vtu 6th sem cse computer networks 2 solved papers of june-2013 june-14 & june-2015

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SOLVED PAPERS OF COMPUTER NETWORKS II (JUNE-2013, JUNE-2014 & JUNE-2015)

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Page 1: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

SOLVED PAPERS

OF

COMPUTER NETWORKS II

(JUNE-2013, JUNE-2014 &

JUNE-2015)

Page 2: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/

Page 3: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

1

1 (a) Differentiate b/w connection oriented and connectionless services. (4 marks)

Ans:

Virtual Circuit Datagram

Circuit setup is required Circuit setup is not required

Each packet contains a short VC number as

address

Each packet contains the full source and

destination address

Route chosen when VC is setup and all

packets follow this route

Each packet is routed independently

In case of router failure, all VC that passed

through the router are terminated

Only crashed packets lost

Congestion control is easy using buffers Difficult congestion control

Complexity in the network layer Complexity in transport layer

1 (b) Explain and derive delays in datagram packet switching. (08 Marks)

Ans: For answer, refer Solved Paper June-2015 Q.No.1a.

1 (c) Define routing algorithm. Explain Bellman-Ford algorithm with e.g. (08 Marks)

Ans:

ROUTING

• Routing means determining feasible paths for packets to follow from each source to each

destination.

BELLMAN-FORD ALGORITHM

• If each neighbor of node-A knows the shortest path to node-Z, then node-A can determine

its shortest path to node-Z by calculating the cost to node-Z through each of its neighbors and

picking the minimum.

• Let Dj = current estimate of the minimum cost from node-j to the destination

Let Cij = link cost from node-i to node-j. (For example C13=C31=2)

The link cost from node-i to itself is defined to be zero (Cii=0).

The link cost between node-i & node-k is infinite if node-i & node-k are

not directly connected.(for example C15=C23=~ )

• If the destination node is node-6, then the minimum cost from node-2 to the destination

node-6 can be calculated in terms of distances through node-1, node-4 or node-5(Fig 7.29):

D2 = min{C21+D1,C24+D4,C25+D5}

= min{3+3,1+3,4+2}

= 4

Algorithm is as follows:

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Page 4: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

2

2 (c) Explain the FIFO and priority queue scheduling for managing traffic at packet

level. (08 Marks)

Ans:

FIFO QUEUEING

• Packets are transmitted in order of their arrival (Figure 7.42a).

• Packets are discarded when they arrive at a full buffer.

• Packet-delay depends on the packet inter-arrival time.

Packet-loss depends on the packet-lengths.

• Disadvantages:

1) Not possible to provide different information flows with different QoS.

2) Hogging occurs when a user sends packets at a high rate and fills the buffers in the

system, thus depriving other users of access to the buffer.

FIFO QUEUEING WITH DISCARD POLICY

• Provide different characteristics of packet-loss performance to different classes of traffic.

• Higher priority packets(Class-1) are discarded when they arrive at a full buffer(Fig 7.42b).

• Lower priority packets(Class-2) are discarded when buffer reaches a certain threshold.

• Disadvantage:

Lower priority packets will experience a higher packet-loss probability.

PRIORITY QUEUEING

• Number of priority classes are defined (Figure 7.43).

• A separate buffer is maintained for each priority class.

• Each time the transmission link becomes available, the next packet for transmission is

selected from the head of the line(HOL) of the highest priority queue.

• The size of the buffers for the different priority classes can be selected to meet different loss

probability requirements.

• Disadvantages:

1) Does not discriminate among users of the same priority.

2) Does not allow for providing some degree of guaranteed access to transmission

bandwidth to the lower priority classes.

3) Fairness problem arises when a certain user hogs the bandwidth by sending an

excessive number of packets.

For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/

Page 5: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

3

2 (b) Suppose that ATM-cells arrive at a leaky bucket policer at time t = 2, 3, 6, 9,

11, 16, 23, 24, 25, 26 and 30. Assume I=4 and L=6. Plot the bucket content and

identify any non-conforming cells. (08 Marks)

Ans:

2 (c) Write a note on traffic management at the flow aggregate level. (04 Marks)

Ans:

TRAFFIC MANAGEMENT AT THE FLOW-AGGREGATED LEVEL

• Traffic management at the flow-aggregated level is called traffic engineering (Figure 7.63).

• Main objective: To map aggregated flows onto the network so that resources are efficiently

utilized.

• The shortest-path routing allows traffic to be forwarded to a destination following the

shortest path.

• Problem: Mapping the traffic according to shortest paths may not result in overall network

efficiency.

Solution: Constraint shortest-path routing is suitable for connection-oriented packet switching

networks.

• Suppose that the traffic demand of bandwidth B between a given source and destination pair

is to be routed.

• First, the algorithm prunes any link in the network that has available bandwidth less than B.

Then, the algorithm runs the shortest path routing to find the paths between the given source

and destination pair.

• Suppose that we wish to set up 3 paths in the following order:

i) node 1 to node 8 (path 1),

ii) node 2 to node 8 (path 2), and

iii) node 3 to node 8 (path 3).

• Initially, path 1 follows the shortest path 1->4->8. Link (1,4) and link (4,8) are then pruned.

Next path 2 follows the shortest path 2->4->5->8 using the pruned network topology.

Now, links (2,4),(4,5) and (5,8) are also pruned.

Path 3 uses the revised pruned topology, which gives 3->6->7->8 as shortest path.

Figure 7.63: Mapping traffic onto the network topology

For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/

Page 6: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

4

3 (a) Explain the format of IPV4 format header. (08 Marks)

Ans:

IP PACKET

1) Version

• This indicates version number used by the packet. (Figure 8.4).

Current version=4.

Version 5 is used for a real-time stream protocol called ST2.

Version 6 is used for IPv6.

2) Internet-Header-Length (IHL)

• This specifies length of header.

Without options field, IHL=5.

3) Type of Service

• This specifies priority of packet based on delay, throughput, reliability & cost requirements.

3 bits are assigned for priority levels.

4 bits are assigned for specific requirement (i.e. delay, throughput, reliability & cost).

4) Total length

• This specifies number of bytes in the packet (including header and data).

Maximum length=65535 bytes.

5) Identification, flags and fragment offset

• These are used for fragmentation and reassembly.

6) Time-to-live(TTL)

• This indicates amount of time(in seconds), the packet is allowed to remain in the network.

• If TTL becomes 0 before packet reaches destination, router discards packet and sends an

error message back to the source.

7) Protocol

• This specifies upper-layer protocol that is to receive the packet at the destination-host.

• Examples of protocols: TCP(protocol=6) UDP(protocol=17).

8) Header checksum

• This is used to verify integrity of header only. If the verification process fails, packet is

discarded.

9) Source IP address and destination IP address

• These contain the addresses of source and destination hosts.

10) Options

• This is of variable length. This allows the packet to request special features such as security

level, route to be taken by packet and timestamp at each router.

11) Padding

• This is used to make the header a multiple of 32-bit words.

For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/

Page 7: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

5

3 (b) With a neat diagram, explain UDP datagram. (08 Marks)

Ans:

USER DATAGRAM PROTOCOL (UDP)

• This is an unreliable, connectionless transport layer protocol.

• This provides only two additional services beyond IP:

i) Demultiplexing and

ii) Error checking

• This can optionally check the integrity of the entire datagram.

• Applications that do not require zero packet loss such as in packet voice systems are well

suited to UDP.

• Applications that use UDP include: DNS, SNMP, RTP & TFTP.

UDP FIELDS

• The destination port allows the UDP module to demultiplex datagrams to the correct

application in a given host.

• The source port identifies the particular application in the source host to receive replies.

• The length field indicates the number of bytes in datagram (including header and Data).

• Checksum field detects errors in the datagram and its use is optional.

• Checksum computation procedure is similar to that in computing IP checksum with 2

exceptions:

1) If the length of the datagram is not a multiple of 16 bits, the datagram will be

padded out with 0s to make it a multiple of 16bits.

2) UDP adds a pseudoheader to the beginning of the datagram when performing the

checksum computation.

• The pseudoheader is also created by the source and destination hosts only during the

checksum computation and is not transmitted.

For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/

Page 8: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

6

3 (c) Write a note on Internet control message Protocol (ICMP). (04 Marks)

Ans:

ICMP(ERROR AND CONTROL MESSAGES)

• This is used to handle error and other control messages (Figure 8.12).

ICMP Fields

1) Type

• This identifies the type of message.

2) Code

• This describes the purpose of the message.

• For example, Type 3 = problem reaching the destinations Possible values for code

field are

0 = network unreachable

1 = host unreachable

2 = protocol unreachable

3 = port unreachable

4 = fragmentation needed and DF set

Type 11 = time exceeded problem. Possible values for code field are

0 = TTL value has been exceeded.

1 = fragment reassembly time has been exceeded.

3) Checksum

• This is used to detect errors in the ICMP message.

4) IP header plus original datagram

• This can be used for diagnostic purposes by matching the information in the ICMP

message with the original data in the IP packet.

Figure 8.12: ICMP basic error message format

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Page 9: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

7

4 (a) With a neat diagram, explain the format of the TCP segment. (08 Marks)

Ans:

TCP SEGMENT

Field Function

Source Indicates the port number at the source node. A port number is the

address on a host when an application makes itself available to

incoming or outgoing data.

Destination Port Indicates the port number at the destination node.

Sequence number Identifies the data segment’s position in the stream of data segments

already sent.

Acknowledgement

number (ACK)

Confirms receipt of the data via a return message to the sender.

