Transcript Cellular Security Overview

Queuing Networks: Burke’s
Theorem, Kleinrock’s
Approximation, and Jackson’s
Theorem
Wade Trappe
Lecture Overview

Network of Queues Introduction
–
–
–
–


Queues in Tandem
Product Form Solutions
Burke’s Theorem
What is reversibility?
Kleinrock’s Approximation
Quick Jackson Theorem
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 queue’s arrival process is Poisson with rate  and
service time is exponential with rate 1  
– Service time at the second queue is exponential with rate  2  
and independent of first server’s service time

How do we model these two queues together?
– What’s the state and state diagram?
Two Queues in Tandem
Queue1
1
Queue2
2
We need to keep track of N1(t) and N2(t) to describe the
state of the system!
Two Queues in Tandem: State Diagram
3
2
1
2

2
2
1
n1=0
1 2
1


n2=0
2

1 2
1 2

1
1 2

1 2
2
2

2

1
1

2
3
Our state diagram must keep track of both N1(t) and N2(t),
and the many transitions that are possible…
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:
p 0  p1
  p j  p j1  p j1
, j  1,2,...

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
Global Balance Equations for 2 Queues, pg 2

Global Balance Equations: Case 1
PN1  0, N2  0  2PN1  0, N2  1

Case 2: n>0, m=0
  1 PN1  n, N2  0  2PN1  n, N2  1  PN1  n 1, N2  0

Case 3: m>0, n=0
  2 PN1  0, N2  m  2PN1  0, N2  m 1  1PN1  1, N2  m 1

Case 4: n>0, m>0
  1   2 PN1  n, N 2  m    2 PN1  n, N 2  m  1
 1PN1  n  1, N 2  m  1
 PN1  n  1, N 2  m 
Steady State PMFs for Two M/M/1 Queues

We may show that the following joint probability mass function
satisfies the global balance equations
PN1  n, N 2  m  1  r1 r1n 1  r2 rm2
where ri=/i
 So, how do we get P(N1=n)?
– Easy, its just an M/M/1!
– So P(N1=n) = (1-r1)r1n for n=0,1,2…

How do we get P(N2=m)?
– Answer: It’s a marginal. Integrate out the joint pmf!
– Sum over all n to get:
PN 2  m  1  r2 rm2
Check
This!
Steady State PMFs for Two Queues, pg 2

This looks “interesting”
PN1  n, N 2  m  1  r1 r1n 1  r2 rm2
means
PN1  n, N2  m  PN1  n PN2  m

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  and exponential service time 2 .

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.
Burke’s Theorem



Burke’s Theorem is the fundamental result describing “product form”
solutions
Burke’s Theorem: Consider an M/M/1, M/M/m, M/M/infinity queuing
system at steady state with arrival rate , then
– The departure process is Poisson with rate ;
– 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 Burke’s Theorem implies:
– Two queue problem follows from Burke’s Thm (arrivals to queue 2 are
Poisson with rate ).
– Arrivals to queue 2 prior to time t are departures from queue 1 prior to
time t, thus Burke’s theorem says queue-1’s departures (queue-2’s
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!
Example Application of Burke’s Thm

Consider the network of queues:
2
1/2
1
1
1/2
2
3

Here Queue 1 is driven by a Poisson process with rate 1, , and
the departures are randomly routed to queues 2 and 3.

Queue 3 has an additional, independent Poisson arrival process
with rate 2.
Example Application of Burke’s Thm, pg 2

Burke’s 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 1/2

Input to Queue 3 is Poisson with rate 1/2 + 2
Thus

PN1 ( t )  k, N 2 ( t )  m, N 3 ( t )  n   1  r1 r1k 1  r 2 r m2 1  r3 r3n
where r1=1/1, r2=1/22 , r3=(1/2 2/3. All queues are
assumed to be stable.
Reversible Markov Processes

In order to prove Burke’s 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
p i p i , j  p j p j, i
where {pi} and {pij} are the stationary probabilities and transition
probabilities of X(t)

This condition can be easily shown for M/M/1 queues…but 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)
Reversible Markov Processes, pg 2

Time-Reversal Theorem: Let {X(t): t>=0} be a stationary
Markov process with (infinitesimal) generator P=[pij], and
with initial distribution equal to stationary distribution. Then
for all T>0, the time-reversed process
X(T  t) :0  t  T
is equivalent to a stationary
~ Markov process with
(infinitesimal) generator Pij given by:
p jp ji
~
pij 
pi
for all state pairs (i,j)
Reversible Markov Processes, pg 3

Proof: Let Q(t)=[qij(t)] denote the transition probabilities of
X(t),
q ij ( t )  PX( t )  j | X(0)  i

We need to show X(T-t) is a Markov process with transition
probabilities
p jq ji ( t )
~
~
Q( t )  qij ( t ) 
pi

Then we obtain the necessary and sufficient condition by
differentiating this and setting t=0. (Its an infinitesimal
generator)
Reversible Markov Processes, pg 4

Consider the interval (0,t+s] and divide it into (0,t] and
(t,t+s], i.e. set T=t+s.

The joint probability of the three random variables
~
X(0)  X( t  s)
is
~
X( t )  X(s)

~
~
~
P X(0)  i 0 , X( t )  i1 , X( t  s)  i 2
~
X( t  s)  X(0)

 PX(0)  i 2 , X(s)  i1 , X( t  s)  i 0 
 p i 2 q i 2 ,i1 (s)q i1 ,i 0 ( t )

Similarly we have


~
~
P X(0)  i 0 , X( t )  i1  PX(s)  i1 , X(s  t )  i 0 
 p i1 q i1 ,i 0 ( t )
Reversible Markov Processes, pg 5

The conditional probability

~
~
~
P X( t  s)  i 2 | X( t )  i1 , X(0)  i 0


p i 2 q i 2 ,i1 ( t )
p i1
~
~
 P X( t  s)  i 2 | X( t )  i1
~
q


i, j

~
Hence, we have shown that X t  is a Markov process with
~
generator P (after differentiation).
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.

Burke’s 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  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)
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 , but all customers
(packets) have deterministic and equal service times, i.e. we
have an M/D/1 queue.


Queue1

Queue2

– Average packet delay for first queue is given by PollaczekKhinchine formula.
Network of Queues, M/D/1 first queue

The interarrival times of the second queue must be at least 1/
– Why?

Now, each packet arriving at either queue takes 1/ time to
process.
– The first packet being finished by first queue is immediately
sent to second server
– It takes at least another 1/ 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!
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!
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.

Kleinrock’s 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.
Quick Look at Jackson’s Theorem

Many queuing networks, a packet/customer may visit a queue
more than once.
Burke’s theorem does not apply!

Typical example: Queue with feedback

p
a



1-p
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).
Jackson’s Theorem, (Open) pg 1.





Consider a queuing network consisting of M separate service stations,
each with its own queue.
Define the vector process:
Nt   N1 (t ), N2 (t ),, NM (t)
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 , and a customer goes to
station i with probability qsi.
– Service time at station k is exponential with rate k .
– Customers are “routed” according to a Markov chain: Probability
that a customer departing station i goes to station j is qij.
Jackson’s Theorem, (Open) pg 2.

Jackson’s Decomposition Theorem: For an open queue as
described, the joint distribution of the queue vector n(t) is
given by:
Pn   PN1  n1 PN2  n 2 PNK  n K 
where P(Nk=nk) is the steady state pmf of an M/M/1 system
with arrival rate k and service rate k, i.e.
PN k  n k   rnk k 1  rk 
Where rk describes the “utilization” factor of system k.