Consider a TCP connection between a client and a server with the following…

2020

Consider a TCP connection between a client and a server with the following specifications; the round trip time is 6 ms, the size of the receiver advertised window is 50 KB, slow-start threshold at the client is 32 KB, and the maximum segment size is 2 KB. The connection is established at time t=0. Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in KB) at time t+60 ms after all acknowledgements are processed is _________ . 

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Correct answer: 44

Key facts: RTT = 6 ms, receiver advertised window = 50 KB, slow-start threshold = 32 KB, MSS = 2 KB, no losses, connection starts at t = 0.

Total time considered: 60 ms, which equals 10 round-trip times (60 / 6 = 10). In slow start the congestion window doubles every RTT until it reaches the slow-start threshold. After reaching the threshold, congestion avoidance increases the congestion window by one MSS per RTT (additive increase).

  • Initial congestion window at t = 0: 1 MSS = 2 KB.

  • After 1 RTT (t = 6 ms): cwnd = 4 KB.

  • After 2 RTTs (t = 12 ms): cwnd = 8 KB.

  • After 3 RTTs (t = 18 ms): cwnd = 16 KB.

  • After 4 RTTs (t = 24 ms): cwnd = 32 KB. This equals the slow-start threshold, so subsequent growth follows congestion avoidance.

  • Remaining RTTs: 10 total RTTs − 4 RTTs = 6 RTTs in congestion avoidance. Congestion avoidance adds 1 MSS (2 KB) per RTT, so increase = 6 × 2 KB = 12 KB.

  • Final congestion window after 10 RTTs (t = 60 ms): 32 KB + 12 KB = 44 KB.

  • Check against receiver advertised window: receiver window = 50 KB, which is larger than 44 KB, so it does not limit the congestion window.

Answer: 44 KB

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