Question:medium
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 10000 segments. Assume that the round-trip time is fixed at 1 millisecond. Assume that the sender always has data to send, the segments are numbered from 1, and no segment is lost. Let \(t\) denote the time (in milliseconds) at which the transmission of segment number 2000 starts.
Which one of the following options is correct?
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 10000 segments. Assume that the round-trip time is fixed at 1 millisecond. Assume that the sender always has data to send, the segments are numbered from 1, and no segment is lost. Let \(t\) denote the time (in milliseconds) at which the transmission of segment number 2000 starts.
Which one of the following options is correct?
Which one of the following options is correct?
Show Hint
The congestion window doubles every RTT in slow start; find which power-of-two doubling round segment number 2000 falls into, then convert the round index to time using RTT = 1 ms.
Updated On: Jul 22, 2026
- \(9 \le t < 10\)
- \(10 \le t < 11\)
- \(11 \le t < 12\)
- \(12 \le t < 13\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Set up a closed-form condition instead of a full table.
With slow start doubling the window every RTT starting from \(cwnd=1\), the cumulative number of segments sent by the end of round \(n\) is \(S(n) = 2^{0}+2^{1}+\cdots+2^{n} = 2^{n+1}-1\). Segment number \(N\) falls in round \(n\), the smallest \(n\) for which \(2^{n+1}-1 \ge N\).
Step 2: Substitute \(N=2000\) and solve for \(n\).
We need the smallest \(n\) with \(2^{n+1} \ge 2001\). Testing powers of two: \(2^{10} = 1024\), so \(2^{10}-1 = 1023\), and \(1023 < 2000\), meaning round 9 is not enough. Next, \(2^{11} = 2048\), so \(2^{11}-1 = 2047 \ge 2000\), meaning round 10 is sufficient. So the smallest valid round is \(n=10\).
Step 3: Convert the round number to time.
Round \(n\) starts at time \(t = n \times \text{RTT} = n \times 1 = n\) ms (round 0 starts the instant transmission begins, at \(t=0\)). So round 10 starts at \(t=10\) ms.
Step 4: Sanity-check the boundary.
Round 10 carries segments \(1024\) to \(2047\); since \(2000\) lies strictly inside this range and is not the first segment of a later round, \(t=10\) ms is confirmed as the start time of segment 2000's transmission. \[ \boxed{10 \le t < 11} \]
With slow start doubling the window every RTT starting from \(cwnd=1\), the cumulative number of segments sent by the end of round \(n\) is \(S(n) = 2^{0}+2^{1}+\cdots+2^{n} = 2^{n+1}-1\). Segment number \(N\) falls in round \(n\), the smallest \(n\) for which \(2^{n+1}-1 \ge N\).
Step 2: Substitute \(N=2000\) and solve for \(n\).
We need the smallest \(n\) with \(2^{n+1} \ge 2001\). Testing powers of two: \(2^{10} = 1024\), so \(2^{10}-1 = 1023\), and \(1023 < 2000\), meaning round 9 is not enough. Next, \(2^{11} = 2048\), so \(2^{11}-1 = 2047 \ge 2000\), meaning round 10 is sufficient. So the smallest valid round is \(n=10\).
Step 3: Convert the round number to time.
Round \(n\) starts at time \(t = n \times \text{RTT} = n \times 1 = n\) ms (round 0 starts the instant transmission begins, at \(t=0\)). So round 10 starts at \(t=10\) ms.
Step 4: Sanity-check the boundary.
Round 10 carries segments \(1024\) to \(2047\); since \(2000\) lies strictly inside this range and is not the first segment of a later round, \(t=10\) ms is confirmed as the start time of segment 2000's transmission. \[ \boxed{10 \le t < 11} \]
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