Question

1. Calculate the total time to transmit a 1000 KB file over a link in the following cases, assuming the one-way delay in either direction is 50ms, and an initial 2 x RTT of "handshaking" before any data is sent.
   1. The bandwidth is 1.5 Mbps, the packet size including the header is 1 KB of which the header is 40 bytes, and data packets are sent continuously and never lost.
   2. Same as (1), but the packet size is 5 KB (incl. header). ·What happens when 15 packets are lost and have to be retransmitted (no time is wasted in detecting which packets have been lost, the last packet in the stream is not retransmitted).
   3. Same as (1), but after we finish transmitting we must wait for one RTT before transmitting the next packet.
   4. Same as (1), but the link bandwidth is "infinite" ( the transmission time is assumed to be zero). We start by sending one packet in the first (2^{(1 – 1)}), during the second RTT we send two ( 2(^2-1) )packets, during the third RTT we send four (2^(3-1)), and so on.

Answer 1

D.

that yields a sum ≥ 1561. When expanded we see that there are 11 terms. (We could also get this directly as [  This means that the total amount of time to send the file is the initial handshake, plus the 11 RTTs to send the groups, plus a final one-way propagation for the last group to reach the destination: 0.08 + 11 ∗ 0.08 + 0.04 = 1.0 sec. (Note: we also accepted an answer of 0.96 sec which left out the final 40 ms of propagation time.) This scheme might be used when we want to maximize the number of packets in flight but we are trying to determine how much the network and receiving host(s) could handle (which it turns out TCP does when starting up).