Skip to content

GATE CS 2012 Q45 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

Explain with AI

Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 22 MSS and the threshold at the start of the first transmission is 88 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

Choose one option
View worked solution

Answer: C

We can trace the congestion window size (extcwndext{cwnd}) and slow-start threshold (extssthreshext{ssthresh}) step-by-step through each transmission:

  • Initial State: extcwnd=2 MSSext{cwnd} = 2\text{ MSS}, extssthresh=8 MSSext{ssthresh} = 8\text{ MSS}.
  • 1st Transmission: extcwnd=2 MSSext{cwnd} = 2\text{ MSS}. Since extcwnd<ssthreshext{cwnd} < \text{ssthresh}, it is in the Slow Start phase. After successful transmission, extcwndext{cwnd} doubles to 4 MSS4\text{ MSS}.
  • 2nd Transmission: extcwnd=4 MSSext{cwnd} = 4\text{ MSS}. Since extcwnd<ssthreshext{cwnd} < \text{ssthresh}, it is in Slow Start. After successful transmission, extcwndext{cwnd} doubles to 8 MSS8\text{ MSS}.
  • 3rd Transmission: extcwnd=8 MSSext{cwnd} = 8\text{ MSS}. Since extcwnd≥ssthreshext{cwnd} \ge \text{ssthresh}, it enters the Congestion Avoidance phase. After successful transmission, extcwndext{cwnd} increases additively by 11 to 9 MSS9\text{ MSS}.
  • 4th Transmission: extcwnd=9 MSSext{cwnd} = 9\text{ MSS}. After successful transmission, extcwndext{cwnd} increases to 10 MSS10\text{ MSS}.
  • 5th Transmission: extcwnd=10 MSSext{cwnd} = 10\text{ MSS}. A timeout occurs during this transmission:
    • New extssthresh=max⁡(cwnd/2,2)=10/2=5 MSSext{ssthresh} = \max(\text{cwnd}/2, 2) = 10/2 = 5\text{ MSS}.
    • extcwndext{cwnd} is reset to the initial slow start value of 2 MSS2\text{ MSS}.
  • 6th Transmission: extcwnd=2 MSSext{cwnd} = 2\text{ MSS}. Since extcwnd<ssthreshext{cwnd} < \text{ssthresh} (2<52 < 5), it is in Slow Start. After successful transmission, extcwndext{cwnd} doubles to 4 MSS4\text{ MSS}.
  • 7th Transmission: extcwnd=4 MSSext{cwnd} = 4\text{ MSS}. Since extcwnd<ssthreshext{cwnd} < \text{ssthresh}, it is in Slow Start. Upon success, doubling would exceed extssthreshext{ssthresh} (55), so extcwndext{cwnd} is capped at extssthresh=5 MSSext{ssthresh} = 5\text{ MSS}.
  • 8th Transmission: extcwnd=5 MSSext{cwnd} = 5\text{ MSS}. Since extcwnd≥ssthreshext{cwnd} \ge \text{ssthresh}, it is in Congestion Avoidance. After successful transmission, extcwndext{cwnd} increases to 6 MSS6\text{ MSS}.
  • 9th Transmission: extcwnd=6 MSSext{cwnd} = 6\text{ MSS}. After successful transmission, extcwndext{cwnd} increases to 7 MSS7\text{ MSS}.
  • 10th Transmission: extcwnd=7 MSSext{cwnd} = 7\text{ MSS}.

Thus, the congestion window size at the start of the tenth transmission (or at the end of the ninth transmission) is 7extMSS7 ext{ MSS}.

Free practice · Your answers on this page are not saved.

Published
Report a correction
Keep exploringMore Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets questions