Multiple Access (1) - York University

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Multiple Access (1)Multiple Access (1)

CSE 3213, Fall 2010Instructor: N. Vlajic

Required reading:Garcia 6.1, 6.2.1, 6.2.2

2Multiple Access Communications

Broadcast NetworksBroadcast Networks – aka multiple access networks – multiple sending & receiving stations share the same transmission medium

• advantages:1) low cost infrastructure2) all stations attached to the medium hear

transmission from any other station ⇒routing not necessary

• disadvantages:access of multiple sending and receiving nodesto the shared medium must be coordinated1) stations should not be transmitting simultaneously

or interrupting each other2) stations should not be able to ‘monopolize’ the

transmission/shared medium

• examples: LAN, cellular and satellite networks

…

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34

5M

Shared multipleaccess medium

…

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34

5M

Shared multipleaccess medium

3Multiple Access Communications (cont.)

Approaches to Approaches to Medium SharingMedium Sharing

(1) StaticStatic Channelization – static & collision free sharingpartition medium into separate channels, which are thendedicated to particular users

advantage: no collisions, perfect fairness – each node getsa dedicated transmission rate R/N during each time interval(suitable for ‘streaming data’, e.g. voice streams)

disadvantage: each user gets only a fraction of the full channel capacity, even when no other station is transmitting

Medium Sharing Techniques

Static Static ChannelizationChannelization

Dynamic Medium Access Control

SchedulingScheduling Random AccessRandom Access


Static Channelization


Scheduling Random Access


Example [ systems employing static channelization ]

uplink f1downlink f2

uplink f3downlink f4

uplink fin downlink fout

satellite network cellular network

two frequency bands: one for uplink and one for downlinkeach station is allocated a channel in the uplink and in the downlink frequency band different approaches can be employed to create uplink/downlink channels (FDMA, TDMA, CDMA)although each station can theoretically transmit to and listen to any channel, stationsremain within their pre-allocated channels to avoid interference

satellite = repeater


(2) DynamicDynamic Medium Access – MAC Schemesthe medium is shared on a ‘per frame’ basisadvantage: transmitting node transmits at the fullrate of the channel

(suitable for ‘bursty data’, e.g. short messages)disadvantage: simultaneous attempts by two or more stations to access the channel result in ‘collision’collision can be minimized through scheduling or random access control

Ring Network

Bus Network










FDMATDMA

ALOHAALOHACSMA

ReservationPollingToken Passing

light burstyload

heavier burstyload

continuous load

7Random Access Techniques: ALOHA

ALOHAALOHA – the earliest random-access method (1970s) – still used in wireless cellular systems for its simplicity• a station transmits whenever it has data to transmit, producing

smallest possible delay – receiver ACKs data

• if more than one frames are transmitted at the same time, they interfere with each other (collide) and are lost

• if ACK not received within timeout (2*propagation delay), the stationpicks random backoff time (to reduce likelihood of subseq. collisions)• station retransmits frame after backoff time

ACK

data

8Random Access Techniques: ALOHA (cont.)

1.1 1.2

TransmissionTime(F)

Station 1

2.1Station 2

3.1 3.1'Station 3

Broadcastchannel

2.2

1.2'

CompleteCollision

PartialCollision

Example [ Aloha throughput ]

R

R

http://www.invocom.et.put.poznan.pl/~invocom/C/P1-4/p1-4_en/p1-4_3_7.htm#pureAapplet

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Vulnerable PeriodVulnerable Period

Random Access Techniques: ALOHA (cont.)

• assume frames of constant length (L) & transmissiontime (X=L/R)

• consider a frame with starting transmission time to –the frame will be successfully transmitted if no otherframe collides with it

any transmission that begins in interval [t0, t0+X], or in the prior X seconds leads to collision

vulnerable period = [ tvulnerable period = [ t00 –– X, tX, t00+ X ]+ X ]

What is the probability of no other transmission, i.e. no collision, in the vulnerable period?!

tt0t0-X t0+X t0+X+2tprop

t0+X+2tprop+B

Vulnerableperiod

Time-out

Backoff period BFirst transmission Retransmission

if necessary

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• definitions and assumptions:S (Sout) – throughput: average # of successful frame transmiss.per X sec (if network operates under ‘stable conditions’ Sout = Sin , where Sin - arrival rate of new frames to the system)

G – load – average # of overall transmission attempts! per X sec

Psucc – probability of a successful frame transmission

How to find Psucc? – suppose a frame is ready for transmissionat time t0 – frame will be transmitted successfully if no other frame attempts transmission X sec before and after t0

random backoff spreads retransmissions so that frame transmission (arrivals) are equally likely at any instant in time –Poisson process!!!


ThroughputThroughput

GPS succ ⋅=

t0 t0 + Xt0 - X

2X

ready for transmission

No, with Psucc

Yes, with (1-Psucc)

new packets

old packets

stationsSin

G channel

Collision?

Sout

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• accordingly, probability of successful transmission (no collision) is:

and throughput


• if general, if frame arrivals are equally likely at any instant in time, and arrivals occur at an average rate of λ [arrivals per sec]

Tk

ek!T)( seconds] T in arrivals P[k λλ −=

• in our case, λ=G/X [arrivals per second] and T=2X, hence

2Gk

ek!

