Poisson Random Process

1

Poisson Random Process

€

Pk = P{k arrivals in (0,t]} =λ t( )

k

k!e−λt

tetPP λ−== ]}(0, in arrivals 0{0

tePtP λ−−=−= 11]}(0, in arrivals some{ 0

2

Mean and Variance Results

You have to memorize these!You should be able to derive any of the above

Exponential: 22 /1 ;/1)( λσλ ==YE

Poisson: tXE λσ == 2)(

Geometric:22 / ;/1)( pqpXE == σ

Binomial: npqnpXE == 2 ;)( σ

3

CMPE 252A: Computer Networks

SET 3:

Medium Access Control Protocols

4

medium access control

logical link control

Medium Access Control Protocols Used to share the use of transmission media that

can be accessed concurrently by multiple users.

PHYSICAL

LINK

NETWORK

TRANSPORT

SESSION

PRESENTATION

APPLICATION

Sharing of link and transportof data over the link

5

Contention-Based MAC Protocols

No coordination: Stations transmit at will when they have data to send (e.g., ALOHA)

Carrier sensing (listen before transmit): Stations sense the channel before transmitting a data packet (e.g., CSMA).

Listen before and during transmission: Stations listen before transmitting and stop if noise is heard while transmitting (CSMA/CD).

Collision avoidance (floor acquisition): Stations carry out a handshake to determine which one can send a data packet (e.g., MACA, FAMA, IEEE802.11, RIMA).

Collision resolution: Stations determine which one should try again after a collision.

6

ALOHA Protocol The first protocol for multiple access

channels; the first analysis of such protocols (Norm Abramson, Univ. of Hawaii, 1970).

Originally planned for systems with a central base station or a satellite transponder.

Two frequency bands;Up link and down link(413MHz, 407MH at 9600bps)

Central node retransmitsevery packet it receives!

7

ALOHA Protocol

Population is a large number of bursty stations.

Each station transmits a packet whenever it receives it from its user; no coordination with other stations!

Central node retransmits all packets (good or bad) on down link.

Stations decide to retransmit based on the information they hear from central node

8

ALOHA Protocol An integral part of the ALOHA protocol is

feedback from the receiverFeedback occurs after a packet is sentNo coordination among sources

Packetready?

transmit

yes

wait for a round-trip time

positiveack?yes

no

compute randombackoff integer kno

delay packettransmission

k times

9

The ALOHA Channel We assume:

An (essentially) infinite population of stations. An ideal perfect down link for the transmission of

feedback to senders. Stations are half duplex; have zero processing delays. Retransmissions are scheduled such that all packets are

statistically independent. Each packet has the same duration P. Stations have the same round-trip delay from one

other; this time can be much longer than P (irrelevant). Packet arrivals are Poisson with rate lambda. Collisions are the only sources of errors.

10

The ALOHA Channel

What percentage of time is the channel sending correct packets? This gives us the throughput of the protocol.

time

user i

time

user j

time

sum...

NEW

NEW

P

NEW NEW

collision

I B I B I B I B I

NEW

NEW

RET.

RET.

RET.

RET.

τ

11

Throughput of ALOHA Protocol is the arrival rate.For convenience, we normalize the arrival rate as: PG λ=

λ

From the definition of throughput: pGS ×=where p is the probability of a successful packet transmission

Because arrivals are Poisson and all packets have equal length, every packet has the same probability of being successful.

