Computer Networks 4 1 May June 13

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QP.1Computer Networks (May/June-2013, R09) JNTU-Hyderabad

B.Tech. IV-Year I-Sem. ( JNTU-Hyderabad )

Code No: 09A70403

Jawaharlal Nehru Technological University Hyderabad

B.Tech. IV Year I Semester ExaminationsMay/June - 2013

COMPUTER NETWORKS

( Common to ECE, EIE, BME, ECM )

Time: 3 Hours Max. Marks: 75

Answer any FIVE Questions

All Questions carry equal marks

- - -

1. (a) Describe the design issues of data link layer of OSI model. (Unit-I, Topic No. 1.4)

(b) Define: protocol, digital signal. [7+8] (Unit-I, Topic No. 1.7)

2. (a) Explain the working of packet switching with an example. (Unit-II, Topic No. 2.5)

(b) What is twisted pair cable? What is the difference between category 3 and category 5 twisted pair cables?

[7+8](Unit-II, Topic No. 2.3)

3. (a) Calculate CRC, if message polynomial (M(x)) = x7+ x5+ 1 and generator polynomial (G(x)) = x3+ 1.

(Unit-III, Topic No. 3.3)

(b) Explain supervisory frame format of HDLC. (Unit-III, Topic No. 3.9)

(c) Describe the working of stop-and-wait ARQ protocol. [15] (Unit-III, Topic No. 3.8)

4. (a) Explain the need for medium access control protocol. Describe the frame format of ethernet.

(Unit-IV, Topic No. 4.5)

(b) List the benefits of a wireless LAN. What is a hidden node in a wireless LAN? [7+8](Unit-IV, Topic No. 4.8)

5. (a) What is a bridge? Explain the working of a spanning tree bridge. (Unit-IV, Topic No. 4.5)

(b) Which is better, a low frequency reuse factor of a high frequency reuse factor? Justify your answer. [7+8]

(Unit-V, Topic No. 5.4)

6. (a) What is UDP? List its uses. (Unit-VII, Topic No. 7.2)

(b) Describe TCP connection establishment using 3-way handshake. [7+8] (Unit-VII, Topic No. 7.2)

7. (a) What is ICMP? Explain its use. (Unit-VI, Topic No. 6.5)

(b) What are the various classes of IPv4 addressing? Give example for each. (Unit-VI, Topic No. 6.1)

(c) Explain integrated services QoS model.[15](Unit-VII, Topic No. 7.8)

8. Write short technical notes on,

(a) DNS (Unit-VIII, Topic No. 8.1)

(b) Multicast routing protocols (Unit-VIII, Topic No. 8.5)

(c) Backbone networks. [15](Unit-V, Topic No. 5.2)

R09Solutions


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QP.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2013


SOLUTIONS TO MAY/JUNE-2013, R09, QP

Q1. (a) Describe the design issues of data link layer of OSI model.

Answer : May/June-13, (R09), Q1(a) M[7]

Data link layer have been designed keeping in mind the type of the task that this layer can perform such as,

1. Providing services to the network layer

2. Grouping the bits from the physical layer into frame (formation of frames)

3. Handling with error that occur during transferring of data (error control)

4. Keeping track on the flow of frames (flow control).

1. Providing Services to the Network Layer

Transferring of data to and from the network layer is the main responsibility of the data link layer. It fetches the data from

an entity called process in the network layer and passes it to the data link layer on the destination machine which in turntransmits it to the network layer as shown in figure (a) it shows the virtual transmission, but in actual transmission data transfer

takes place through the physical layer of both source and destination machine as shown in figure (b).

Application

Presentation

Session

Transport

Network

Datalink

Physical

Virtual path for

data transmission

Application

Presentation

Session

Transport

Network

Datalink

Physical

Application

Presentation

Session

Transport

Network

Datalink

Physical Actual path for

data transmission

Application

Presentation

Session

Transport

Network

Datalink

Physical

Application

Presentation

Session

Transport

Network

Datalink

Physical

Virtual path for

data transmission

Application

Presentation

Session

Transport

Network

Datalink

Physical

Application

Presentation

Session

Transport

Network

Datalink

Physical Actual path for

data transmission

Application

Presentation

Session

Transport

Network

Datalink

Physical

Figure (a): Virtual Transmission Figure (b): Actual Transmission

Data Link Layer Services

(i) Unacknowledged connectionless service

(ii) Acknowledged connectionless service

(iii) Acknowledged connection-oriented service.