TCP header length Indicates the length of the TCP header.

Reserved A field reserved for later use

Flags A collection of six 1-bit fields that signal special conditions through

flags. The following flags are available for the sender’s use:

URG-If set to 1, the Urgent pointer field contains information for the

receiver.

ACK-If set to 1, the Acknowledgment field contains information for the

receiver. (If set to 0, the receiver will ignore the Acknowledgment Field.

PSH-If set to 1, it indicates that data should be sent to an application

without buffering.

RST- If set to 1, the sender is requesting that the connection be reset.

SYN-If set to 1, the sender is requesting a synchronization of the

sequence numbers between the two nodes. This code is used when TCP

requests a connection to set the initial sequence number.

FIN-IF set to 1, the segment is the last in a sequence and the

connection should be closed.

Sliding-window size(or

window)

Indicates how many bytes the sender can issue to a receiver while

acknowledgment for this segment is outstanding. This field performs

flow control, preventing the receiver from being deluged with bytes.

Checksum Allows the receiving node to determine whether the TCP segment

became corrupted during transmission.

Urgent pointer Indicates a location in the data field where urgent data resides.

Options Specifies special options, such as the maximum segment size a network

can handle.

Padding Contains filler information to ensure that the size of the TCP header is a

multiple of 32 bits.

Data Contains data originally sent by the source node.

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Page 10: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

8

4 (b) Explain the Border Gateway Protocol (BGP). (08 Marks)

Ans:

BGP (Border Gateway Protocol)

• The default standard for inter-AS routing.

• BGP: glues the whole Internet together.

• Path Vector protocol uses TCP (service port: 179).

• Similar to Distance Vector protocol. Each Border Gateway broadcast to neighbors (peers)

entire path (i.e., sequence of AS’s) to destination.

• This is an exterior gateway protocol (EGP), which means that it performs routing between

multiple autonomous systems or domains and exchanges routing and reachability information

with other BGP systems.

• The information exchanged among BGP routers allow a router to construct a graph of AS’s

and each AS is identified by Autonomous System Number (ASN).

• This routes to networks (ASs), not individual hosts.

• The messages exchanged using TCP and BGP messages are:

-> OPEN: opens TCP connection to peer and authenticates sender.

-> UPDATE: advertises new path (or withdraws old).

-> KEEPALIVE keeps connection alive in absence of UPDATES; also ACKs.

-> OPEN request.

-> NOTIFICATION: reports errors in previous msg; also used to close connection.

• Session is established between two BGP speakers to exchange information regarding:

-> New active routes

-> Old inactive routers or

-> To report error conditions

• The below figure illustrates core routers using BGP to route traffic between autonomous

systems:

• All BGP message types use the basic packet header. Open, update, and notification

messages have additional fields, but keep-alive messages use only the basic packet header.

BGP Packet Fields

1) Marker: contains an authentication value that the message receiver can predict.

2) Length: indicates the total length of the message in bytes.

3) Type: specifies the message type as one of the following:

i) Open

ii) Update

iii) Notification and

iv) Keep-alive

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Page 11: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

9

4 (c) Write a note on Network Address Translation (NAT). (04 Marks)

Ans:

NAT (Network Address Translator)

• This is the translation of an IP address used within one network to a different IP address

known within another network.

• One network is designated the inside network and the other is the outside.

• Typically, a company maps its local inside network addresses to one or more global outside

IP addresses and unmaps the global IP addresses on incoming packets back into local IP

addresses.

• This helps ensure security since each outgoing or incoming request must go through a

translation process that also offers the opportunity to qualify or authenticate the request.

• This also conserves on the number of global IP addresses that a company needs and it lets

the company use a single IP address in its communication with the world.

• This is included as part of a router and is often part of a corporate firewall.

• Network administrators create a NAT table that does the global-to-local and local-to-global

IP address mapping.

• This can be statically defined or it can be set up to dynamically translate from and to a pool

of IP addresses.

• This lets an administrator to create tables that map:

-> A local IP address to one global IP address statically

-> A global IP address to any of a pool of local IP addresses on a round-robin basis.

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Page 12: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

10

5 (a) Explain the remote login protocols. (08 Marks)

Ans:

REMOTE LOGIN PROTOCOLS

• Using client/server model, a user can establish a session on the remote-machine and then

run its applications. This application is known as remote login.

• For example, an employee working at home can log into his work-server to access

application programs for doing a project.

• Two remote login protocols: 1) TELNET & 2) SSH.

1) TELNET

• This is a TCP/IP standard for establishing a connection to a remote-machine.

• This allows a user to log into a remote-machine across the Internet.

• This makes a TCP connection and then passes the detail of the application from the user to

the remote-machine.

Logging to Remote Servers

• TELNET has the following properties:

1) Client-programs are built to use the standard client/server interfaces without

knowing the details of server-programs.

2) A client and a server can negotiate data format options.

3) Once a connection is established, both ends of the connection are treated

symmetrically.

• When a user logs into a remote-server, the client's terminal-driver accepts the

keystrokes and interprets them as characters by its operating-system.

• Characters are typically transformed to a universal character set called NVT (network

virtual terminal).

• The client then establishes a TCP connection to the server.

• Texts in the NVT formal are transmitted using a TCP session and are delivered to the

operating-system of the remote-server.

• The server converts the characters back from NVT to the local client machine's

format.

2) SSH (SECURE SHELL) PROTOCOL

• This is based on UNIX programs.

• This uses TCP for communications.

• This is more powerful and flexible than TELNET.

• This allows the user to more easily execute a single command on a remote client.

• This has the following advantages over TELNET

1) Security: SSH provides a secure communication by encrypting and authenticating

messages. Security is implemented by using public-key encryption.

2) Multiplexing: SSH provides several additional data transfers over the same

connection by multiplexing multiple channels.

SSH Packet Fields

1) Length: This indicates the size of the packet (Figure 9.7).

2) Padding: This causes an intrusion to be more difficult.

3) Type: This identifies the type of message.

4) CRC: This is an error detection field.

Figure 9.7. SSH packet format

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Page 13: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

11

5 (b) Explain the RSA algorithm with an example. (08 Marks)

Ans:

RSA

• Assume that a plaintext m must be encrypted to a ciphertext c. This has three phases

i) Key Generation Algorithm:

1) Choose two prime numbers p and q.

2) Compute n=p*q.

3) Compute φ(n) = (p - 1) * (q - 1)

4) Choose e such that 1 < e < φ(n) and e and n are coprime.

5) Compute a value for d such that (d * e) % φ(n) = 1.

6) The public key = {e, n}

7) The private key = {d, n}

ii) Encryption:

1) Both sender and receiver must know the value of n.

2) The sender knows the value of e and only the receiver knows the value of d.

3) Ciphertext c is constructed by c=memod n

iii) Decryption:

1) Given the ciphertext c, the plaintext m is extracted by m=cd mod n.

• Example for RSA:

1) Choose p = 3 and q = 11

2) Compute n = p * q = 3 * 11 = 33

3) Compute φ(n) = (p - 1) * (q - 1) = 2 * 10 = 20

4) Choose e such that 1 < e < φ(n) and e and n are coprime. Let e = 7

5) Compute a value for d such that (d * e) % φ(n) = 1.

One solution is d = 3 [(3 * 7) % 20 = 1]

6) Public key is (e, n) => (7, 33)

7) Private key is (d, n) => (3, 33)

The encryption of m = 2 is c = 27 % 33 = 29

The decryption of c = 29 is m = 293 % 33 = 2

5 (c) Write a note on firewalls. (04 Marks)

Ans:

FIREWALL

• This is placed between hosts of a certain network and the outside world (Figure 10.8).

• This is used to protect the network from unwanted web sites and potential hackers.

• Main objective: To monitor and filter packets coming from unknown sources.

• This can be a software program or a hardware device.

1) Software firewalls can be installed in home computers by using an Internet

connection with gateways.

2) Hardware firewalls → are more secure than software firewalls

→ are not expensive.

• A firewall controls the flow of traffic by one of the following 3 methods:

1) Packet filtering: A firewall filters those packets that pass through.

If packets can get through the filter, they reach their destinations:

otherwise, they are discarded

2) A firewall filters packets based on the source IP address.

3) Denial of Service(DOS). This method controls the number of packets entering a

network.

Figure 10.8. A simple configuration of a secured network using

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Page 14: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

12

6 (a) With a neat diagram, explain the integrated services QoS. (08 Marks)

Ans:

INTEGRATED SERVICES

• This provides QoS to individual applications and flow records.

• This consists of 2 service classes:

1) Guaranteed service class: is used for applications that cannot tolerate a delay

beyond a particular value.

• This can be used for real-time applications such as video communications.

2) Controlled-load service class: is used for applications that can tolerate some

delay and loss.

• This is designed such that applications run very well when the network is not heavily

loaded or congested.

• Four common categories of processes providing QOS:

1) Traffic-shaping: regulates turbulent traffic (Figure 12.1).

2) Admission-control: governs whether the network can admit or reject the flow

based on given information about an application's flow,.

3) Resource allocation: allows network-users to reserve bandwidth on neighboring

routers.

4) Packet-scheduling: sets the timetable for the transmission of packet flows.

• Any involving router needs to queue & transmit packets for each flow appropriately.

• This approach has been deterred owning to scalability issues.

• As the network size increases, routers need to handle larger routing tables and switch larger

numbers of bits per second. In such situations, routers need to refresh information

periodically.

Figure 12.1. Overview of QoS methods in integrated services

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Page 15: VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

13

6 (b) Explain MPLS operation & packet format. (08 Marks)

Ans:

MPLS (MULTI PROTOCOL LABEL SWITCHING)

• MPLS transmission is a special case of tunneling.

