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Homework Answers
Given,
Initial MSS ( Maximum Segment Size ) = 1 MSS
The TCP connection here uses AIMD and not slow start which means the congestion window ( cwnd ) will increase by 1.
a) It takes 1 RTT ( Round Trip Time ) to increase the congestion window size by 1 MSS.
So, during the transmission, the following communication takes place :
1. It takes 1 RTT to increase the cwnd from 1 MSS to 2 MSS.
2. It takes 1 RTT to increase the cwnd from 2 MSS to 3 MSS.
3. It takes 1 RTT to increase the cwnd from 3 MSS to 4 MSS.
4. It takes 1 RTT to increase the cwnd from 4 MSS to 5 MSS.
5. It takes 1 RTT to increase the cwnd from 5 MSS to 6 MSS.
6. It takes 1 RTT to increase the cwnd from 6 MSS to 7 MSS.
7. It takes 1 RTT to increase the cwnd from 7 MSS to 8 MSS.
8. It takes 1 RTT to increase the cwnd from 8 MSS to 9 MSS.
9. It takes 1 RTT to increase the cwnd from 9 MSS to 10 MSS.
10.It takes 1 RTT to increase the cwnd from 10 MSS to 11 MSS.
11. It takes 1 RTT to increase the cwnd from 11 MSS to 12 MSS.
So, the transmission takes place as :
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12
The number of bars is equal to the number of round trip times.
Here, the number of round trip times are 11.
So, it takes 11 RTTs for CWND to increase to 12 MSS.
b.
The transmission till 5 RTT takes place as follows :
1 MSS -----> 2 MSS -------> 3 MSS -------> 4 MSS ---------> 5 MSS ---------> 6 MSS
So, after 5 RTT, the cwnd is 6 MSS.
Average throughput = Segment Size / Total round trip time
= 6 MSS / 5 RTT
= 1.2 MSS / RTT .
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