mcs-042 solved assignment 2014-15

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Course Code : MCS-042Course Title : Data Communication and Computer NetworkAssignment Number : MCA (4)/042/Assign/2014-15Maximum Marks : 100Weightage : 25%Last Dates for Submission : 15thOctober,

2014 (For July 2014 Session)15thApril, 2015 (For January 2015 Session)

This assignment has nine questions. Answer all questions. Rest 20 marks are forviva voce. You may use illustration and diagrams to enhance the explanations.Please go through the guidelines regarding assignments given in the ProgrammeGuide for the format of presentation.

Q.1.Derive expressions for throughput in Aloha and slotted Alohaprotocols. Show the vulnerability period of Aloha and alsoload vs. through graph for Aloha and slotted Aloha.(10 marks)

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Q.2.Explain functioning of the following protocols with the helpof a diagram. 3 bit sequence numbers are used.Go back N








Selective RepeatHow these two protocols are different from stop and waitprotocols?(10 marks)

1.Go Back NIn this method, many frames can be transmitted during the process without waiting foracknowledgement. In this, we can send n frames without making the sender wait foracknowledgements. At the same time, the sender will maintain a copy of each sent frame tillacknowledgement reaches it safely. As seen earlier in the Stop & Wait ARQ where, we used numbermechanism, here also, we will use number method for transmission and receipt of frames. Eachframe will have a sequence number that will be added with the frame. If, the frame can have k bitsequence number then the numbers will range between 0 to 2k-1. For example if k is 2 bit thennumbers will be 0,1,2,3,0,1,2,3,0,.The sender can send 4 frames continuously without waiting for acknowledgment. But, the receiverwill look forward to only one frame that must be in order. If, the frame received is not in order, itwill simply keep on discarding the frame till, it receives the desired sequence number frame. Thereceiver is not bound to send an individual acknowledgment for all frames received; it can send acumulative acknowledgment also. The receiver will send a positive acknowledgement if, thereceived frame is the desired sequence number frame. Otherwise, it will keep on waiting if, theframe received is corrupted or out of order. As soon as the timer expires for the frame sent by thesender, the sender will GoBack and retransmit all frames including the frame for which the timerexpired, till, the last sent frame. Hence, it is named as Go-Back-N ARQ. The process ofretransmission for an error frame is shown in Figure.6.Go-Back- N is used in HDLC protocol.








Selective Repeat ARQThis method increases the efficiency of the use of bandwidth. In this method, the receiver has a windowwith the buffer that can hold multiple frames sent by the sender. The sender will retransmit only thatframe which has some error and not all the frames as in Go-Back-N ARQ. Hence, it is named as selectiverepeat ARQ. Here, the size of the sender and the receiver window will be same. The receiver will notlook forward only to one frame as in Go-Back-N ARQ but it will look forward to a continuous range offrames. The receiver also sends NAK for the frame which had the error and required to be retransmittedby the sender before the time out event fires. As in the earlier section we discussed, each frame willhave a sequence number that will be added with the frame. If, the frame can have k bit sequencenumber then the sequence number of frames will range between 0 to 2k-1. For example, if k is 2 bit thennumbers will be 0,1,2,3,0,1,2,3,0,.Size of sender and receiver window would be 2k/2 i.e 4/2=2 or it canbe written as 2k-1. If the window size is 2 and acknowledgement for frame0 and frame1 both gets lostduring transmission then, after timer expires the sender will retransmit frame0 though the receiver isexpecting frame2 after frame1 was received without any error. Hence, frame0 will be discarded by thereceiver as a duplicate frame. If the receiver window size is more than two, the receiver will acceptduplicate frame0 as a new frame and hence, the size of window should be 2k/2. The process is shown infigure 7:








The problem with stop and wait is that only one frame can be transmitted at a time and this leads toinefficiency of transmission. To deal with this, we have another error and flow control method i.e. GoBack N and Selective Repeat ARQ.

