Question

A TCP sender successfully establishes a connection with a TCP receiver and starts the transmission of segments. The TCP congestion control mechanism's slow-start threshold is set to 1000010000 segments. Assume that the round-trip time is fixed at 11 millisecond. Assume that the sender always has data to send, the segments are numbered from 11, and no segment is lost. Let tt denote the time (in milliseconds) at which the transmission of segment number 20002000 starts.

Which one of the following options is correct?

Choose one option
Official answer and solution
Official answer
B
From
Official GATE 2026 answer key

Solution

Track the exponential growth of the congestion window (cwnd=2k\text{cwnd} = 2^k segments) at each millisecond t=kt = k during TCP slow start.

Since the slow-start threshold is 10000>200010000 > 2000 and no loss occurs, TCP remains in slow start throughout, doubling its congestion window every RTT=1 ms\text{RTT} = 1\text{ ms}:

  • At t=0 mst = 0\text{ ms} (cwnd=1\text{cwnd} = 1): sends segment 11.
  • At t=1 mst = 1\text{ ms} (cwnd=2\text{cwnd} = 2): sends segments 22 to 33.
  • At t=2 mst = 2\text{ ms} (cwnd=4\text{cwnd} = 4): sends segments 44 to 77.
  • In general, in the round starting at t=k mst = k\text{ ms} (cwnd=2k\text{cwnd} = 2^k), the sender transmits segments 2k2^k through 2k+1−12^{k+1} - 1.

For k=10k = 10 (t=10 mst = 10\text{ ms}), the sender transmits segments 210=10242^{10} = 1024 through 211−1=20472^{11} - 1 = 2047. Since 1024≤2000≤20471024 \le 2000 \le 2047, segment 20002000 is transmitted during the interval 10≤t<1110 \le t < 11.

Final answer

Option (B)

Common trap

Being off by 1 RTT due to starting the first transmission at t=1t = 1 instead of t=0t = 0.

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

Source

Paper
GATE 2026 Computer Science and Information Technology - Session 1 (CS1)
Question
44 of 65
Section
Computer Science and Information Technology

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