Question

consider sending a large file from a host to another over a tcp connection that has no loss. a) suppose tcp uses aimd for its congestion control without slow start. assuming congwin increases by 1 mss every time a batch of acks is received and assuming approximately constant round-trip times, how long does it take for congwin to increase from 1 mss to 6 mss (assuming no loss events)? b) what is the average throughout (in terms of mss and rtt) for this connection up through time

Answer 1

It takes 5 RTTs for the congestion window to increase from 1 MSS to 6 MSS in a TCP connection using AIMD without slow start because the window increases by 1 MSS each RTT. The average throughput, in this case, can be calculated as 21 MSS divided by 5 RTTs.

Step-by-step explanation:

The scenario involves analyzing the behavior of TCP congestion control with Additive Increase Multiplicative Decrease (AIMD) without the slow start phase. Since there is no slow start, congestion window (congwin) increases linearly.

To calculate the time taken for congwin to increase from 1 Maximum Segment Size (MSS) to 6 MSS, we consider that the round-trip time (RTT) is approximately constant and congwin increases by 1 MSS for each RTT after a batch of acknowledgements (ACKs) is received. Therefore, it takes 1 RTT to increase to 2 MSS, another RTT to increase to 3 MSS, and so on, until it reaches 6 MSS after 5 RTTs.

For part b), to find the average throughput, we note that throughput is the rate of successful message delivery over a communication channel. We can calculate the average throughput as the sum of congwin sizes divided by the number of RTTs considered. Using the given increase, the sum of sizes from 1 MSS to 6 MSS is 1+2+3+4+5+6 = 21 MSS. Since it takes 5 RTTs to reach 6 MSS, the average throughput is then 21 MSS / 5 RTTs.