Q.3.A bit stream 101010101101 transmitted using standard CRCmethod. The generator polynomial is x4+1. Show the actualbit string transmitted. Suppose the fourth bit from the left isinverted during transmission. Show that this error is detectedat the receivers end.

-----------------------send me if you have solved------------------------Q.4.Sketch the Manchester and differentiate Manchesterencodings for the bit stream: 100111010101(10 marks)(5 marks)








Q.5.Explain the Ethernet frame format. Why there is arequirement of maximum and minimum Ethernet framelength?(10 marks)Ans:

Ethernet protocol is a MAC sublayer protocol. Ethernet stands for cable and IEEE 802.3 Ethernetprotocol was designed to operate at 10 Mbps. Here, we will begin discussing the Ethernet with varioustypes of cabling. With the help of Figure 9, we will try to summarise the cabling used for baseband 802.3LANs.








IEEE 802.3 Ethernet accesses the channel using 1-persistent CSMA/CD method in LAN. Now we willdiscuss MAC frame structure for IEEE 802.3 with the help of Figure 10.

Each frame has seven fields explained as follows:Preamble: The first field of 802.3 frame is 7 byte (56 bits) long with a sequence of alternate 1 and 0 i.e.,10101010. This pattern helps the receiver to synchronise and get the beginning of the frame.

Starting Delimiter (SD): The second field start delimiter is 1 byte (8 bit) long. It has pattern 10101011.Again, it is to indicate the beginning of the frame and ensure that the next field will be a destinationaddress. Address, here, can be a single address or a group address.

Destination Address (DA): This field is 6 byte (48 bit) long. It contains the physical address of thereceiver.

Source Address (SA): This filed is also 6 byte (48 bit) long. It contains the physical address of thesender.

Length of Data Field: It is 2 byte (16 bit) long. It indicates the number of bytes in the information field.The longest allowable value can be 1518 bytes.Data: This field size will be a minimum of 46 bytes long and a maximum of 1500 bytes as will beexplained later.








Pad: This field size can be 0 to 46 bytes long. This is required if, the data size is less than 46 bytes as a802.3 frame must be at least 64 bytes long.Frame Checksum (FCS): This field is 4 bytes (32 bit) long. It contains information about errordetection. Here it is CRC-32.Minimum and Maximum Length of FrameMinimum frame length = 64 bytes = 512 bitsMaximum frame length= 1518 bytes = 12144 bitsMinimum length or lower limit for frame length is defined for normal operation of CSMA/CD. This isrequired so that, the entire frame is not transmitted completely before its first bit has been received by thereceiver. If, this happens then the probability of the occurrence of collision will be high (the same hasbeen explained earlier in the previous section CSMA/CD).Hence, Ethernet frame must be of 64 bytes long. Some of the bytes are header and trailer parts of theframe. If, we consider 6 bytes destination address, 6 bytes source address, 2 bytes length and 4 bytes FCS(6+6+2+4=18) then, the minimum length of data will be 64-18= 46 bytes. If, frame is less than 46 bytesthen, padding bits fill up this difference.As per 802.3 standard, the frames maximum length or upper limit of frame is = 1518 bytes (excludingpreamble and SD). If we subtract 18 bytes of header and trailer then, the maximum length will be 1500bytes.

Qns 6:How is OSPF implemented in IP network? Illustrate.Ans:

OSPF: The Interior Gateway Routing ProtocolOpen Shortest Path First (OSPF) has become standard interior gateway routing protocol. It supports manyadvanced features to meet a long list of requirements.

The protocol specification should be publicly available. O in OSPF stands for open. It means thatOSPF is not a proprietary protocol. It is a dynamic algorithm one that adapts to changes in the topology automatically and quickly. Load distribution: When multiple paths to a destination have the same cost OSPF allowsmultiple paths to be used.

Support for hierarchy within a single routing domain. By 1988, the internet had grown so largethat no router was expected to know the entire topology. OSPF was designed so that no routerwould have to do so. Security: All exchanges between OSPF routers are authenticated and allows only trusted routersto participate. OSPF supports three kinds of connections and networks:(i) Point-to-point lines between two routers only.(ii) Multi-access networks with broadcasting (e.g., LANs).