(i) Unacknowledged Connectionless Service

In this type of service, the source machine does not establish any connection with the destination machine. The

source continuously send the frames without waiting for the acknowledgment from the receiver. The major drawbackof this service is that, the sender will have no knowledge about which frames are being received by the receiver or

which frames are lost due to some transmission impairments like attenuation, noise etc. So, if frames are lost, the

sender does not take any responsibility to recover the lost frames. This service is used when the rate of error is low

or for real time traffic.

(ii) Acknowledged Connectionless Service

Source machine does not establish any connection with the destination machine, but here the source machine

receive feedback from the destination in the form of acknowledgment. This help the sender who have an idea of

which frames are lost or damaged. This service increases the reliability. If the timer expires and still the sender did not

receive the acknowledgments, it assumes that the frame have lost and retransmit the frame.

This type of service is used mainly in the wireless systems.


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(iii) Acknowledged Connection-oriented Service

As, the name itself specifies that there will be a connection between the source machine and destination. Before

starting the transmission the sender first establishes a connection, and then start transmitting the frames. The

receiver in turn send the positive acknowledgment for correct frames and negative acknowledgment for the lost or

damaged frames. Each frame is given by the sequence number that helps the data link layer to discard any duplicate

frame or acknowledgment. This is most difficult service of the datalink layer.

Steps for making a connection,

(a) Establishing a connection

(b) Transmission of frames

(c) Releasing the connection.

2. Formation of Frames

Grouping the bits together from the physical layer and constructing a frame is called framing. The main responsibility

of the data link layer is to detect and correct the errors. The error arises when the bits on the physical layer get scattered or

grabbled while reaching at the destination. In order to perform this the data link layer creates frames. The frames are createdby adding a header and a trailer to the packet which specify the source and the destination address. Destination address tell

where the frame is to be delivered and the source address used to help the destination in sending the acknowledgment. All

data is not sent as a single frame, because the error recovery process can take long time, instead it is sent by fragmenting the

single frame into number of unequal frames which makes the error detection and correction simpler.

Framing are of two types,

(i) Fixed-size Framing (Static Framing)

The frames are of fixed size specifying the start of frame is not necessary.

(ii) Variable-sizes Framing (Dynamic Framing)

The size of frames changes. Here it is necessary to specify end of the frame.

Framing is done using one of the following four methods,(a) Counting the character

(b) Character stuffing

(c) Bit stuffing

(d) Error control.

(a) Counting the Character

In this method, a field called the header is appended to each frame which specify the character following it i.e., it tells

the number of characters in that frame. Using this header field, destination can easily recognize the start and end of

the frame.

Example

4 1 2 3 4 1 2 3 5 1 2 3 4 7 1 2 3 4 5 6 8 1 2 3 4 5 6 7 4 1 2 3

Character countheader



4 1 2 3 4 1 2 3 5 1 2 3 4 7 1 2 3 4 5 6 8 1 2 3 4 5 6 7 4 1 2 3




In the above example, each header specifies the number of characters following it.

Disadvantages

1. If the bits in the count become corrupted during transmission, the receiver will think that the frame contain

fewer (or more) bits than it actually does.

2. Although check sum will detect the incorrect frames the receiver will have difficulty in resynchronizing the

start of new frame.


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4 1 2 3 4 1 2 3 5 1 2 3 4 5 1 2 3 4 5 6 8 1 2 3 4 5 6 7 4 1 2 3

This becomes thestart of new frame

4 1 2 3 4 1 2 3 5 1 2 3 4 5 1 2 3 4 5 6 8 1 2 3 4 5 6 7 4 1 2 3

This becomes thestart of new frame

(b) Character Stuffing

Character stuffing is a technique of inserting the ASCII character DLE STX (Data Link Escape Start of Text) and DLE

ETX (Data Link Escape End of Text) that mark the start and end of the frame. In certain situations it may also happen

that DLE character may appear in the data section of the frame. In order to avoid the mistake of understanding the

data and the start/end of the frame, the ASCII character DLE is stuffed before each DLE character in the data. Then,

this DLE character is removed after monitoring the data stream on the receiver side. If two DLE DLE follows, then the

receiver think that it is stuffed and therefore it is deleted.

The major disadvantage of character stuffing is that character is the smallest unit that can be operated on, but all the

architectures are not byte-oriented.