A packet is successful if no packets arrive within P seconds before it starts or while it is being transmitted; accordingly, GP eep 2)2( −− == λ

Therefore,

GGeS 2−= G

S

0.18

0.5

Throughput of ALOHA Protocol

time

node i frame

t0

All packets have the same length

packet overlaps with start of packet from node i

t0 - 1 t0 + 1

packet overlaps with end of packet from node i

interfering frame

interfering frame

Node i’s frame is vulnerable from any arrival in the time interval (t0-1, t0+1]

Throughput of ALOHA Protocol

Highest throughput when we have one packet for each 2-packet time period

time

node i frame

t0

packet overlaps with start of packet from node i

t0 - 1 t0 + 1

packet overlaps with end of packet from node i

interfering frame

interfering frame

Node i’s frame is vulnerable from any arrival in the time interval (t0-1, t0+1]

14

Slotted ALOHA The throughput of ALOHA can be improved by

reducing the time a packet is vulnerable to interference from other packets.

Slotted ALOHA works in a “slotted channel” providing discrete time slots.

Stations can start transmitting only at the beginning of time slots.

The time synchronization needed for slotting is accomplished at the physical layer, and some synchronization is required in many cases anyway.

15

Slotted ALOHA Protocol

Packetready?

transmit

yes

wait for a round-trip time

quantized in slots

positiveack?

yes

no

compute randombackoff integer k

no

delay packettransmission

k times

Wait for start of next slot

16

Throughput of Slotted ALOHA

The vulnerability period of a packet is a slot time:

Any arrivals in prior slotcollide with packet i

time

arrivals

i

If T is the duration of a time slot and G is the normalized packet arrival rate, then PT GeSGeS GT === −− for ;λ

We can obtain the same result by computing the likelihood and average length of utilization, idle and busy periods.

17

Slotted ALOHA

...

P τ

collision

time

user i

time

user j

time

sum

time slot

I B I B B I B

NEW

NEW

NEW

NEW

NEW

NEW RET

RET

IMPORTANT:The starting point of a busy period is a “renewal point”! System is busy

18

Renewal Theory Recall the Poisson random process:

N(t) = number of arrivals in (0, t] Inter-arrival times are exponentially distributed N(t) is a counting process with exponential inter-arrival times.

Definition of Renewal Process: A counting process N(t) for which inter-arrival times X1, X2, …, Xn are an independent identically

distributed (iid) random sequence.

19

Poisson Random Variable

A sequence of n independent Bernoulli trials;

with X being the number of arrivals in (0, t]

arrivals. 1or 0 having ofy probabilit the tocompared

negligible is in arrival one than more ofy probabilit The

:0 and Make

t

tn

Δ→Δ∞→

By assumption, whether or not an event occurs in a subinterval is independent of the outcomes in other subintervals.We have:

tΔ

…. time

t0

arrival

1 2 3 4 n

1 2 3 k

20

Renewal Theory Example

At each time t = 1, 2, …, a Bernoulli process N(t) has an arrival with probability p, and this is independent of the occurrence of arrivals at any other times.

Is N(t) a renewal process?

…. time

0

arrival

1 2 3 4 n

1 2 3 k

21

Renewal Theory Example

Answer: For any inter-arrival period n, the inter-arrival time

Xn equals x if there were x-1 Bernoulli failures followed by a success.

…. time

0 (1-p)(1-p) (p)

1 2 3

Xn = 3 if we have 2 failures followed by a success!

22

Renewal Theory Example Therefore, each inter-arrival time Xn has the Geometric PMF:

Because each Bernoulli trial is independent, Xn is independent of the previous inter-arrival times X1, X2 ,…Xn-1.

This implies that a Bernoulli arrival process is a renewal process!⎩⎨⎧ =−

=−

otherwise 0

1,2,... x)1()(

1 ppxP

x

X n

23

Renewal Theory

After an arrival (in a renewal process), the subsequent inter-arrival times are distributed identically to the original inter-arrival times.

Effectively, the process restarts, or has a renewal, whenever an arrival occurs!

24

Renewal Theory

Suppose that N(t) has n arrivals by time t1, the additional time until the next arrival is denoted by Sn+1 - t1, and the subsequent inter-arrival times are Xn+2 , Xn+3 ,… and so on.

