Question

Exercise 3. Suppose you are sending a large file from your laptop to another host over a TCP connection and assume that the connection have no loss. Round-trip times for this problem is constant 1 RTT) (a) Suppose TCP uses slow start and then uses Additive-Increase and Multiplicative-Decrease (AIMD) for its congestion control after cund reaches 16 MSS. Assuming cwnd increases by 1 MSS every time a ACK is received during the slow start and by 1 MSS every time a batch of ACKs is received during the AIMD how long does it take for cund to increase up to 20 MSS from 1 MSS? b) What is the average throughput of this connection (in terms of MSS and RTT) from timet0 up to time t 10 RTT?

Answer 1

1. In Slow Start phase cwnd increases by 1 MSS every time an ACK is received.

Therefore, when we send 1^st MSS, we receive 1 ACK. Hence cwnd is increased by 1MSS.

Hence cwnd becomes (1+1) 2MSS.

Then, 2MSS will be sent and 2 ACK will be received.

For 2 ACK cwnd will be increased by 2MSS.

Hence cwnd becomes (2+2) 4MSS. Means cwnd is doubled after every RTT

This will happen until cwnd becomes 16MSS.

Then, cwnd I increased by 1MSS for every batch of ACK.

So, cwnd increases as follows:

1 (1 RTT) 2 (2 RTT) 4 (3 RTT) 8 (4 RTT) 16 (5 RTT) 17 (6 RTT) 18 (7 RTT) 19 (8 RTT) 20

It takes 8 RTT to increase up to 20 MSS from 1MSS.

2. 1 (1 RTT) 2 (2 RTT) 4 (3 RTT) 8 (4 RTT) 16 (5 RTT) 17 (6 RTT) 18 (7 RTT) 19 (8 RTT) 20 (9 RTT) 21 (10 RTT) 22

From t=0 to t=10, in 10 RTT total no. of MSSs send is : 1+2+4+8+16+17+18+19+20+21 = 126

Therefore, average throughput : 126 MSS/ 10 RTT = 12.6 MSS/RTT

After 10RTT, 22MSS is sent but not acknowledged. Hence it will not be counted.