Example

DLE STX A B DLE DLE D E DLE DLE DLE ETX

Stuffed character

A B DLE D E DLE Original data

A B DLE D E DLE Data passed to thenetwork layer

Figure: Character Stuffing

(c) Bit Stuffing

Bit-stuffing is same as that of character stuffing but the former one is bit oriented and the later one is byte-oriented.

It is the process of adding an extra-bit in the data-unit of the frame to prevent sequence of bits from looking like a flag.

Basically, bit stuffing is used to differentiate between the flag and data to achieve data transparency. In this process,

the sender add extra 0 bit to the outgoing data stream after the occurrence of five consecutive 1s and the receiver

remove the bit 0. If it is followed after five consecutive 1s.

Example

0111 00111110 111111110 1111110

Original data

0111 001111100 11111 01110 11111010

Stuffed bits

0111 0011111011111111 01111110 data passed to the network layer.

3. Error Control

Error control is a mean for protecting data from errors. The main function of the data link layer is error control.

Basically, there are two steps in the error control,

Error control

Error detection Error correction

Error control

Error detection Error correction


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Error occur when the bits from the physical layer are

misplaced or grabbled during transmission, or when there is

hardware malfunctioning that result in the lost of frames.

There are different kinds of error.

(a) Single-bit Error

Only a single-bit in the data unit is replaced.

(b) Multiple-bit Error

Two or more non-consecutive bits in the data unit

have changed.

(c) Burst Error

Two or more consecutive bits in the data unit have

changed. Burst error are very difficult to detect.

Methods Adopted for Controlling the Error

(i) Feedback (ii) Timer

(iii) Sequence number.

(i) Feedback

Feedback is nothing but sending either the positive

acknowledgment when the frame is correct or the

negative acknowledgment when the frame is damaged

to the sender. So, that the sender can retransmit the

damaged frame.

(ii) Timer

If the link is noisy, then there is a chance that the

frame could be lost, neither the sender nor the receiverhave the knowledge about the frame. If the timer

expires and still the sender have not received the

acknowledgment then it will retransmit the frame that

is lost.

(iii) Sequence Number

They are used to discard any duplicate frame at both

sender and receiver side i.e., each frame is received

exactly once.

The method that are used for detecting the error are

the CRC, LRC, Check sum.

Method used for Error Correction, Hamming code.

4. Flow Control

Flow control means controlling the overflow of data

on the receiver side. It is used when the receiver is not having

the capacity to process the incoming data or when it doesnt

have enough space in its memory to store the data. Flow

control mechanism is used to control the speed of the sender,

so that it can transmit the data with a speed which is

acceptable by the receiver. If the sender sends the data with

greater speed then the receiver tend to loose the data i.e., it

discards the frame. This situation make the receiver flooded

with the data, due to which it is not able to process the data.

Receiver is overwhelmed when its processing speed is much

slower than the processing speed of sender. There are two

things that are to be considered while transmitting data from

sender to receiver.1. Capacity of link or communication channel.

2. Capacity of the receiver buffer.

If the receiver have a capacity of n, then the sender

should transmit data with the speed which is less than are

equal to n.

(b) Define: protocol, digital signal.

Answer : May/June-13, (R09), Q1(b) M[8]

Protocol

For answer refer Unit-I, Q10.

Digital SignalFor answer refer Unit-I, Q28, Topic: Digital Signal.

Q2. (a) Explain the working of packet switchingwith an example.


Switching

Switching is a technology that links the intended

machines for data transfer by using the special type of device

called switch. The two different techniques employed by

switching are,

(i) Circuit switching

(ii) Packet switching.

(i) Circuit Switching

In this switching technique, a dedicated line (path) is

established to transmit the data among various devices.

Example: Telephone

Following are few steps of how two devices

communicates with each other by using circuit switching,

(i) Setup a connection

(ii) Transfer of data

(iii) Terminate the connection

(i) Setup a Connection

Before sending any signal or data from one point to

another, a circuit must be established. The figure (1) shows

four switching nodes. Each and every node consist of few

incoming and few outgoing lines. When a telephone user

dials a number, the various switching offices establishes a

connection among themselves. The elapsed time between

the end of dialing a number at one end and start ringing the

bell at the other end is approximately 10 seconds for local

calls, during this time it will establish a copper path i.e.,

Microwave link.


