Ele4Net Communication Networks Answers | Assessment Answer

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).

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

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.

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

References

Fall, K. R., & Stevens, W. R. (2012). TCP/IP Illustrated, Volume 1: The Protocols (2 ed.). New York: addison-Wesley.

Blanchet , M. (2010, 02 02). Patent No. 7,657,642. Washington.