Answer:
The answer is "6 RTT and [tex]\frac{5\ MSS}{3 \ RTT}[/tex]"
Explanation:
TCP uses AIMD without a sluggish start for congestion control.
Sliding windows original = 4 MSS
Every RTT is increased by 1 MSS throughout the congestion window(cwnd).
In (a):
Its City is rising between 6 and 12 MSS
Each RTT cwnd window increases by 1 MSS
[tex]6 MSS | \ 7 MSS | \ 8 MSS | \ 9 MSS | \ 10 MSS | \ 11 MSS | \ 12 MSS[/tex]
We can see that 6 RTTs from 12 MSS city sizes, so it takes time to achieve 6 MSSs = 6 RTT
In (b):
Time = 6 RTT
Each RTT cwnd window will up by 1 MSS
Window Size Original = 4 MSS
[tex]4 MSS \ | 5 MSS \ | 6 MSS \ | 7 MSS \ | 8 MSS \ | 9 MSS \ | 10 MSS[/tex]
It is possible to view the window after 6 RTT (cwnd) = 10 MSS
[tex]\text{Average throughput} = \frac{window \ size}{time}[/tex]
[tex]= \frac{10 \ MSS}{6 \ RTT} \\\\= \frac{5\ MSS}{3 \ RTT}[/tex]