Solutions to Select Exercises
Chapter 1
4. We will count the transfer as completed when the last data bit arrives at its destination
(a) 1.5 MB = 12582912 bits. 2 initial RTTs (160 ms) + 12,582,912/10,000,000 bps (transmit) + RTT/2 (propagation) ≈ 1.458 seconds.
(b) Number of packets required = 1.5 MB/1KB = 1536. To the above we add the time for 1535 RTTs (the number of RTTs between when packet 1 arrives and packet 1536 arrives), for a total of
seconds.
seconds.(c) Dividing the 1536 packets by 20 gives 76.8. This will take 76.5 RTTs (half an RTT for the first batch to arrive, plus 76 RTTs between the first batch and the 77th partial batch), ...
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