Question: Suppose a client is connecting with the server through TCP . In the last packet of the three - way handshake, the client adds a

Suppose a client is connecting with the server through TCP. In the last packet of the three-way handshake, the client adds a message to ask server send a file with L bytes. There are some
assumptions:
The transmission speed between client and server is R bytes per second. The round trip time
between them is a fixed value, RTT.
The payload of a TCP packet is M bytes, and the TCP window size is M (e.g. n packets per TCP window).
There is no packet loss on the link and all ACKs arrive at the client before the timeout.
You can ignore the header construction, header length and other header related time overhead. Let's denote T is the time duration from the client sends the TCP connection request to the client successfully receives the file. Then, we have,
T=2RTT+L/R, when nM > R*RTT+M
T ={2RTT + L/R+(K -1)(M/R+ RTT nM/R},when nM < Rx RTT + M
Where K =[L/nM], round down of L/nM
1. What does 2RTT mean in the equation?
2. What does the condition, nM > Rx RTT + M, mean?
3. Please proof the equation when nM > Rx RTT + M. Hint: you can draw a figure.

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