Chapter 3

Chapter 3

Delay models in Data Networks

Section 3.2

Little`s Theorem

3.2 Little`s Theorem : average number of customers

in system : mean arrival rate T:mean time a customer spends in

system

N

N

T

Little`s Theorem Proof

N(t) = number of customers in system at time t

(t) = number of customers who arrived in interval [0,t]

Ti = time spent in system by the i-th customer

Little`s Theorem

Little`s Theorem

TN

take

t

T

t

tT

tdN

t

t

t

iit

ii

t

lim

)(

)(1)(

1

)(

1)(

10

3.2.3 Application of Little`s Theorem

Ex3.1 : arrival rate in a transmission line NQ : average number of packets

waiting in queue W : average waiting time spent by a

packet in queue

NQ = W

Application of Little`s Theorem

If = average Tx time =

: Average number of packets under Tx I.e. fraction of time that s busy utilization fact

or

XX

Application of Little`s Theorem Ex3.2

N : average number packets in network T : average delay per packet

also Ti : average delay of packets arriving

at node i

n

iiTN

1

iii TN

3.3 M/M/1 Queuing System M/M/1

First M : arrival , Poisson Second M : service , Exponential 1 : server number

M/M/1 Queuing System Arrival Poisson process

A(t) : number of arrivals from 0 to time t

Number of arrivals that occur in disjoint intervals are independent

Number of arrivals in any interval of length is Poisson distributed with parameter , ,1,0,

!

)()()(

nn

entAtAP

M/M/1 Queuing System Properties of Poisson process

1. Inter arrival times are independent and exponentially distributed with parameter

tn : time of the n-th arrival

2

1

1var,

1)( :

0,1

iancemeanePpdf

SeSP

tt

nn

Sn

nnn

M/M/1 Queuing System

2. For every t0, 0

!2

)(1:

0)(

)(2)()(

)(1)()(

)(10)()(

2

0lim

eNote

owhere

otAtAP

otAtAP

otAtAP

M/M/1 Queuing System3. A = A1+A2++AK is also Poisson with

rate = 1+ 2++ K

Poissonmerge

A1

A2

AK…

……

..

M/M/1 Queuing System

4.

Poisson split

P

1-P Poisson with (1-P)

Also Poisson with

P

M/M/1 Queuing System Service time : Exponential distribution

with parameter Sn : service time of n-th customer

nS

n

Sn

eSPpdf

SeSSP

)(:

0,1

M/M/1 Queuing System Properties of Exponential : memoryles

s

0, ),(|

0, ),(|

trforrSPtStrSP

trforrPttrP

nnn

nnn

Markov chain formulation Let's focus at the times,0,,2,…,k,…

Nk = number of customers in system at time k = N(k)

Where N(t) is continuous-time Markov Chain Nk is discrete-time

Let Pij : transition probabilities = P{Nk+1=j|Nk=i}

Markov chain formulation

1,1, ),(

0),(

0),(

1),(1

)(1

,

1,

1,

00

iiijandioP

ioP

ioP

ioP

oP

ji

ii

ii

ii


Note During any time interval, the total number

of transitions from state n to n+1 must differ from the total number of transitions from n+1 to n by at most 1

I.e. frequency of transitions from n+1 to n = frequency of transitions from n to n+1

ntNPnNPP

iesprobabilitstatesteady

tk

kn

)(

limlim



Take ->0 Pn=Pn+1

Pn+1=Pn, n=0, 1, … 等比數列where = / utilization Pn+1= n+1P0, n=0,1,…

Since <1, and

)23.3,...(1,0)1(

110

0

nP

PP

nn

nn


1,

11

)25.3(1

)1(

)24.3(1)1(

1)1(

)1()(lim

2

2

00

WNW

NT

nnPtNEN

Q

n

n

nn

t

M/G/1 System

Let Ci ： customer I Wi = waiting time of Ci

Xi = service time of Ci

Ni = # of customers found waiting in queue when Ci arrives Ri = residual service time of the customer in service when Ci arrives

M/G/1 System

Ci start serviceCi arrives

Ri

Ni

Xi-1Xi- Ni

)48.3(1

)()()()(

1lim

1

RW

XENEREWE

WRu

NRW

XRW

jiii

Qi

i

Nijjii

i

In steady-state,

M/G/1 System

To calculate R, by graphical approach:

Time

Residual service time r()

Ci starts service

M(t)=# of service completion in [0, t]

tXM(t)

XM(t)

X2

X2

X1

X1

M/G/1 System

Time avg of r() in [0, t]

