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Title: Queuing Networks: Burke


1
Queuing Networks Burkes Theorem, Kleinrocks
Approximation, and Jacksons Theorem
  • Wade Trappe

2
Lecture Overview
  • Network of Queues Introduction
  • Queues in Tandem
  • Product Form Solutions
  • Burkes Theorem
  • What is reversibility?
  • Kleinrocks Approximation
  • Quick Jackson Theorem

3
Network 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?

4
Two 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!
5
Two 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
6
Global 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

7
Global Balance Equations for 2 Queues, pg 2
  • Global Balance Equations Case 1
  • Case 2 ngt0, m0
  • Case 3 mgt0, n0
  • Case 4 ngt0, mgt0

8
Steady 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

9
Steady 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.

10
Burkes 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!

11
Example 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
12
Example 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.

13
Reversible 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)

14
Reversible 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)

15
Reversible 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)

16
Reversible 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

17
Reversible Markov Processes, pg 5
  • The conditional probability
  • Hence, we have shown that is a Markov
    process with generator (after
    differentiation).

18
B-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)

19
Step 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
20
Network 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!

21
Two 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!

22
The 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.

23
Quick 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
24
Jacksons 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.

25
Jacksons 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.
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