MCS-042 ASSIGNMENT SOLUTION (2018-19)

If you have any queries please leave a message here
Your Message
×


Question 1 :

(a) What are the main differences between connectionless and connection oriented communication?

Answer : -

  1. In connection-less communication there is no need to establish connection between source (sender) and destination (receiver), but in connection-oriented communication connection must established before data transfer.

  2. In conection-oriented system a HANDSHAKING process is used i.e the process starts from the sender computer by sending a request to start the transfer that is to be acknowledged by the destination device before the data can be send.
    While in connection-less system no handshaking process is used. if one computer wants to send data,it just sends the data packet to the end system.

  3. In conection-oriented system communication between the sender and receiver continues untill the transmission has been verified.thus it requires higher overhead and places greater demand on bandwidth.
    While the connection-less system requires less overhead and bandwidth.

  4. There is guranteed delivery of data in conection-oriented system. Any packet that is not received by the destination is resend by the sending device.
    There is no confirmation that the data has been received, thus there is no resending of data in connection-less system. Thus transmission is not guranteed.

  5. Example of connectiov-oriented protocol is - TCP/IP and example of connection-less protocol is - UDP




(b) What are the essential differences between packet switching and message switching? Explain with the help of a suitable diagram.

Answer : -

Packet Switching

Packet switching is a digital network transmission process in which data is broken into suitable-sized pieces or blocks for fast and efficient transfer via different network devices. When a computer attempts to send a file to another computer, the file is broken into packets so that it can be sent across the network in the most efficient way. These packets are then routed by network devices to the destination.


Message Switching

Message switching was a technique developed as an alternate to circuit switching, before packet switching was introduced. In message switching, end users communicate by sending and receiving messages that included the entire data to be shared. Messages are the smallest individual unit.

Message Switching provide 2 distinct and important characteristics :




Question 2 :

system has 𝑛-layer protocol hierarchy. Applications generate message of length M bytes. At each of the layers, an h byte header is added. What fraction of the network bandwidth is filled with headers?

Answer : - From the given data, a system contains n-layers protocol hierarchy and “h” bytes of data are added at each layer.

Therefore, the fraction of the network bandwidth filled with header is - {(n-1) * h} / [{(n-1) * h} + M]




Question 3 :

State the Nyquist theorem. For what kind of physical medium it is applicable? Will it work for a noisy Channel?

Answer : - According to Henry Nyquist, a perfect channel has a finite transmission capacity. He derived an equation expressing the maximum data rate for a finite bandwidth noiseless channel. Nyquist said that if an arbitrary signal has been run through a low-pass filter of bandwidth B, the filtered signal can be completely reconstructed by making only 2Bsamples per second. Sampling the line faster than 2B times per second is pointless because the higher frequency components that such sampling could recover have already been filtered out. If the signal consists of L discrete levels, Nyquist’s theorem states :

Maximum bit rate =2* Bandwidth log2 L

For noisy channel :

Capacity = bandwidth * log2 (1+SNR), where SNR is signal to noise ratio.

Nyquist theorem is valid for both fibre optics and copper wires this is because the Niquist theorem is purely depend upon the mathematic function and has not any concern with the transmission media for communication.




Question 4 :

Define the throughput expressions of Pure Aloha and Slotted Aloha. Also draw throughput Vs load graphs for the above protocols.

Answer : -

Pure Aloha

The probability of k frames transmission in t seconds and is given by the poisson distribution as follows :
P[k] = (2G)k * e-2G / k! , { k=0,1,2,3,.... }

The throughput for Pure ALOHA is : S = G * e-2G
G = One frame time
e-G = Probability of zero frames in t seconds

The maximum throughput S = 0.184 when G = 1/2


Slotted Aloha

The probability of k frames transmission in t seconds and is given by the poisson distribution as follows :
P[k] = (G)k * e-G / k! , { k=0,1,2,3,.... }

The throughput for slotted ALOHA is: S = G * e-G
G = One frame time
e-G = Probability of zero frames in t seconds

The maximum throughput S = 0.368 when G = 1




Question 5 :

(a) Explain hidden station and exposed station problems in WLAN protocols with the help of an illustration. What is the limitation of CSMA protocol in resolving the above problems? Explain the use of virtual channel sensing method as a proposed solution.

Answer : - In a formal way hidden terminals are nodes in a wireless network that are out of range of other node or a collection of nodes.
Consider the scenario of wireless networking with three wireless devices -



The transmission range of access point A reaches at B, but not at access point C, similarly transmission range of access point C reaches B, but not at A. These nodes are known as hidden terminals. The problem occurs when nodes A and C start to send data packets simultaneously to the access point B. Because the access point A and C are out of range of each other and resultant they cannot detect a collision while transmitting, Carrier Sense Multiple Access with Collision Detection (CSMA/CD) does not work and collisions occur, which then corrupt the data received by the access point B due to the hidden terminal problem.