Renewal Point: For a renewal process N(t), time t1 with N(t1) = n is a renewal point if Sn+1 - t1, Xn+2, Xn+3,… is an iid random sequence statistically identical to X1, X2, X3,…

Every instant of time is a renewal point for a Poisson process!

25

Renewal Theory

Alternating renewal process: System is on and off (that is, or busy and idle).

time

system collision

...

success success

off on off on off on off ...

Y1 X1 Y2 X2 Y3 X3 Y4 ...

X1, X2, X3,…. are i.i.d. with mean E(X)Y1, Y2, Yx,…, are i.i.d. with mean E(Y)

P(t) = P{system is ON at time t in steady state} = E(x)/{E(x)+E(Y)}Average cycle length = E(X) + E(Y)

26

Evaluating Throughput We assume that the system is “stationary,” i.e., system behaves in cycles that are

statistically equivalent Average cycle consists of an idle period (I ) and a busy period (B ). The busy portion of a cycle has good and part parts.

The portion of time used to send user data is called the utilization period (U )

€

Throughput is defined as : S =U

I + B=

U

I +(PgBg + PbBb )

timesuccess failure

gB I bB

good busy period

idle period bad busyperiod

27

Evaluating Throughput The expression for S amounts to simply taking averages. What we need to do now is compute the probability that

I, B (good a bad parts), and U happen in an average cycle, and their average duration.

Ideally, these probabilities are based on independent events, and we can express S based on knowledge of the present state of the system.

then of component each of occurrence ofy probabilit

andduration average thedenote and Let

Si

PT ii

BBII

UU

PTPT

PT

BI

US

+=

+=

28

Throughput of Slotted ALOHAIdle, busy and utilization periods are multiples of time slots.

We need to count the time slots in each average period and we are done.Average length of idle period:

I = number of slots in idle period

time

arrivals transmissionsstart

no arrivals

idle periodstarts

29

Idle Period in Slotted ALOHA

TenoP

somePslotoneIPλ−−=−=

==

1 }slot in arrive pkts {1

}slot idle in the arrive pkts {} {

I has one slot:

time

there is a prior busy period

...at least one arrival!

time

I has two slots:

...at least one arrival!no arrivals

)1(

2}slot in pkts some{1}slot in pkts no{

slot} 2ndin pkts some andslot first in pkts no{} {

TT ee

PP

PslotstwoIP

λλ −− −=

=×====

30

Idle Period in Slotted ALOHA

time

I has k slots:

...

no arrivals

1 k-1 k

… some arrivals

2}slot in pkts some{}1slot in pkts no{...1}slot in pkts no{

slot}last in pkts some and slots 1first in pkts no{} {

PkPP

kPslotskIP

×−××=−==

€

P{I = k slots} = e−λT( )

k−1(1− e−λT )

This corresponds to the Geometric r.v., and we know its average value to be 1/p, with p being the probability of success.

Success now consists of ending the idle period! Therefore:

)1(

1Te

I λ−−=

31

Busy Period in Slotted ALOHA We follow the same approach:

Solve the problem with the Geometric random variable

time

B has k slots:

...

some arrivals

1 k-1 k

… no arrivals

2}slot in pkts no{}1slot in pkts some{...1}slot in pkts some{

slot}last in pkts no and slots 1first in pkts some{} {

PkPP

kPslotskBP

×−××=−==

€

P{B = k slots} = 1− e−λT( )

k−1(e−λT )

TT

ee

B λλ == −

1

prior slot consideredin idle period

32

Utilization Period Here we have to make use of conditional probability! A busy period has good and bad time slots

(transmission periods).

collision

time

sum

The probability that a slot (transmission period) in the current busy period is successful is the probability that only one packet arrives in the prior slot, given that there is a busy period

Arrivals are Poisson, so we make use of the definition of thatrandom variable as follows….

