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241-460 Introduction to Queueing
Netw orks : Engineering Approach
Assoc. Prof. Thossaporn KamolphiwongCentre for Network Research (CNR)
Department of Computer Engineering, Faculty of EngineeringPrince of Songkla University, Thailand
Email : [email protected]
Outline
M/ M/ 1 queue
Birth-Death rocess for M M 1
Average Number of customer in System
Average Number of Customer in Queue Waiting Time
Example
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M / M / 1
M/M/1
Interarrival Time Service Time #Servers
Memoryless/Memoryless/1 Server
Poisson arrival process
Exponential service time distribution
1 server
infinite population
FCFS
Chapter 9 : M/M/1 Queue
M / M / 1
exponentialservice(Infinite buffer)
PoissonArrival
n
Queue Server
!)(
n
enP
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Birth-Death Process : M/ M/ 1
Let n : arrival rate of customer
Stage : # of customers in the system at time t
2 n10
1 n0
n+1n-1
n-12 n-2
1 2 n n+1n-13
Chapter 9 : M/M/1 Queue
Balance Equ. : M/ M/ 1
n-11 n0 2 n-2
State Rate In = Rate Out0 1P1 = 0P0
n-12 n
n-1
10
1
n+1
n2 n+1
3
0 0 + 2 2 = 1 + 1 12 1P1 + 3P3 = (2 + 2)P2.... ...................
n n-1Pn-1 + nPn+1 = (n + n)Pn
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Steady State Process : M/ M/ 1
Finding Steady State Process:
0: P1 = (0/1)P0
1: P2 = (1/2)P1 + (1P1 - 0P0)/2
= (1/2)P1 + (1P1 - 1P1)/2
= (1/2)P1
0
12
01 P
Chapter 9 : M/M/1 Queue
Steady State Process : M/ M/ 1
State
- n n-1 n n-1 n-1 n-1- n-2 n-2 n
= (n-1/n)Pn-1 + (n-1Pn-1- n-1Pn-1)/n
= (n-1/n)Pn-1
0
11-nn
02-n1-n PPn
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Example : M/ M/ 1
The customers arrive according to a Poissonprocess with rate
The time it takes to serve every customer is anexponential r.v. with parameter .
There is only one server
The system can hold infinite customers (nobuffer overflow)
Chapter 9 : M/M/1 Queue
Example : M/ M/ 1
At equilibrium:P0 = P1P1 = P2
-
P2 = P3................
Pn-1 = Pn
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Example : M/ M/ 1
(P0)(P1)(P2)(Pn-1) = (P1)(P2)(P3).(Pn)n
0 = n n
Let= / then
n
00 PPP nn
Chapter 9 : M/M/1 Queue
Example : M/ M/ 1
P0 + P1 + P2 + + Pn + = 1
10
n
nP
10
0
n
nP 10
0
n
nP
0PPn
n
< 1, P0 must not be zero
0
0
1
n
n
P
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Example : M/ M/ 1
Because
1 1nn i
01
n
P
If || < 1 then
1||if1
1
1
1limlim
1
00
n
n
n
i
i
nn
n
0n
So
11
1
10P
Chapter 9 : M/M/1 Queue
(Continue)
0PPn
n
0
155P
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Example : M/ M/ 1
Prob. that system is empty
= 1 Prob. that system is busy
0
Chapter 9 : M/M/1 Queue
Average Number of customer (N)
in System
N(t) : average number of customer in system attime t
1)( nntN
0)( n nnPtN
n
0
11)(n
nntN
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Average Number of customer (N)
in System
1 nn
d
00 nn d
1
1
0 d
d
d
d
n
n
,n
201
11
d
nn
n
Chapter 9 : M/M/1 Queue
Average Number of customer (N)
in System
2
11)( tN
1
)(tN
average number of customer in system
at time t =
1
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Average Number of customer (N)
in SystemM/M/1
10
20
30
40
Averagenumberof
customerinsystem
0 0.2 0.4 0.6 0.8 1
Utilization
Chapter 9 : M/M/1 Queue
Average Number of Customer inQueue (Nq)
Nq : average number of customer in queue at time t
11 n
n
n
nq PnPN
1
1n
nq PnN
111
2
0PNNq
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Average Number of Customer in
Queue (Nq)
11)(tN
Customer in WaitinServer customer
(Nq)
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Waiting Time (T)
Littles LAW
11
T
1
11
T
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Waiting time in Queue (Wq
)
Wq : Waiting time in Queue
Wq = Waiting time in system service time
11
qW
qW
Chapter 9 : M/M/1 Queue
Waiting time in Queue (Wq)
M/M/1
2
3
4
5
6
ingtimeinQ
ueue
(sec)
0
0 0.2 0.4 0.6 0.8 1
Utilization
Wait
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M/ M/ 1 queueing model Summary
1
T
10P TN
1nnP
qW
2
qq WN
Chapter 9 : M/M/1 Queue
Example
Router A send 8 packets per seconds, on theavera e to router B. The mean size of a acketis 400 byte (exponentially distributed). The linespeed is 64 kbit/s. How many packets are there
on the average in router A waiting fortransmission or being transmitted andwhat isthe probability that the number is10 or more?
