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 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. 9 ≤ 𝑡 < 10
• B. 10 ≤ 𝑡 < 11
• C. 11 ≤ 𝑡 < 12
• D. 12 ≤ 𝑡 < 13

Hint:

In slow start, cwnd doubles every RTT starting from 1. Cumulative segments sent after round $k$ equal $2^k - 1$. Find the round where segment 2000 falls, then multiply (round number - 1) by the RTT to get the start time.

Answer 1

The Correct Option is B

This question tests the TCP slow-start phase of congestion control, where the congestion window (cwnd) grows exponentially every round-trip time (RTT) until it reaches the slow-start threshold (ssthresh).

Step 1: Setup. ssthresh = 10000 segments, RTT = 1 ms, cwnd starts at 1 segment and doubles every RTT (slow start). Since 10000 is very large, we stay in slow start for a long time, so we only need to track cwnd growth until segment 2000 is reached.

Step 2: Segments sent per round. In round \(k\) (starting from round 1, \(k = 1, 2, 3, \ldots\)), cwnd = \(2^{k-1}\), so \(2^{k-1}\) new segments are transmitted in that round.

Step 3: Cumulative segments after each round. The total number of segments sent by the end of round \(k\) is:

\[ 1 + 2 + 4 + \cdots + 2^{k-1} = 2^k - 1 \]

Round	cwnd	Segments sent	Cumulative total
1	1	1	1
2	2	2-3	3
3	4	4-7	7
4	8	8-15	15
5	16	16-31	31
6	32	32-63	63
7	64	64-127	127
8	128	128-255	255
9	256	256-511	511
10	512	512-1023	1023
11	1024	1024-2047	2047

Step 4: Locate segment 2000. After round 10, cumulative segments sent = 1023. In round 11 (cwnd = 1024), segments numbered 1024 through 2047 are transmitted. Since \(1024 \le 2000 \le 2047\), segment number 2000 is sent during round 11.

Step 5: Find the start time of round 11. Round 1 starts at \(t = 0\). Each subsequent round begins exactly one RTT (1 ms) after the previous one. Round 11 therefore starts at:

\[ t = (11 - 1) \times 1\ \text{ms} = 10\ \text{ms} \]

So the transmission of segment number 2000 starts at \(t = 10\) ms, which satisfies \(10 \le t < 11\).

Final Answer:

\[ \boxed{10 \le t < 11 \ \text{(Option B)}} \]