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

An alternate way to solve this is by directly working out at which power-of-two 'boundary' segment 2000 falls, rather than building the full round table.

Idea: In TCP slow start with an initial cwnd of 1 segment and doubling every RTT, after \(k\) full RTTs (i.e. at the start of round \(k+1\)), the sender has already transmitted \(2^k - 1\) segments, so the next segment sent is number \(2^k\).

Step 1: Since ssthresh = 10000 is large, the connection is still in slow start well past segment 2000, so exponential growth applies throughout.

Step 2: Find the smallest \(k\) such that \(2^k - 1 < 2000 \le 2^{k+1} - 1\), i.e. locate where 2000 sits between consecutive cumulative totals.

Compute powers of 2: \(2^{10} = 1024\), so cumulative segments after round 10 is \(2^{10} - 1 = 1023\). Next, \(2^{11} = 2048\), so cumulative segments after round 11 is \(2^{11} - 1 = 2047\).

Since \(1023 < 2000 \le 2047\), segment 2000 falls inside round 11 (segments numbered 1024 to 2047 are all sent in that round).

Step 3: Each round takes exactly one RTT = 1 ms, and round 1 begins at \(t = 0\). The start time of round \(n\) is \((n-1) \times \text{RTT}\). For round 11:

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

Step 4: Check the given ranges: \(10 \le t < 11\) is satisfied exactly, matching option B.

This confirms the answer without needing to individually track every round's segment range - only the doubling pattern of cumulative totals (\(2^k - 1\)) is needed.

Final Answer: \(10 \le t < 11\) (Option B)