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TCP Flow and Congestion Control

GATE CSE 2020 · Q55

NAT2 marksMedium

Time

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=0t = 0. Assume that there are no timeouts and errors during transmission. Then the size of the congestion window (in KB) at time t+60 mst + 60\text{ ms} after all acknowledgements are processed is ________.

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Official answer: 44

TCP Flow and Congestion Control

Given parameters:

  • Round Trip Time (RTT) = 6 ms6\text{ ms}

  • Maximum Segment Size (MSS) = 2 KB2\text{ KB}

  • Slow-start threshold (ssthresh) = 32 KB32\text{ KB}

  • Receiver advertised window (rwnd) = 50 KB50\text{ KB}

  • Total time elapsed = 60 ms=606=10 RTTs60\text{ ms} = \frac{60}{6} = 10\text{ RTTs}

Congestion Window Progression:

At t=0t = 0, the initial congestion window is:

cwnd=1 MSS=2 KB\text{cwnd} = 1\text{ MSS} = 2\text{ KB}

Phase 1: Slow-Start (cwnd doubles each RTT until it reaches ssthresh = 32 KB32\text{ KB}):

  • After 1st1^{\text{st}} RTT (t=6 mst = 6\text{ ms}): cwnd=2×2=4 KB\text{cwnd} = 2 \times 2 = 4\text{ KB}

  • After 2nd2^{\text{nd}} RTT (t=12 mst = 12\text{ ms}): cwnd=2×4=8 KB\text{cwnd} = 2 \times 4 = 8\text{ KB}

  • After 3rd3^{\text{rd}} RTT (t=18 mst = 18\text{ ms}): cwnd=2×8=16 KB\text{cwnd} = 2 \times 8 = 16\text{ KB}

  • After 4th4^{\text{th}} RTT (t=24 mst = 24\text{ ms}): cwnd=2×16=32 KB=ssthresh\text{cwnd} = 2 \times 16 = 32\text{ KB} = \text{ssthresh}

Phase 2: Congestion Avoidance (cwnd increases linearly by 1 MSS=2 KB1\text{ MSS} = 2\text{ KB} per RTT):

  • After 5th5^{\text{th}} RTT (t=30 mst = 30\text{ ms}): cwnd=32+2=34 KB\text{cwnd} = 32 + 2 = 34\text{ KB}

  • After 6th6^{\text{th}} RTT (t=36 mst = 36\text{ ms}): cwnd=34+2=36 KB\text{cwnd} = 34 + 2 = 36\text{ KB}

  • After 7th7^{\text{th}} RTT (t=42 mst = 42\text{ ms}): cwnd=36+2=38 KB\text{cwnd} = 36 + 2 = 38\text{ KB}

  • After 8th8^{\text{th}} RTT (t=48 mst = 48\text{ ms}): cwnd=38+2=40 KB\text{cwnd} = 38 + 2 = 40\text{ KB}

  • After 9th9^{\text{th}} RTT (t=54 mst = 54\text{ ms}): cwnd=40+2=42 KB\text{cwnd} = 40 + 2 = 42\text{ KB}

  • After 10th10^{\text{th}} RTT (t=60 mst = 60\text{ ms}): cwnd=42+2=44 KB\text{cwnd} = 42 + 2 = 44\text{ KB}

Since cwnd=44 KB≤rwnd=50 KB\text{cwnd} = 44\text{ KB} \le \text{rwnd} = 50\text{ KB}, the congestion window size is 44 KB.

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This TCP Flow and Congestion Control previous year question is Q55 of the GATE CSE 2020 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. 6 of them are 2-mark NAT questions like this one, and it is the only one in GATE CSE 2020.

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GateAI solution cardGATE CSE 2020 · Q55 · NAT

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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