Computer Networking

Computer Ne bothrking1. Data Link Layer Protocols eer put CRC in there trailer rather than in the header. Why?Ans.The CRC is exercised during transmitting and app eradicateed to the bug output stream as soon as the last objet dart goes out onto the wire. But If we put CRC in the header, before transmitting it is necessary to compute CRC modus operandis and and so add it. This approach will result in handling of from each one byte twice once for check heart and soulming and once for transmitting. While do-gooder of CRC in trailer results is the easy way and less effort. Thats the reason wherefore we add CRC at the trailer part of the frame.2. Slotted aloha is a ameliorate version on aloha protocol. On what factors the improvement is implemented in slotted aloha.Ans.In reason of slotted ALOHA concept of conviction slot has been introduced. During the transmittance system we make use of this time slot to commove a frame. A frame will be enrapture only at the rise of the time slot. As a result of this method the determines of collision decreases considerably. compromising time in case of slotted ALOHA is also less than the insecure time in pure aloha. Or in other lyric poem we jackpot say that the vulnerable time is just half in the slotted ALOHA as compargond to that of ALOHA.As the probability of collisions has been decreased this results in the c aren magnitude efficiencyof the transmittal.3.When bit stuffing is use, is it possible for divergence, insertion or revision of a single bit to cause an fault not spy by checksum? If not why not? If so how? Does checksum play a role there?Ans.Yes. There is a possibility of modification a single bit which will result in an wrongdoing.But this computer flaw can good be detected by the checksum fallacy sleuthing method.This is so because checksum is the sum of the information elements to be transmitted.If a bit in frame get modified therefore it will result in the change of sum of th e selective information elements.Which in turn will change the checksum. And if the chechsum dose not match at both the sides because error will be detected by the pass catcher.Also there is chance of loss of a bits.Lets suppose if there change occurs in the bit which was stuffed in the data part means stuffed 0 bit becomes 1. in this case receiver will take it as flag and conseder it as the end of the frame which will finally result in the loss of some data bits. This loss of data bits can easily be detected by the checksum method. As due to the loss of data bits the sum will change and it will result in change in checksum itemise. And the error will be detected easily.Insertion of a bit is not possible because flags are utilise to indicate the start and end of a frame. And also the size of the frame is fixed. So we can not add one extra bit to the frame.4. Give two reasons why network might use an error correcting code instead of error detection and retransmission?Ans.Sending data with error correcting code and detecting error consequently ask for retransmission of data are two different methods used for data transmission.The inaugural option have some advantages everyplace the second one.1 Fast transmission of data takes place.If any frame is received with error accordingly receiver can regenerate the correct frame utilize the error control information. While in 2nd case receiver will 1st send request for retransmission of that particular frame so sender will again send that frame, which is a very time consuming process.2 Efficiency of transmission increases.If we use 1st option then the whole bandwidth of the stockpile will be used to send the data only in one direction which will automatically increases the data transmission efficiency. While in case of 2nd option bandwidth of the channel is divided in two parts which results in loss of efficiency.5. radio set transmission and wired transmission use different set of six-fold channel allocatio n strategies. Why there was a need of scheme when detection was already available?Ans.In case of wired transmission we use wires for the transmission. If data have to send over a puny distance it is easily transferred but if data is to be send over a long distance then we makes use of repeaters. Because of the repeaters the energy of the frames are maintained. So if any collision occurs in wired connection then it is easy to recover the data.While in case of wireless transmission data travels through the air which results in loss of energy of the bundles. During transmission energy of the data packets decreases. So if collision occurs then the packets destroy easily and completely. So we need to overturn collisions in case of wireless transmission.6. racy tooth supports two figures of links between a slave and master .What are they and why is each one used for?Ans.There are two types of link that can be created between a primary ( rule ) and a second-string ( Slave ) station s.1 SCO, Synchronous connection oriented.This connection is used when it is much important to deliver data in time ( to avoid latency) than integrity ( i. e. error-free delivery ).The basic units of connection is two slots, one for each direction. At regular intervals specific slots are reserved for primary and secondary stations to establish the connection. If any error occurs ( such as packet lost) then it is never retransmitted.This type of connection is used in real-time applications.2 ACL, Asynchronous connectionless link.This type of connection is used when data integrity is more important than avoiding delay in data delivery.If any error occurs, then the damaged frame is retransmited.The number of slots are not fixed for ACL, it can use one, two or more number of slots.After the comer of the data frame secondary station sends ACL frame if and only if the prior slot has been talked to it.7. Using 5-bit sequence numbers, what is the maximum size of the send and receive wind owpanepanes for each of the following protocols?a. Stop-and-Wait ARQAns. In this protocol sequence number is based on modulo-2 arithmeetic.Send window Size - 1 ( always )Receive Window Size - 1 ( always )b. Go-Back-N ARQAns.Send window size - 32 ( frames numbering from 0 to 31 )receive window size- 1 ( always )c. Selective-Repeat ARQAns. In this protocol the size of send window is equal to receive window.Send Window size - 16Receive Window size - 168. If an Ethernet conclusion address is 070102030405, what is the type of theaddress (unicast, multicast, or broadcast)?Ans.A source address is always a unicast address as frame comes only from one station.Now destination address can be unicast, multicast or broadcast.To identify a address wheather it is a unicast or a multicast we conseder least significant bit of the 1st byte.If this bit is 0 then address is Unicast.If this bit is 1 then address is Multicast.While broadcast address is a special case of the multicast address.If all the bits in the this 6 byte address are 1 then its a broadcast address.