COMMUNICATION NETWORKS ECE 422 DATA COMMUNICATION & COMPUTER - - PowerPoint PPT Presentation

MULTIPLE ACCESS IN COMMUNICATION NETWORKS

ECE 422 – DATA COMMUNICATION & COMPUTER NETWORKS Wednesday, 12 February 2020

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Course Content: Introduction: Overview of Data Communications and Networking. Physical Layer: Analog and Digital, Analog Signals, Digital Signals, Analog versus Digital, Data Rate Limits, Transmission Impairment, More about signals. Digital Transmission: Line coding, Block coding, Sampling, Transmission mode. Analog Transmission: Modulation of Digital Data; Telephone modems, modulation of Analog signals. Multiplexing: FDM, TDM, CDM. Transmission Media: Guided Media, Unguided media (wireless). Data Link Layer: Error Detection and correction - Types of Errors, Detection, Error Correction; Data Link Control and Protocols-Flow and Error Control, Stop-and-wait ARQ. Go-Back-N ARQ, Selective Repeat ARQ, HDLC. Point-to-Point Access- Point–to-Point Protocol (PPP), PPP Stack, Multiple Access Random Access, Controlled Access, Channelization. Network Layer: Host to Host Delivery: Internetworking, addressing and Routing Network Layer Protocols: ARP, IPV4, ICMP, IPV6 and ICMPV6 Transport Layer: Process to Process Delivery: UDP; TCP congestion control and Quality of service. Application Layer: Client Server Model, Socket Interface, Domain Name System (DNS): Electronic Mail (SMTP) and file transfer (FTP) HTTP and WWW. Local area Network: Ethernet - Traditional Ethernet, Fast Ethernet, Gigabit Ethernet; Token bus, token ring; Wireless LANs - IEEE 802.11, Bluetooth virtual circuits: Frame Relay and ATM. Industrial Communication and Control Networks: Transmission methods, Network topology, Contemporary networks – Profibus, Controller Area Network (CAN), DeviceNet, CANopen, Actuator Sensor Interface (AS-1),Industrial Ethernet.

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WHERE ARE WE IN THE SYLLABUS?

DATA LINK

The Data Link Layer

• 1. Makes the physical layer appear error-

free to the upper layer.

• 2. Functions include:

a) Physical addressing b) Flow-control c) Error Control d) Access Control Suppose the Physical Media is Multipoint (Ethernet)

• r Broadcast (Terrestrial Radio or Satellite)?

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NEED FOR RULES FOR MEETINGS OR ASSEMBLIES & APPLICATION TO COMPUTER NETWORKS

For effective Communication, we require that no two people: 1. Should speak at the same time, 2. Interrupt each other, 3. Monopolize the discussion. We can therefore introduce control measure as follows: a) Have a chairman selecting who speaks and limiting speech time, b) Use a microphone automated switch-off to place a limit on speech time, The alternative, is to have a free for all meeting (Shouting match!)

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Computer Networks Apply the Same Rules when many Computers/Devices share the same communication Media

CATEGORIES OF MULTIPLE ACCESS PROTOCOLS

Multiple Access Protocols Random Access

• 1. Aloha
• 2. CSMA
• 3. CSMA/CD
• 4. CSMA/CA

Controlled Access

• 1. Reservation
• 2. Polling
• 3. Token Passing

Channelization

• 1. FDMA
• 2. TDMA
• 3. CDMA

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Covered in ECE 416

RANDOM ACCESS

• 1. Each station has access to the media without

being controlled by another station.

• 2. If more than one station transmit at the

same time, then there is an access conflict- collision and the frames are either discarded

• r modified.
• 3. A random access protocol must address the

following questions:

a) When can the station access the medium? b) What can the station do if the medium is busy? c) How can the station determine the success or failure of the transmission? d) What can the station do if there is an access conflict?

A B C E F

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HISTORY OF RANDOM ACCESS

1. ALOHA was the earliest random access method. 2. It was developed at the University of Hawaii in early 1970. 3. It is a system for coordinating and arbitrating access to a shared communication Networks channel. 4. It was designed for a radio (wireless) LAN, but it can be used on any shared medium. 5. There are two different types of ALOHA: a) Pure ALOHA b) Slotted ALOHA

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ILLUSTRATION OF PURE ALOHA

• 1. Assume each station sends two

frames.

• 2. In the diagram, there are a total of

eight frames send on the shared medium.

• 3. Some of these frames collide

because multiple frames are in contention for the shared channel.

• 4. The Receiver acknowledges each

correctly received frame.

• 5. If the Transmitting station does not

receive ACK before timeout, it retransmits the frame.

