Advanced Computer Networks Rab Nawaz Jadoon · Department of Computer Science 28 DATA Link Layer For block coding we use modulo-2 arithmetic's. In this arithmetic, the modulus N is

Department of Computer Science

DCS

COMSATS Institute of Information Technology

Advanced Computer Networks

Rab Nawaz JadoonAssistant Professor

COMSATS University, Lahore

Pakistan

Advanced Computer Networks

Department of Computer Science 22

Course Plan

Lectures

Seminars

Reading of various Research Papers

Presentations by the students on selected topics

Quizzes and Reports

Recommended Books

Data Communication & Networks by Behrouz A. Forouzan

Computer Networks by Andrew Tannenbaum


Grading

Percentage Grade GP

90% and above A 4

80% to 89% B 3.5

70% to 79% C 3

60% to 69% D 2.5

Below 60% F 0


Marks Distribution

Description Weight age

Quiz/Assignment 25%

Midterm 25%

Final 50%


Layer Task

The Letter is written, put in

envelop and dropped in

mailbox.

The Letter is carried from

mailbox to a post office.

The Letter is delivered to a

carrier by the post office.

The Letter is picked up,

removed from the envelop

and read.

The Letter is carried from

the post office to the

mailbox.

The Letter is delivered from

the carrier to the post office


OSI layers


Internet Model

H5 D5 (Application Layer)

H4 D4 ( Transport Layer)

H3 D3 ( Network Layer)

H2 D2 ( Data Link Layer)

Physical Layer

Layered Architecture


Organization of layers

TCP/IP protocol suite can be divided into threesubgroups

Network support layers

Deals with the physical aspect of the moving data from onedevice to other device

Electrical specification, physical connections, addressingtiming and reliability

User support layer

Transport layer links the two subgroups


Functionality of layers

Physical Layer

Responsible for transmitting individual bits from onenode to another.

It defines the physical characteristics of the interfacebetween the device and transmission media.

Representation of bits into signals.

Data rate such as duration of bits.

Clock Synchronization of bits.

Multiplexing



Data link Layer

Responsible for transmitting frames from one nodeto another.

Framing: it divides stream of bits received fromnetwork layer into manageable data units calledframes.

Physical addressing

Flow Control

Error Control: it adds reliability to the physical layerby adding mechanism to detect and retransmitdamaged or lost frames. Trailer

Access control



Network Layer

Responsible for original source to final destinationdelivery of packet across multiple networks.

It oversees the delivery of the packets between twosystems on the same networks.

Logical addressing.

Routing



Transport Layer

Responsible for delivery of a message from processto process delivery.

Also maintaining the logical connections betweensender and receiver

Port Addressing.

Segmentation and reassembly.

Connection control

Flow control

Error Control



Application Layer

Responsible for providing services to user.

Mail services

File Transfer and access.

Remote login

Accessing www


OSI Model

Presentation Layer

It is concerned with the syntax and semantics of theinformation exchange between two systems.

Translation: sender dependent format to commonformat.

Encryption

Compression


Summary

Node-to-Node Delivery

(At MAC Layer)

Host-to-Host Delivery

(At Network Layer)

Process-to-Process Delivery

(At Transport Layer)


TCP/IP Suit


Comparison of Network models


Functioning of Layers


Data Link Layer

Common responsibilities and functionalities provided by DATA Link Layer,

Hop to Hop Delivery

Also include framing, addressing, flow control, error control, and media access control.

The data link layer divides the stream of bits received from the network layer into manageable data units called frames.

The data link layer adds a header to the frame to define the addresses of the sender and receiver of the frame.


Data Link Layer

Flow Control

If the rate at which the data are absorbed by the receiver is less than the rate at which data are produced in the sender, the data link layer imposes a flow control mechanism to avoid overwhelming the receiver.

Error Control

It also adds reliability to the physical layer by adding mechanisms to detect and retransmit damaged, duplicate, or lost frames.


DATA Link Layer

Access Control

When two or more devices are connected to the same link, data link layer protocols are necessary to determine which device has control over the link at any given time.

Due to this functionality it is also called MAC layer.


Error Detection and Correction

A system must guarantee that the data received should be identical to the data transmitted.

Data can be corrupted during transmission. Some applications require that errors be detected and corrected.

Many factors can alter one or more bits of a message.


