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For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/
VTU
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COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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1 (a) Differentiate b/w connection oriented and connectionless services. (4...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
2
2 (c) Explain the FIFO and priority queue scheduling for managing traffic a...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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2 (b) Suppose that ATM-cells arrive at a leaky bucket policer at time t = 2...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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3 (a) Explain the format of IPV4 format header. (08 Marks)
Ans:
IP PACKET
1...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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3 (b) With a neat diagram, explain UDP datagram. (08 Marks)
Ans:
USER DATAG...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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3 (c) Write a note on Internet control message Protocol (ICMP). (04 Marks)
...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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4 (a) With a neat diagram, explain the format of the TCP segment. (08 Marks...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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4 (b) Explain the Border Gateway Protocol (BGP). (08 Marks)
Ans:
BGP (Borde...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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4 (c) Write a note on Network Address Translation (NAT). (04 Marks)
Ans:
NA...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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5 (a) Explain the remote login protocols. (08 Marks)
Ans:
REMOTE LOGIN PRO...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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5 (b) Explain the RSA algorithm with an example. (08 Marks)
Ans:
RSA
• Ass...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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6 (a) With a neat diagram, explain the integrated services QoS. (08 Marks)...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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6 (b) Explain MPLS operation & packet format. (08 Marks)
Ans:
MPLS (MULTI ...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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7 (a) List and explain the compression methods without loss. (08 Marks)
An...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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LEMPEL-ZIV ENCODING
• Lempel-Ziv codes are independent of the source stati...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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7 (c) Write a note on real-time media transport protocols. (04 Marks)
Ans:...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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8 (b) Explain the DEEP clustering algorithm. (06 Marks)
Ans:
DEEP (Decentr...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013
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8 (c) Explain the intracluster and intercluster routing protocols. (08 Mar...
For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/
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COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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1 (a) Differentiate b/w connection oriented and connectionless services. (5...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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1 (c) Explain the Dijkstra's routing algorithm, with an example. (9 Marks)
...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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2 (b) Define congestion control with graph. Explain the leaky bucket algori...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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3 (a) Explain:
i) IP address classification
ii) Subnet addressing. (10 Mark...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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Ans: (ii)
SUBNET ADDRESSING
• To allow a single network address to span mul...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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3 (b) Give the format of IPV6 basic header. Compare IPV6 with IPV4. (10 Mar...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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4 (a) Explain OSPF protocol and its operation. (10 marks)
Ans:
OSPF (OPEN S...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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Step 3: Propagation of OSPF Link State Request and building routing tables
...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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Ans: (ii)
MOBILE IP
• This allows location-independent routing of IP datagr...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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5 (a) Write a note on only Two:
i) Remote login protocols
ii) File transfe...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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5 (b) Define network management & explain SNMP and SNMP messages. (6 Marks...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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5 (c) Compare secret key and public key cryptography systems. (6 Marks)
An...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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6 (a) Explain the differentiated services QoS with a neat diagram. (8 Mark...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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6 (b) Explain VPN and its types based on tunnelling. (8 Marks)
Ans:
VPN (V...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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6 (c) Explain the need for overlay networks. (4 Marks)
Ans:
OVERLAY NETWOR...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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7 (a) Briefly explain the MPEG standards and frame types of compression. (...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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7 (c) With a neat diagram, explain the H.23 components and list the steps ...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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8 (b) Briefly explain the direct and multi-hop routing of intra-cluster ro...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014
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8 (c) Write short notes on:
i) Clustering in sensor networks
ii) Security ...
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TE...
For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/
VTU
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TE...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
1
1 (a) Explain datagram and virtual circuit packet switching with delay calc...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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1 (b) With neat diagram, explain the generic packet switch. (04 Marks)
Ans:...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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1 (c) Consider the network, in the following Fig. Q1 (c):
i) Use the Dijkst...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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2 (a) Discuss the (i) Priority queuing & (ii) Weighted fair queuing. (06 Ma...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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2 (c) Give the classification congestion control algorithms. Explain the le...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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3 (a) Describe the various fields of IP version 4 header. (06 Marks)
Ans: F...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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4 (a) Explain TCP 3 way handshake for establishing a TCP connection. (06 Ma...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
8
4 (b) What are the classification of internet routing protocols? Explain in...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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4 (c) Write note on: i) Mobile IP ii) Reverse path multicasting. (06 Marks)...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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5 (b) What are the elements of network management? Discuss the interaction...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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6 (a) Describe the various types of resource allocation schemes. (06 Marks...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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7 (a) Explain the jpeg compression for still images. (06 Marks)
Ans:
STILL...
COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015
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8 (a) Explain DSDV, TORA routing protocols for mobile adhoc networks. (08 ...
VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015
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VTU 6TH SEM CSE COMPUTER NETWORKS 2 SOLVED PAPERS OF JUNE-2013 JUNE-14 & JUNE-2015

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

  1. 1. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  2. 2. 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: For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  3. 3. 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/ VTU N O TESBYSR I
  4. 4. 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/ VTU N O TESBYSR I
  5. 5. 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/ VTU N O TESBYSR I
  6. 6. 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/ VTU N O TESBYSR I
  7. 7. 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  8. 8. 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  9. 9. 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  10. 10. 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  11. 11. 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  12. 12. 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=me mod 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  13. 13. 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  14. 14. 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  15. 15. 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  16. 16. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013 15 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  17. 17. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013 16 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  18. 18. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013 17 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  19. 19. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2013 18 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  20. 20. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  21. 21. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 1 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  22. 22. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 2 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  23. 23. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 3 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  24. 24. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 4 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  25. 25. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 5 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  26. 26. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 6 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  27. 27. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 7 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). For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  28. 28. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 8 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  29. 29. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 9 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  30. 30. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 10 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  31. 31. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 11 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  32. 32. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 12 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  33. 33. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 13 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  34. 34. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 14 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  35. 35. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 15 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  36. 36. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 16 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  37. 37. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 17 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  38. 38. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 18 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  39. 39. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2014 19 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  40. 40. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  41. 41. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  42. 42. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 1 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  43. 43. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 2 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  44. 44. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 3 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: For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  45. 45. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 4 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 For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  46. 46. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 5 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  47. 47. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 6 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). For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  48. 48. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 7 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  49. 49. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 8 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  50. 50. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 9 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  51. 51. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 10 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  52. 52. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 11 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  53. 53. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 12 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I
  54. 54. COMPUTER NETWORKS II SOLVED PAPER JUNE- 2015 13 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. For Solved Question Papers of UGC-NET/GATE/SET/PGCET in Computer Science, visit http://victory4sure.weebly.com/ VTU N O TESBYSR I

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