AppliedCourse MOCK

Question

Consider a TCP connection between two machines (A and B) in an environment with 0% packet loss. Assume the round trip time (RTT) between the two machines is 4 [seconds], and the segment size is 3 [Kbytes]. The bandwidth of the connection is 500 [kbps]. What is the minimum possible TCP window size for which there will be no stalling?(We say a TCP connection experiences no stalling if the acknowledgments arrive back to the sender before the sliding window over the send buffer close to zero. I.e., TCP packets are continuously, back-to-back, sent out of the sender before the first ack arrives back.)

Answer 1

There will be no stalling if
time-to-send-entire-window ≤ time-for-first-ack-to-arrive-back
That is:
W*S / B ≤ RTT + S / B
Hence: W ≤ RTT * B / S + 1
In this particular case: W ≤ 4 sec * 500 kbps / 24 kbits + 1 = 83.3 + 1 = 84.3

(3Kbytes = 24 Kbits)
That is: W ≤ 84

So the smallest TCP window size is 84.

Answer 2

WS/R > RTT + S/R. In this case, the server receives an acknowledgement for the first segment in the first window before the server completes the transmission of the first window.

WS/R < RTT + S/R. In this case, the server transmits the first window's worth of segments before the server receives an acknowledgement for the first segment in the window.

The above figure follows the case WS/R > RTT + S/R

where W = Window size, S = Segment size, R = Bandwidth, RTT = round trip time.

Reference : http://www2.ic.uff.br/~michael/kr1999/3-transport/3_07-congestion.html