courtesy of r.martin, rutgers univ introduction to queueing theory comp5416 advanced network...

Courtesy of R.Martin, Rutgers Univ

Introduction to Queueing TheoryCOMP5416Advanced Network Technologies

M/M/1-2School of Information Technologies

Queuing Theory

• View network as collections of queues– FIFO data-structures

• Queuing theory provides probabilistic analysis of these queues

• Examples:– Average length– Probability queue is at a certain length– Probability a packet will be lost


Little’s Formula (aka. Little’s Law)

• Little’s Law:– Mean number tasks in system = arrival rate x mean

residence time• Observed before, Little was first to prove• Applies to any system in equilibrium, as long as

nothing in black box is creating or destroying tasks

SystemArrivals Departures


Example using Little’s Formula

• Consider 40 customers/hour visit Hungry Jack Restaurant– Average customer spend 15 minutes in the restaurant

• What is average number of customers at at given time?

10/25.0/40 custhrhrcustTr r


Model Queuing System

• Use Little’s formula on complete system and parts to reason about average time in the queue


Kendal Notation

• Six parameters in shorthand (x/x/x/x/x/x)– First three typically used, unless specified

1. Arrival Distribution2. Service Distribution3. Number of servers4. Total Capacity (infinite if not specified)5. Population Size (infinite)6. Service Discipline (FCFS/FIFO)


Distributions

• M: Exponential• D: Deterministic (e.g. fixed constant)• Ek: Erlang with parameter k

• Hk: Hyperexponential with param. k• G: General (anything)

• M/M/1 is the simplest ‘realistic’ queue


Kendal Notation Examples

• M/M/1:– Exponential arrivals and service, 1 server, infinite capacity

and population, FCFS (FIFO)• M/M/n

– Same as above, but n servers• G/G/3/20/1500/SPF

– General arrival and service distributions, 3 servers, 17 queue slots (20-3), 1500 total jobs, Shortest Packet First


M/M/1 Queue

Single Isolated Link

•Assume messages arrive at the channel according to a Poisson process at a rate messages/sec.

•Assume that the message lengths have a negative exponential distribution with mean 1/ bits/message.

•Channel transmits messages from its buffer at a constant rate c bits/sec

•The buffer associated with the channel can considered to be effectively of infinite length.

=> an M/M/1 queue.


M/M/1

• The first M says that the interarrival times to the queue are negative-exponential (ie Markovian or memoryless);

• The second M says that the service times are negative-exponential (ie Markovian or memoryless again);

• The 1 says that there is a single server at the queue.

• Shorthand notation for "a queue with Poisson arrivals, negative exponentially distributed message lengths, a single server, and infinite buffer space".


• Arrival rate messages/sec,

• Message lengths neg exp, mean 1/ bits/message,

• Transmission rate c bits/sec, – ie messages transmitted at rate = c messages/sec.

• Define the state of the system as the total number of messages at the link (waiting + being transmitted).

• The system is therefore a Markov chain, since both the arrival and message length distributions are memoryless.

0 1 2


M/M/1 queue model

Tw 1/

Tr

r

w


Solving queuing systems

• Given: : Arrival rate of jobs (packets) : Service rate of the server (output link)

• Solve:– r: average number of items in queuing system– Tr: avg. time an item spends in whole system

– Tw: avg. time an item waits in the queue


Equilibrium conditions

0 1 2

Boundary for flux balance

Boundary for global balance

,2,11 ipp ii


• Solving the flux balance equations recursively, we get

02

2

1 ppppi

iii

0pp ii

Define occupancy


(normalising condition)

(geometric distribution)

)1(

1

00

i

ip

,2,1,0)1( ip ii

Note:Solution valid only for < 1 (stability condition)For 1, there is no steady state!