• Features:

1) Connection-oriented forwarding mechanism

2) Has layer 2 label-based lookups

3) Enables traffic engineering to implement peer-to-peer VPNs effectively

4) Supports other applications, such as IP multicast routing and QoS extension.

• This uses a small label appended to packets and typically makes efficient routing decisions.

MPLS OPERATION

• MPLS network consists of nodes called label-switch-routers (LSR).

• An LSR switches labeled packets according to particular switching tables (Figure 16.5).

• An LSR has 2 functional components: i) Control component & ii) Forwarding component.

1) The control component: uses routing protocols such as OSPF and BGP.

• The control component also facilitates the exchange of information with other LSRs

to build and maintain the forwarding table.

• A label is a header used by an LSR to forward packets.

• When a packet arrives, the forwarding component uses the label of the packet as an

index to search the forwarding table for a match.

2) The forwarding component: then directs the packets from the input interface to

the output interface through the switching fabric.

• Key to scalability of MPLS: Labels have only local significance between two devices that

communicate.

MPLS Packet Fields

1) Label value: This is significant only locally (Figure 16.6).

2) Exp: This is reserved for future experimental use.

3) S is set to 1 for the oldest entry in the stack and to 0 for all other entries.

4) TTL: This is used to encode a hop-count value to prevent packets from looping

forever in the network.

Figure 16.5. An MPLS network

Figure 16.6. MPLS header encapsulation for an IP packet

6 (c) Write a note on virtual private networks. (04 Marks)

Ans: For answer, refer Solved Paper June-2014 Q.No.6b.

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COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013

14

7 (a) List and explain the compression methods without loss. (08 Marks)

Ans:

RUN-LENGTH ENCODING

• This technique is fairly effective for compression of plaintext and numbers , especially for

facsimile systems.

• Repeated letters can be replaced by a run length, beginning with Cc to express the

compression letter count.

• The longer the text, the smaller the compression ratio becomes.

Example

Assume a system that represents b as a blank. Find the compressed version of the

following sentence :

THISSSSSS b IS bbbb AN b EXAMPLE b OF b RUNLENGTH b CODE

Solution

According to the conventions stated, the compressed version of that sentence turns

into:

THIS Cc 6 Sb IS Cc b 4 b ANbEXAMPLE b OF b RUN Cc -5LENGTH b CODE

HUFFMAN ENCODING

• This is an efficient frequency-dependent coding technique.

• Source values with smaller probabilities appear to be encoded by a longer word.

• This technique reduces the total number of bits, leading to an efficient compression of data.

Begin Huffman Encoding Algorithm

1) Sort outputs of the source in decreasing order of their probabilities. For example, 0.7, 0.6,

0.6, 0.59, ..., 0.02, 0.01.

2) Merge the two least probabilistic outputs into a single output whose probability is the sum

of corresponding probability, such as 0.02 + 0.01 = 0.03.

3) If the number of remaining outputs is 2, go to the next step; otherwise, go to step 1.

4) Assign 0 and 1 as codes on the diagram.

5) If a new output is the result of merging two outputs, append the code word with 0 and 1;

otherwise, stop.

Example

Design a Huffman encoder for a source generating { a1 , a2 , a3 , a4 , a5 } and with

probabilities {0.05, 0.05, 0.08, 0.30, 0.52}.

Solution

Following the algorithm, the output of the information source shown in Figure 17.11,

the information related to { a1 , a2 , a3 , a4 , a5 } is compressed to 1100, 1101, 111,

10, 0, respectively.

Figure 17.11. Huffman encoding

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LEMPEL-ZIV ENCODING

• Lempel-Ziv codes are independent of the source statistics.

• This coding technique is normally used for UNIX compressed files.

Begin Lempel-Ziv Encoding Algorithm

1) Any sequence of source output is passed in a phrase of varying length. At the first step,

identify phrases of the smallest length that have not appeared so far. Note that all phrases are

different, and lengths of words grow as the encoding process proceeds.

2) Phrases are encoded using code words of equal length. If k1 = number of bits are needed to

describe the code word and k2 = the number of phrases, we must have

k1 = log2|k2 |2

3) A code is followed by the last bit of parser output to double-check the last bit.

Example

For the following string of bits, find the encoded Lempel-Ziv words:

11110111011000001010010001111010101100

Solution

Implementing step 1 on the string, there are 14 phrases, as follows:

1 – 11 - 10 - 111- 0 - 110 - 00 - 001- 01- 0010 - 0011- 1101- 010 – 1100

Thus, k2 = 14 and k1 = log2 |14|2 = 4.

7 (b) With a neat diagram, explain the session initiation protocol (SIP). (08 Marks)

Ans:

SIP (SESSION INITIATION PROTOCOL)

• This is a VoIP signaling protocol.

• This can

-> perform both unicast and multicast sessions

-> support user mobility

-> handles signals & identifies user location, call setup & call termination

SIP Components

• A call is initiated from a user agent: the user's IP telephone system (which is similar

to a conventional phone).

• A user agent assists in initiating or terminating a phone call in VoIP networks.

Figure 18.3. Overview of SIP

SIP has following 5 servers

1) DNS server: The Domain Name System (DNS) server maps the domain name to an

IP address in the user information database (UID).

• The UID database contains such user information as preferences and the services to

which it has subscribed.

2) Proxy server: The proxy server forwards requests from a user agent to a different

location and handles authorizations by checking whether the caller is authorized to

make a particular call.

3) Location server: This server is responsible for UID management.

• The location server interacts with the database during call setup.

4) Redirect server: This server performs call forwarding and provides alternative

paths for the user agent.

5) Registrar server: This server is responsible for registering users in the system and

updating the UID that the location server consults.

• Requests for registration must be authenticated before the registration is granted.

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7 (c) Write a note on real-time media transport protocols. (04 Marks)

Ans:

REAL TIME MEDIA TRANSPORT PROTOCOLS

• Real-Time Transport Protocol (RTP) & Real-Time Control Protocol (RTCP) are the most widely

used protocols for real-time transmission.

RTP

• This provides some basic functionalities to real-time applications and includes some specific

functions to each application.

• This runs on top of the transport protocol as UDP.

• This provides application-level framing by adding application-layer headers to datagrams.

• The application breaks the data into smaller units, called application data units (ADUs).

• Lower layers in the protocol stack, such as the transport layer, preserve the structure of the

ADU.

RTCP

• This also runs on top of UDP.

• This performs several functions, using multicasting to provide feedback about the data

quality to all session members.

• The session multicast members can thus get an estimate of the performance of other

members in the current active session.

• Senders can send reports about data rates and the quality of data transmission.

• Receivers can send information about packet-loss rate, jitter variations, and any other

problems they might encounter.

8 (a) Briefly explain the classification of routing protocol. (06 Marks)

Ans:

CLASSIFICATION OF ROUTING PROTOCOLS

1. Centralized versus distributed

• In centralized routing protocols, the routing decision is made at a central node.

• In distributed routing protocols, the routing decision is made by all the network

nodes.

2. Static versus adaptive

• In static routing protocols, a route of a source/destination pair does not change

because of any traffic condition or link failure.

• In adaptive routing protocols, routes may change because of any congestion.

3. Table driven (or proactive) vs. Source initiated (or reactive)

Table-driven Routing Protocols

• These protocol finds routes to all possible destinations ahead of time.

• The routes are recorded in the nodes’ routing tables and are updated within the

predefined intervals.

• Proactive routing protocols are faster in decision making, but cause problems if the

topology of the network continually changes.

• These protocols require every node to maintain one or more tables to store updated

routing information from every node to all other nodes.

Source-initiated Routing Protocols

• These protocols are on-demand procedures and create routes only when requested to

do so by source nodes.

• A route request initiates a route-discover process in the network and is completed

once a route is discovered.

• If it exists, at the time of request, a route is maintained by a route-maintenance

procedure until either the destination node becomes irrelevant to the source or the

route is no longer needed.

• Control overhead of packets is smaller than of proactive protocols.

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8 (b) Explain the DEEP clustering algorithm. (06 Marks)

Ans:

DEEP (Decentralized Energy Efficient Propagation)

• The protocol that establishes clusters with uniformly distributed cluster heads.

• This protocol balances the load among all the cluster heads by keeping the clusters' radii

fairly equal.

• This protocol is completely decentralized, and there is no need for any location-finder device

or hardware.

• The protocol starts with initial cluster head and forms new cluster-head candidates gradually

by controlling the relative distance between a pair of cluster heads and the circular radius of

each cluster.

Clustering Algorithm

1) Initialize: Initial cluster head finds cluster members by sending "cluster-head

declaration."

2) Initial cluster head finds new cluster-head candidates by sending "cluster-head

exploration signal."

3) Repeat: Cluster-head candidates that are placed on the (dr1,dr2) ring find cluster

members.

4) Nodes that receive more than one cluster-head declaration choose the closest

cluster head, based on the received signal energy.

5) Cluster-head candidates that receive a cluster-head declaration signal negotiate with

the sender, and one of them gets eliminated.

6) Confirmed cluster heads send "cluster-head exploration" signals to find new cluster-

head candidates (Go to step 4).

7) Finalize: If the number of members in a cluster is less than mn , all the members

find new clusters by sending the membership-search signal.

8) At the end, a node that has not received any control signal sends the membership-

search signal.

