Chapter 3, Problem P47P

Explanation of Solution

Description of the TCP link to always be busy with sending data:

• Consider the maximum window size is “W”.
• Buffer size is “S”.
• Round trip time “RTT” to send data packets through TCP connection.

If the window size reaches “W” then loss occurs on packet. When loss occurs the sender changes the size from “W” to “W/2” which means the sender will cut its congestion window size by half.

After reducing the size, the sender waits for the acknowledgement (ACK) of “W/2” outstanding packets before it starts sending data segments again.

The link can be busy, if it sends the data in the period “W/ (2×C)” which means the time period is equal to interval where the sender is waiting for the ACK of the W/2 outstanding packets. Thus, “S/C must be no less than W/ (2×C)”.

Therefore, $\text{S}\ge \frac{\text{W}}{\text{2}}$

Now, consider the one way propagation delay between sender and receiver is T_p