25-08-2017, 09:32 PM
Computer Networks: A Systems Approach Fourth Edition Solutions Manual
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Success here depends largely on the ability of one’s search tool to separate out
the chaff. The following are representative examples
We will count the transfer as completed when the last data bit arrives at its destination.
An alternative interpretation would be to count until the last ACK arrives
back at the sender, in which case the time would be half an RTT (50ms) longer.
(a) 2 initial RTT’s (200ms) + 1000KB/1.5Mbps (transmit) + RTT/2 (propagation)
≈ 0.25 + 8Mbit/1.5Mbps = 0.25 + 5.33 sec = 5.58 sec. If we pay more
careful attention to when a mega is 106 versus 220, we get
8,192,000 bits/1,500,000bits/sec = 5.46 sec, for a total delay of 5.71 sec.
(b) To the above we add the time for 999 RTTs (the number of RTTs between
when packet 1 arrives and packet 1000 arrives), for a total of 5.71+99.9 = 105.61.