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 congwinincreases by 1 mss every time a batch of acks is received and assuming approximately constantround-trip times, how long does it take for congwin to increase from 1 mss to 6 mss (assumingno loss events)?b) what is the average throughout (in terms of mss and rtt) for this connection up through time

Answer 1

The time it takes for the TCP congestion window to increase from 1 MSS to 6 MSS without any loss under AIMD congestion control is 5 RTTs. The average throughput for this timeframe is 3 MSS/RTT.

Step-by-step explanation:

The student question pertains to the time it takes for the TCP congestion window to grow from 1 Maximum Segment Size (MSS) to 6 MSS, under Additive Increase Multiplicative Decrease (AIMD) congestion control with no slow start and no loss. Let's work through each congestion window increase:

• Rounds to go from 1 MSS to 2 MSS: 1 round
• Rounds to go from 2 MSS to 3 MSS: 1 round
• Rounds to go from 3 MSS to 4 MSS: 1 round
• Rounds to go from 4 MSS to 5 MSS: 1 round
• Rounds to go from 5 MSS to 6 MSS: 1 round

Total rounds for the congestion window to grow from 1 MSS to 6 MSS is 5 rounds.

Assuming no loss and constant round-trip time (RTT), each round is one RTT long. Therefore, the time to increase from 1 MSS to 6 MSS is 5 RTTs.

For part b), the average throughput in terms of MSS and RTT can be calculated as the sum of the congestion window sizes divided by the total time (in RTTs).

Average throughput = (1 + 2 + 3 + 4 + 5) MSS / 5 RTT = 15 MSS / 5 RTT = 3 MSS/RTT