33

Utilization Period

T

T

S

e

Te

TsomeP

ToneP

TsomeP

TsomeandToneP

TsomeTPP

λ

λλ−

−

−==

=

=

1} in arrivals {} in arrival {

} in arrivals {} in arrivals in arrival {

} in arrivals | in arrival one{

The probability that a given slot within a busy period is successful is:

The portion of an average busy period used to send useful data equals the length of the average busy period in slots, times the probability that any given slot is successful.

TT

TT

S e

T

e

TeePBU λλ

λλ λλ

−−

−

−=⎟⎟

⎠

⎞⎜⎜⎝

⎛

−=×=

11

We can use the Binomial random variable to proof the above!

34

Throughput of Slotted ALOHA We now just substitute B,I, and U in S:

10-3

10-2

10-1

100

101

102

103

104

105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Analytical Results

Offered Load: G

S (Throughput)

Pure CSMA

Slotted CSMA

Pure Aloha

Slotted Aloha

T

TT

TTe

ee

e

T

BI

US λ

λλ

λ λλ −

−

− =+

−

⎟⎠

⎞⎜⎝

⎛−

=+

=

11

11

Maximum throughput is twice that of ALOHA.This occurs when G = 1

TGwith

GeS G

λ== −

35

Average Delay of MAC Protocols

We want to measure or compute the average time from the instant the first bit of a packet is first transmitted to the moment the last bit is received correctly at the destination.

Assume that arrivals (of new and retransmitted data or control packets) to the channel are Poisson.

Assume fully-connected networks.

36

Average Delay in ALOHA

Direct method: The average number of transmissions needed for a packet to be received correctly is GeSG 2/ =

Therefore, the number of retransmissions is 11/ 2 −=− GeSG

Assumptions:

A satellite channel with propagation delay NxP, where P is the packet length and NxP >> PA retransmission is sent after an average backoff time of BxP seconds.

A packet is transmitted (G/S-1) times in error (due to collisions) and each such transmission wastes P+NxP +BxP seconds.

The last transmission is successful and must take P+NxP seconds.Therefore, the average delay incurred is:

))(1( 2 PBPNPePNPD G ×+×+−+×+=

37

Average Delay in ALOHAIndirect Method:

Based on the fact that the success of a transmission is independent of others, and knowing how many times we have retransmitted does not change the likelihood of success in the next transmission!We use a diagram showing possible states, probabilities of transition, and delay incurred in that transition.

START END

BACKOFF

PNP , ×+SP

PBPNP

,1

×+×+− SP

PBPNP

,1

×+×+− SP

PNP , ×+SP

From the diagram. we obtain a number of simultaneous equations that we solve to obtain delay from START to END.

38

Average Delay in ALOHAFrom the diagram we have:

))(1()( RPBPNPPPNPPD SS +×+×+−+×+=

))(1()( RPBPNPPPNPPR SS +×+×+−+×+=Solving these two equations:

)()1(

PBPNPP

PPNPR

S

S ×+×+−

+×+=

))(1( 1 PBPNPPPNPD S ×+×+−+×+= −

Substituting GS eP 2−= we obtain the same result expected from the

Geometric r.v.

The same method can be applied on the other MAC protocols!

39

Average Delay of ALOHA The delay increases exponentially with heavy load,

which is not acceptable for real-time applications.

40

CSMA: Carrier Sense Multiple Access

The capacity of ALOHA or slotted ALOHA is limited by the large vulnerability period of a packet.

By listening before transmitting, stations try to reduce the vulnerability period to one propagation delay.

This is the basis of CSMA (Kleinrock and Tobagi, UCLA, 1975)

Many of the assumptions made for ALOHA are made now for CSMA.

41

CSMA ProtocolAssume non-persistent carrier sensing.Requires a maximum propagation delay much smaller than packet lengths!

transmit

no

wait for a round-trip time

positiveack?yes

compute randombackoff integer kno

delay packettransmission

k times

Packetready

ChannelBusy?

yes

42

CSMA Throughput

A virtual secondary channel used to send ACKs reliable and in 0 time!Same assumptions made for pure ALOHA analysis.All stations are at one propagation delay from each other and that equals: τ

Arrivals are Poisson with average rate λ

Peer-to-peer communicationNo base station or transponderExplicit feedback to sender!