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Switching node

Microwave link

Telephonecaller

Telephonereceiver

Switching node

Microwave link

Telephonecaller

Telephonereceiver

Switching node

Microwave link

Telephonecaller

Telephonereceiver

Figure (1): Circuit Switching Network

(ii) Transfer for Data

Once the connection has been established, the data

can be transferred or transmitted through that connection.

The data can be either in analog or digital form, depending

on the nature of the network used.(iii) Terminating the Connection

Once the transmission of data is completed, then the

connection is dropped or terminated. A typical example of

circuit switching is the telephone conversation between two

telephone devices as shown in the a box figure.

The arrangement of circuit switched network can be

done by using two types of special devices.

(a) Circuit Switch

It is a special type of device where different

machines are arranged in such a way that

depending on the number of inputs given, same

number of outputs are obtained.

Circuitswitch

Input 1

Input 2

Output 1

Output 2

Circuitswitch

Input 1

Input 2

Output 1

Output 2

Figure (2): Circuit Switch

(b) Folded Switch

The arrangement of machines or terminals in

this type of device can be done in such a way

that all the machines are connected to oneanother. Mostly arrangement of telephone can

be done in folded switch.

F.SF.S

Figure (3): Folded Switch

Circuit switching consists of the two main methods,

Space division circuit switching

Time division circuit switching.

(ii) Packet Switching

In this technique messages to be transmitted are

divided into packets. These packets are then transmitted

individually to the destination. This technique can be

employed in two ways,

(a) Datagram packet switching

(b) Virtual circuit packet switching.

(a) Datagram Packet Switching

In this approach packets that has to be transmitted

are independent without maintaining the addresses

of the transmitted packets. The packets can be

transmitted along any rout.Working of Datagram Packet Switching

Suppose as source device-1 wants to send three

datagrams to the destination device-4. When source

transmits the datagrams, they are transmitted

individually. Network assumes that only one packet

is to be transmitted, but actually each datagram will

be transmitted through different routes i.e., there will

be no fixed route for each datagram.

The figure (4) is the example for the above method.

Q

R

P

R

R

P

Q P QR P

QP

Device-2

Device-1

R

R

Q

P

R

P Q

R

R

Q P

QP

Q PR

Receiver

Intermediate

Node

Sender

Q

Q

R

P

R

R

P

Q P QR PQR P

QP

QP

Device-2

Device-1

R

R

Q

P

R

P Q

R

R

Q P

QP

Q PR

Receiver

Intermediate

Node

Sender R

Q

P

R

Q

P

R

P Q

R

R

Q PQ P

QP

QP

Q PR Q PR

Receiver

Intermediate

Node

Sender

Q

Figure (4): Datagram Method

In the figure (4), we can see that each datagram is

transmitted through different nodes with different routes.

(b) Virtual Circuit Packet Switching

In this method the routs along which the packets has

to be transmitted are specified. It employs several

routing methods for specifying the routs such as

fined routing, flooding, adaptive routing and random

routing.


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Q

R

P

R

P

Q PQR P

QP

Q

R

P

R

P

Q PQR PQR P

QP

QP

Figure (5): Example of a Virtual Circuit Method

(b) What is twisted pair cable? What is the difference between category 3 and category 5 twistedpair cables?


Twisted Pair Cable

For answer refer Unit-II, Q24, Topic: Physical Description.

Differences between Category-3 and Category-5 Twisted Pair Cables

Category-3 Twisted Pair Cable Category-5 Twisted Pair Cable

1. It is an unshielded twisted pair cable consisting of 1. It is an unshielded twisted pair cable consisting of

4-pair of wires with a maximum data rate of 10 Mbps. 4-pair of wires with a maximum data rate of 100 Mbps.

2. Number of twists are less when compared to 2. Number of twists are relatively more. That is,

category-5. That is, cables of category-3 can be category-5 cables can be twisted 12 times per foot.

twisted only 3 times per foot.

3. Supports the transmission of upto 10 Mbps. 3. Supports the transmission of upto 100 Mbps.

4. Used effectively at lengths of upto 100 mts. 4. Used effectively at lengths of upto 100 Mbps.

5. Maximum frequency is 16 MHz. 5. Maximum frequency is 100 MHz.

6. Can be used in 10 BASE-T networks. 6. Used in 100 BASE-T and 1000 BASE-T networks.

Q3. (a) Calculate CRC, if message polynomial (M(x)) = x7+ x5+ 1 and generator polynomial (G(x)) =x3+ 1.

