Communication Networks - Assignment 1 Solutions

This article provides solutions to Assignment 1 of Communication Networks. It covers topics like TCP/IP protocol architecture, differences between IPv4 and IPv6, fundamental frequency and period of a signal, Shannon capacity in noisy channel and Nyquist formula.

Running ead C MM CA R SH : O UNI TION NETWO K 0
Communication etworksN
Assignment 1
Student ameN
8/27/2018

Communication etworksN 1
Contents
Communication etworksN .............................................................................................................................1
Assignment - 1.............................................................................................................................................1
Solution-1:.....................................................................................................................................................1
Solution-2:.....................................................................................................................................................1
Solution-3:.....................................................................................................................................................2
Solution-4:.....................................................................................................................................................2
Solution a- .................................................................................................................................................. 2
Solution b- ..................................................................................................................................................2

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Communication Networks
Assignment - 1
Solution-1:
In TCP/IP protocol architecture, four parts of protocol data unit at application later it is Data, at
transport layer known as Segment, Internet layer known as Packet, at data link layer it is frame,
and at physical layer, it is known as bits.
(Source: ccnablog, 2018)
As in diagram above, different form of the data at each layer of TCP/IP model is called protocol
data unit. Data bind in next packet with header information of particular layer. Each layer have
own header. Headers have small information about that data. Therefore, at the end all header and
data converted in row bits and transfer from physical media ( Fall & Stevens, 2012).

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Solution-2:
These are the following differences in IPv4 and IPv6 Protocol versions:
S. No. IPv4 IPv6
1. IPv4 have 32 bit address Length of IPv6 address is 128 bits
2. Addresses are represented in decimals. Addresses are represented in hexadecimals.
3. IPsec is optional in the IPv6 protocol IPsec is inbuilt in the IPv6 protocol
4. Address Representation In decimal Address Representation In hexadecimal
5. It supports Manual and DHCP
configuration.
It supports Auto-configuration and
renumbering.
6. Encryption and Authentication is not
Provided
Encryption and Authentication is Provided
7. Fragmentation performed by Sender and
forwarding routers
Fragmentation performed by Only by the
sender
8. Broadcast messages are available Broadcast messages are not available.
9. Checksum field is available in IPv4
header
Checksum field is not available in IPv6
header

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Solution-3:
The greatest common divisor of all the frequency components contained in a signal is called
fundamental frequency and least common multiple of all individual periods of the components is
known as fundamental period of the signal.
Fundamental frequency is 1f, where f=1.
The period of the Fundamental is 1/f, so period is = 1.
Solution-4:
I have a channel with a 1-MHz bandwidth.
SNR for this channel is 63.
Solution-a
An ideal noiseless channel never exists.
Shannon capacity in Noisy Channel:
The maximum data rate for any noisy channel is C = B* log2 (1+SNR)
Where,
C= Channel capacity in bits per second
B= bandwidth of channel
SNR= signal to noise ratio
First, I use the Shannon formula to find the upper limit.

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C= 106 log2 (1+63)
C= 6 Mbps
Solution-b
The Shannon formula gives us 6 Mbps, the upper limit to the data rate.
I choose 4 Mbps, which is 2/3 of the maximum theoretical limit.
Then we use the Nyquist formula to find the number of signal levels.
The formula for maximum bit rate in bits per second (bps) is:
Maximum bit rate = 2*BW* log2 L
Where, BW =bandwidth at channel
L= number of signed levels used to represent data.
So
4 Mbps = 2*1MHz * log2 L
L = 4
Works Cited
all R Stevens RF , K. ., & , W. . (2012). CP P llustrated olume he ProtocolsT /I I , V 1: T ed ew ork(2 .). N Y :
addison esley-W .
lanchet MB , . (2010, 02 02). Patent oN . 7,657,642. ashingtonW .