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Chapter 5

Multiple Access Networks

Main reference: Copyright © The McGraw-Hill Companies Inc.



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PART 1:PART 1:

5.1: RANDOM ACCESS5.1: RANDOM ACCESS

5.2 CONTROLLED ACCESS5.2 CONTROLLED ACCESS



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Introduction

Local Area Network [LAN]

- used to interconnect distributed communities of computer-

based DTEs located within a single building or localizedgroup of buildings

- also referred to as private data networks

Advantage:

i. Inexpensive and fast interconnections of minicomputers,

PCs, workstations, etc., in business, education andresearch environments

ii. Allow users to share expensive resources, such as

- repository data [e.g., file servers]

- repository knowledge [e.g., database servers]- service provider [e.g., printer, plotter]



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Characteristics:

i. Moderate-size geographic area [diameter of no more than a few

km]

ii. Located within a single building or localized group of buildings

[warehouse, campus, hostel]

iii. Total data rate of at least several Mbps

iv. Complete ownership by a single organization

The link layer is divided into 2 sub layers:i. Logical link control (LLC)

ii. Medium access control (MAC)

ApplicationPresentation

SessionTransportNetwork

LLCMAC

Physical

Linklayer



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Medium Access Control [MAC]

- regulates the access to the shared link

Logical Link Control [LLC]

- implements reliable packet transmission [flow and errorcontrol]

LLC sublayer was originally designed to be the same for all LAN for

interoperability

Standards- MAC: IEEE 802.3-12

- LLC: IEEE 802.2

Most common protocols

- ALOHA, Ethernet, Token Ring, WIFI

Two main classes of LANs:i. Wired LANs

ii. Wireless LANs



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Main performance parameters of interest

- Throughput: maximum bit transmission rate when the

LAN is heavily loaded- Efficiency: fraction of throughput to channel rate

- Delay: typical time taken to transmit a packet

between two nodes



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LAN protocols are concerned principally with lower layers of the OSI

Model.

Thus, higher-layer protocols are independent of network architectureand are applicable to LANs, MANs and WANs.

IEEE 802 LAN/MAN Standards Committee (www.ieee802.org) has

developed the IEEE 802 reference model for LAN specifications.

ApplicationPresentation

SessionTransportNetwork

Physical

Data link layer

( ) ( ) ( )

Logical link control

Medium accesscontrol

Upper layerprotocol

OSI reference model IEEE 802 referencemodel

Physical

MediumMedium

LLC serviceaccess point

[LSAP]

Scope ofIEEE 802standards

IEEE 802



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The lowest layer of the IEEE 802 reference model corresponds to

the physical layer of the OSI model and includes functions:

- Encoding/decoding signal- Preamble generation/removal [for synchronization]

- Bit transmission/reception

In addition, the physical layer of 802 model also includes

specification of transmission medium and topology, as the choice of

transmission medium and topology is critical in LAN.

The data link layer in OSI model is implemented as 2 sublayers in

IEEE 802 model:

- logical link control [LLC]- medium access control [MAC]

Logical link control layer provides interface to higher layers and

perform flow and error control.



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Medium access control layer provides functions such as:

- assemble data into frame with address and error-detectionfields on transmission

- dissemble frame and perform address recognition and error

detection on reception

- Govern access to the LAN transmission medium

The separation is required because:- The logic required to manage access to a shared-access

medium is not found in traditional layer 2 data link control

- For the same LLC, several MAC options may be provided



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IEEE 802.1 Higher Layer Interfaces

Bridging , Management , etc.

IEEE 802.2

1. Unacked CL

2. Acked CL

3. CO

NL

etc.

LLC

MAC

PHY

CSMA/CD Token BusTokenRing

TokenRing

DQDB Wireless100(Base)VG-AnyLAN

I E E E

8 0 2 . 3

I E E E

8 0 2 . 4

I E E E

8 0 2 . 5

F D D I

I E E E

8 0 2 . 6

I E E E 8 0 2 . 1 1

I E E E 8 0 2 . 1 2

U n s h i e l d e d t w i s t e d p a

i r

B r o a d b a n d c o a x i a l

O p t i c a l f i b e r

B r o a d b a n d c o a x i a l


U n s h i e l d e d t w i s t e d p a

i r

S

h i e l d e d t w i s t e d p a i r

O p t i c a l F i b e r

O p t i c a l F i b e r

R a d i o

I n f r a r e d

S

h i e l d e d t w i s t e d p a i r


The architecture of IEEE 802 LAN standards

CL – connectionlessCO – connection oriented



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Taxonomy of multiple-access protocols

. .



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5.1 RANDOM ACCESS5.1 RANDOM ACCESS

In In random access random access or or contention contention methods, no methods, no

station is superior to another station and none is station is superior to another station and none is assigned the control over another. assigned the control over another.