Error Detection and Correction

Error vs. Fault Fault leads to error and error leads to failure.

Types of errors Single bit error

Single-bit errors are the least likely type of error in serial data transmission.

Burst error The term burst error means that 2 or more bits in the data

unit have changed from 1 to 0 or from 0 to 1.


DATA Link Layer

Burst Error

In figure below, 0100010001000011 was sent, but 0101110101100011 was received.

Burst error does not occur in consecutive bits.

The length of the burst is measured from the first corrupted bit to the last corrupted bit.

Burst error of length 8


Data link layer

Redundancy

The central concept in detecting or correcting errors is redundancy.

This is done through by sending extra bits with the data to be sent.

These redundant bits are added by the sender and removed by the receiver.

It helps the receiver to detect and then correct the corrupted bits.


Data link layer

Coding

Redundancy is achieved through various coding schemes.

The sender adds redundant bits through a process that creates a relationship between the redundant bits and the actual data bits.

The receiver checks the relationships between the two sets of bits to detect or correct the errors.

Block coding is used to do the stuff.


Data Link Layer

Message

Generator

Message +

redundancy

Message

Checker

Received DATA

Sender Receiver

Unreliable transmission

Accepted or discarded

Encoder Decoder


DATA Link Layer

For block coding we use modulo-2 arithmetic's.

In this arithmetic, the modulus N is 2.

Mean that use only 0 and 1.

Operations in this arithmetic are very simple.

The following shows how we can add or subtract 2 bits.

In this arithmetic we use the XOR (exclusive OR) operation for both addition and subtraction.

Adding: 0+0=0 0+1=1 1+0=1 1+1=0 Subtracting: 0 -0=0 0- 1=1 1 - 0 =1 1-1 =0


BLOCK Coding

In this coding the actual message is divided into blocks each of k-bits, called data words.

Then add r redundant bits to each block to make the length n = k + r.

The resulting n-bit blocks are called codewords. E.g. The 4B/5B block coding


Block Coding

How can errors be detected by using block coding?

Two ways,

1. The receiver has (or can find) a list of valid codeword.

2. The original codeword has changed to an invalid one.


Example

Let us assume that k = 2 and n = 3. Table below shows the list of datawords and codewords.

A code for error detection


Solution

Assume the sender encodes the dataword 01 as 011 and sends it to the receiver.

Consider the following cases, The receiver receives 011. It is a valid codeword.

The codeword is corrupted during transmission, and 111 is received. This is not valid codeword so it is discarded.

The codeword is corrupted during transmission, and 000 is received. This is a valid codeword. The receiver incorrectly extracts the dataword 00. Two

corrupted bits have made the error undetectable.


Error Correction Mechanism

Structure of encoder and decoder in error correction,


Hamming Distance

It is a central concept in coding for error control.

The Hamming distance between two words (of the same size) is the number of differences between the corresponding bits.

It is depicted as d(x, y).

It can easily done by applying the XOR operation on the two words and count the number of 1’s in the result.

The Hamming distance d(10101, 11110) is 3 because 10101 xor 11110 is 01011 (three 1’s).


Minimum Hamming Distance

The minimum Hamming distance is the smallest Hamming distance between all possible pairs in a set of words.

For example, d(00000, 01011) = 3

d(01011, 10101) = 4

d(00000, 10101) = 3

d(01011, 11110) = 3

d(00000, 11110) = 4

d(10101, 11110) = 3

The dmin in this case is 3.


LINEAR BLOCK CODES

Almost all block codes used today belong to a subset called linear block codes.

Non linear block codes are not used widely b/c their structure makes the theoretical analysis and implementation difficult.

A linear block code is a code in which the exclusive OR (addition modulo-2) of two valid codewords creates another valid codeword.


Some Linear Block Codes

Following are some linear block codes are,

Simple Parity-Check Code

Most familiar error-detecting code is the simple parity-check code.

A k-bit dataword is changed to an n-bit codeword where

n = k + 1.

The extra bit, called the parity bit, is selected to make the total number of 1’s in the codeword even.

The minimum Hamming distance for this category is

dmin = 2

The code is a single-bit error-detecting code; it cannot correct any error.


Simple parity-check code

Simple parity-check code C(5, 4)


Simple parity-check code

Encoder and decoder for simple parity-check code.


