Successfully reported this slideshow.
Upcoming SlideShare
×

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

5,638 views

Published on

Published in: Engineering
• Full Name
Comment goes here.

Are you sure you want to Yes No

Are you sure you want to  Yes  No

Are you sure you want to  Yes  No

### 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
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
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
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
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
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
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