No station permits, or does not permit, another No station permits, or does not permit, another

station to send. At each instance, a station that station to send. At each instance, a station that

has data to send uses a procedure defined by the has data to send uses a procedure defined by the

protocol to make a decision on whether or not to protocol to make a decision on whether or not to

send. send.



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First Multiple Access Protocol using contention scheme

Precursor to CSMA/CD

Applicable to any shared transmission medium: radiotransmitters, coaxial cable, twisted pair or optical fiber

Two versions:

Pure ALOHA

Slotted ALOHA

Pure ALOHA is the original ALOHA

Each station starts transmitting whenever it has data to

send After transmitting a data frame, the sender expects the

receiver to send an acknowledgment.

ALOHA



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If the acknowledgement does not arrive after a timeout

period, the sender assumes that the data frame has been

destroyed and resends the frame. If all stations experience collision try to resend their data

frames after the timeout period, the data frames will

collide again.

To overcome this, each station waits a random amount of

time before resending its data frame.

This random waiting time is called the backoff time (T B)

and it helps avoid more collisions. To avoid congesting the channel with retransmitted

frames, a station must give up after a maximum number of

retransmission attempts

PURE ALOHA



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Frames in a pure ALOHA network

As there is only one channel to share, there is a possibility of

collision if more stations are sending at about the same time.

Due to its simplicity, the number of collisions rises rapidly withincreased load.



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Procedure for pure ALOHA protocol



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The stations on a wireless ALOHA network are a maximum

of 600 km apart. If we assume that signals propagate at 3 ×

108 m/s, we find

T p = (600 × 10 5 ) / (3 × 108 ) = 2 ms.

Now we can find the value of T B for different values of

K .

a. For K = 1, the range is {0, 1}. The station needs to|

generate a random number with a value of 0 or 1.

This means that T B is either 0 ms (0 × 2) or 2 ms (1 × 2), based on the outcome of the random variable.

Example 1



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b. For K = 2, the range is {0, 1, 2, 3}. This means that T B

can be 0, 2, 4, or 6 ms, based on the outcome of the

random variable.

c. For K = 3, the range is {0, 1, 2, 3, 4, 5, 6, 7}. This

means that T B can be 0, 2, 4, . . . , 14 ms, based on the

outcome of the random variable.

d. We need to mention that if K > 10, it is normally set to

10.

Example 1(continued)



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Vulnerable time for pure ALOHA protocol

Vulnerable time is the duration with a possibility of collision

Consider 3 stations (A, B & C) with each station sending a fixed

length data frame taking transmission time T fr

Pure ALOHA vulnerable time= 2 x T fr



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A pure ALOHA network transmits 200-bit frames on a

shared channel of 200 kbps. What is the requirement to

make this frame collision-free?

Example 2

Solution

Average frame transmission time T fr is 200 bits/200 kbps or 1 ms. The vulnerable time is 2 × 1 ms = 2 ms. This means no

station should send later than 1 ms before this station starts

transmission and no station should start sending during the

one 1-ms period that this station is sending.



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The throughput for pure ALOHA isS = G × e −2G .

The maximum throughputSmax = 0.184 when G= (1/2).

Note



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A pure ALOHA network transmits 200-bit frames on a

shared channel of 200 kbps. What is the throughput if the

system (all stations together) produces a. 1000 frames per second b. 500 frames per second

c. 250 frames per second.

Example 3

SolutionThe frame transmission time is 200/200 kbps or 1 ms.

a. If the system creates 1000 frames per second, this is 1

frame per millisecond. The load is 1. In this case

S = G× e−2 G or S = 0.135 (13.5 percent). This means

that the throughput is 1000 × 0.135 = 135 frames. Only

135 frames out of 1000 will probably survive.



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Example 3 (continued)

b. If the system creates 500 frames per second, this is

(1/2) frame per millisecond. The load is (1/2). In this

case S = G × e −2G or S = 0.184 (18.4 percent). This means that the throughput is 500 × 0.184 = 92 and that

only 92 frames out of 500 will probably survive. Note

that this is the maximum throughput case,

percentagewise.

c. If the system creates 250 frames per second, this is (1/4)

frame per millisecond. The load is (1/4). In this case

S = G × e − 2G or S = 0.152 (15.2 percent). This means

that the throughput is 250 × 0.152 = 38. Only 38

frames out of 250 will probably survive.



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Frames in a slotted ALOHA network

Pure ALOHA has a vulnerable time of 2T fr as there is no rule that

defines when a station can send (stations may send at arbitrary

time)

Slotted ALOHA was invented to improve the efficiency

In slotted ALOHA, time is divided into slots of T fr each

Each station can only transmit at the beginning of the time slot



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The throughput for slotted ALOHA isS = G × e−G .