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8 (c) Explain the intracluster and intercluster routing protocols. (08 Marks)

Ans:

INTRA CLUSTER ROUTING

• If the routing is happening within a cluster, then the protocol is called as intra cluster

routing.

• In intra cluster routing, the packets are transmitter with in a cluster.

• It can be of 2 types.

1) Direct Routing

• The cluster head as the destination for all cluster nodes is located in the center of the

cluster, so all nodes can communicate with the cluster head directly.

2) Multihop Routing

• A node can face multiple hops in order to reach the destination.

INTER CLUSTER ROUTING

• If the routing is happening between the nodes of different clusters it is called as inter cluster

routing.

• These protocols are not typically different from the multihop ones for intradomain cases.

• Interdomain protocols are available for

1) Intercluster energy conscious routing (ICR)

2) Energy-aware routing (EAR)

3) Direct diffusion

ICR (Intercluster Energy-Conscious Routing)

• It is a destination initiated reactive routing algorithm.

• The destination is called as local base station [LBS] it will start the route discovery by

creating interest signal and following them.

• ICR works in two phases, Route discovery and data acquisition.

1. Route Discovery Phase: In this phase, the LBS initiates route discovery by

sending an interest signal within the range Ri,

1. All the nodes which are in the range Ri will receive the interest signal.

2. Upon receiving the interest signal, it will be stored and flooding

continues.

3. If an intermediate node receive already processed interest signal, it

will be discarded

4. Before flooding the interest signal, the cost value will be updated.

2. Data-acquisition phase: occurs after each cluster head collects the

requested information from sensor nodes and compresses it into a packet with

fixed length, searches for the neighbor's address in memory, and relays the

packet to that neighbor.

Figure 20.11. LBS starts route discovery by generating interest signals.

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1 (a) Differentiate b/w connection oriented and connectionless services. (5 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.1a.

1 (b) Compare datagram packet switching vs. virtual packet switching. (6 Marks)

Ans:

Circuit Switching Packet Switching

1. Call set up is required 1. Call setup is not required

2. Dedicated connection b/w two Hosts 2. No dedicated connection b/w 2 Hosts

3. Connection/Communication is lost, if any

link in the path between the Hosts is broken

3. Connection/Communication could continue

between the Hosts since data have many

routes between the Hosts

4. Information take the same route

between the connected Hosts

4. Information could take different routes to

reach the destination Host

5. Information always arrives in order 5. Information could arrive out of order to the

destination

6. Bandwidth available is fixed 6. Bandwidth available is variable

7. Congestion is call based 7. Congestion is packet based

8. Bandwidth utilization is partial 8. Bandwidth utilization is full

9. It does not use store-and forward

transmission

9. It uses store-and forward transmission

10. It is Transparent 10. Not transparent

11. Charging is time based 11. Charging is packet based

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1 (c) Explain the Dijkstra's routing algorithm, with an example. (9 Marks)

Ans:

DIJKSTRA'S ALGORITHM

• This is used to find the shortest paths from a source node to all other nodes in a network

• Main idea: Progressively identify the closest nodes from the source node in order of

increasing path cost (Figure 7.32).

Algorithm is as follows:

2 (a) Explain the FIFO and priority queue scheduling for managing traffic at packet

level. (8 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.2a.

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2 (b) Define congestion control with graph. Explain the leaky bucket algorithm for

policing the traffic at flow level. (12 Marks)

Ans:

CONGESTION CONTROL

• The process of managing traffic-flow in order to control congestion is called congestion-

control.

LEAKY BUCKET ALGORITHM

• The process of monitoring & enforcing the traffic-flow is called the policing.

• When traffic-flow violates agreed-upon contract, the network may choose to tag (or discard)

the nonconforming traffic.

• Tagging essentially lowers priority of nonconforming traffic.

• When network resources are exhausted, tagged traffic is the first to be discarded.

• Policing-device can be implemented based on the concept of a leaky bucket.

• Imagine the traffic-flow to a policing-device as water being poured into a bucket that has a

hole at the bottom.

• Bucket leaks at a constant rate (Figure 7.53).

• When bucket is full, a new portion of water is said to be nonconforming and the water can be

discarded.

When water is poured into bucket & overflow does not occur, a new portion of water (i.e.

packet) is said to be conforming.

• The hole ensures that bucket will never overflow as long as drain-rate is higher than rate

water is being poured in.

• The inverse of I is called the sustainable rate, which is the long-term average rate allowed

for the conforming traffic.

• Suppose the peak rate of a given traffic flow is denoted by R and its inverse is T, that is,

T=1/R. Then, the maximum burst size is given by

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3 (a) Explain:

i) IP address classification

ii) Subnet addressing. (10 Marks)

Ans: (i)

IP ADDRESS CLASSIFICATION

• This is a numeric identifier assigned to each machine on an IP network.

• This consists of network ID(NID) and host ID(HID).

• HID identifies the network-connection to the host rather than the actual host.

• NID identifies the network to which the host is connected. All the hosts connected to the

same network, have the same NID.

• HID is assigned by the network administrator at the local site.

NID for an organization may be assigned by the ISP(Internet Service Provider).

• Class D addresses are used for multicast services (Figure 8.5).

• Multicast means a host sends message to a group of hosts simultaneously.

• IP addresses are usually written in dotted-decimal notation. The address is broken into four

bytes.

For example, an IP address of

10000000 10000111 01000100 00000101

is written as

128.135.68.5

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Ans: (ii)

SUBNET ADDRESSING

• To allow a single network address to span multiple physical networks is called subnet

addressing

• The main feature of the subnet addressing scheme is that it is oblivious to the network

outside the organization (Figure 8.6).

• Inside the organization the local network administrator is free to choose any combination of

lengths for the subnet and host ID fields.

• For the subnet address scheme to work, every machine on the network must know which

part of the host address will be used as the subnet address. This is accomplished by assigning

each machine a subnet mask.

• In the subnet mask,

i) 1's represent the positions that refer to the network (or subnet address)

ii) 0's represent the positions that refer to the host part of the address.

Question: If a packet with a destination IP address of 150.100.12.176 arrives at site from the

outside network, which subnet should a router forward this packet to? Assume subnet mask is

255.255.255.128(Fig 8.7).

Solution: The router can determine the subnet number by performing a binary AND between

the subnet mask and the IP address.

IP address: 10010110 01100100 00001100 10110000 (150.100.12.176)

Subnet mask: 11111111 11111111 11111111 10000000 (255.255.255.128)

Subnet number: 10010110 01100100 00001100 10000000 (150.100.12.128)

This number(150.100.12.128) is used to forward the packet to the correct subnet work inside

the organization.

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3 (b) Give the format of IPV6 basic header. Compare IPV6 with IPV4. (10 Marks)

Ans:

IPV6 BASIC HEADER

CHANGES FROM IPV4 TO IPV6

1) Longer address fields

• Length of address field is extended from 32 bits to 128 bits.

• The address space can support up to 3.4X1038 hosts.

2) Simplified header format

• Some of the header fields in IPv4 such as identification, flags and fragment offset do

not appear in the IPv6 header.

3) Flexible support for options

• The options in IPv6 appear in optional extension headers.

• The options are encoded in a more efficient and flexible fashion than they are in IPv4.

4) Flow label capability

• IPv6 adds a "flow label" to identify a certain packet "flow" that requires a certain QoS.

5) Security

• IPv6 supports build-in authentication and confidentiality.

6) Large packets

• IPv6 supports payloads that are longer than 65 Kbytes called jumbo payloads.

7) Fragmentation at source only

• Routers do not perform packet fragmentation.

• If a packet needs to be fragmented, the source should check the minimum MTU along

the path and perform the necessary fragmentation.

8) No checksum fields

• The checksum field has been removed to reduce packet processing time in a router.

• Packets carried by the physical network (such as Ethernet, token shop) are typically

already checked.

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4 (a) Explain OSPF protocol and its operation. (10 marks)

Ans:

OSPF (OPEN SHORTEST PATH FIRST)

• This is a Link State protocol based on cost rather than hops.

• This is an Interior Gateway Protocol (IGP) Protocol, uses flooding of link state information

and Dijkstra’s least-cost path algorithm.

• Router constructs a complete topological map of the entire autonomous system.

• The router then locally runs the Dijkstra’s shortest-path algorithm to determine shortest-

path tree to all networks with itself as root node.

• The router’s routing table is then obtained from this shortest-path tree.

• At steady state: All routers have same LS database, Know how many routers in network,

interfaces & links between routers, Cost of each link.

• Occasional Hello messages (10 sec) & LS updates sent (30 min).

The operation of OSPF can be explained in 3 steps:

Step1: Discovery of neighbors can be done by sending hello packets in point-to-point links

and designated routers in multi-access networks (Figure 8.35).

• To discover, establish and maintain relationships, the OSPF transmit hello packet to

each interface periodically, typically for every 10 to 30 sec.

• When a router receives a hello packet, it replies with a hello packet containing router

ID of each neighbor it has seen.

• When a router receives a hello packet containing its router ID in one of the neighbor

fields, the router is assured that communication to sender is bidirectional.

Step 2: Establishment of adjacent and synchronization of link-state databases

• Once the connection is established between two neighbor routers, the database

description packet is used to synchronize their link-state databases

• One router acts as master and other as slave (Figure 8.36).

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Step 3: Propagation of OSPF Link State Request and building routing tables

• When a router wants to update the link-state database, it sends a LS request packet

to neighbor to update part of its link-state database (Figure 8.38).

• Each LSA request is specified by the link state type, link state ID, and the advertising

router.