(iii) Multi-access networks without broadcasting (e.g., Packet Switched WANs).

OSPF identifies four types of routers:(i) Internal routers (within one area).(ii) (ii) Area border routers (connects two or more areas).(iii) (iii) Backbone routers (performs routing within the backbone).(iv) (iv) AS boundary routers (exchanges routing information with routers in other ASes).(v)

OSPF allows a longer AS to be divided into smaller areas. Topology and details ofone area is not visible to others. Every AS has a backbone area (called Area 0) asshown in Figure3.

All areas are connected to the backbone possibly by a tunnel, so it is possible to move from one areatoanother area through a backbone. The primary role of the backbone area is to route traffic between the








other areas in the AS. Inter area routing within the AS requires the follows movement of packets asshown below by arrows.

After having described all the components of OSPF, let us now conclude the topic by describing itsoperation.

Q.7.Consider the following network with the indicated link cost.Use Dijkstras and Bellmans shortest path algorithm to findthe shortest path from source node A to all other nodes.

(15 marks)

------------------send me if you have solved------------------

Q.8.What are the benefits of using digital certificate? (5 marks)








A Digital Certificate is an electronic card, or the digital equivalent of an identification card, that certifies

online transactions, e-commerce and other authentications on the Internet. A third-party Certificate

Authority (CA) issues a public key on high-volume e-commerce sites, in emails send by financial

services, banks and other online privacy-based businesses. These certificates are meant to

authenticate the legitimacy of the communications, interactions and possible transactions between the

sender and the receiver. These certificates also authenticate casual surfing of portals, sites and

forums.

1. Minimal involvement needed from end users2. No additional hardware needed3. Easy to manage4. Certificates are natively supported by manyenterprise applications and networks

Q.9.How does TCP 3-way handshake mechanism handle thefollowing:(i) Delayed arrived and SYN packets(ii) Delayed ACK Packet

The Transmission Control Protocol (TCP) level of the TCP/IP transport protocolis connection-oriented. That means that, before any data can be transmitted, a reliablecommunication channel must be created. To establish a connection, TCP uses a 3-way handshake. Before a client attempts to connect with a server, the server must firstbind to a port to open it up for connections: this is called a passive open. Once thepassive open is established, a client may initiate an active open. To establish aconnection, the 3-step handshake occurs (Explanation of the Three-Way Handshakevia TCP-IP)








1. The active open is performed bysending a SYN packet to the server. In this packet Synchronize sequencenumber flag set to 1. This make it the request to the server to synchronize thesequence numbers. It specifies its initial sequence number (ISN), which isincremented by 1, For example 8221821+1=8221822, and that is sent to theserver.

2. In response, the server replies with a SYN-ACK packet in whichboth Acknowledgement field and Synchronize sequence number field set to 1.the server is acknowledging the request of the client for synchronization. Atthe same time, the server is also sending its request to the client forsynchronization of its sequence numbers. There is one major difference in thissegment. The server transmits an acknowledgement number (8221823) to theclient. The acknowledgement is just proof to the client that the ACK is specificto the SYN the client initiated. The process of acknowledging the client'srequest allows the server to increment the client's sequence number by oneand uses it as its acknowledgement number.

3. Finally the client sends a packet with ACK bit (Acknowledgement field) setback to the server. In this packet, the client is acknowledging the request fromthe server for synchronization. The client uses the same algorithm the serverimplemented in providing an acknowledgement number. The client'sacknowledgment of the server's request for synchronization completes theprocess of establishing a reliable connection, thus the three-way handshake.

Please note that for each connection there is a send sequence number and a receivesequence number. The initial send sequence number (ISN or ISS) is chosen by thedata sending TCP, and the initial receive sequence number (IRS) is learned duringTCP handshake.






mcs-042 solved assignment 2014-15

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