IEEE 802.11 uses 802.11 RTS/CTS acknowledgment and handshake techniques over wireless networks to transferring packets that partly overcome the hidden node problem. RTS/CTS is not a proper and permanent solution and may decrease throughput even further, but adaptive acknowledgments from the base station can help too.

Request To Send (RTS)/Clear To Send (CTS) Process
When A has to send data to B, it begins by sending an RTS frame to B to request permission to send it a frame. When B receives this request, it may decide to grant permission, in which case it sends the CTS frame back. Upon receipt of the CTS, A now sends its frame and starts an ACK timer. Upon correct receipt of the data frame, B responds with an ACK frame leading to the closure of data transfer operation between A & B. In case A's ACK timer expires before the ACK gets back to it, the whole protocol is run again.

Network Allocation Vector (NAV)
As an alternative to carrier sensing, the network allocation vector (NAV) is used to inform other nodes how long the current node will need the channel. Any nodes overhearing the NAV know that they have no need of sensing the channel for the time indicated. Since idle sensing of the channel is one of the biggest uses of energy in a network, the NAV reduces the amount of idle sensing required at any nodes which can overhear it, thus saving energy at all nodes in the network.




(b) Sketch the differential Manchester encoding for the bit stream: 0011111010111. Assume the line is initially in the low state.

Answer : -




Question 6 :

Draw the Ethernet frame format and explain its fields. Is there any limitation on a maximum and minimum frame size of Ethernet frame? Explain.

Answer : -

Field Name
Size (bytes)
Description
Preamble
7
This is a stream of bits used to allow the transmitter and reciever to synchronize their communication. The preamble is an alternating pattern of binary 56 ones and zeroes.
Start Frame Delimiter
1
This is always 10101011 and is used to indicate the beginning of the frame information.
Destination Address
6
It contain the physical address (MAC Address) of the receiver.
Source Address
6
It contain the physical address (MAC Address) of the sender.
Length of Data Field
2
It indicates the number of bytes present in "Data" field.
Data
0 To 1454
The data is inserted here. This is where the IP header and data is placed if you are running IP over Ethernet.
Payload
46
This field size can be 0 to 46 bytes long. This is required if, the data size is less than 46 bytes.
Frame Checksum
4
This field contains the Frame Check Sequence (FCS) which is calculated using a Cyclic Redundancy Check (CRC). The FCS allows Ethernet to detect errors in the Ethernet frame and reject the frame if it appears damaged.

Minimum frame length = 64 bytes
Maximum frame length = 1518 bytes
Minimum length or lower limit for frame length is defined for normal operation of CSMA/CD.
An ethernet frame has a minimum size because anything that is shorter than the 64 byte minimum is interpreted by receiving stations as a collision.




Question 7 :

How does the Border Gateway Protocol work? Explain it with the help of a diagram. How does it resolve the count to infinity problem that is caused by other distance vector routing-algorithms?

Answer : - Border Gateway Protocol (BGP) is designed to exchange routing and reachability information between Autonomous Systems (AS) on the Internet.

The term Autonomous System (AS) refers to a network that operates separately from other networks and usually operates within a single administrative domain. Each AS is represented by an AS number.

Each BGP speaker, which is called a “peer”, exchanges routing information with its neighboring peers in the form of network prefix announcements. This way, an AS doesn’t need to be connected to another AS to know its network prefix.

The BGP decision-making mechanism analyzes all the data and sets one of its peers as the next stop, to forward packets for a certain destination.

Each peer manages a table with all the routes it knows for each network and propagates that information to its neighboring autonomous systems.

In this way, BGP allows an AS to collect all the routing information from its neighboring autonomous systems and “advertise” that information further. Each peer transfers the information internally inside its own autonomous system.

Just like in real life, usually more than one route exists to reach a given destination. BGP is responsible for determining the most suitable route according to the information collected and an organization’s routing policy, which is based on cost, reliability, speed, etc.


In distance vector routing, count to infinity problem/routing loops usually occur when an interface goes down. It can also occur when two routers send updates to each other at the same time.




Question 8 :

Draw the header format of TCP and explain the followings fields: ACK bit, RST bit & PSH bit, and Flags.

Answer : -


Control Flags Table

Subfield Name
Size (bits)
Description
Urgent Pointer (URG)
1 bit
When set to 1, indicates that the current segment contains urgent (or high-priority) data
Acknowledgement (ACK)
1 bit
When set to 1, indicates that this segment is carrying an acknowledgment, and the value of the Acknowledgment Number field is valid and carrying the next sequence expected from the destination of this segment.
Push (PSH)
1 bit
The sender of this segment is using the TCP push feature, requesting that the data in this segment be immediately pushed to the application on the receiving device. It is useful for transmitting small units of data.
Reset (RST)
1 bit
The sender has encountered a problem and wants to reset the connection.
Synchronize (SYN)
1 bit
This segment is a request to synchronize sequence numbers and establish a connection; the Sequence Number field contains the Sequence Number of the sender of the segment.
Finish (FIN)
1 bit
The sender of the segment is requesting that the connection be closed.