4 (b) Write a note on:

i) IGMP protocol

ii) Mobile IP. (10 marks)

Ans: (i)

IGMP (INTERNET GROUP MANAGEMENT PROTOCOL)

• This runs between hosts and their immediately-neighboring multicast routers.

• The mechanisms of the protocol allow a host to inform its local router that it wishes to

receive transmissions addressed to a specific multicast group.

• Also, routers periodically query the LAN to determine if any group members are still active.

• If there is more than one IP multicast router on the LAN, one of the routers is elected

"querier" and assumes the responsibility of querying the LAN for the presence of any group

members.

• Based on the group membership information learned from the IGMP, a router is able to

determine which (if any) multicast traffic needs to be forwarded to each of its "leaf"

subnetworks.

• "Leaf" subnetworks are those that have no further downstream routers; they either contain

receivers for some set of groups, or they do not.

• Multicast routers use the information derived from IGMP, along with a multicast routing

protocol, to support IP multicasting across the MBone.

IGMP Message Fields

1) Version: This field identifies the version number.

2) Type: This field identifies the message type. There are two types:

1) Type 1 indicates a query message sent by router and

2) Type 2 indicates a report sent by host.

3) Unused: This field must be set to zero.

4) Checksum: This field contains checksum for all eight bytes of IGMP message.

5) Group Address: This address is the class D IPV4 address.

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Ans: (ii)

MOBILE IP

• This allows location-independent routing of IP datagrams on the Internet.

• Each mobile node is identified by its home address disregarding its current location in the

Internet.

• While away from its home network, a mobile node is associated with a care-of address which

identifies its current location and its home address is associated with the local endpoint of a

tunnel to its home agent.

• Mobile IP allows portable devices called mobile hosts (MHs) to roam from one area to

another.

Mobile IP operates as follows

• When a correspondent host (CH) wants to send a packet to MH, the CH transmits the

standard IP packet with its address as the source IP address and MH’s address as

destination IP address.

• This packet will be intercepted by the mobile host’s router called home agent (HA).

The HA keeps track of the current location of the MH. The HA manages all MHs in its

home network that use the same address prefix.

• If the MH is located in the home network, the HA simply forwards the packet to its

home network.

• When an MH moves to a foreign network, the MH obtains a care of address from the

foreign agent (FA) and registers the new address with its HA. The care-of-address

reflects the MH’s current location and is typically the address of FA.

• Once the HA knows the care-of-address of the MH, the HA can forward the

registration packet to the MH via the FA.

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5 (a) Write a note on only Two:

i) Remote login protocols

ii) File transfer and FTP

iii) World wide web and HTTP. (8 Marks)

Ans: (i) For answer, refer Solved Paper June-2013 Q.No.5a.

Ans: (ii)

FILE TRANSFER PROTOCOLS

FTP (File Transfer Protocol)

• FTP is used to transfer files from one host to another host over the internet.

• TELNET provides broader access to a user, whereas FTP allows access only to certain files.

Begin File Transfer Protocol

1) A user requests a connection to a remote-server.

2) The user waits for an acknowledgement.

3) Once connected, the user must enter a user ID, followed by a password.

4) The connection is established over a TCP session.

5) The desired file is transferred.

6) The user closes the FTP connection.

Ans: (iii)

WORLD WIDE WEB (WWW) AND HTTP (HYPER TEXT TRANSFER PROTOCOL)

• Web is a global network of servers.

• The servers are linked by a common protocol such as http, ftp & so on.

• When a client-host requests an object, a web-server responds by sending the requested-

object.

• A browser is a user-agent which displays the requested web-page.

• HTTP transfers web-page at the application layer.

• HTTP uses TCP rather than UDP, since reliable delivery of web-pages with text is important.

• The TCP connection-establishment delay is one of the main contributing delay factors

associated with downloading web-documents.

• Web-page consists of HTML (Hypertext Markup Language) files that can be addressed by a

single URL (Uniform Resource Locator).

• A URL is a global address of an HTML document.

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5 (b) Define network management & explain SNMP and SNMP messages. (6 Marks)

Ans:

NETWORK MANAGEMENT

• The purpose of network management is → to monitor, test and analyze the h/w, s/w and human elements of a network and

→ then to configure & control these elements to meet the operational performance

requirements of the network.

SNMP (SIMPLE NETWORK MANAGEMENT PROTOCOL)

• This runs on top of UDP and uses client/server configuration.

• PDUs(Protocol Data Unit) are carried in the payload of a UDP datagram, and so its delivery

to a destination is not guaranteed.

• Managed-devices (such as routers and hosts) are objects, and each object has a formal

ASN.1 definition.

• The task of SNMP is to transport MIB information among managing-centers and agents

executing on behalf of managing-centers.

• For each managed MIB object, an SNMP request is used to retrieve (or change) it’s

associated value.

• If an unsolicited message is received by an agent (or when an interface/device goes down),

the protocol can also inform the managing-center.

• SNMPv2 has seven PDU's( or messages) as follows.

1) GetRequest: This is used to obtain the value of a MIB object.

2) GetNextRequest: This is used to obtain the next value of a MIB object.

3) GetBulkRequest: This is used to get multiple values, equivalent to multiple

GetRequests but without using multiple overheads.

4) InformRequest: This is a manager-to-manager message that two communicating

management centers are remote to each other.

5) SetRequest : This is used to set the value of a MIB object.

6) Response: is a reply message to a request-type PDU.

7) Trap: This notifies a managing-center that an unexpected event has occurred.

• Get or Set PDU Fields

1) PDU type: This indicates one of the seven PDU types.

2) Request ID: This is used to verify the response of a request.

3) Error status: This indicates types of errors reported by an agent.

4) Error index: This indicates to a n/w administrator which name has caused an error.

• Trap PDU Fields

1) Enterprise: This is for use in multiple networks

2) Timestamp: This is used for measuring up time.

3) Agent address:This indicates address of managed agent is included in PDU header.

Figure 9.12. SNMP PDU format

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5 (c) Compare secret key and public key cryptography systems. (6 Marks)

Ans:

SECRET KEY ENCRYPTION PROTOCOLS

• This is also called as symmetric encryption or single-key encryption.

• Sender and receiver conventionally use the same key for an encryption process.

• This consist of

→ an encryption-algorithm

→ a key and

→ a decryption-algorithm

• Two popular protocols are: 1) DES (Data Encryption Standard)

2) AES (Advanced Encryption Standard)

• A shared secret-key between a transmitter and a receiver is assigned at the transmitter and

receiver points.

• At the receiving end, the encrypted information can be transformed back to the original data

by using

→ decryption algorithm and

→ secret key

PUBLIC KEY ENCRYPTION PROTOCOLS

• This is also called as asymmetric or two key encryption.

• A sender/receiver pair use different keys.

• This is based on mathematical functions rather than on substitution or permutation.

• Two popular protocols are: 1)RSA protocol 2)Diffie-Hillman key-exchange protocol.

• Either of the two related keys can be used for encryption; the other one for decryption.

• Each system publishes its encryption key by placing it in a public-register & sorts out key as

public one. The companion key is kept private.

• If A wishes to send a message to B, A encrypts the message by using B's public key.

• At receiving end, B decrypts the message by using its private key.

• No other recipients can decrypt the message, since only B knows its private key.

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6 (a) Explain the differentiated services QoS with a neat diagram. (8 Marks)

Ans:

DIFFERENTIATED SERVICES QOS

• This provides QoS support to a broad class of applications.

• This provides a simpler and more scalable QoS.

• This minimizes the amount of storage needed in a router by processing traffic flows in an

aggregate manner.

• The traffic-classifier routes packets to specific outputs, based on the values found inside

multiple fields of a packet header (Figure 12.14).

• The traffic-conditioner detects and responds if any packet has violated any of the rules

specified in the TCA(Traffic-Conditioning Agreement).

• The traffic-conditioner has 4 major components:

1) Meter

• This measures traffic to make sure that packets do not exceed their traffic profiles.

2) Marker

• This marks or unmarks packets in order to keep track of their situations in DS node.

3) Shaper

• This delays any packet that is not complaint with the traffic profile.

4) Dropper

• This discards any packet that violates its traffic profile.

• A bandwidth-broker is needed to allocate & control the available bandwidth within the DS

domain,

• In order to process traffic flows in an aggregate manner, a packet must go through

SLA(Service Level Agreement) that includes a TCA

• An SLA indicates the type of forwarding service, and

A TCA presents all the detailed parameters that a customer receives.

• An SLA can be either static or dynamic

1) Static SLA is a long-term agreement.

2) Dynamic SLA uses the bandwidth-broker that allows users to make changes more

frequently.

Figure 12.14. Overview of DiffServ operation

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6 (b) Explain VPN and its types based on tunnelling. (8 Marks)

Ans:

VPN (VIRTUAL PRIVATE NETWORKS)

• This is a networking infrastructure whereby a private-network makes use of the public-

network.

• VPN part of public network is set up "virtually" by a private-sector entity to provide public

networking services to small entities.

• This maintains privacy by using tunneling protocols and security procedures.

• Benefits to an organization by using VPN

1) Extended geographical communication

2) Reduced operational cost

3) Enhanced organizational management

4) Enhanced network management with simplified local area networks

5) Improved productivity and globalization.

• There are 2 types of VPNs: 1) remote access and 2) site-to-site (Figure 16.2).

Figure 16.2. Three types of VPNs to and from a headquarter organization

1) REMOTE ACCESS VPN

• This is a user-to-LAN connection.

• An organization uses VPN to connect its users to a private network from various remote

locations.

• This allows encrypted connections between an organization's private network and remote-

users through a third-party service provider.