)(

)(

2

1

2

11)(

1

)(

1

2

2)(

10

tM

X

t

tM

Xt

drt

tM

ii

i

tM

i

t

M/G/1 System

)1(2

)(

1

)(2

1

)(lim

)(lim

2

1

)(1

lim

2

2

)(

1

2

0

XERW

XE

tM

X

t

tM

drt

R

tTake

tM

ii

tt

t

t

P-K Formula(3.53)

Ex3.15

Consider a go back n ARQ:

Assume that error in the forward channel is p, return channel is error-free

Packet arrives as a Poisson process with rate packets/frame

1 2 3 … n-1 n 1

1 2 3

time

time

sender

receiver

Timeout (n-1) frames

Prop. delay

Ex3.15

Service time X : from when a packet transmitted until it is successfully received

1 , if 1st tx is successful (1-p)X={ 1+n, if 1st tx is un- successful; 2nd is su

ccessful p(1-p)1+kn, if 1st k is un- successful;(k+1)th successful Pk(1-p)

Ex3.15

0

22

0 0

0

)1()1(][

)1(

)1()1()(

k

k

k k

kk

k

k

ppknXE

kpnpp

ppknXE

Ex3.15

WXETXE

XEW

p

ppn

p

npXE

p

npXE

)(,))(1(2

)(

)1(

)(

1

21][

11)(

2

2

222

3.5.1 M/G/1 Queue with vacations

1. When the server has served all customers, it goes on vacation

2. If the system is still idle after a vacation interval, go on another vacation interval

3. If a customer arrives during a vacation, customer waits until the end of vacation. Chapter 1 section 1.3.1 page 34in Network or Transport Layer

M/G/1 Queue with vacations


Assume vacation intervals v1, v2… are iid and are independent of customers arrival & service times.→A customer must wait for the completion

of the current service or vacation interval, and then the service of all customers waiting before it.


Where R is the mean residual time for completion of service or vacation when the customer arrives.

1

RW


Let L(t) = # of vacations completed by tM(t) = # of services completed by t

2)(

1

2)(

10 2

11

2

11)(

1i

tL

ii

tM

i

tV

tX

tdr

t

)(

)(

2

1

2

11

)(

1

2

2)(

1 tL

V

t

tLX

t

tL

ii

i

tM

i

)( 2VE

)(2

)(

)1(2

)(

1

22

VE

VEXERW

)(

)()1(

2

1

2

)( 22

VE

VEXER

)(

1)(),(

)(

)1(lim

VEt

tLVE

tL

tt

Because Fraction of time occupied with vacation = 1-

Ex3.16 : FDM, SFDM, TDM m streams of traffic with rate /m(Poi

sson) FDM system – Divide available bandwi

dth into m subchannels. Transmission time for a packet on each of these subchannels is m.

FDM

mm

XE )(

2222 0)()( mmXEXVarXE

)56.3()1(2)1(2

mmWFDM

Slotted FDM System Packet trans starts only at time m,

2m,…When the queue is idle, server takes a vacation of m. (if idle again after vacation, take another)

)57.3(2)1(2

mW

mW FDMSFDM

TDM System

Look at SFDM queue, ->same queue WTDM=WSFDM

Summary

mm

T

mm

T

mifbetterm

Tm

T

FDM

SFDM

FDMTDM

)1(2

)1(2

2),12

(1)1(2

Service time

Reservations & Polling

Satellite

Collision -> solution:polling or reservation

S1 D1 D1 S2 D2

Cycle

S1 D1 D1 S2 D2…

Reservation & Polling M Poisson traffic streams with rate /m1. Gated System – only those packets

which arrive prior to the user’s preceding reservation period are transmitted.

2. Exhaustive system – all packets are transmitted including those that arrive during this data period

3. Partially gated – all packets that arrive up to the beginning of the data interval.

Single-User

Gated system:

S D D S D D … D

m=1

time

Di arrives Di startsWi

Ri

Vl(I)

Di ends tx

l(i)-th reservation intervalNi : # of packets arrive in front of Di

Single-User A reservation(vacation)starts when the

system has served all packets which arrive prior to the preceding reservation interval.

A vacation(M/G/1 queue with vacation) starts when the system has served all packets which have arrived.(corresponds to exhaustive system)

Single-User

WXWEXENE

WNEN

VE

VEXER

itake

RE

i

i

i

)()()(

)(

)(2

)()1(

2

)(

)(

22

與以前一樣

Single-User

exhaustiveVE

VEXEW

VE

VE

VEXEW

)(2

)(

)1(2

)(

)61.3(1

)(

)(2

)(

)1(2

)(

22

22

Single-user

gated

Multi-User

Ni is redefined as # of packets which must be transmitted before packet i

S D D D … S D D time

Packet i arrives Packet i startsWi

Ri

Ni

Pakcet i ends

… S D D

Sum=Yi

Multi-user

Where Yi : includes all reservation intervals packet I must want for.

