Assignment 2 - Comp Networks.docx - Name: Simon Frank...

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Name: Simon FrankAssignment 2Comp 3271 -Computer Networks2017
Q10-8. K = 5N = 8Code word size is same as dividend size which is 8 bitsWe need to augment the data with three 0’sSize of the remainder r = n-k =3 bitsSize of the divisor = r+1 =4 bitsQ10-12.The error cannot be detected because the sum of items is not affected in thisswapping.P10-10.P10-12. Data word = 101001111Divisor = 10111...here divisor is 5 bits so pad 4 0’s in the data word10111) 101001111 000010111---------------------------------000111111 000010111----------------------------------010001 000010111--------------------------------00110 0000101 11----------------------------------011 110010 111----------------------------------01 00101 0111-----------------------------------0 0101Codeword would be1010011110101.P10-18. a.Q11-12.The simple protocol is designed for error-free medium (which nevermaterialized). The Ethernet protocol is an example of Stop-and-Wait protocol. If apacket is lost or corrupted, it should be resent. However, there is no explicitacknowledgement in the Ethernet. If a packet is lost or corrupted, the upper-layerprotocols force the data-link to resend the frame.Q11-16.In one-way communication, the data frames are moving from the sender tothe receiver, the acknowledgements are moving from the receiver to the sender. Sincewe require that the data frames be acknowledged, we need a timer at the sender site
to resend lost or corrupted data frames. Since acknowledgements frame are notrequired to be acknowledged, we do not need a timer at the receiver site. If anacknowledgement frame is lost, the sender interprets at the situation as the loss ofthe data frame and resend the data frame.

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Term
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Transmission Control Protocol

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