CN-TCP

Question

Assume a new TCP connection starts by sending 1 segment and then increases it congestion windows by 1 segment each time it receives an acknowledgement i,e after 1RTT congestion window is 2 segments in slowstart.Assume connection never leaves slow start.Find the number of RTT's it takes to send N segments

Given ans is Log₂(N+1) 

Answer 1

$$\begin{align*} &\text{They are saying packets instead of MSS } \\ &\text{So, According to the question : } \\ \\ &\begin{matrix} &\text{1st RTT} &\text{packet sent}&=1=2^0 \\ \\ &\text{2nd RTT} &\text{packet sent}&=2=2^1 \\ \\ &\text{3rd RTT} &\text{packet sent}&=4=2^2 \\ \\ &\text{4th RTT} &\text{packet sent}&=8=2^3 \\ \\ &\text{5th RTT} &\text{packet sent}&=16=2^4 \\ \\ &\text{6th RTT} &\text{packet sent}&=32=2^5 \\ \\ &---&---&---\\ \\ &\text{kth RTT} &\text{packet sent}&={\color{red}{\bf 2^{k-1}}} \\ \\ \hline\\ &\text{at the end of kth RTT}&\text{Total packet sent} &= \sum_{x=0}^{k-1}\left [ {\color{red}{\bf 2^{x}}} \right ] \end{matrix} \\ \\ &\text{Total packet sent} = \sum_{x=0}^{k-1}\left [ {\color{red}{\bf 2^{x}}} \right ] = {\color{red}{\bf 2^k - 1}} \\ & \Rightarrow N = {\color{red}{\bf 2^k - 1}} \\ &\Rightarrow k = {\color{red}{\bf \log_2\left [ N+1 \right ]}} \\ \end{align*}$$

Comments

you are forgetting size of congestion window bro, congestion window size si currently 2 and increases only by 1 each time you recieve an ack .