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PURE ALOHA FLOWCHART

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𝑶𝒄𝒈 = 𝟏 is is 𝑶𝒄𝒈 ≥ 𝑶𝑴𝒋𝒏? 𝑶𝒄𝒈 = 𝑶𝒄𝒈 + 𝟐 Transmitter Sends Receiver Listens

VULNERABLE TIME FOR PURE ALOHA

Vulnerable time = 2 × 𝑢

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If a station wants to transmit data, its transmission may damage the transmission of another packet which commenced transmission up to t seconds before. Once the station has started transmitting data, its transmission may be disrupted

• ver the t seconds if another

station starts transmitting.

EXAMPLE - PURE ALOHA VUNERABILITY

A pure ALOHA network transmits 500-bit frames on a shared channel

• f 1 Mbps. Under what circumstances will this frame be collision-free?

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SOLUTION

• Frame Transmission time, 𝑢 =

𝐺𝑠𝑏𝑛𝑓 𝑇𝑗𝑨𝑓 𝑈𝑠𝑏𝑜𝑡𝑛𝑗𝑡𝑡𝑗𝑝𝑜 𝑇𝑞𝑓𝑓𝑒 = 500 1,000,000 = 500𝜈𝑡𝑓𝑑

• The Vulnerable time = 2𝑢 = 2 × 500 = 1,000𝜈 sec = 1 𝑛𝑡𝑓𝑑
• Therefore there will be no collision if no station transmits in a period of 1 msec,

i.e 0.5 msec before or 0.5 msec after the station starts transmitting.

B C A t t 0.5 ms 0.5 ms

PURE ALOHA THROUGHPUT

• 1. Let us call G the average number of

frames generated by a pure ALOHA system during a frame transmission time i.e attempts per packet.

• 2. Then the average number of successful

transmissions, S is given by: 𝑇 = 𝐻𝑓−2𝐻

• 3. The maximum throughput occurs when

G = 0.5, i.e 𝑇𝑛𝑏𝑦 = 0.5𝑓−1= 0.184

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EXAMPLE

QUESTION: A pure ALOHA network transmits 200-bit frames on a shared channel

• f 200 kbps. What is the throughput if the system -all stations together-

produces the average of: a) 1000 frames per second b) 500 frames per second c) 250 frames per second

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The frame transmission time is

200 200,000 or 1 ms.

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

The load, G = 1 Or S = Ge-2G = 1xe-2 = 0.135 or 13.5% Throughput = 1000 x 0.135 = 135 frames per second

SOLUTION (a)

SOLUTION (b)

The frame transmission time is

200 200,000 or 1 ms.

(a) If the system creates 500 frames per second, this is

1 2 frame per

second.

The load, G =

1 2

Or S =

1 2e-1 = 0.184 or 18.4%

Throughput = 500 x 0.184 = 93 frames per second

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SOLUTION (c)

The frame transmission time is

200 200,000 or 1 ms.

(a) If the system creates 250 frames per second, this is

1 4 frame per

second.

The load, G =

1 4

Or S =

1 4e-0.5 = 0.152 or 15.2%

Throughput = 250 x 0.152 = 36 frames per second

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SLOTTED ALOHA

• 1. The efficiency of the pure ALOHA system can be improved by

“slotting” time into segments whose duration is exactly equal to the transmission time of a single packet (assuming constant-length packets).

• 2. In slotted ALOHA, each user starts transmitting only at the

beginning of a slot.

• 3. As a result, when two packets conflict, they will only overlap

completely rather than partially, thus increasing channel efficiency.

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B C A t t Slot-1 Slot-2 Slot-3

COLLISSIONS IN SLOTTED ALOHA

• xxx

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VULNERABILITY OF SLOTTED ALOHA

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Vulnerable time = 𝑈

THROUGHPUT IN SLOTTED ALOHA

• 1. Let us call G the average number of frames

generated by a pure ALOHA system during

• ne frame transmission time.
• 2. Then the average number of successful

transmissions, S is given by: 𝑇 = 𝐻𝑓−𝐻

• 3. The maximum throughput occurs when G =

1, i.e 𝑇𝑛𝑏𝑦 = 1𝑓−2= 0.368

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PURE ALOHA Vs SLOTTED ALOHA THROUGHPUT

EXAMPLE

A slotted ALOHA network transmits 200-bit frames using a shared channel with a 200-kbps bandwidth. Find the throughput if the system (all stations together) produces a) 1000 frames per second b) 500 frames per second c) 250 frames per second

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The frame transmission time is

200 200,000 or 1 ms

(a) System produces 1000 frames per second Let us call G the average number of frames generated In this case G = 0.001 x 1,000 = 1 S =G x e-G or S =0.368 (or 36.8 percent) (was 13.5% for pure Aloha!) Throughput is therefore 1000 x 0.0368 =368 frames. Therefore only 368 out of 1000 frames will probably survive.

