M/M/1 queue

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1 M/M/1 queue λ n = λ, (n >=0); μ n = μ (n>=1) λ: arrival rate μ: service rate λ μ 1 1 ...) 1 ( 1 ... ... ; ... ... 0 2 0 1 0 0 0 0 1 0 1 1 0 P P P P P P P P P P P n n n n n n n n n

description

M/M/1 queue. λ n = λ , (n >=0); μ n = μ (n>=1). λ. μ. λ : arrival rate μ : service rate. Traffic intensity. rho = λ / μ It is a measure of the total arrival traffic to the system Also known as offered load Example: λ = 3/hour; 1/ μ =15 min = 0.25 h - PowerPoint PPT Presentation

Transcript of M/M/1 queue

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M/M/1 queue

λn = λ, (n >=0); μn = μ (n>=1)

λ: arrival rateμ: service rate

λ μ

11...)1(

1......

;

......

02

0

10

0

0010

110

PP

PPP

PP

PPPP

n

nn

n

n

nn

nn

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Traffic intensity rho = λ/μ

It is a measure of the total arrival traffic to the system Also known as offered load

Example: λ = 3/hour; 1/μ=15 min = 0.25 h

Represents the fraction of time a server is busy In which case it is called the utilization factor

Example: rho = 0.75 = % busy

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Queuing systems: stability λ<μ

=> stable system

λ>μ Steady build up of customers => unstable

Time 1 2 3 4 5 6 7 8 9 10 11

123

busy idleN(t)

Time 1 2 3 4 5 6 7 8 9 10 11

123

N(t)

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Example#1 A communication channel operating at 9600 bps

Receives two type of packet streams from a gateway Type A packets have a fixed length format of 48 bits

Type B packets have an exponentially distribution length With a mean of 480 bits

If on the average there are 20% type A packets and 80% type B packets

Calculate the utilization of this channel Assuming the combined arrival rate is 15 packets/s

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Performance measures L

Mean # customers in the whole system

Lq

Mean queue length in the queue space

W Mean waiting time in the system

Wq

Mean waiting time in the queue

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Mean queue length (M/M/1)

L

n

nnPnEL

n

n

n n

nn

n

n

nn

1)'

11)(1(

)'()1(

)'()1()()1(

)1(][

0

0 0

1

00

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Mean queue length (M/M/1) (cont’d)

q

nn

nn

nnq

LLLL

PL

PnP

PnL

))1(1()1(

)1(

0

11

1

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Little’s theorem This result

Existed as an empirical rule for many years And was first proved in a formal way by Little in 1961

The theorem Relates the average number of customers L

In a steady state queuing system

To the product of the average arrival rate (λ) And average waiting time (W) a customer spend in a system

WL .

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LITTLE’s Formula

: average number of messages in system : average delay λ: arrival rate

Little’s relation holds for any Service discipline Arrival process Holding area

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Graphical Proof A(t)

Cumulative arrival process L(t)

Nb. of customers that left system up to t

=> N(t) = A(t) – L(t) Nb. of customers in system at time t

di : interval between ith arrival and its departure

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Graphical Proof (continued)

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Graphical Proof (continued)

Now, let

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Mean waiting time (M/M/1) Applying Little’s theorem

1

1.1

.LW

WL

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Z-transform: application in queuing systems X is a discrete r.v.

P(X=i) = Pi, i=0, 1, … P0 , P1 , P2 ,…

Properties of the z-transform g(1) = 1, P0 = g(0); P1 = g’(0); P2 = ½ . g’’(0)

, +

0

)(i

ii zPzg

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M/M/1 Queue – Infinite Waiting Room Probability generating function

Mean

Variance

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M/M/S

0

01

110

.!

1.

...3.2.......

;;

Pn

nPP

snsnssnn

n

n

n

nn

n

n

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M/M/S (cont’d)

.1

1.!

1.!

1.

1

;!.

1

;!

1

!.1

........3.2.

1

0

0

0

0

0

0

SSn

P

SnPSS

SnPn

P

PSS

PSSS

P

Sn

S

n

Sn

Sn

n

n

n

Sn

n

n

n

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M/M/S

SnPSS

SnPn

P

PSS

PSSS

PSn

Pn

Pn

PPSn

SnSSnn

Sn

n

n

n

Sn

nn

n

nn

n

nn

n

n

;!.

