-Khinchine formula.
l
m
m
Queue2
Queue1
20Network of Queues, M/D/1 first queue
- The interarrival times of the second queue must
be at least 1/m - Why?
- Now, each packet arriving at either queue takes
1/m time to process. - The first packet being finished by first queue is
immediately sent to second server - It takes at least another 1/m amount of time for
first queue to get and finish the next
packet/customer. - So, first packet will be finished by second
server at or before the next packet arrives to
second server. - Result No queue (waiting) at second system!
21Two Queues, correlated service times
- Earlier we considered the service times
independent of each other and independent of the
arrival times. - Reality A big packet at the first system is
probably still a big packet at the second system! - Interarrival times at the second queue are
strongly correlated with packet lengths! - Long packets at first system will typically find
the queue at the second server more empty - Shorter packets from the first system will
typically find the queue at the second server
more busy because the second server is processing
some prior big packet - It is tough to find an analytical solution for
joint pmf under dependence assumptions!
22The Kleinrock Independence Approximation
- We have argued that in practice there is
dependence upon the interarrival times and
service times. - Independence is lost after the first system!
- Reality hurt us, but reality provides us one more
gift - Reality Real networks typically involve more
than one stream of packets merging at a node The
combination of multiple streams helps restore
independence in many cases! - This observation is due to Kleinrock.
- Kleinrocks Approximation It is often
appropriate to use M/M/1 queues for each
communication link when the arrivals at entry
points are Poisson, packet lengths are roughly
exponentially distributed, network is dense and
traffic is heavy.
23Quick Look at Jacksons Theorem
- Many queuing networks, a packet/customer may
visit a queue more than once. - Burkes theorem does not apply!
- Typical example Queue with feedback
- If the arrival rate is much less than departure
rate, then net arrival process has a few,
isolated external arrivals followed by a burst of
feedback arrivals (dependent on packet length).
p
m
l
a
1-p
24Jacksons Theorem, (Open) pg 1.
- Consider a queuing network consisting of M
separate service stations, each with its own
queue. - Define the vector process
- In an open queuing network, customers may arrive
from an external source and eventually may
leave the network (as depicted in previous
slide). - A closed network has no arrivals or departures
from the system (total customers is fixed just
they may move around) - Assumptions
- The rate of the source (birth process) is l, and
a customer goes to station i with probability
qsi. - Service time at station k is exponential with
rate mk . - Customers are routed according to a Markov
chain Probability that a customer departing
station i goes to station j is qij.
25Jacksons Theorem, (Open) pg 2.
- Jacksons Decomposition Theorem For an open
queue as described, the joint distribution of the
queue vector n(t) is given by - where P(Nknk) is the steady state pmf of an
M/M/1 system with arrival rate lk and service
rate mk, i.e. -
- Where rk describes the utilization factor of
system k.