)63.3()()()()()( iiii YEXENEREWE

Multi-User If number the users 0, 1, 2,…,m-1, the

l-th reservation interval is used to make reservation for user l mod m

1

0

1

0

22

)(2

)()1(

2

)(m

ll

m

ll

VE

VEXE

R

Multi-user

?),65.3(1

YYR

W

YWRW

Multi-user For an exhaustive systemLet lj=E ( Yj | packet i arrives in user l’s

reservation or data intervals and belongs to user (l+j) mod m)

0,...

0,0

mod)(mod)2(mod)1( jVVV

j

mjlmlml

Packet i belongs to each user with same prob. = 1/m

Multi-user

)66.3(

...

1

1

1

1

mod)(

mod)1(mod)2(mod)1(

mod)2(mod)1(

mod)1(

1

1

1

1

m

j

mjl

mmlmlml

mlml

ml

m

jlj

m

jlj

Vm

jm

VVV

VV

V

m

m

Multi-user

All users have equal average data length in steady state.

P(packet i arrives during user l’s data interval)

P(packet i arrives during user l’s reservation interval)

m

1

0

)1( m

k

k

l

V

V

Multi-user

1

1

m

j

E (Yi|pkt i arrives in user l’s data or reser. int.) X P(pkt i arrives in user l’s reser. Or data int. )

1

0

1

0

2

1

0

1

11

0

mod)(

1

)68.3(2

)1(

2

)(

))1(

(

m

l

l

m

ll

m

l

m

jm

k

k

l

mjl

Vm

Vwhere

Vm

VVm

V

V

mV

m

jm

Multi-user

If Vl’s have same dist.

m

VVwhere

V

VmXW

m

l

ll

V

V

1

0

2

2

22

)(

2)1(2

)(

)1(2

2

Exhaustive system(3.69)

- The partially gated system is the same as the exhaustive system except that if a packet arrives in its own user’s data interval (with prob. /m), it is delayed an extra cycle of reservation periods(mV)

Y is increased by

Multi-user

Vm

Vmm

0)1(

Multi-user- The fully gated system is the same as

partially gated system except if a pkt arrives during a user’s own reservation interval (prob. (1-)/m)

- It is delayed by an additional mV

- Y is increased by Vmm

)1

(

Priority Queuing N classes of customers class i

arrives a Poisson process with rate I

service time

Each class joins a separate queue

2,

1i

i

i Xu

X

1

2

Server

Priority Queuing Single server will server customers from

the highest priority queue first1. Non-preemptive

- a lower priority customer, once started, is allowed to finish, when a high priority customer arrives.

2. Preemptive resume- Service for a low priority customer is interrupted when a high priority customer arrives and is resumed from the point of interruption when all higher priority customers have been served

Non-preemptive

Let NQk=avg. # in queue for priority k

Wk= avg. queueing time for priority k k = k/k = system utilization for

priority k R = mean residual service time.

Non-preemptive

2112

2

1

12

11

111

11

111

1

1

WNNRW

RW

WRNRW

QQ

Q

Where 1W2 is the avg. # of higher priority customers that arrives while you are waiting

Non-preemptive

)...1)(...1(

)1)(1(

21121

2112

kkk

RW

RW

Similarly,

Non-preemptiveR=the residual time

2

1

)(2

1X

n

ii

Where =2nd moment of the service time avg. over all priority

2X

221

1 ... nn XX

Non-preemptive

)81.3()(2

1 2

1

XRn

ii

代入

)83.3(1

)82.3()...1)(...1(2 21121

2

1

kkk

kk

i

n

ii

k

WTand

XW

Preemptive

k

kT 1

Note that Tk will not be affected by customers from class k+1 to n

Unfinished work of Class 1 to k (A)

Work due to class 1to k-1 who arrives whenthis customer is waiting (B)

)84.3(2

)1

(Re...1

)(

1

2

1

k

iii

k

k

k

XRwhere

RWcall

RA

Preemptive

)86.3()1(

)1(1

,1

)87.3()...1)(...1(

)...1(1

...1

1

1)(

1

111

1

111

11

1

11

11

RTkfor

RT

TR

T

TTB

kk

kkk

k

k

iki

k

k

kk

k

iki

k

iki

i

Documents

Chapter 3