43

CSMA Protocol

The big difference compared to ALOHA is that busy periods are bounded!

time

user i

time

user j

time

sum...

I B I B I B I

RET.

RET.

RET

RET

NEW

NEW

P

NEWNEW

collision

τ2+≤ PB

τ

44

CSMA ThroughputP

timeRET.RET.NEWNEW

collisionτ

I B I B I B I

τ2+≤ Pfailed period

τ+Psuccessful period

Length of average idle period (exponential interarrivals)? λ/1=IThe probability that a packet is successful is?

λτ−=ePS

The average length of a utilization period is?

(no packets can arrive within tau sec. after the start of the packet!)λτ−×= ePU

We can approximate: τ2+= PB

45

CSMA Throughput

Substituting we have:

λτ

λτ

12 ++

≈−

P

PeS Pretty accurate for << P

More accurate estimation of S requires finding the average length of B.

timeFIRST

LAST

P

START END

Y

Y Y is a random variable!τ++= YPB

€

τ

€

τ

€

τ

46

CSMA Throughput

)()sec in arrivals 0()()( yY eyPyFyYP −−=−==≤ τλτ

timeFIRST

LAST

START

Y =y

no arrivals, no more arrivals occur after LASTy−τ

)1(1

0)1())(1()(0

)(

0

λττ

τλ

λτ −−−

∞

−−=+−=−= ∫∫ edyedyyFYE yY

)1(1

2 λτ

λτ −−−+= ePB

Note that the average length of B is determined by the time between the start of the first and the last packet in the busy period.

τ

47

CSMA ThroughputSubstituting we get:

10-3

10-2

10-1

100

101

102

103

104

105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Analytical Results

Offered Load: G

S (Throughput)

Pure CSMA

Slotted CSMA

Pure Aloha

Slotted Aloha

aG

aG

eGa

GeS −

−

++=

)21(

Pa

PG

/

,with

τλ

==

€

S =Ge−aG

(1+ 2a)G +1

Approximate:

48

Slotted CSMA Non-persistent strategy. A slot lasts one maximum propagation delay.

no

wait for a round-trip timequantized in slots

positiveack?yes

compute randombackoff integer k

no

delay packettransmission

k times

Packetready

ChannelBusy?

yes

transmit

wait for start of next slot

49

Computing the Throughput of Slotted CSMA

NEW

P

time

collision

τI B I B I

success

RET.

time

I has k slots:

...

no arrivals

1 k-1 k

… some arrivals

Just as in slotted ALOHA, with slot duration equal to τ

τ

sec )1(

; slots )1(

1λτλτ

τ−− −

=−

=e

Ie

I

50

Throughput of Slotted CSMA We follow the same approach as in slotted ALOHA B has k transmission periods, each of P + τ sec What happens in a transmission period depends only on the

last time slot of the prior transmission period!

€

P{B = k trans. periods} = P{B = k(P + τ ) sec} = 1− e−λ τ( )

k−1(e−λ τ )

time...

some arrivals in last slot of each transmission period

1k-1

k… no arrivals in last slot of last transmission period

τ+PSlotted CSMA:

51

Throughput of Slotted CSMA

sec )( λτλτ ττ

ePeP

B +=+

= −

so ,1

}in arrivals {

}in arrival { λτ

λτλττ

τ−

−

−==

ee

somePoneP

PS sec SPP

BPU ×

+×=

τ

A transmission period is successful with probabilityand the period has useful data for P seconds

SP

sec 1

λτ

λτ−−

=eP

U

periodsion transmiss1λτ−=

eB

We use the Geometric r.v. to obtain the average number of transmission periods in B :

52

Throughput of Slotted CSMA We now substitute U, B and I in S:

PaPGea

aGeS

aG

aG

/ , with ;)1(

τλ ==−+

= −

−

53

CSMA Throughput Because prop. delay is much smaller than pkt length,

slotted and pure CSMA have very similar performance. When MAC protocol requires small prop delays,

we can use slotted version to predict performance of unslotted version.

