At n = 10 we have thus been able to send all 1,536 packets the last batchĪrrives 0.5 RTT later. At n RTTs past the initial handshaking we have sent 1+2+4+ ・・・ +2 n = 2 n+1 −1 packets. (d) Right after the handshaking is done we send one packet. 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), plus the initial 2 RTTs, for 6.28 seconds. (c) Dividing the 1,536 packets by 20 gives 76.8. To the above we add the time for 1,535 RTTs (the number of RTTs between when packet 1 arrives and packet 1,536 arrives), for a total of 1. We will count the transfer as completed when the last data bit arrives at its destination. (d) Zero transmit time as in (c), but during the first RTT we can send one packet, during the second RTT we can send two packets, during the third we can send four = 23−1, and so on. (c) The link allows infinitely fast transmit, but limits bandwidth such that only 20 packets can be sent per RTT. (b) The bandwidth is 10Mbps, but after we finish sending each data packet we must wait one RTT before sending the next. (a)The bandwidth is 10Mbps, and data packets can be sent continuously. 1.Calculate the total time required to transfer a 1.5MB file in the following cases,assuming a RTT of 80 ms, a packet size of 1 KB data, and an initial 2×RTT of“handshaking” before data is
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