EBK COMPUTER NETWORKING
EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Chapter 1, Problem P33P

Explanation of Solution

Given,

Number of bits present in the file is F

Number of segments present in the file is S

Therefore, the length of the packets is S + 80 bits

The transmission rate of each link is R bps

Number of packets present in the packet can be calculated by using the following formula is given below:

Numberofpackets=NumberofbitsNumberofsegments

Therefore, the number of packets present in the file is FS.

The time required to receive the last packet at first router is S+80R×FSsec.

At S+80R×FS time, the packet FS-2 will be at destination whereas the packet FS-1 will be at second router.

The last packet will be transmitted by first and second routers with S+80Rsec time delay

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Consider a simple protocol for transferring files over a link. After some initial negotiation, A sends data packets of size 1 KB to B; B then replies with an acknowledgment. A always waits for each ACK before sending the next data packet; this is known as stop-and-wait. Packets that are overdue are presumed lost and are retransmitted. (a) In the absence of any packet losses or duplications, explain why it is not necessary to include any "sequence number" data in the packet headers. (b) Suppose that the link can lose occasional packets, but that packets that do arrive always arrive in the order sent. Is a 2-bit sequence number (that is, N mod 4) enough for A and B to detect and resend any lost packets? Is a 1-bit sequence number enough? (c) Now suppose that the link can deliver out of order and that sometimes a packet can be delivered as much as 1 minute after subsequent packets. How does this change the sequence number requirements?
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Chapter 1 Solutions

EBK COMPUTER NETWORKING

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