(2G) seconds] 2X in onstransmissi P[k −=

2Gsucc e seconds] 2X in onstransmissi P[0P −==

Poisson process

t0 t0 + Xt0 - X

2Gsucc e GPGS −⋅=⋅=


S vs. GS vs. Gin Purein PureALOHAALOHA

• NOTE: our analysis assumed that many nodes share a common channel& have comparable transmiss. rates (if only 1 node uses the medium, S=1)

• initially, as G increases S increases until it reaches Smax – after that point the network enters ‘unstable operating conditions’ in which collisions become more likely and the number of backlogged stations increases (consequently, Sin > Sout )

• max throughput of ALOHA (Smax = 0.184) occurs at G=0.5, which correspondsto a total arrival rate of ‘one frame per vulnerable period’

• Smax = 0.184 ⇒ max ALOHA throughput = 18% of channel capacity

18% of channel utilization, with Aloha, is not encouraging. But, with everyone transmitting at will we could hardly expected a 100% success rate.

00.020.040.060.080.1

0.120.140.160.180.2

00.0

0781

250.0

1562

50.0

3125

0.062

50.1

25 0.25 0.5 1 2 4

G - attempts per frame time

S - throughputper frame time

stable operating area(Sin~Sout)

unstable operating area(Sin≠Sou)


Example [ Aloha ]

a) What is the vulnerable period (in milliseconds) of a pure ALOHA broadcast system with R=50-kbps wireless channel, assuming 1000-byteframes.

b) What is the maximum throughput S of such a channel (system), in kbps?

a) (frame transmission time =) X = 1000 bytes / 50 kbps = 8000 bits / 50 kbps⇒ X = 160 milliseconds

vulnerable period = 2*X = 320 milliseconds

b) From theory, max throughput = 0.18 * R = 0.18 * 50 kbps = 9.179 kbps

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Slotted ALOHASlotted ALOHA – “improved ALOHA”, with reduced probabilityof collision• assumptions:

time is divided into slots of size X=L/R (one frame time)nodes start to transmit only at the beginning of a slotnodes are synchronized so that each node knows whenthe slots begin

• operation:1) when node has a fresh frame to send, it waits until next

frame slot and transmits3) if there is a collision, node retransmits the frame after a

backoff-time (backoff-time = multiples of time-frames)

Random Access Techniques: Slotted ALOHA

http://www.invocom.et.put.poznan.pl/~invocom/C/P1-4/p1-4_en/p1-4_3_8.htm

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A

B

C

D

E

tic tic tic tic tic tic tic tic tic tic tic tic tic

Slotted ALOHA

Random Access Techniques: Slotted ALOHA (cont.)

A

B

C

D

E

Pure ALOHA

Example [ Aloha vs. Slotted Aloha ]

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Vulnerable PeriodVulnerable Periodof Slotted ALOHAof Slotted ALOHA

Random Access Techniques: Slotted ALOHA (cont.)

• consider one arbitrary packet P that becomes readyfor transmission at some time t during the time slot[k, k+1]

• packet P will be transmitted successfully if no otherpacket becomes available for transmission duringthe same time slot

vulnerable period = [ tvulnerable period = [ t00 –– X, tX, t00 ]]

Gsucc e seconds] X in arrivals P[0P −==

Gsucc e GPGS −⋅=⋅=

t(k+1)XkX t0 +X+2tprop+ B

Vulnerable periodperiod in which packet P becomes

ready for transmission

Time-out

Backoff period B

t0 +X+2tprop

packet P

17Random Access Techniques: Slotted ALOHA (cont.)

S vs. G inS vs. G inSlotted ALOHASlotted ALOHA

• max throughput of Slotted ALOHA (Smax = 0.36) occurs at G=1, which corresponds to a total arrival rate of ‘one frame per vulnerable period’

• Smax = 0.36 ⇒ max Slotted ALOHA throughput = 36% of actual channel capacity

0 0.5 1 1.5 2 2.5 30

0.1

0.2

0.3

0.4

0.5

R

Thro

ughp

ut (A

LOH

A)

Slotted ALOHA: Re-R

Pure ALOHA: Re-2R

Ideal (no collisions): R

G

S

ideal no collision

Slotted ALOHA: S=G*e-G

Slotted ALOHA: S=G*e-2G

Slotted ALOHASlotted ALOHAvs. Pure ALOHAvs. Pure ALOHA

• slotted ALOHA reduces vulnerability to collision, but also addsa waiting period for transmission

• if contention is low, it will prevent very few collisions, & delaymany of the (few) packets that are sent


Example [ slotted Aloha ]

Measurements of slotted ALOHA channel with an infinite number of users show that 10% of the slots are idle.a) What is the channel load, G?b) What is the throughput, S?c) Is the channel underloaded or overloaded?

a) 10% of slots idle ⇒frame will be successfully transmitted if sent in those 10% of slots ⇒Psucc = 0.1

According to theory, Psucc = e-G ⇒ G = -ln(Psucc) = - ln(0.1) = 2.3

b) According to theory, S = Psucc*G = G*e-G

as G=2.3 and e-G=0.1 ⇒ S = 0.23

c) Whenever G>1, the channel is overloaded, so it is overloaded in this case.


http://www.tti.unipa.it/mat_bianchi/rm/04.pdf

Example [ slotted Aloha in cellular (GSM) systems ]

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average arrival rate = λ [passenger / sec]

average # of arrived passengers in T [sec]:

Arrival: passengers arrive randomly and independently – a Poisson process.

probability of having exactly kpassengers in line after T [sec]:

λT [sec] T in passengers of #average =

Tk


λ [passenger / sec]

Departure: passengers arrive in batches – NOT a Poisson process.What is the probability that exactly 1 passenger arrivesto the station, off the buss, in T sec?

Passenger arrivals are equally likely at any instant in time.

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http://commons.wikimedia.org/wiki/File:Poisson_distribution_PMF.png

k

T=1

Tk


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Multiple Access (1) - York University