• Tunneling uses mainly the Point-to-point protocol(PPP).

• PPP is the carrier for other Internet protocols when communicating over the network

between a host computer and a remote point.

• Besides IPsec, other types of protocols associated with PPP are

i) L2F (Layer 2 forwarding) protocol uses authentication scheme supported by PPP.

ii) PPTP (Point to Point Tunneling Protocol) supports 40-bit and 128-bit encryption and

uses the authentication scheme supported by PPP.

iii) L2TP (Layer 2 tunneling protocol) combines features of both PPTP & L2F.

2) SITE TO SITE VPN

• An organization uses VPN to connect multiple fixed sites over a public-network.

• VPNs can be classified as either intranets or extranets.

i) Intranet VPNs: connect an organization's remote site LANs into a single private

network.

ii) Extranet VPNs: allow two organizations to work in a shared environment through a

tunnel built to connect their LANs.

• The main benefit of using a VPN is scalability with a reasonable cost.

• GRE (Generic Routing Encapsulation) is normally the encapsulating protocol. It provides the

framework for the encapsulation over an IP-based protocol.

• L2TP can also be used. This fully supports IPSec regulations and can be used as a tunneling

protocol for remote access VPNs.

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6 (c) Explain the need for overlay networks. (4 Marks)

Ans:

OVERLAY NETWORK

• This is an application specific computer network built on top of another network (Fig 16.11).

• This creates a virtual topology on top of the physical topology of the public network.

• This type of network is created to protect the existing network structure from new protocols

whose testing phases requires Internet use.

• These have no control over how packets are routed in the underlying network between a pair

of overlay source/destination nodes.

• However, these can control a sequence of overlay nodes through a message passing function

before reaching destination.

• These are self-organized. When a node fails, the overlay-network algorithm should provide

solutions that let the network recover and recreate an appropriate network structure.

• These permit routing messages to destinations when the IP address is not known in

advance.

Figure 16.11. An overlay network for connections between two LANs associated with routers

R1 and R4

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7 (a) Briefly explain the MPEG standards and frame types of compression. (6 Marks)

Ans:

MPEG STANDARDS

1) MPEG-1: primarily for video on CD-ROM

2) MPEG-2: for multimedia entertainment and high-definition television (HDTV) and

the satellite broadcasting industry

3) MPEG-4: for object-oriented video compression and videoconferencing over low-

bandwidth channels

4) MPEG-7: for a broad range of demands requiring large bandwidths providing

multimedia tools

5) MPEG-21: for interaction among the various MPEG groups

Frame Types of Compression

1) Interimage (I) frames: An I frame is treated as a JPEG still image and

compressed using DCT.

2) Predictive (P) frames: These frames are produced by computing differences

between a current and a previous I or P frame.

3) Bidirectional (B) frames: A B frame is similar to a P frame, but the P frame

considers differences between a previous, current, and future frames.

• In any frame sequence, I frames appear periodically as the base of the scene.

Normally, there is a P frame between each two groups of B frames (Figure 17.9).

Figure 17.9. Snapshot of moving frames for MPEG compression

7 (b) Explain the Huffman encoding, with an example. (6 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.7a.

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7 (c) With a neat diagram, explain the H.23 components and list the steps in

signaling. (8 Marks)

Ans:

H.323 COMPONENTS

1) Multimedia Terminal: A multimedia terminal is designed to support video and data

traffic and to provide support for IP telephony.

2) DNS Server: As in SIP, a DNS server maps a domain name to an IP address.

3) Gateway: The gateway is a router that serves as an interface between the IP

telephone system and the traditional telephone network.

4) Gatekeeper: The gatekeeper is the control center that performs all the location and

signaling functions. The gatekeeper monitors and coordinates the activities of the

gateway. The gateway also performs some signaling functions.

5) Multicast or Multipoint Control Unit (MCU): This unit provides some multipoint

services, such as conference calls.

Figure 18.5. Overview of H.323 protocol connection

STEPS IN SIGNALING

1) Call setup: When user1 dials user 2's telephone number, the first set of signals are

exchanged between these two users in conjunction with opening a TCP connection.

2) Initial communication capability: All the end points' communication capabilities

available over TCP are exchanged.

3) Audio/video communication establishment: Step 3 implements the

establishment of a logical channel, which in H.323 is unidirectional; therefore, a logical

channel must be established in either direction in order to have two-way

communications.

4) Communications: Step 4 comprises the communications between the two users.

This phase is handled using RTP over UDP.

5) Call termination: The call is terminated by either user.

8 (a) Explain the wireless routing protocol for AD-HoC networks. (5 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.8a.

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8 (b) Briefly explain the direct and multi-hop routing of intra-cluster routing

protocol, with the help or relevant diagrams. (6 Marks)

Ans:

DIRECT ROUTING

• The cluster head as the destination for all cluster nodes is located in the center of the

cluster, so all nodes can communicate with the cluster head directly (as shown in Figure 20.8).

• Note that in this figure, 2 nodes cannot reach the destination, as they are located far from it.

• The number shown in each node indicates the level of energy the corresponding node has.

• The number associated with each node indicates a normalized value of the remaining energy

in that node.

Figure 20.8. Direct routing in a cluster.

MULTIHOP ROUTING

• The destination is reached through multiple hops. (Figure 20.9).

• If there are many paths, then only the path which is energy efficient will be considered.

• The sensor node will be at different distances apart from other nodes.

• A packet from a node is routed to a neighbor node that exhibits high energy.

• The number in the node indicates the remaining energy in the node

• The algorithm aims to choose the appropriate next neighbor for each node, using a central

command node.

• Typically, a central command node collects the information about direct paths' costs and

geographical positions of the nodes and finds the best path.

Figure 20.9. Multihop routing in a cluster in which the number associated with each node

indicates a normalized value of the remaining energy in that node

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8 (c) Write short notes on:

i) Clustering in sensor networks

ii) Security vulnerabilities of AD-HoC networks. (9 Marks)

Ans: (i)

CLUSTERING IN SENSOR NETWORKS

• The region being sensed is normally partitioned into equally loaded clusters of sensor nodes

(as shown in Figure 20.1).

• A cluster in a sensor network resembles a domain in a computer network.

• In other words, nodes are inserted in the vicinity of a certain predefined region, forming a

cluster.

• Different types of sensors can also be deployed in a region.

• The most effective routing scheme in sensor networks is normally based on the energy

(battery level) of nodes.

• In such routing schemes, the best path has the highest amount of total energy.

• The network of such sensing nodes is constructed with identical sensor nodes, regardless of

the size of the network.

• In Figure 20.1, each cluster contains a cluster head responsible for routing data from its

corresponding cluster to a base station.

Figure 20.1. A sensor network and its clusters

• Communicating nodes are normally linked by a wireless medium, such as radio.

• The wireless sensor node is equipped with a limited power source, such as a battery or even

a solar cell, where there is enough sunlight exposure on the node.

• In some application scenarios, sensor-node lifetime depends on the battery lifetime.

• Removal of dead nodes can cause significant topological changes and may require packet

rerouting.

Ans: (ii)

SECURITY IN ADHOC NETWORKS

• Because of dynamic topological changes, ad-hoc networks are vulnerable at the physical

link, as they can easily be manipulated.

• An intruder can easily attack ad-hoc networks by loading available network resources, such

as wireless links and energy (battery) levels of other users, and then disturb all users.

• Attackers can also disturb the normal operation of routing protocols by modifying packets.

• The intruder may insert spurious information into routing packets, causing erroneous routing

table updates and thus misrouting.

• Some other security vulnerabilities of ad-hoc networks follow.

1) Limited computational capabilities: the nodes in the mobile adhoc network are

modular, independent and will have limited computational capability. It becomes a

source of vulnerability when they handle public key cryptography.

2) Limited power supply: since nodes have limited power supply attacker can

exhaust batteries by giving excessive computations to be carried out.

3) Challenging key management: the key management becomes extremely difficult

as the mobile devices will be under movement.

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1 (a) Explain datagram and virtual circuit packet switching with delay calculation

diagrams. (08 Marks)

Ans:

DATAGRAM PACKET SWITCHING

• Let transmission delay of message = p seconds.

• Let transmission time of message = T seconds.

• Let the message is broken into 3 separate packets.

Let the 3 packets follow the same path and are transmitted in succession.

Let each packet requires P=T/3 seconds to transmit.

• As shown in figure 7.16, the first packet arrives at the switch after p+P seconds.

The first packet is received at the second switch at time 2p+2P.

The first packet is received at the third switch at time 3p+3P.

The final packet will arrive at the destination at time 3p+3P+2P=3p+5P=3p+t+2p.

• In general, if the path followed by a sequence of packets consists of L hops with identical

propagation delays and transmission speeds, then the delay incurred by a message that

consists of k packets is given by

Lp+LP+(k-1)P

VIRTUAL CIRCUIT PACKET SWITCHING

• A modified form of virtual-circuit packet switching, called cut-through packet stitching, can

be used when retransmissions are not used in the underlying data link control (Figure 7.22).

• The minimum delay in transmitting the message is approximately equal to the sum of the

propagation delays in the various hops plus the one-message transmission time.

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1 (b) With neat diagram, explain the generic packet switch. (04 Marks)

Ans:

PACKET SWITCH

• A packet-switch performs 2 main functions:

1) Routing function uses algorithms to find a path to each destination and store the

result in a routing table.

2) Forwarding function

-> processes each incoming packet from an input port and

-> forwards the packet to the appropriate output port (based on the information

stored in the routing table).

