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TCP Flow and Congestion Control

GATE CSE 2012 · Q45

MCQ2 marksMedium

Time

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.

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TCP Flow and Congestion Control

Let us trace the congestion window (cwndcwnd) and slow-start threshold (ssthreshssthresh):

  • Initial Conditions:

    • Starting cwnd=2 MSScwnd = 2\text{ MSS}

    • Initial ssthresh=8 MSSssthresh = 8\text{ MSS}

  • Transmission Schedule:

    • Transmission 1: cwnd=2 MSScwnd = 2\text{ MSS} (Slow Start). Successful   ⟹  cwnd\implies cwnd doubles to 4 MSS4\text{ MSS}.

    • Transmission 2: cwnd=4 MSScwnd = 4\text{ MSS} (Slow Start). Successful   ⟹  cwnd\implies cwnd doubles to 8 MSS8\text{ MSS} (reaches ssthreshssthresh).

    • Transmission 3: cwnd=8 MSScwnd = 8\text{ MSS} (Congestion Avoidance). Successful   ⟹  cwnd\implies cwnd increases by 11 to 9 MSS9\text{ MSS}.

    • Transmission 4: cwnd=9 MSScwnd = 9\text{ MSS} (Congestion Avoidance). Successful   ⟹  cwnd\implies cwnd increases by 11 to 10 MSS10\text{ MSS}.

    • Transmission 5: cwnd=10 MSScwnd = 10\text{ MSS}. A timeout occurs.

  • Reaction to Timeout:

    • ssthresh=max⁡(2,⌊cwnd2⌋)=102=5 MSSssthresh = \max\left(2, \left\lfloor\frac{cwnd}{2}\right\rfloor\right) = \frac{10}{2} = 5\text{ MSS}.

    • cwndcwnd resets to the slow-start window (typically 1 MSS1\text{ MSS}, or 2 MSS2\text{ MSS} as stated at the start of the slow-start phase).

  • Subsequent Transmissions (assuming reset to 2 MSS2\text{ MSS}):

    • Transmission 6: cwnd=2 MSScwnd = 2\text{ MSS} (Slow Start)   ⟹  cwnd\implies cwnd becomes 4 MSS4\text{ MSS}.

    • Transmission 7: cwnd=4 MSScwnd = 4\text{ MSS} (Slow Start)   ⟹  cwnd\implies cwnd reaches threshold 5 MSS5\text{ MSS}.

    • Transmission 8: cwnd=5 MSScwnd = 5\text{ MSS} (Congestion Avoidance)   ⟹  cwnd\implies cwnd becomes 6 MSS6\text{ MSS}.

    • Transmission 9: cwnd=6 MSScwnd = 6\text{ MSS} (Congestion Avoidance)   ⟹  cwnd\implies cwnd becomes 7 MSS7\text{ MSS}.

    • Transmission 10: cwnd=7 MSScwnd = 7\text{ MSS} (Congestion Avoidance)   ⟹  cwnd\implies cwnd at the end of 10th transmission becomes 8 MSS8\text{ MSS}.

Due to ambiguities in whether cwndcwnd resets to 1 MSS1\text{ MSS} (RFC standard) or 2 MSS2\text{ MSS}, and the interpretation of window size "during" vs "at the end of" the tenth transmission, this question was awarded Marks to All in the official answer key.

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This TCP Flow and Congestion Control previous year question is Q45 of the GATE CSE 2012 paper. Find more GATE CSE PYQs on TCP Flow and Congestion Control in Computer Networks, or browse all GATE CSE PYQs.

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TCP Flow and Congestion Control in GATE CSE

TCP Flow and Congestion Control has 23 GATE CSE PYQs across 21 papers from 2012 to 2026. 2 of them are 2-mark MCQs like this one, and it is the only one in GATE CSE 2012.

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GateAI solution cardGATE CSE 2012 · Q45 · MCQ

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.

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