M/M/1 Queue Length Distribution

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4 5 6 7 8 9

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 5 5 6 7 8 9


M/M/1 Mean Number in the System

1

)1()1(

)1(}{

2

00

i

ii

i

ipiRE


M/M/1 Mean Delay

1

11}{

111}{1

}|{}{delayMean

0

0

R

ii

iiRR

TE

REp

i

piTETE

0.00

2.00

4.00

6.00

8.00

10.00

E{T

}

0 0.25 0.5 0.75 1

M/M/1 queue


M/M/1 Mean Waiting Time

• The waiting time is defined as the time in the queue, excluding service time:

1

11

1

111}{}{ RW TETE


Summary of M/M/1 (infinite buffer)

1

1}{

1

11}{

1}{

1

W

R

TE

TE

RE


Little’s Formula

• If messages/sec enter a “system”, and each spends, on average, W seconds there, then the total number in the “system” must be given by W by a simple flow argument.

• Proof for M/M/1:

}{}{

}{1

1

/1

1

11}{

R

R

TERE

RETE


M/M/1/n – Finite Buffer

0 1

2

n

nipp ii ,,1,00

1

1

00

1

1

n

n

i

ip

nip ini ,,1,0

1

11


M/M/1/n

• Probability that the buffer will be blocked:

nnnB pP

11

1}blocked is messagePr{


M/M/1/n exampleExample : = 0.5• Assume that we want

• Choose n so that

• i.e. choose n = 9

310BP

96.91101

105.0

105.01

5.05.0

5.01

5.01

3

31

31

1

1

n

n

n

nn

n


M/M/1/n – validity of solution

• Solution for the finite buffer queue is valid for all – not just for < 1 as was the case for the infinite

buffer.

• This is because the finite buffer copes with overloads by blocking messages, rather than by creating a backlog of messages.


M/M/1/n – Little’s formula

and it is only these messages which contribute to the buffer size. This implies that:

)1( BP

}{)1(}{}{ RBR TEPTERE

To use Little's formula for the finite buffer case, note that the rate of non-blocked messages is given by:


Summary of M/M/1/n

n

i

in

n

ii

n

n

B

RB

iipRE

P

TEPRE

01

0

1

1

)1(}{

1

)1(

}{)1(}{

1) <or 1 be(can


State dependent Queues

Let i= arrival rate when the system is in state i

i = service rate when the system is in state i

0 1 2

Balance Equations:,2,111 ipp iiii

,2,1021

110 ippi

ii

with solution


M/M/2

• State dependent with

,3,22

1

,1,0

i

i

i

i

i

0 1 2

0101 2)2(ppp

ii

i

i

i

i


M/M/1 vs M/M/2

1

1

2

1}{ 2 rate server, 1RTE

)1)(1(

11}{ rateeach servers, 2

RTE

12

1

}{

}{

rateeach servers, 2

2 rate server, 1

R

R

TE

TE

Compare M/M/2 (2 servers – each at rate ) to that of a single link at rate 2 .

(M/M/1 result)

(M/M/2 result)

Conclusion: Single fast server is always more efficient!


M/M/1 vs M/M/2M/M/1 vs M/M/2

0

1

2

3

4

Occupancy

Mea

n D

elay

(u

nit

s o

f si

ng

le p

acke

t se

rvic

e ti

mes

)

M/M/1: 1 server, rate 2

M/M/2: 2 servers, each rate


M/M/

(a Poisson distribution)

0 1 2

Ai

Aep

ip

iA

i

i

i with !!

0


M/M/1 delay distribution

(an exponential distribution)

tT etp

R

)1()1()(

• Probability of delay within certain limit:

tR etT )1(1)Pr(


Example

Suppose that packets arrive at a link at a Poisson rate of 250 packets/second, and the packet lengths are exponentially distributed with mean length 1000 bits/packet. Assume that the buffer length is sufficiently large that it can be assumed to be effectively infinite.

Find the required capacity of the link in bits/second so that the following criteria are satisfied:

(1) Mean delay 10 msec

(2) Pr{delay 20 msec} 0.10


Solution

bps000,2502500001000250

:Occupancy

ccc

bps000,350

010.0250000

1000

2500001

11000

1

11}{ :DelayMean

c

c

cc

TE R


Solution

In order to satisfy both criteria, we need to choose capacity >365,129 bps.

(The next highest standard rate is 384 kbps.)

tR etT )1(}Pr{ :Delay of Percentile

10.0}msec 20Pr{ 1000

20

1000

250000

c

R eT

bps 129,365c

courtesy of r.martin, rutgers univ introduction to queueing theory comp5416 advanced network...

Documents