Question 9 :

Explain the window management scheme in TCP through an illustration.

Answer : - Window management in TCP is an important concept that ensures reliability in packet delivery as well as reduce the wastage of time in waiting for the acknowledge after each packet.

Window size determines the amount of data that you can transmit before receiving an acknowledgment. Sliding window refers to the fact that the window size is negotiated dynamically during the TCP session.




Question 10 :

Discuss the Silly Window Syndrome which cause degradation of TCP performance. What is the proposed solution? Explain.

Answer : - Silly window syndrome is a problem in computer networking caused by poorly implemented TCP flow control. A serious problem can arise in the sliding window operation when the sending application program creates data slowly, the receiving application program consumes data slowly, or both. If a server with this problem is unable to process all incoming data, it requests that its clients reduce the amount of data they send at a time (the window setting on a TCP packet). If the server continues to be unable to process all incoming data, the window becomes smaller and smaller, sometimes to the point that the data transmitted is smaller than the packet header, making data transmission extremely inefficient. The name of this problem is due to the window size shrinking to a "silly" value.

Solution - When the silly window syndrome is created by the sender, Nagle's algorithm is used and when it is created by the receiver, David D Clark's solution is used.




Question 11 :

Explain the terms: bit rate, baud rate and bandwidth with the help of an example. Also draw modulation schemes for the followings: (i) QPSK (ii) QAM-16 and describe.

Answer : -

Bit Rate - Bit rate can be defined as the number of bit intervals per second. And bit interval is referred to as the time needed to transfer one single bit. In simpler words, the bit rate is the number of bits sent in one second, usually expressed in bits per second (bps). For example, kilobits per second (Kbps), Megabits per second (Mbps), Gigabits per second (Gbps), etc.

Bit Rate can be expressed by the given equation :
Bit rate = baud rate x the number of bits per signal unit

Baud Rate - Baud rate is expressed in the number of times a signal can change on transmission line per second. Usually, the transmission line uses only two signal states, and make the baud rate equal to the number of bits per second that can be transferred.

An example can illustrate it. For example, 1500 baud rate illustrates that the channel state can alter up to 1500 times per second. The meaning of changing state means that channel can change its state from 0 to 1 or from 1 to 0 up to 1500 times per second (in the given case).

Baud Rate can be expressed by the given equation :
Baud rate = bit rate / the number of bits per signal unit

Bandwidth - Bandwidth refers to how much digital information we can send or receive across a connection in a certain amount of time. Sometimes it's called data transfer rate too.


QPSK


QAM-16




Question 12 :

What are the differences between leaky bucket and token bucket algorithm? How is token bucket algorithm is implemented? Explain.

Answer : -

Leaky Bucket
Token Bucket
When the host has to send a packet , packet is thrown in bucket.In this leaky bucket holds tokens generated at regular intervals of time.
Bucket leaks at constant rate.Bucket has maximum capacity.
Bursty traffic is converted into uniform traffic by leaky bucket.If there is a ready packet , a token is removed from Bucket and packet is send.
In practice bucket is a finite queue outputs at finite rate.If there is a no token in bucket, packet can not be send.


The token bucket algorithm can be conceptually understood as follows :




Question 13 :

How does DES work? Explain.

Answer : - DES is a block cipher which encrypts data in 64-bit blocks. A 64-bit block of plaintext goes in one end of the algorithm and a 64-bit block of ciphertext comes out the other end. DES is a symmetric algorithm : The same algorithm and key are used for both encryption and decryption (except for minor differences in the key schedule).

The key length is 56 bits. (The key is usually expressed as a 64-bit number, but every eighth bit is used for parity checking and is ignored. These parity bits are the least-significant bits of the key bytes.) The key can be any 56-bit number and can be changed at any time.

At its simplest level, the algorithm is nothing more than a combination of the two basic techniques of encryption : confusion and diffusion. The fundamental building block of DES is a single combination of these techniques (a substitution followed by a permutation) on the text, based on the key. This is known as a round. DES has 16 rounds; it applies the same combination of techniques on the plaintext block 16 times.

DES operates on a 64-bit block of plaintext. After an initial permutation, the block is broken into a right half and a left half, each 32 bits long. Then there are 16 rounds of identical operations, called Function f, in which the data are combined with the key. After the sixteenth round, the right and left halves are joined, and a final permutation (the inverse of the initial permutation) finishes off the algorithm.



ABOUT US

QuestionSolves.com is an educational website that helps worldwide students in solving computer education related queries.

Also, different software like Visual Studio, SQL Server, Oracle etc. are available to download in different versions.

Moreover, QuestionSolves.com provides solutions to your questions and assignments also.


MORE TOPIC


Windows Command

UNIX Command

IGNOU Assignment Solution

IGNOU Question Paper Solution

Solutions of Different Questions


WHAT WE DO


Website Devlopment

Training

Home Learning

Provide BCA, MCA Projects

Provide Assignment & Question Paper Solution


CONTACT US


Follow Us