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Assignment 2

Question 1: What is the drawback of VRC

Vertical Redundancy Check is a way of error checking by attaching a parity bit to each byte

of data to be transmitted, which is then tested to determine if the transmission is correct.

The resulting parity bit is constructed by XORing the word. The result is a "1" if there is an odd

number of 1s, and a "0" if there is an even number of 1s in the word.

Draw Back:

This method is unreliable, because if there are even number of errors in the data, the error

will go undetected.

Question 2: Calculate the LRC of the following data: [HEX] 864C9BF3

In telecommunication, a longitudinal redundancy check (LRC) or horizontal redundancy

check is a form of redundancy check that is applied independently to each of a parallel group

of bit streams. The data must be divided into transmission blocks, to which the additional

check data is added.

Data: 1000 0110 0100 1100 1001 1011 1111 0011

Arranging data:

1 0 0 0

0 1 1 0

0 1 0 0

1 1 0 0

1 0 0 1

1 0 1 1

1 1 1 1

0 0 1 1

1 0 0 0 (Parity bit calculated for every column)

Transmitted Data: 1000 0110 0100 1100 1001 1011 1111 0011 1000

VRC:

Solution:

LRC:

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Assignment 2

Question 3: A: For P=110101 and M=1011010001, Find CRC?

B: In CRC error Detection scheme, choose P(x)=x4+x+1. Encode the bits

10111011011?

1 1 0 1 1 0 1

110101 / 1 0 1 1 0 1 0 0 0 1 0 0 0 0 0

1 1 0 1 0 1 : : : : : : : : :

0 1 1 0 0 0 0 : : : : : : : :

1 1 0 1 0 1 : : : : : : : :

0 0 0 1 0 1 0 0 1 : : : : :

1 1 0 1 0 1 : : : : :

0 1 1 1 0 0 0 : : : :

1 1 0 1 0 1 : : : :

0 0 1 1 0 1 0 0 : :

1 1 0 1 0 1 : :

0 0 0 0 0 1 0 0 <= FCS

The FCS is: X2

Transmitted bits: 1 0 1 1 0 1 0 0 0 1 0 0 1 0 0

1 1 0 1 0 1 1

10011 / 1 0 1 1 1 0 1 1 0 1 1 0 0 0 0

1 0 0 1 1 : : : : : : : : : :

0 0 1 0 0 0 1 : : : : : : : :

1 0 0 1 1 : : : : : : : :

0 0 0 1 0 1 0 1 : : : : :

1 0 0 1 1 : : : : :

0 0 1 1 0 1 0 : : :

1 0 0 1 1 : : :

0 1 0 0 1 0 : :

1 0 0 1 1 : :

0 0 0 0 1 0 0 <= FCS

Predetermined Divisor: 10011

FCS = X3

Transmitted bits: 1 0 1 1 1 0 1 1 0 1 1 0 1 0 0

Part A:

Part B:

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Assignment 2

Question 4: Calculate the hamming pair wise distance among the following

codewords:

A: 00000, 10101, 01010

B: 000000, 010101, 101010, 110110

Let V1 = 0 0 0 0 0

V2 = 1 0 1 0 1

V3 = 0 1 0 1 0

D(V1,V2) = 3 D(V1,V3) = 2 D(V2,V3) = 5

Let V1 = 0 0 0 0 0 0

V2 = 0 1 0 1 0 1

V3 = 1 0 1 0 1 0

V4 = 1 1 0 1 1 0

D(V1,V2) = 3 D(V1,V3) = 3 D(V1,V4) = 4 D(V2,V3) = 6 D(V2,V4) = 3

D(V3,V4) = 3

Question 5: Suppose a file of 8,000 bytes is to be sent over a line of 5600 bps.

A: Calculate the overhead in bits and time in using asynchronous communication. Assume one

start bit and a stop bit element of length one bit, and 8bits to send the byte itself for each

character. The 8 –bit character consists of all data bits with no parity bit.

B: Calculate the overhead in bits and time using synchronous communication. Assume that the

data are sent in frames. Each frame consists of 1000 characters = 8000 bits and overhead of 48

control bits per frame.

Part A:

Part B:

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Assignment 2

Number of Bytes = 8000

Number of bits = 8 x 8000 = 64000

Line capacity = 5600 bps

Start bit length = 1 bit

Stop bit Length = 1 bit

Number of bits for 1 character = 8 bits

Bits apart from data = 2 bits per character (i.e. 2 bits for every byte)

Total bits apart from data = 2 x 8000 = 16000 bits

Total bits transmitted = 64000 + 16000 = 80000

Therefore, the overhead ratio = (Number of extra bits) / (Total Number of bits Transmitted)

= 16000 / 80000

= 0.2

Percentage overhead = 0.2 x 100 % = 20 % (in bits)

Time overhead = 16000/5600 = 2.85 sec

Number of data bits per frame = 8000

Total frames in 8000 bytes = 64000/8000 = 8 frames

Control bits per frame = 48

Total number of overhead bits = 48 x 8 = 384 bits

Therefore, the overhead ratio = (Number of control bits) / (Total Number of bits Transmitted)

= 384 / 64384

= 0.00596 = 0.006

Percentage overhead = 0.006 x 100 % = 0.6 % (in bits)

Time overhead = 384/5600 = 68 ms

Question 6:

A: Consider the use of 1500-bit frames on a 1-Mbps satellite channel with a 270-ms delay.

What is the maximum link utilization for?

a: Stop-and-wait flow control?

Part A:

Part B:

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University:
Birla Institute of Technology and Science

Subject:
Computer Networks

Upload date:
30/07/2012