The maximum throughputSmax = 0.368 when G = 1.

Note



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Vulnerable time for slotted ALOHA protocol

There is still possibility of collision as two or more stations may

send in the same time slot

Nonetheless, the vulnerable time is reduced to only T frc



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Efficiency ratio (ratio of throughput achieved to channel rate)

o Pure ALOHA = 0.184

o Slotted ALOHA=0.368

Both versions exhibit poor utilization as they fail to take advantage of short propagation delay (w.r.t frame transmission time) in LANs

For short propagation delay, when a station launches a frame, a;; other

station will know it almost immediately⇒CSMA



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A slotted ALOHA network transmits 200-bit frames on a

shared channel of 200 kbps. What is the throughput if the

system (all stations together) produces a. 1000 frames per second b. 500 frames per second

c. 250 frames per second.

Example 4

Solution

The frame transmission time is 200/200 kbps or 1 ms.

a. If the system creates 1000 frames per second, this is 1

frame per millisecond. The load is 1. In this caseS = G× e−G or S = 0.368 (36.8 percent). This means

that the throughput is 1000 × 0.0368 = 368 frames.

Only 386 frames out of 1000 will probably survive.



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Example 4 (continued)

b. If the system creates 500 frames per second, this is

(1/2) frame per millisecond. The load is (1/2). In this

case S = G × e−G or S = 0.303 (30.3 percent). This

means that the throughput is 500 × 0.0303 = 151.

Only 151 frames out of 500 will probably survive.

c. If the system creates 250 frames per second, this is (1/4)

frame per millisecond. The load is (1/4). In this case

S = G × e −G or S = 0.195 (19.5 percent). This means

that the throughput is 250 × 0.195 = 49. Only 49 frames out of 250 will probably survive.



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CSMA is a polite version of ALOHA

With CSMA, a station wishing to transmit first listens to the medium

(carrier sense) and obeys the following rules:

1) If the medium is idle, transmit; otherwise, go to step 2

2) If the medium is busy, continue to listen for idle medium; when

medium becomes idle, transmit whole frame immediately

CSMA reduces the possibility of collision, but cannot eliminate it

Even though each station listens to the medium before transmitting,collision still exists due to propagation delay

When a frame is sent, it takes a while(though very short) for every

station to sense it

A station may sense the medium and find it idle, only because the

frame has not yet reach the station

Carrier Sense Multiple Access (CSMA)



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Space/time model of the collision in CSMA

CSMA reduces the possibility of collision, but cannot eliminate it

Collision still exists because of propagation delay

At time t 1 , B senses that the

medium is idle and send a

frame.

At time t 2(t 2 > t 1), C sensesthat the medium is idle(as

the frame from B has not

arrived at C yet) and sends

a frame

Both frames collide and are

destroyed



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Vulnerable time in CSMA

The vulnerable time for CSMA is the propagation time, T p (the time

time for a signal to propagate from one end to another end of the

medium)

If the first bit of the frame reaches the end of the medium, every

station will already have heard the bit and will refrain from sending Consider the worst case where A at one end is sending a frame

This frame is susceptible for collision until its first bit arrives at the

other end.



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Behavior of three persistence methods

Persistence Strategy defines the procedure for a station that senses a

busy medium.

Two strategies:

1) Non-persistent

o A station with frame to send senses the mediumo If the medium is idle, it sends immediately

o If the medium is busy, it waits a random time before sensing the

medium again

o Reduce the chance of collision as stations are unlikely to wait the same

amount of timeo Lower efficiency as the medium may be idle while the stations are

waiting



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2) Persistento A station with frame to send senses the medium

o If the medium is idle, it sends immediately

o If the medium is busy,

a) 1-persistent- The station sends a frame immediately ( with a probability of 1) if

the medium is sensed idle

-Increase the chance of collision as 2 or more stations may send their

frames simultaneously after finding the medium idle.



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b) p-persistent

- The station send with probability p (i.e. may or may not send) if the

medium is sensed idle

- Reduce chance of collision and improve efficiency



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Flow diagram for three persistence methods



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With CSMA, the medium remains unusable for the duration of frame

transmission when two frames collide

For long frames, the amount of wasted capacity can be significant

[wasted time equal to frame transmission time]

This waste can be reduced if a station continues to listen to the

medium while transmitting [collision detection]

How to detect:

i. The station sends frame and senses the mediumii. Collision detected if

- Coax:station senses power exceeding transmitted signal

strength [maximum length must be restricted to limit

signal attenuation]

- UTP: there is signal on more than one port [sensed by

hub and collision presence signal is generated and sent

to all stations]

Carrier Sense Multiple Access with Collision Detection (CSMA/CD)