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ANSWER (a)

The frame transmission time is

200 200,000 or 1 ms

(b) System produces 500 frames per second In this case G = 500 x 0.001 =

1 2

S =G x e-G =

1 2 𝑓−1

2 =0.303 ( or 30.3 percent)

(was 18.4 % for pure Aloha!) Throughput is 500 x 0.0303 =151. Only 151 frames out of 500 will probably survive.

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ANSWER (b)

• 1. Carrier sense multiple access

(CSMA) requires that each station first listen to the medium (or check the state of the medium) before sending.

• 2. It is based on the principle "sense

before transmit" or "listen before talk.“

• 3. Compared with previous cases of

ALOHA, it reduces the possibility of collision, but it does not eliminate it.

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CARRIER SENSE MULTIPLE ACCESS

EFFECT OF DELAY IN CSMA/CD

• 1. The possibility of collision still exists

CSMA/CD because of propagation delay.

• 2. When a station sends a frame, it still

takes time (although very short) for the first bit to reach every station and for every station to sense it.

• 3. Therefore a station may sense the

medium and find it idle, only because the first bit sent by another station has not yet been received.

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∆t

VULNERABLE TIME IN CSMA/CD

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PERSISTENCE METHODS IN CSMA

1. Persistence methods were developed to answer the question .. ‘What should a station do if it senses the channel to be busy?’

• 2. There are three persistence methods, i.e

a) Non-Persistence b) 1-persistence c) p-Persistence

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}

NON-PERSISTENCE CSMA

• 1. All stations sense the channel

continuously.

• 2. If a station finds the line idle, it

sends its frame immediately.

• 3. Of the 3 methods, It has the

highest chance of collision because two or more stations may find the line idle and send their frames simulteneously.

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1. The p-persistent method is used if the channel has time slots with a slot duration equal to or greater than the maximum propagation time. 2. After the station finds the line idle it follows these steps:

a) With probability p, the station sends its frame. b) With probability q = 1 - p, the station waits for the beginning of the next time slot and checks the line again. I. If the line is idle, it goes to step (a). II. If the line is busy, it acts as though a collision has occurred and uses the backoff procedure.

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p-PERSISTENCE CSMA

ADVANTAGES/DISADVANTAGES OF NON-PERSISTENCE

Advantage of non-persistent

• It reduces the chance of collision because the stations wait a random

amount of time.

• It is unlikely that two or more stations will wait for same amount of

random time and will retransmit at the same time. Disadvantage of non-persistent

• It reduces the efficiency of network because the channel remains idle

when there may be stations with frames to send as stations wait a random amount of time after the collision

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CARRIER SENSE MULTIPLE ACCESS WITH COLLISION DETECTION (CSMA/CD) - 1

• 1. Carrier sense multiple access with collision detection

(CSMA/CD) adds the algorithm to CSMA to handle the collision.

• 2. A station monitors the medium after it sends a

frame to see if the transmission was successful.

a) If successful, the station is finished. b) If there is a collision, the frame is sent again.

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• CSMA/CD uses the following procedure:
• 1. Host A checks for the presence of a digital

signal on the wire.

• 2. If no other hosts are transmitting packets,

the it begins sending the frame.

• 3. It also monitors the medium to make sure

no other hosts begin transmitting.

• 4. if another host begins transmitting at the

same time a a collision occurs.

• 5. The transmitting host sends a jam signal

that causes all hosts on the network segment to stop sending data.

• 6. After a random period of time, hosts

retransmit their packets.

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CARRIER SENSE MULTIPLE ACCESS WITH COLLISION DETECTION (CSMA/CD) - 2

• 1. Before sending the last bit of the frame, the sending

station must detect a collision, if any, and abort the transmission.

• 2. This requirement imposes restrictions on the frame size

to conform to the maximum round-trip propagation delay.

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CARRIER SENSE MULTIPLE ACCESS WITH COLLISION DETECTION (CSMA/CD) - 3

EXAMPLE

Assume a network using CSMA/CD operates at 10 Mbps. If the maximum propagation time (including the delays in the devices and ignoring the time needed to send a jamming signal) , is 25.6 µsec, what is the minimum size of the frame?

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SOLUTION The frame transmission time is Tfr = 2 x Tp =2X25.6 =51.2 µsec. This means, in the worst case, a station needs to transmit for a period of 51.2 µsec to detect the collision. The minimum size of the frame is 10x106 x 51.2x10-6 =512 bits or 64 bytes. 512 bits (or 512/8 = 64 bytes) is actually the minimum size of the frame for Standard Ethernet, i.e 10baseT.