Answer : May/June-13, (R09), Q3(a)For answer refer Unit-III, Q16.

(b) Explain supervisory frame format of HDLC.

Answer : May/June-13, (R09), Q3(b)For answer refer Unit-III, Q33, Topic: Supervisory Frame (S-Frame).

(c) Describe the working of stop-and-wait ARQ protocol.

Answer : May/June-13, (R09), Q3(c)

For answer refer Unit-III, Q26, Topic: Stop and Wait ARQ.


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Q4. (a) Explain the need for medium access control protocol. Describe the frame format of ethernet.


Need for Medium Access Control Protocol

Medium access control is one of the two sub-layers of data link layer of OSI model. A MAC protocol is used to

provide an efficient use of the channel capacity by controlling access to the transmission medium. It is even used in

transferring the data packet from one network interface card to the other and vice versa, across a shared channel. MAC

protocols prevents collisions between various stations moving across the same channel.

However, this protocol is specific to the LAN being used like token ring, token bus, etc.

Frame Format of Ethernet

For answer refer Unit-IV, Q21.

(b) List the benefits of a wireless LAN. What is a hidden node in a wireless LAN?

Answer :

May/June-13, (R09), Q4(b) M[8]Benefits of a Wireless LAN

The benefits of wireless LANs are as follows,

1. Fast and easy installations without the use of any physical cables.

2. Convenience in accessing the internet irrespective of the location.

3. Productive for the users who prefer to be connected even while on move i.e., while travelling.

4. Provides scalability in configuring the systems to various topologies so as to meet the change in needs of applications

and installations.

5. Handle and serve various connections at a time using an access point. Access point is nothing but a box used for

setting up a wireless LAN.

6. Low cost when compared to wired hardware. Though the initial investment might be higher than the cost of wired

hardware, but the overall expenses costs much lower.

Hidden Nodes

A set of nodes which are unreachable to other nodes in a wireless network are called hidden nodes. Consider a

physical star topology that has an access point along with the nodes enclosing it in circular fashion. That is, all the nodes

are within the range of access point, but are unable to interact with each other, since they are not connected physically. A

node which is at the far edge of the range of access point, called P, in a wireless network can view the access point. But, it

cannot see the node R which is at the opposite end of the access point range. Such nodes are referred as hidden. When

both of these nodes start sending packets simultaneously to another node Q, the problem occurs. The carrier sensing

multiple access without collisions will not work because PandRare unable to sense the carrier. Therefore, collisions occur

because of which the data gets corrupted at the access point.

This problem can be overcome by handshaking along with CSMA/CA scheme.

Q5. (a) What is a bridge? Explain the working of a spanning tree bridge.


Bridge

A bridge is an electronic device used to connect multiple LANs. The bridges operate in the data link layer.

Suppose, we have a pair of ethernet LANs and we want to interconnect. One approach is to connect them by using

a repeater, but this would not be a workable solution. An alternative way to connect two LANs is by bridge. With the help

of this bridge device we can forward frames from one ethernet LAN to another. Usually, bridges operate in both the physical

and the data link layers of the OSI model. Bridges can divide a large network into smaller segments.


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7

6

5

4

32

1

7

6

5

4

32

1

PhysicalLayer

Bridge

7

6

5

4

32

1

7

6

5

4

32

1

PhysicalLayer

Bridge

Figure: A Bridge in the OSI Model

A bridge operates at the data link layer, giving it access to the physical addresses of all stations connected to it.

When a frame enters a bridge, the bridge not only regenerates the signal but also checks the address of its destination and

forwards the new copy only to the address to which it belongs. As a bridge encounter a packet, it reads the address

contained in the frame and compares that address with a table of the stations on both segments. When it finds a match, it

sends the packet to that segment.

Working of Spanning Tree Bridge

Spanning tree bridges are also called as transparent bridges. This bridge run under the promiscous (immoral) mode.

It accepts every frame which is send on all the LANs to which it is connected.

A C

B1

E

D B2

F G

4

LAN2LAN1

LAN3

Bridge

B

A C

B1

E

D B2

F G

4

LAN2LAN1

LAN3

Bridge

B

Figure (1): A Configuration with 4 LANs and Two Bridges

Figure (1) consists of 3 LANs (8 stations) with 2 bridges.

When the frame arrives, the bridge must take decision whether to forward or discard it and after sometime the bridge

must put the frame on another LAN. This decision is made by seeing the destination address in a hash table of the bridge.

