ReviewCh9 M M 1 Queue

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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

Chapter 9 : M/M/1 Queue


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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


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


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


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)


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


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


(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


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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in System

1 nn

d

00 nn d

1

1

0 d

d

d

d

n

n

,n

201

11

d

nn

n



in System

2

11)( tN

1

)(tN

average number of customer in system

at time t =

1



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in SystemM/M/1

10

20

30

40

Averagenumberof

customerinsystem

0 0.2 0.4 0.6 0.8 1

Utilization


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)


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


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


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


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


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


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


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


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


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


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


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


(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


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.


Documents

ReviewCh9 M M 1 Queue