• A line card contains several input/output ports (Figure 7.10).

The line card is concerned with symbol timing, line coding, framing, physical layer

addressing & error checking.

The line card also contains some buffers and the associated scheduling algorithms.

• Programmable network processor performs packet-related tasks such as table lookup and

packet scheduling.

• A controller contains a general-purpose processor which is used for control & management

functions depending on the type of packet switching.

The controller acts as a central coordinator, as it communicates with each line card and

the interconnection fabric.

• Interconnection fabric is used to transfer packets between the line cards.

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1 (c) Consider the network, in the following Fig. Q1 (c):

i) Use the Dijkstra's algorithm to find the set of shortest path from node 4 to other

nodes.

ii) Find the set of associated routing table entries. (12 Marks)

Ans: (i)

Solution:

Ans: (ii)

Solution:

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2 (a) Discuss the (i) Priority queuing & (ii) Weighted fair queuing. (06 Marks)

Ans: (i) For answer, refer Solved Paper June-2013 Q.No.2a.

Ans: (ii)

WEIGHTED FAIR QUEUEING

• Each user flow has its own buffer, but each user flow also has a weight that determines its

relative share of the bandwidth (Figure 7.49).

• If buffer 1 has weight 1 and buffer 2 has weight 3, then buffer 1 will receive 1/4 of the

bandwidth and buffer 2 will receive 3/4 of the bandwidth.

• This is also easily approximated in ATM. In each round, each non-empty buffer would

transmit a number of packets proportional to its weight.

• Packet by packet weighted fair queueing is also easily generalized from fair queueing.

• Weighted fair-queueing systems are means for providing QoS guarantees.

2 (b) Explain the concept of Random Early Detection. (04 Marks)

Ans:

RED (RANDOM EARLY DETECTION)

• This is a buffer management technique that attempts to provide equitable access to a FIFO

system by randomly dropping arriving packets before the buffer overflows (Figure 7.48).

• A dropped packet provides feedback information to the source and informs the source to

reduce its transmission rate.

• This algorithm uses an average queue length rather than instantaneous queue length to

decide how to drop packets.

• Specifically, two thresholds are defined: minth and maxth.

• When the average queue length is below minth, RED does not drop any arriving packets.

• When the average queue length exceeds maxth, RED drops any arriving packets.

• When the average queue length is between minth and maxth, RED drops an arriving packet

with an increasing probability as the average queue length increases.

Figure 7.48: Packet drop profile in RED

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2 (c) Give the classification congestion control algorithms. Explain the leaky bucket

and token bucket traffic shaper with neat diagram. (10 Marks)

Ans:

• Congestion control algorithms can be classified into 2 types:

1) Open-Loop Control

• This prevents congestion from occurring by making sure that the flow generated by

source will not degrade network-performance to a level below the specified QoS.

• If QoS cannot be guaranteed, network rejects flow before it enters the network.

• The function that makes the decision to accept or reject a new traffic-flow is called an

admission-control.

2) Closed-Loop Control

• This reacts to congestion when it is already happening or is about to happen.

• This regulates traffic-flow according to state of network.

• This does not use any reservation.

LEAKY-BUCKET TRAFFIC-SHAPER

• Packets are served periodically so that the stream of packets at the output is smooth.

• Buffer is used to store momentary bursts of packets.

• If buffer is full, incoming packets are discarded.

• A policing-device checks and passes each packet on the fly (Figure 7.59).

• A traffic shaping device needs to introduce certain delays for packets that arrive

earlier than their scheduled departures.

• Drawback: The leaky-bucket traffic-shaper is very restricted, since the output rate is

constant. Many applications produce variable rate traffic. If the traffic flows from such

applications have to go through the traffic-shaper, the delay through the buffer can be

unnecessarily long.

TOKEN BUCKET TRAFFIC SHAPER

• This regulates only the packets that are not conforming (Figure 7.60).

• Packets that are deemed conforming are passed through without further delay.

• Tokens are generated periodically at a constant rate.

• Tokens are stored in a bucket.

• If the bucket is full, arriving tokens are discarded.

• A packet from the buffer can be taken out only if a token in the bucket can be drawn.

• If bucket is empty, arriving packets have to wait in the buffer.

• The backlogged packets have to wait for new tokens to be generated before they can

be transmitted out.

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3 (a) Describe the various fields of IP version 4 header. (06 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.3a.

3 (b) i) A small organization has a class C address for seven networks each with 24

hosts. What is an appropriate subnet mask?

ii) Perform CIDR aggregation on the /24 IP address 200.96.86.0/24,

200.96.87.0/24, 200.96.88.0/24, 200.96.89.0/24. (06 Marks)

Ans: (i)

Solution:

• A Class C address requires 21 bits for its network ID, leaving 8 bits for the host ID and

subnet ID to share.

• One possible scheme would assign 4 bits to the host and 4 to the subnet ID, as shown

below. The number of bits assigned to the host can be increased to 5 as well.

Ans: (ii)

Solution:

200.96.86.0/20 = 11001000.01100000.01010110.00000000

200.96.87.0/20 = 11001000.01100000.01010111.00000000

200.96.88.0/20 = 11001000.01100000.01011000.00000000

200.96.89.0/20 = 11001000.01100000.01011001.00000000

Mask = 11111111.11111111.11110000.00000000

The resulting prefix is 200.96.80.0/20.

3 (c) (i) Why transition from IPV4.0 to IPV6.0 is required? (ii) Explain the IPV6

network addressing. (08 Marks)

Ans: (i) For answer, refer Solved Paper June-2014 Q.No.3b.

Ans: (ii)

NETWORK ADDRESSING

• IPv6 addresses are divided into three categories:

1) Unicast addresses: These identify a single network interface.

2) Multicast addresses: These identify a group of network interfaces, typically at

different location. A packet will be sent to all network interfaces in the group.

3) Anycast addresses: These also identify a group of network interfaces.

However, a packet will be sent to only one network interface in the group, usually

nearest one.

SPECIAL PURPOSE ADDRESSES

• The address 0::0 is called the unspecified address and is never used as a destination

address.

• The address ::1 is used for a loopback.

• IPv4 compatible addresses are needed during the transition period where an IPv6

packet needs to be "tunneled" across an IPv4 network.

• IP mapped addresses are used to indicate IPv4 hosts and routers that do not support

IPv6.

• Provides based unicast addresses are identified by the prefix 010. These addresses

will be mainly used by the Internet service providers to assign addresses to their

subscribers. (Figure 8.11).

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4 (a) Explain TCP 3 way handshake for establishing a TCP connection. (06 Marks)

Ans:

TCP CONNECTION ESTABLISHMENT

• To establish a connection, TCP uses a three-way handshake.

• Before a client attempts to connect with a server, the server must first bind to a port to open

it up for connections: this is called a passive open.

• Once the passive open is established, a client may initiate an active open.

• To establish a connection, the three-way (or 3-step) handshake occurs:

1) The active open is performed by the client sending a SYN to the server.

2) In response, the server replies with a SYN-ACK.

3) Finally the client sends an ACK back to the server.

• At this point, both the client and server have received an acknowledgment of the connection.

• Each SYN message during connection establishment can specify options such as maximum

segment size (MSS), window scaling and time stamps.

• The three way handshake procedure ensures that both host’s agree on their initial sequence

numbers.

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4 (b) What are the classification of internet routing protocols? Explain in detail

routing information protocol (RIP) (08 Marks)

Ans:

CLASSIFICATION OF INTERNET ROUTING PROTOCOLS

1) IGP (Interior Gateway Protocols)

• Used for routing within an AS. It is also referred to as intra-AS routing.

• Companies, organizations, and even service providers use an IGP on their internal networks.

• IGPs include RIP, EIGRP, OSPF, and IS-IS.

2) EGP (Exterior Gateway Protocols)

• Used for routing between autonomous systems.

• It is also referred to as inter-AS routing.

• Service providers and large companies may interconnect using an EGP.

• The Border Gateway Protocol (BGP) is the only currently viable EGP and is the official routing

protocol used by the Internet.

RIP (ROUTING INFORMATION PROTOCOL)

• This is a dynamic routing protocol used in LAN and WAN.

• It is classified as an interior gateway protocol (IGP) using the distance-vector routing

algorithm.

• A RIP run on top of UDP, port number 520 is used.

• RIP is a distance-vector routing protocol, which employs the hop count as a routing metric.

• Suitable for small networks (local area environments).

• The maximum number of hops allowed with RIP is 15, and the hold down time is 180

seconds.

• Value of 16 is reserved to represent infinity, i.e. node is not reachable.

• Small number limits the count-to-infinity problem.

• Originally each RIP router transmits full updates every 30 seconds by default.

• A router expects to receive an update message from each of its neighbors within 180

seconds in the worst case.

• If router does not receive update message from neighbor X within this limit, it assumes the

link to X has failed and sets the corresponding minimum cost to 16 (infinity).

RIP Message Fields

1) Command: specifies the purpose of the message, two values are defined value 1

requests the system to send its routing information and values 2 indicates a response

containing the routing information.

2) Version: two versions, RIPV1 and RIPV2

3) Address Family Identifier: identifies type of address used currently only IP

address is defined

4) IP address: indicates the address of destination, which can be network or host

address.

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4 (c) Write note on: i) Mobile IP ii) Reverse path multicasting. (06 Marks)

Ans: (i) For answer, refer Solved Paper June-2014 Q.No.4b.

Ans: (ii)

RPM (REVERSE PATH MULTICASTING)

• Reverse Path Multicasting (RPM) is an enhancement to Reverse Path Broadcasting and

Truncated Reverse Path Broadcasting.