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In CSMA/CD, a station wishing to transmit obey these rules:

i. If the station senses that the channel is idle (for 96 bit times), it

start to transmit the data frame and go to step (iii), otherwise go

to step (ii).

ii. If the channel is busy, the station waits (1-persistent) until it

senses that the channel is idle for 96 bit times and start to

transmit the frame and go to step (iii).

iii. While transmitting, it monitors the channel for possiblecollision; if it transmits the entire frame without collision, the

transmission is considered as successful.

iv. If collision is detected, the station stops transmitting its frame,

transmit a 48-bit jam signal and go to step (v)v. After aborting, the station enters an exponential backoff phase

and waits a random time before returns to step (i)



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Binary Exponential Backoff

When transmitting a given frame, after experiencing the nth collision

in a row for this frame, the station chooses a value K at random from

{0, 1, 2, …2m-1}, where m = min(n, 10).

The station then waits K slot times before it attempts to transmit

1st collision waits 0 or 1 slot time2nd collision waits 0, 1, 2 or 3 slot times3rd collision waits 0, 1, 2 … 7 slot timesmth collision waits 0 … 2m-1 slot times

After 10th collision, m fixed at 10After 16th collision, frame is discarded and reportfailure to upper layer

Note: Slot time = worst-case round-trip propagation time



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In the presence of collisions, the mean value of the random delay is

doubled after each collision

As congestion increases, stations back off by larger and largeramount to reduce the probability of collision

After 16 unsuccessful attempts, the station gives up and report an

error to the upper layer



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CSMA/CD with exponential backoff



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Collision Detection & Frame Transmission time

In CSMA/CD, the transmitting station senses the voltage levels before

and during transmissions

A collision results in a change of voltage level, and it takes time to

propagate back to the station

A station will only sense collision while transmitting; if a very short

frame is transmitted, the station might stop transmitting/sensing before

the collision signal arrives.



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Consider the worst case scenario :

i. At t = 0, A transmits a frame onto the medium.

ii. At t = Tp − ε, frame from A almost arrives at D.

iii. At t = Tp, D just starts to transmit frame onto medium. Itimmediately detects collision and transmit a jamming signal

iv. At t = 2Tp, A detects the collision.

Note: Tp is the end-to-end propagation time.

A B C D

t = 0

t = tp

t = 2tp



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The amount of time required to detect a collision is no greater than

twice the end-to-end propagation delay

A frame must take at least 2T p to send to prevent the sender from

incorrectly concludes that the transmission was successfully

A starts to transmit frame

A has been completelytransmitted the frame

B starts to transmit frame priordetecting frame from A

Collision occurs

Collision propagates throughtthe network

A does not detect the collision

as it has finished transmission



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A B

1. A starts totransmit frame 3. B starts to transmit frame

just before frame from A arrives

5. Collision propagatesthrough the network

2. A continue tosense mediumwhile transmitting

4. Collisionoccurs

6. A detects collisionand stop transmitting

7. A transmit 48-bit jam signal



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A network using CSMA/CD has a bandwidth of 10 Mbps. If

the maximum propagation time (including the delays in the

devices and ignoring the time needed to send a jamming

signal, as we see later) is 25.6 µ s, what is the minimum size

of the frame?

Example 5

SolutionThe frame transmission time is T fr = 2 × T p = 51.2 µ s. This

means, in the worst case, a station needs to transmit for a

period of 51.2 µ s to detect the collision. The minimum size

of the frame is 10 Mbps × 51.2 µ s = 512 bits or 64 bytes.This is actually the minimum size of the frame for Standard

Ethernet.



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Major difference between ALOHA and CSMA/CD

- ALOHA: transmit as soon as there is a frame to be sent

- CSMA/CD:wait first for the channel to be idle before a frame istransmitted

Efficiency of CSMA/CD (empirical):

where

CSMA/CD will be more efficient than

- pure ALOHA for a < 0.89- slotted ALOHA for a < 0.34

a

U

51

1

+

=

timeontransmissi frame

ndelay propagatioend toend a

−−=



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5.2 CONTROLLED ACCESS5.2 CONTROLLED ACCESS

In In controlled access controlled access , the stations consult one , the stations consult one

another to find which station has the right to send. another to find which station has the right to send.

A station cannot send unless it has been authorized A station cannot send unless it has been authorized

by other stations. by other stations.



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Reservation access method

A station needs to make a reservation before sending

data.

Time is divided into intervals.

In each interval, a reservation frame frame precedesthe data frames sent in that interval.



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Select and poll functions in polling access method

All data exchanges must be made through the primary device

even when the ultimate destination is a secondary device.

The primary device controls the link; the secondary devicesfollow its instruction



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Logical ring and physical topology in token-passing access method

The stations in a network are organized in a logical ring.

Each processor has a predecessor and a successor.

A special packet called a token circulates through the ring

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