The hash table consists of address of all possible destinations and inform which output does LAN belongs to. The

main algorithm used in transparent bridge is backward learning. By seeing the address they can inform which station is

accessible on which LAN.

The routing procedure for an incoming frame depends on the source LAN and the destination LAN, which are as

follows,

(i) If the source and destination stations are on the same LAN then bridge just discards the frame.

(ii) If the source and destination stations are on the different LANs then the bridge must forward the frame.

(iii) Use flooding algorithm, if the destination LAN is not known.

B1

B2 LAN2

1

LAN1

2

B1

B2 LAN2

1

LAN1

2

Figure (2): Parallel Connection of Bridge

In figure (2) two or more bridges can be connected in parallel between the LANs to increase the reliability.

To construct the spanning tree the following steps are implemented,

(i) One bridge will be selected as the root of the tree. The root can be selected when each of the bridge broadcast its

serial number (given by manufacturer). If the serial number is small then that bridge will become the root.

(ii) Construct a tree of shortest paths from the root to every bridge and LAN. This tree is considered as a spanning tree.

(iii) Now, a unique path is developed from every LAN to the root.


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B1 B2

B3 B4

LAN4 LAN5

LAN3LAN2LAN1

B1 B2

B3 B4

LAN4 LAN5

LAN3LAN2LAN1

Figure (3): Spanning Tree, Interconnected LANs

1 2 3

4 5

B4

B2B1

B3

Bridge that is not a part

of spanning tree

1 2 3

4 5

B4

B2B1

B3

Bridge that is not a part

of spanning tree

Figure (4): Spanning Tree

(i) Figure (3) shows, 5 LANs connected by 4 bridges. By

looking at this figure we can construct the spanning

tree.

(ii) In spanning tree only path exists from every LAN to

every other LAN.

(iii) Spanning tree will not allow closed loops which is

shown by a dotted line in figure (4).(b) Which is better, a low frequency reuse

factor of a high frequency reuse factor?Justify your answer.


Low Frequency Reuse Factor Vs High Frequency Reuse

Factor

In most of the cases neighboring cells are not suppose

to use same set of frequencies to perform communication it

is so because the adjacent cell signal may create interference

for the users who are located near to it. But it is difficult to

implement because of the availability of a limited set offrequencies. Therefore, the frequency reuse principle can

be implemented by using a set of patterns in which

frequencies can be reused.

The reuse factor, here is significantly used in the

capacity of cells. This factor is divided into two parts namely,

low frequency reuse factor and high frequency reuse factor.

The low frequency reuse factor maximizes the capacity of

cells using the interference consideration. These interference

considerations are used for balancing the capacity. Whereas,

high frequency reuse factor lowers the interferences since it

provides much distance among the cells without the use of

other channels. Therefore, it can be said that, a high

frequency reuse factor is much better than the low frequency

reuse factor as the cells that make use of the identical set of

frequencies are separated by more cells.Q6. (a) What is UDP? List its uses.


UDP

For answer refer Unit-VII, Q6 (only 1stparagraph)

Uses of UDP

For answer refer Unit-VII, Q8.

(b) Describe TCP connection establishmentusing 3-way handshake.



Q7. (a) What is ICMP? Explain its use.

Answer : May/June-13, (R09), Q7(a)

ICMP

For answer refer Unit-VI, Q22, Topic: Internet Control

Message Protocol (ICMP).

Use of ICMP

ICMP is used for two purposes,

(i) To send error reporting messages.

(ii) To relay query messages.

Error Reporting Messages

For answer refer Unit-VI, Q22, Topic: Error Reporting

Messages, Q24.Query Messages

For answer refer Unit-VI, Q22, Topic: Query

Messages, Q25.

(b) What are the various classes of IPv4addressing? Give example for each.

Answer : May/June-13, (R09), Q7(b)

For answer refer Unit-VI, Q5.

(c) Explain integrated services QoS model.

Answer : May/June-13, (R09), Q7(c)


Q8. Write short technical notes on,

(a) DNS

(b) Multicast routing protocols

(c) Backbone networks.

Answer : May/June-13, (R09), Q8 M[15]

(a) DNS

For answer refer Unit-VIII, Q1.

(b) Multicast Routing Protocols

For answer refer Unit-VIII, Q20(c).

(c) Backbone Networks

For answer refer Unit-V, Q4, Topic: Backbone Network.

Documents

Computer Networks 4 1 May June 13