# Vertical Redundancy Check-Computer Networks-Assignment, Exercises for Computer Networks. Birla Institute of Technology and Science

## Computer Networks

Description: This is an introductory course in Data Communication and Computer Networks. The course is designed with objectives: Provide solid foundation in the field of data communication and computer networks, give practical experience on networks and networking devices, introduce the cutting edge technologies. This assignment main points are: Drawback, Vertical, Redundancy, Check, Data, Transmission, Errors, Longitudinal, Horizontal, Arranging
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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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