A8 packets/s
400 bytes64 kbps
B
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Solution
A8 packets/s B
This is system is M/M/1 because
- one server
400 bytes64 kbps
- arrival is Poisson
- service is exponential
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Solution
A8 packets/s B
= 8 packets/s
Packet size = 400 bytes
400 bytes64 kbps
Link capacity = 64 kbits/s
= 64x1000/(8x400) = 20 packets/s
= / = 8/20 = 0.4
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Solution
How many packets are there on the average inrouter A waitin for transmission or beintransmitted ?
Average packets waiting for transmission are
N= /(1-)
enN= /(1-) = 0.4/0.6 = 2/3
Chapter 9 : M/M/1 Queue
Solution
What is the probability that the number is10 ormore?
Pn = n(1-)
1010
10 1n
n
nnn PP
lim1
110
10
n
nP n
4101010
10 104.01
1
nP
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(Continue)
n
i
n
n lim
nn
i
iS
...1110
10
11211 ... nnS
1101 nSSS
1
110 n
S
1lim
110
10
n
nn
n
Chapter 9 : M/M/1 Queue
More Question
How long are the packets waiting for transmissionin router A?
T=N/ = (2/3)/8 = 1/12 second
How long are the packets waiting for transmissionin queue?
q = = - = secon
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Finite Storage (M/ M/ 1/ K)
Customer arrive according to a Poisson processwith rate
The system has a finite capacity ofKcustomers
including the one in service
Service times are exponential with rate
n-12 n
10
n+1
Chapter 9 : M/M/1 Queue
Solution
The arrival rate is
n = , n < K
n-12 n
10
n = 0, n > K
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M/M/1 /K
At equilibrium: P0 = P1P1 = P2
n-12 n
10
P2 = P3...................
Pn-1 = Pn
Chapter 9 : M/M/1 Queue
M/M/1 /K
(P0)(P1)(Pn-1) = (P1)(P2)(P3).(Pn)n n
0 n
where ,n < K00PPP
n
n
n
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M/M/1 /K
Since 10
K
n
nP
10
0
K
n
nP
10
0
K
n
nP
K
n
n
P
0
01
< 1, P0 must not be zero
Chapter 9 : M/M/1 Queue
M/M/1 /K
Because
1 1KK n
So
0n
10 11
KP
KnPn
Kn,...,2,1,0,
1
11
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M/M/1 /K
The probability Kof state Kis the probability
that an arrivin customer finds the s stem full(the buffer overflows).
When K= 1, we have a single server loss system
n
1,0,1 nPn
Chapter 9 : M/M/1 Queue
Average Number of Customer inSystem
N(t) : average number of customer in system at time t
K
K
nntN
11
1)(
K
n
nnPtN
0
)(
K
n
n
Kn
011
1
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Average Number of Customer in
System
11 KK
0 11n
Kn
1
111)(
K
KKKK
tN
Chapter 9 : M/M/1 Queue
Average Number of customer inQueue
K
nq PnN 1n
K
n
n
K
n
n PnP
11
01 PN
11
1
1
1
11
11
KK
KK
q
KKN
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Litt le s result for M/ M/ 1/ K
We shall use Littles formula to find Tand
q
Recall that was the arrival rateBut if there are Kentities in the system, any
arrivals find the system full, cannot arrive
So of the arrivals per time unit, some
proport on are turne awayPK is the probability of the system being full
Chapter 9 : M/M/1 Queue
(Infinite buffer)
Actual rate of arrival
exponential
service
PoissonArrival
Queue
Server
Actual rate of Arrival eff= (1 PK)
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Litt le s result for M/ M/ 1/ K
Average time spend in the system per customer
The average waiting time per customer
KP
T
1
K
qq
PW
1
Chapter 9 : M/M/1 Queue
References
1. Robert B. Cooper, Introduction to QueueingTheor 2nd edition North Holland 1981.
2. Donald Gross, Carl M. Harris, Fundamentals ofQueueing Theory, 3rd edition, Wiley-Interscience Publication, USA, 1998.
3. Leonard Kleinrock, Queueing Systems Volumn-
Canada, 1975.
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4. Georges Fiche and Gerard Hebuterne,Communicatin S stems & Networks: Traffic &Performance, Kogan Page Limited, 2004.
5. Jerimeah F. Hayes, Thimma V. J. Ganesh Babu,Modeling and Analysis of TelecommunicationsNetworks, John Wiley & Sons, 2004.
Chapter 9 : M/M/1 Queue