Calculating parity mechanism

This is normally done by adding the 4 bits of the dataword (modulo-2); the result is the parity bit. In other words,

r0 = a3 + a2 + a1 + a0 (modulo-2)

If the number of 1’s is even, the result is 0; if the number of 1’s is odd, the result is 1.

At the reciever end, the addition is done over all 5 bits.

The result which is called the syndrome, is just 1 bit.

so = b3 + b2 +b1 + b0 + q0 (modulo-2)

If so is equal to 0 then no error, if 1 then the dataword is not extracted and hence the dataword is not created.


Hamming Codes

It is a category of error-correcting codes.

These codes were originally designed with

dmin = 3, which means that they can detect up to two errors or correct one single error.

First let us find the relationship between n and k in a Hamming code.

We need to choose an integer m >= 3.

The values of n and k are then calculated from m as n = 2m - 1 and k = n - m.

The number of check bits r = m.


Hamming Code Mechanism



A copy of a 4-bit dataword is fed into the generator that creates three parity checks r0, r1, and r2, r0 = a2 + a1 + a0 modulo-2

r1 = a3 + a2 + a1 modulo-2

r2 = a1 + a0 + a3 modulo-2

Each of the parity-check bits handles 3 out of the 4 bits of the dataword.

The total number of 1’s in each 4-bit combination (3 dataword bits and 1 parity bit) must be even.

Above three equations are not unique.

A combination of two equation cannot create the third one.


Hamming code C(7, 4)



The checker in the decoder creates a 3-bit syndrome (S2S1SO) in which each bit is the parity check for 4 out of the 7 bits in the received codeword.

s0 = b2 + b1 + b0 +q0 modulo-2

s1 = b3 + b2 + b1 +q1 modulo-2

s2 = b1 + b0 + b3 +q2 modulo-2

The equations used by the checker are the same as those used by the generator with the parity-check bits added to the right-hand side of the equation.



The 3-bit syndrome creates eight different bit patterns (000 to 111) that can represent eight different conditions.

Syndrome 000 001 010 011 100 101 110 111

Error None q0 q1 b2 q2 b0 b3 b1

Generator is not concerned with the four cases shaded b/c there is either no error or an error in the parity bit.

In the other four cases, 1 of the bits must be flipped (changed from 0 to 1 or 1 to 0) to find the correct dataword.


Example

Let us trace the path of three datawords from the sender to the destination: Dataword Codeword Received Syndrome

0100 0100011 0100011 000

Result: No error

0111 0111001 0011001 011

Result: b2 is in error, after flipping b2 we get 0111

1101 1101000 0001000 101

Result: b0 is in error, after flipping b0 we get 0000, the wrong dataword.

Note: This code cannot correct two errors.


Burst Error detection and correction

The key is to split a burst error between several codewords.

Normally we send a packet or a frame of data.

Make the hamming code of size N codeword out of our frame to be sent.

Instead of sending one frame, arrange codeword in table and send the bits in the table a column at a time.

Bits are sent column by column. (from the left).


Burst error correction using H. Code


Cyclic Codes

Cyclic codes are special linear block.

In a cyclic code, if a codeword is cyclically shifted (rotated), the result is another codeword.

For example if 1011000 is a codeword, then 0110001 is also a codeword after left shifting the dataword.


Cyclic Redundancy Check

A CRC is an error-detecting code.

Its computation resembles a polynomial long division operation in which the quotient is discarded and the remainder becomes the result.

The coefficients are calculated according to the carry-less arithmetic of a finite field.

The length of the remainder is always less than the length of the divisor.


Computation of CRC

To compute an n-bit binary CRC

Line the bits representing the input in a row.

Put the divisor underneath the left-hand end of the row.

Here is the first calculation for computing a 3-bit CRC,

11010011101100


CRC

Direct method calculation,

If the input bit above the leftmost divisor bit is 0, do nothing and move the divisor to the right by one bit.

If the input bit above the leftmost divisor bit is 1, the divisor is XORed into the input.

The divisor is then shifted one bit to the right, and the process is repeated until the divisor reaches the right-hand end of the input row.

Complete calculation is on the next slide.


CRC


2nd Method of Calculating CRC

In the encoder, the dataword has k bits(4 here).

The codeword has n bits(7 here).

The size of the dataword is augmented by adding n–k (3 here).

