GATE CS Mastery
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GATE CS 2012 · Computer Networks

Question 45

MCQPrinted as CS Q.452 marksNo negative markingMedium (estimate)Marks to all

Question

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 2 MSS and the threshold at the start of the first transmission is 8 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
Official answer and solution
Official answer
Marks to all
From
Official GATE 2012 answer key
Why the marks went to every candidate

Official GATE answer key awarded Marks to All for this question.

Solution

Trace TCP congestion window evolution during slow start and congestion avoidance up to and after timeout.

  1. Initial parameters:
  • Start of slow start window: W1=2 MSSW_1 = 2\text{ MSS}.
  • Initial threshold: ssthresh=8 MSS\text{ssthresh} = 8\text{ MSS}.
  1. First 5 transmissions:
  • Trans 1: W=2 MSSW = 2\text{ MSS} (slow start). Successful →W=4 MSS\to W = 4\text{ MSS}.
  • Trans 2: W=4 MSSW = 4\text{ MSS} (slow start). Successful →W=8 MSS\to W = 8\text{ MSS}.
  • Trans 3: W=8 MSSW = 8\text{ MSS} (equals threshold, enters Congestion Avoidance). Successful →W=9 MSS\to W = 9\text{ MSS}.
  • Trans 4: W=9 MSSW = 9\text{ MSS}. Successful →W=10 MSS\to W = 10\text{ MSS}.
  • Trans 5: W=10 MSSW = 10\text{ MSS}. Timeout occurs!
  1. Timeout handling:
  • New threshold is set to max⁡(W/2,2)=10/2=5 MSS\max(W/2, 2) = 10/2 = 5\text{ MSS}.
  • The congestion window is reset.
  1. Reason for Marks to All: Standard RFC TCP resets the congestion window to 1 MSS, while some textbooks/interpretations reset it to the stated initial slow-start window of 2 MSS. Furthermore, ambiguity in whether 'end of tenth transmission' referred to the window size applied during transmission 10 or updated after transmission 10 led to different answers (7 MSS or 8 MSS). Due to these conflicting interpretations, GATE awarded Marks to All.
Final answer

Marks to All (Ambiguity in reset window and round numbering led to Marks to All; mathematically 7 or 8 MSS depending on interpretation).

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2012 Computer Science and Information Technology (CS)
Question
45 of 65, printed as CS Q.45
Section
Computer Science and Information Technology

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