1

;!

1

!.1

........3.2.,

.!

1.....3.2....

...,

;;

0

0

00

0001

110

λ

μS servers

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M/M/S: normalizing equations

1!.

1!

1

1!.

1!

1

1.........

1

00

0

1

00

110

SnSn

nS

n

n

SnSn

nS

n

n

nSS

SSnP

PSS

Pn

PPPPP

...1.1.1!

1.

...1.1.!

1

1!

1!.

1

2

21

2

21

SSS

SSS

SSSS

S

SSS

SnSn

n

SnSn

n

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M/M/S: stable queue is λ/Sμ < 1 ?

Otherwise you will not get a stable queue, as such

.1

1.!

1.!

1.

1

1

1.!

1.

...1.1.1!

1.

1

0

0

2

21

SSn

P

SS

SSS

S

n

Sn

S

S

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M/M/S: performance measures Mean queue length

Mean waiting time in the queue (Little’s theorem)

Mean waiting time in the system

Mean # of customers in the whole system

02 .

1!.

)/(.)( P

SS

SPSnLS

SnSq

q

qqq

LWWL .

1

qWW

qWLWL ..

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Erlang C formula A quantity of interest

Probability to find all s servers busy

Ratio between Lq and Pc

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M/M/S: stability revisited Stable

If λ/Sμ < 1

Arrival rate to an individual server

Utilization of a server

Utilization of all servers

S

1.S

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M/M/1/N

Birth and death equations

λ μ% loss

N

NnPPPP

NnNn

n

n

n

n

n

nn

n

n

,...,1,0,....

...

;01;

0,

00021

110

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M/M/1/N: normalizing constant Let ρ=λ/μ

As such

10

1

0

0

00

10

111

1)1(

1)...1(

1......

1...

N

N

N

NN

N

PP

P

PPP

PPP

)1()1(

1)1(. 110

N

N

NN

nn

n PPP

Probability of arrivingto a full waiting room

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M/M/1/N: what percent of λ gets into the queue? Percentage of time the queue is full

is equal to PN

Rate of lost customers = λ.PN

Rate of customers getting in : λ.(1-PN) Often referred to as effective customer arrival rate

Utilization of server

.75 .25

fullNot full

)1.( NP

)1.( NP

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M/M/1/N: performance measures Mean # of customers in the system

Mean queue length

Waiting time in system: W = L/λ

Waiting time in queue: Wq = Lq/λ

1

1

1)1(

1

N

NNL

LM/M./1

)1( 0PLLq

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M/M/1/N: equivalent systems When an M/M/1/N queue is full

Continuous arrival A system with loss

is equivalent to shutting up the service For the duration during which the queue is full

And starting it up again when system no longer ful

This system is called a shut down system

This equivalence holds only when the inter-arrival is exponential

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Proof: rate diagrams M/M/1/N system with loss

Consider the special case where N = 5

0 1 2 3 4 5λ λ λ λ λ

μ μ μ μ μ

λ

45545

201

10

....).(..

..).(..

PPPPP

PPPPP

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Proof: rate diagrams (cont’d) M/M/1/N shut down system

Consider the special case where N = 5

0 1 2 3 4 5λ λ λ λ λ

μ μ μ μ μ

45

201

10

....

..).(..

PP

PPPPP

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M/M/infinity: birth and death equations

.

.

λ μ

000021

110 .!

.!

1..!

1....

...

.

Pn

Pn

Pn

PP

nnn

n

n

n

nn

n

n

Infinite number ofservers

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M/M/infinity: normalizing constant

en

P

ePePP

Pn

PPP

PPP

Pn

P

n

n

n

n

n

n

.!

1.1...!2!1

1

1....!

....!2!1

1......

.!

00

2

0

00

2

00

10

0

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Erlang system: M/M/S/S

.

.

λ μ

Finite number ofServers = S

0

0

2

0

0

!

1

1!

...!2!1

1

.!

n

n

S

n

n

n

P

SP

Pn

P

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Erlang loss formula What percent gets in and

What percent gets lost

PS = prob S customers in system

Effective arrival rate

Rate of lost customers = λ.PS

)1.( SP

S

n

n

S

S

n

SP

0 !

!/

Erlang loss formula

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Erlang B formula Probability of finding all s servers busy

In an iterative form: