Queuing Systems
4/13/2015
© 2007 Raymond P. Jefferis III
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Basic Queuing Model
M/M/1 Queue
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© 2007 Raymond P. Jefferis III
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Note
• The queue forms because the instantaneous
arrival rate exceeds the service rate. If
arrivals and service completions were
synchronized, there would be no queue.
• The average arrival rate must be less than
the service rate, or the length of the queue
will grow without limit.
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© 2007 Raymond P. Jefferis III
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Notation
M/M/1
1= number of servers
M=Service:
M=exponential
G=general
D=deterministic (constant)
M= Arrivals: Poisson (Markov) process
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© 2007 Raymond P. Jefferis III
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Input Source - Arrivals
• Generates arrivals according to a statistical
distribution (random interarrival times)
• Arrivals may require random service times
• Average interarrival time is the mean
(expected value) of the model probability
density function
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Expected Value (Mean)
The expected value E(Xi) of a discrete
random variable Xi (i=1,2, …, n) is defined
as,

E( X i )   x j p X i ( X i  x j )
j 1
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© 2007 Raymond P. Jefferis III
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Variance
• The variance of a distribution function
having mean E(k) is given by,
  E[k  E( k )]  E( k )  E ( k )
2
k
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2
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Example Arrival Distributions
• Weibull distribution (component failures)
(a and b are shape and scale parameters)
b1  ax b
P( x )  abx e
0< x<
P( x )  0
x0
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Example Arrival Distribution
• Poisson distribution (telephone call arrivals)
• Assumptions:
– the probability of a single arrival in Dt is lDt,
where l is the average arrival rate [packets/s]
– the probability of no arrivals in Dt is 1-lDt
– arrivals are independent of previous events
[Markov (memoryless) system]
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Poisson Arrival Distribution
• The probability of k arrivals during interval
T is given by,
 lT
e
p( k )  ( lT )
k!
k
• The mean and variance are,
E( k )  lt
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  lt
2
k
© 2007 Raymond P. Jefferis III
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Mean of Poisson Distribution
 2k  E[k  E( k )]2

E ( k )   kp( k )
k 0
( lT )k e  lT
E( k )   k
k!
k 0

( lT )k 1 e lT
E ( k )   lT
( k  1)!
k 0

E ( k )  lTe
 lT
 2k  E( k 2 )  E 2 ( k )
 2k  E( k )
 2k  lT
( lT )k 1

k  0 ( k  1)!

E ( k )  lTe  lT elT  lT
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Poisson (l=0.5)
0 .5
e
k
ll
k
0
0
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5
k
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Poisson (l=0.8)
0 .5
e
k
ll
k
0
0
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5
k
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Poisson (l=2)
0 .6
0 .5
e
k
ll
k
0
0
0
0
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5
k
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10
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Poisson (l=5)
0 .5
e
k
ll
k
0
0
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5
k
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Observations
• The mean increases with lT, as expected.
• The variance increases with lT, as derived.
• If only m arrivals can be serviced in time T,
there is a probability that k>m. In this case
k-m arrivals must be queued. This
probability increases with lT.
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Example
A student at a terminal generates a call to the
network server l=2 times per hour, on average.
What is the probability of two or more such calls
in the next hour, assuming Poisson statistics?
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Answer to example
• The probability of two or more calls in a
one-hour interval is,
p(k  2)  1  p(k  0)  p(k  1)
20 2
p(0)  e  0.1352
0!
21 2
p(1)  e  0.2707
1!
p  2    1  [ p(0)  p(1)]  1  0.4060  0.5940
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© 2007 Raymond P. Jefferis III
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Cumulative Distribution
• The cumulative distribution F(x<k) is the
integral of the probability density function.
Thus, for the previous example:
k
0
1
2
3
4
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p(x=k)
0.13534
0.27067
0.27067
0.18045
0.090224
F(x<k)
0.13534
0.40601
0.67668
0.85713
0.947354
F(X³k)
0.86466
0.59399
0.32332
0.14287
0.052646
© 2007 Raymond P. Jefferis III
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Queue
•
•
•
•
•
holds arrivals until serviced
has maximum length parameter
state is length at given time
arrivals experience waiting time in queue
multiple queues are possible
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Service Mechanisms
•
•
•
•
One or more servers (single or parallel)
servers may also be queued
a service discipline is imposed (FIFO, etc.)
a holding time is required for service
(reciprocal of the serving rate)
• server is busy until service is complete
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Example Service Times
• Exponential distribution (tel. call service)
P( x ) 
1

e

x

mean  
var   2
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Erlang (gamma) Service Times
• Distribution
( k )k k 1  kt
F (t ) 
t e
( k  1)!
• k is integer
• m, k positive
mean  1 / 
std.dev.  1 /  k
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Summary of System Properties
•
•
•
•
•
•
interarrival times randomly distributed
service times randomly distributed
service discipline imposed
number of servers
length of queue (has maximum)
Markov process if no influence of queue
length on arrivals (memoryless)
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Queuing Analysis
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Example
• A certain communications channel delivers
data at its capacity of C=14,400 [bits/sec].
The average packet (message) length, l is
450 bits. The packet arrival rate is thus:
l C/l
l  14400 / 450
l  32 [packets / sec]
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© 2007 Raymond P. Jefferis III
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Utilization Factor
• The ratio of arrival rate to service rate is the
utilization factor.
l/
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Example
• If a channel transmits 35 packets/second and
its service rate is 40 packets/second, the
utilization factor can be calculated as:
l/
l  35 [packets / sec]
 = 40 [packets / sec]
 = 35 / 40 = 0.875
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Total Time Delay
• The average total time delay before
transmission of an arriving packet, W, is
equal to the waiting time in the queue, wq,
plus the time to service the packet, ws. That
is (mean values):
W  wq  ws
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© 2007 Raymond P. Jefferis III
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Delay if n Packets Are Queued
• If n packets are in the queue, the average
time delay for an incoming packet will be:
W  nws  ws  ( n  1)ws
Since ws=1/, then,
W
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( n  1)

© 2007 Raymond P. Jefferis III
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Result
• The expected time delay depends upon the
state of the queue and the mean service rate.
• Note that the service rate may depend upon
the average length of the message serviced.
• For message length li and channel
capacity C,
ws  i / C
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© 2007 Raymond P. Jefferis III
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Number of Packets in System
• The expected (mean) number of packets in
the system will be,
L  lW
where
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l is the average arrival rate
W is the expected waiting time,
including service
© 2007 Raymond P. Jefferis III
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Packets in System vs Traffic
• The expected number of packets in the
system is:
L
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
1 

l
l
© 2007 Raymond P. Jefferis III
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Graphical Representation
1 00

1

50
0
0
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0 .5

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Number of Packets in Queue
• The expected (mean) number of packets in
the queue will be,
Lq  lWq
where
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l is the average arrival rate
Wq is the expected waiting time in
the queue
© 2007 Raymond P. Jefferis III
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Queued Packets vs Traffic
• The expected number of packets in the
queue is:
2
l2
Lq 

1 
(  l )
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Graphical Representation
1 00

1
2

50
0
0
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0 .5

© 2007 Raymond P. Jefferis III
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Total Expected Waiting Time
• If the service time is constant at its
average rate 1/, then
W  Wq 1/ 
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© 2007 Raymond P. Jefferis III
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Wait in Queue vs Traffic
• The waiting time in the queue is:
Wq 
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1
l
Lq 
1


2
l 1 
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Total Waiting Time vs Traffic
• The total waiting time will be:
1

2
1
W 



l 1  
l
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Little’s Result
• There are 4 variables and three independent
equations, so if one is known the rest can
be calculated
L  lW 

1 

l
 l
2
l2
Lq  lWq 

1 
(  l)
Wq 
1
l
Lq 
W  Wq 
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1

1
l


1
l
2
1 


2
1 
l
(  l )

1


© 2007 Raymond P. Jefferis III
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 l
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