USE OF ENERGY LEVEL

• 1. The level of energy in a channel can

have three values:

a) Zero Level: the channel is idle b) Normal Level: A station has successfully captured the channel and is sending its frame. c) Abnormal Level: There is a collision and the level of the energy up to twice the normal level.

• 2. A station in a CSMS/CD system is

therefore designed to monitor the energy level to determine if the channel is idle, busy, or in collision mode.

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APPLICATIONS OF CSMA/CD

• CSMA/CD is commonly used in networks with repeaters and hubs

because these devices run in the half-duplex mode and all of their ports are in the same collision domain.

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CATEGORIES OF MULTIPLE ACCESS PROTOCOLS

Multiple Access Protocols Random Access

• 1. Aloha
• 2. Slotted Aloha
• 3. CSMA
• 4. CSMA/CD
• 5. CSMA/CA

Controlled Access

• 1. Reservation
• 2. Polling
• 3. Token Passing

Channelization

• 1. FDMA
• 2. TDMA
• 3. CDMA

CONTROLLED ACCESS

• 1. In controlled access, the stations consult
• ne another to find which station has the

right to send.

• 2. A station cannot send unless it has been

authorized by other stations.

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RESERVATION METHOD

• 1. In the reservation method, a station needs to make a reservation

before sending data.

• 2. Time is divided into intervals (time slots).
• 3. In each interval, a reservation frame precedes the data frames sent

in that interval.

• 4. A systems with N stations will have exactly N reservation minislots

in the reservation frame. Each minislot belongs to a station.

• 5. When a station needs to send a data frame, it makes a reservation

in its own minislot.

• 6. The stations that have made reservation can send their data frames.

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RESERVATIONS METHOD

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• 1. Every frame consists of N

mini-slots and N*k data- slots.

• 2. Every station has got its
• wn mini-slot and can

reserve up to k data slots.

• 3. Any station can send data

in unused data-slots.

POLLING

• 1. In polling systems, one device is

designated as a primary station and the other devices are secondary stations.

• 2. All data exchanges are made through

the primary device even when the ultimate destination is a secondary device.

• 3. The primary device :

a) Controls the link and determines which device is allowed to use the channel at a given time. b) Initiates all sessions.

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POLL AND SELECT FUNCTIONS

Poll Function

• The poll function is used by the primary device to solicit transmissions

from the secondary devices. Select Function

• The select function is used whenever the primary device has

something to send.

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TOKEN PASSING

• In the token-passing method, the stations in a network are organized

in a logical ring.

• For each station, there is a predecessor and a successor.

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LOGICAL RING

• In logical token-passing network, stations are not physically connected in a

ring.

• Each node knows its predecessor and successor.

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DUAL RING TOPOLOGY

• 1. The dual ring topology uses a second (auxiliary) ring which operates

in the reverse direction compared with the main ring.

• 2. The second ring is for emergencies only (such as a spare tire for a

car).

• If one of the links in the main ring fails, the system automatically

combines the two rings to form a temporary ring.

• After the failed link is restored, the auxiliary ring becomes idle

again.

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BUS RING / TOKEN BUS

• 1. In the bus ring topology, also called a token

bus, the stations are connected to a single cable called a bus.

• 2. They, however, make a logical ring, because

each station knows the address of its successor (and also predecessor for token management purposes).

• 3. When a station has finished sending its data,

it releases the token and inserts the address

• f its successor in the token.

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A B C E F

STAR RING TOPOLOGY

• In a star ring topology, the physical topology is a star. There is a hub,

however, that acts as the connector.

• The wiring inside the hub makes the ring;

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CHANNELIZATION

• 1. Channelization is a multiple-access method in which the available

bandwidth of a link is shared in time, frequency, or through code, between different stations.

• 2. Each station can use the time slot or frequency channel or code division

to send its data in the common transmission medium.

• 3. The available methods are:

a) TDMA b) FDMA c) CDMA

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Channels Channels Link

• Frequency division multiplexing

(FDM) is a technique of multiplexing which combines more than one signal of different frequencies over a shared medium.

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FREQUENCY-DIVISION MULTIPLEXING (FDM)

TIME-DIVISION MULTIPLEXING (TDM)

• Time-division multiplexing (TDM) is a

method of transmitting and receiving independent signals over a common signal path by means of synchronized switches at each end of the transmission line so that each signal appears on the line only a fraction of time in an alternating pattern.

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• Code Division Multiplexing

(CDM) allow multiple signals from different users to share a common communication channel using

• codes. Individual conversations

are encoded in a digital sequence. Data from all users is available on the shared channel, but only those users associated with a particular code can access the data.

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CODE-DIVISION MULTIPLEXING (CDM)

Unique codes Data streams