EE450_HW1_solns

EE450_HW1_solns - EE450: Computer Networks, Fall 2007...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
EE450: Computer Networks, Fall 2007 Homework #1, Solutions (100 points) 5. (20 points, 5 points each) 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 10 6 versus 2 20 , 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 sec. (c) This is 49.5 RTTs, plus the initial 2, for 5.15 seconds. (d) Right after the handshaking is done we send one packet. One RTT after the handshaking we send two packets. At n RTTs past the initial handshaking we have sent 1 + 2 + 4 + ··· + 2 n = 2 n+1 - 1 packets. At n = 9 we have thus been able to send all 1,000 packets; the last batch arrives 0.5 RTT later. Total time is (2+9.5)*RTTs, or 1.15 sec
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

EE450_HW1_solns - EE450: Computer Networks, Fall 2007...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online