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. \(9 \le t < 10\)
• B. \(10 \le t < 11\)
• C. \(11 \le t < 12\)
• D. \(12 \le t < 13\)

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.

Answer 1

The Correct Option is B

Step 1: Recall the TCP slow-start model.
In TCP slow start, the sender begins with congestion window \(cwnd = 1\) segment. All segments admitted by the current window are treated as sent at the same instant (the idealized GATE model), and after one round-trip time (RTT) the sender receives the acknowledgements for that batch, which doubles the congestion window. Since the slow-start threshold, 10000 segments, is far larger than any window size reached in this problem, the window keeps doubling every RTT throughout -- congestion avoidance never kicks in.

Step 2: Set up the round-by-round timeline.
Let round \(n\) (for \(n = 0, 1, 2, \ldots\)) start at time \(t = n\) ms, since RTT \(=1\) ms and round \(0\) begins the instant data transmission starts (\(t=0\)). In round \(n\), \(cwnd = 2^n\) segments are transmitted, with segment numbers running from \(2^n\) to \(2^{n+1}-1\) (because segments are numbered from 1, and all earlier rounds together already sent \(1+2+4+\cdots+2^{n-1} = 2^n - 1\) segments).

Step 3: Build the round table.

Round n | Start time t=n ms | cwnd=2^n | Segment numbers sent
0       | 0                 | 1        | 1
1       | 1                 | 2        | 2 - 3
2       | 2                 | 4        | 4 - 7
3       | 3                 | 8        | 8 - 15
4       | 4                 | 16       | 16 - 31
5       | 5                 | 32       | 32 - 63
6       | 6                 | 64       | 64 - 127
7       | 7                 | 128      | 128 - 255
8       | 8                 | 256      | 256 - 511
9       | 9                 | 512      | 512 - 1023
10      | 10                | 1024     | 1024 - 2047

Step 4: Locate segment number 2000.
By the end of round 9, a cumulative total of \(2^{10}-1 = 1023\) segments have been sent (segments 1 through 1023). Round 10 starts at \(t = 10\) ms and sends segments numbered \(1024\) through \(2047\) (since \(2^{11}-1 = 2047\)). Since \(1024 \le 2000 \le 2047\), segment number 2000 is transmitted in round 10, and every segment of round 10 is treated as starting transmission at \(t=10\) ms.

Step 5: Match to the given range.
\(t = 10\), which satisfies \(10 \le t < 11\). \[ \boxed{10 \le t < 11} \]