10-3

10-2

10-1

100

101

102

103

104

105

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1Analytical Results

Offered Load: G

S (Throughput)

Pure CSMA

Slotted CSMA

Pure Aloha

Slotted Aloha

Reminder: These results are only an upper bound on performance, because we did not take into account the effect of ACKs sent from receivers!

54

CSMA/CD:CSMA with Collision Detection

CSMA improves on the performance of ALOHA tremendously. The remaining limitation is that, once a packet is sent,

feedback occurs a roundtrip time after the entire packet is transmitted.

The solution to improve on the performance of CSMA is to listen to the channel while a packet is being sent.

This is called collision detection. R.M Metcalfe and D.R. Boggs, “Ethernet: Distributed Packet Switching for

Local Computer Networks,” Comm. ACM, Vol. 19, 1976 (Xerox PARC).

55

CSMA/CD ProtocolNon-persistent transmission strategyCollision detection serves as a NACK!

transmit

no

abort transmission

Collision detected?

no compute randombackoff integer k

yes

delay packettransmission

k times

Packetready

Channelbusy?

yes

send jammingsignal

Assumption are:• All stations hear one another • Propagation delay is much smaller than packets

Station listens to channel while transmitting; Collision is detected when signals sent and heard differ.Jamming signal sent to ensure all stations know of the collision.

56

Throughput of CSMA/CD

time

JJZC +≤++= ττ 32collision interval: successful packet:

τ+P

first packet starts (A)

τ ττ≤Z

J

first interfering packet starts (B)

A starts hearing B and jams

B starts hearing A and jams

idleperiod

τP

NEWAB

ττ ++ J

average idle period:

λ/1=I

)(e2

e ; with )(]2)[1( -

-

ττ

τττλτ

λτ

−−+++=

=≤+++++−=

JPJZB

PZPZPJZPB SSS

57

Throughput of CSMA/CD Notes:

The average length of a bad busy period is much smaller than in CSMA because J<<P.

This length is determined by the time between the first packet in the busy period and the first packet that interferes (in contrast, in CSMA, it is the last interfering packet that counts)

For

However, we can derive an exact value of the average value of Y.

P<<τ we can approximate: )(3 ττ λτ −−++≈ − JPeJB

The utilization period is only that portion of a packet transmission that has no overhead, that is:λτ−=PeU

58

Throughput of CSMA/CD

Substituting we get:)(3

1 ττλ

λτ

λτ

−−+++≈

−

−

JPeJ

PeS

59

Throughput of CSMA/CD

λτ

τλλ

λτ

τλλ

ττ

ττ

ττ

−

−−

−

−−−

−−

=−−

=

=

=>

=>≤≤>−=≤=

e

ee

e

ee

P

zPzP

P

zPzZP

zPzZP

zzZPzZPzF

zzz

Z

11

)1(

)},0(in arrivals some{

)},(in arrivals some {)},0(in arrivals no{

)},0(in arrivals some{

)},0(in arrivals some and ),0(in arrivals no{}{

)},0(in arrivals some|),0(in arrivals no{}{

0for };{1}{)(

)()(

Therefore:

λτ

λ

λτ

λτλ

−

−

−

−−

−−

=−−

−=>−=e

e

e

eezZPzF

zz

Z 1

1

11 }{1)(

⎥⎦

⎤⎢⎣

⎡

−−+=−= −

−

∫ λτ

λττ

λττ

e

edttFZE Z 1

12))(1()(

0