• RPM creates a delivery tree that spans only

1) Subnetworks with group members, and

2) Routers and subnetworks along the shortest path to those subnetworks.

• RPM allows the source-based "shortest-path" tree to be "pruned" so that datagrams are only

forwarded along branches that lead to active members of the destination group.

• When a multicast router receives a packet for a (source, group) pair, the first packet is

forwarded following the TRPB algorithm across all routers in the internetwork.

• The TRPB algorithm guarantees that each leaf router will receive at least the first multicast

packet.

• If there is a group member on one of its leaf subnetworks, a leaf router forwards the packet

based on this group membership information.

• If none of the subnetworks connected to the leaf router contain group members, the leaf

router may transmit a "prune" message on its parent link.

• An upstream router receiving a prune message is required to store the prune information in

memory.

• If the upstream router has no recipients on local leaf subnetworks and has received prune

messages from each downstream neighbor on each of the child interfaces for this (source,

group) pair, then the upstream router does not need to receive any more packets for this

(source, group) pair.

5 (a) Explain the remote login protocols. (06 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.5a.

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5 (b) What are the elements of network management? Discuss the interaction

between SNMP management station and SNMP agent. (08 Marks)

Ans:

ELEMENTS OF NETWORK MANAGEMENT

1) Managed devices: A managed device is a node that has an SNMP agent and

resides on a managed network. These devices can be routers, switches and bridges or

hubs.

2) Agents: An agent is a software module residing within a device. The agent

translates information into a compatible format with SNMP.

3) Network management systems (NMSs): An NMS runs monitoring applications.

They provide the bulk of processing and memory resources required for network

management.

Figure 5.1: SNMP managed configuration

• The manager provides the interface b/w the network manager and the management

system.

• The agent provides the interface b/w the manager and the physical device(s).

• The manager and agent use an MIB (Management Information Base) and a relatively small

set of commands to exchange information.

• The MIB is organized in a tree structure with individual variables, such as point status or

description, being represented as leaves on the branches.

• A long numeric tag or object identifier (OID) is used to distinguish each variable uniquely in

the MIB and in SNMP messages.

• SNMP uses 5 basic messages to communicate between the manager and the agent (GET

GET-NEXT, GET-RESPONSE, SET, and TRAP)

1) The GET and GET-NEXT messages allow the manager to request information for a

specific variable.

2) The agent will then respond with a GET-RESPONSE message to the manager with

either

-> the information requested or

-> an error indication as to why the request cannot be processed.

3) A SET message allows the manager to request a change be made to the value of a

specific variable in the case of an alarm remote that will operate a relay.

4) The TRAP message allows the agent to spontaneously inform the manager of an

"important" event.

5 (c) Write RSA algorithm. For RSA algorithm of 4 bit message 1001, choose p = 3

and q =11, find the public keys and private keys for this and show the cipher text.

(06 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.5b.

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6 (a) Describe the various types of resource allocation schemes. (06 Marks)

Ans:

Router based Host based

Routers have primary responsibility for

congestion control.

End hosts have primary responsibility for

congestion control.

A router selectively forwards packets and

drops them, if required, to manage the

allocation of existing resources.

The end host observes the network

conditions and adjusts their behavior (such

as throughput, delay and packet losses) and

adjust rate at which they generate and send

packets.

A router also sends information to end host

on the amount of traffic it can generate &

send.

Fixed based Adaptive based

The end host request the resource at the

router level before a flow is established.

The end host sends data without first

reserving any resources at the router.

Each router then allocates enough resource

to satisfy the request.

The end host then adjusts their sending rate

according to feedback they receive.

Window based Rate based

A receiver chooses a window size. This

window size is dependent on the buffer

space available to the receiver.

A receiver specifies the maximum rate of bits

per second it can handle.

The receiver then sends the window size to

the sender.

A sender sends traffic in compliance with the

rate advertised by the receiver.

6 (b) Define VPN. Discuss the concept of tunneling & PPP. (06 Marks)

Ans:

VPN (VIRTUAL PRIVATE NETWORKS)

• This is a networking infrastructure whereby a private-network makes use of the public-

network.

TUNNELING AND PPP (POINT-TO-POINT PROTOCOL)

• A tunnel is a connection that forms a virtual-network on top of a physical-network(Fig 16.3).

• Tunneling is a process of encapsulating packets and sending them over the public-network.

• A tunnel is a relatively inexpensive connection, since it uses the Internet as its primary form

of communication.

• Tunneling requires two other types of protocols:

1) Carrier protocols are used for carrying information over the public network..

2) Encapsulating protocols are used for wrapping, encapsulating and securing data.

• VPNs can put a packet that uses a non-routable IP address inside a packet to extend a

private-network over the Internet.

PPP

• This is a Tunneling protocol.

• This allows an organization to establish secure connection from one point to another while

using public resources.

• A PPP connection is a serial connection between a user and an ISP.

Figure 16.3. A customized protocol packet tunneling through the Internet

6 (c) What is MPLS network. Explain MPLS operation. (08 Marks)

Ans: For answer, refer Solved Paper June-2014 Q.No.1a.

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7 (a) Explain the jpeg compression for still images. (06 Marks)

Ans:

STILL IMAGES AND JPEG COMPRESSION

• The Joint Photographic Experts Group (JPEG) is the compression standard for still images.

• It is used for gray-scale and quality- color images.

• JPEG is a lossy process i.e. an image obtained after the decompression at a receiving end

may not be the same as the original.

• A typical JPEG process consists of three processes: discrete cosine transform (DCT),

quantization, and compression, or encoding (Figure 17.4).

Figure 17.4. A typical JPEG process for production and compression of still images

• The DCT process is complex and converts a snapshot of a real image into a matrix of

corresponding values.

•The quantization phase converts the values generated by DCT to simple numbers in order to

occupy less bandwidth.

• The compression process makes the quantized values as compact as possible.

• The compression phase is normally lossless and uses standard compression techniques.

7 (b) Explain: i) Huffman encoding ii) Lempel-Ziv-Wetch encoding (08 Marks)

Ans: For answer, refer Solved Paper June-2013 Q.No.7a.

7 (c) Briefly explain with neat diagram, how Constant Distribution Network (CDN)

interaction with DNS(Domain Name System). (06 Marks)

Ans:

Figure 18.12. Video streaming provided to a user in ISP 3 domain using DNS servers

• CDNs use Domain Name System (DNS) servers to direct browsers to the correct server.

• When a user requests the movie.mpg video object, its browser sends its request for the base

object to the origin server, www.filmmaker.com.

• Then the browser finds the object at

www.cdnco.com/www.filmmaker.com/movies/movie.mpg.

• Now, the browser requests the corresponding DNS server for the location of the CDN

company.

• The DNS server is set up to transfer all movie queries about www.contentco.com to the

corresponding DNS server.

• Finally, the corresponding DNS server extracts the IP address of the requesting browser and

returns the IP address of the closest CDN server to the browser, located in the ISP 3 domain in

the previous example.

• This way, the user communicates with the CDN server located nearby with minimal

congestion.

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8 (a) Explain DSDV, TORA routing protocols for mobile adhoc networks. (08 Marks)

Ans:

DSDV (DESTINATION SEQUENCED DISTANCE VECTOR)

• This is table driven based routing algorithm.

• This is improved version of Bellman Ford routing algorithm.

• Each DSDV node maintains 2 routing tables:

1) Table for forwarding packets, and

2) Table for advertising incremental updates.

• The nodes will maintain a routing table that consists of a sequence number.

• The routing table periodically exchanged so that every node will have latest information.

• This is suitable for small networks.

The algorithm works as follows:

• A node or a mobile device will make an update in its routing table and send the information

to its neighbor upon receiving the updated information and make an update in its own routing

table.

• The update is made by comparing the sequence number received is greater than present

sequence number than the new one will be used.

• If there is a link failure in one of the node will change the metric value to infinity and

broadcast the message.

TORA (TEMPORARY ORDERED ROUTING ALGORITHM)

• It is also a source initiated routing algorithm, creates multiple routes for any source/

destination pair.

• The advantage of multiple routes is that route discovery is not required for every alteration

in the network topology.

• This consists of 3 phases:

1) Route Creation/discovery

2) Route maintenance

3) Route erasure

• This uses 3 types of packets:

1) Query Packets for route creation,

2) Update Packets for both creation and maintenance

• The route will be discovered from the source to destination only when a request is made for

the transmission.

The algorithm works as follows:

• The source will generate a query packet which will be broadcasted in the network. This

continues as long as a node that is directly connected to the destination is identified.

• When the destination is identified an update packet will be generated and sent back to the

source.

• The update packet will have the path information if there are more than one update packet

received by the source, it means there are multiple paths to the destination, the source has to

choose best path available.

8 (b) (i) What are the classification of sensor networks?

(ii) Explain with relevant diagram DEEP clustering in sensor network. (06 Marks)

Ans: i)

• Sensor networks can be used for target tracking, environmental monitoring, system control,

and chemical or biological detection.

• In military applications, sensor networks can enable soldiers to see around corners and to

detect chemical and biological weapons long before they get close enough to cause harm.

• Civilian uses include environmental monitoring, traffic control, and providing health care

monitoring for the elderly while allowing them more freedom to move about.

Ans: ii) For answer, refer Solved Paper June-2013 Q.No.8b.

8 (c) Describe the direct or multihop intracluster routing protocols with neat

diagram. (06 Marks)

Ans: For answer, refer Solved Paper June-2014 Q.No.8b.

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