The n-bit result is fed into the generator.

The generator uses a divisor of size n - k + 1(4 here), predefined and agreed upon i.e. 1011.

The generator divides the augmented dataword by the divisor (modulo-2 division).

The quotient of the division is discarded; the remainder (r2r1ro) is appended to the dataword to create the codeword.


2nd Method of Calculating CRC

CRC encoder and decoder


CRC

At decoder end,

Receives the corrupted codeword.

All n bits are fed into checker which is replica of the generator.

Remainder produced by the checker is a syndrome of n-k bits, which is fed into decision logic analyzer.

If all bits of the syndrome is zero then no error take the 4 leftmost bits of the codeword are accepted, else discarded.


At Encoder side


At Decoder Side


At decoder end

If the code word changes during transmission, then,


Simulation of division in CRC encoder


Polynomial representation

A pattern of Os and 1 s can be represented as a polynomial with coefficients of 0 and 1.

The power of each term shows the position of the bit; the coefficient shows the value of the bit.


Polynomial representation

X6 + x + 1

1 0 0 0 0 1 1

We have seven bit number, the degree of the polynomial is 6 one less then the number of bits in the pattern.


Polynomial operation

Adding and subtracting Adding or subtracting is done by combining terms

and deleting pairs of identical terms.

For example adding x5 + x4 + x2 and x6 + x4 + x2

gives just x6 + x5. The terms x4 and x2 are deleted.

If we add three polynomial of the same degree then two of them deleted, one is kept left.

Multiplying or Dividing Terms In multiplications, the powers are added.

For division subtract the power of the second term from the power of first.


Polynomial operation

Multiplying Two Polynomials Each term of the first polynomial must be multiplied

by all terms of the second.

The result, of course, is then simplified, and pairs of equal terms are deleted.

Shifting Shifting number of bits to the right or left of the

given pattern.

Shifting left 3 bits: 10011 becomes 10011000

x4 +x+ 1 becomes x7 +x4+x 3

Shifting right 3 bits: 10011 becomes 10

x 4 + x + 1 becomes x


CRC division using polynomials


CHECKSUM

The checksum is used in the Internet by several protocols.

The checksum is based on the concept of redundancy.

Basic Idea Suppose our data is a list of five 4-bit numbers.

send the sum of the numbers along with the data.

If the set of numbers is (7, 11, 12, 0, 6),

we send (7, 11, 12, 0, 6, 36).

The receiver adds the five numbers and compares the result with the sum. If the result is same then no error else data is not accepted.


Checksum

Another heuristic is the negative (complement) of the sum called the checksum.

We send (7, 11, 12, 0, 6, -36).

Receiver add all the numbers, if result is 0 then no error otherwise an error.


One's Complement

The previous example has one major drawback.

The data is of 4 bit word (0 to 15) except checksum.

One solution is to use the 1’s complement.

Using unsigned numbers between 0 and 2 n - 1 using only n bits.

If the number has more than n bits, then wrapping the bits.

For example, 21 in binary is 10101.

We can wrap the leftmost bit and add it to the four rightmost bits.

We have (0101 + 1) = 0110 or 6.


Checksum

Problem

How can we represent the number -6 in one's complement arithmetic using only four bits?


Checksum using 1’s complement

Let the data to be sent is (7, 11, 12, 0, 6)

Figure below show the process at the sender and receiver.


Checksum computation

Sender site:

The message is divided into 16-bit words.

The value of the checksum word is set to 0.

All words including the checksum are added using one's complement addition.

The sum is complemented and becomes the checksum.

The checksum is sent with the data.



Receiver Site:

The message (including checksum) is divided into 16-bit words.

All words are added using one's complement addition.

The sum is complemented and becomes the new checksum.

If the value of checksum is 0, the message is accepted; otherwise, it is rejected.



Let us calculate the checksum for a text of 8 characters ("Forouzan").

The text needs to be divided into 2-byte (16-bit) words.

We use ASCII to change each byte to a 2-digit hexadecimal number.

F is represented as 0x46 and o is represented as 0x6F.


Solution


ASCII Chart

Documents

Advanced Computer Networks Rab Nawaz Jadoon · Department of Computer Science 28 DATA Link Layer For block coding we use modulo-2 arithmetic's. In this arithmetic, the modulus N is