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GATE CS 2014 Set 1 Q27 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

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Let the size of congestion window of a TCP connection be 3232 KB when a timeout occurs. The round trip time of the connection is 100100 msec and the maximum segment size used is 22 KB. The time taken (in msec) by the TCP connection to get back to 3232 KB congestion window is _________.

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Answer: 1100

When a TCP timeout occurs, the congestion control parameters are updated as follows:

  • The slow start threshold (ssthreshssthresh) is set to half of the current congestion window size:
    ssthresh=32 KB2=16 KB=8 MSSssthresh = \frac{32\text{ KB}}{2} = 16\text{ KB} = 8\text{ MSS}
  • The congestion window (cwndcwnd) is reset to 1 MSS=2 KB1\text{ MSS} = 2\text{ KB}.

Now, we trace the growth of cwndcwnd over successive Round Trip Times (RTTs):

  1. Slow Start Phase (exponential growth until cwndcwnd reaches ssthresh=8 MSSssthresh = 8\text{ MSS}):
    • Initially (at 00 RTT): cwnd=1 MSScwnd = 1\text{ MSS}
    • After 11 RTT: cwnd=2 MSScwnd = 2\text{ MSS}
    • After 22 RTT: cwnd=4 MSScwnd = 4\text{ MSS}
    • After 33 RTT: cwnd=8 MSScwnd = 8\text{ MSS} (reaches ssthreshssthresh)
  2. Congestion Avoidance Phase (additive growth of 1 MSS1\text{ MSS} per RTT):
    • After 44 RTT: cwnd=9 MSScwnd = 9\text{ MSS}
    • After 55 RTT: cwnd=10 MSScwnd = 10\text{ MSS}
    • …\dots
    • After nn RTT: cwnd=8+(n−3) MSScwnd = 8 + (n - 3)\text{ MSS}

To reach the target congestion window of 32 KB32\text{ KB} (16 MSS16\text{ MSS}):

16=8+(n−3)  ⟹  n=11 RTTs16 = 8 + (n - 3) \implies n = 11\text{ RTTs}

Given that 1 RTT=100 msec1\text{ RTT} = 100\text{ msec}, the total time taken is:

Total Time=11×100 msec=1100 msec\text{Total Time} = 11 \times 100\text{ msec} = 1100\text{ msec}

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