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?

The correct option is 10 ≤ t < 11.

Step-by-Step Explanation:

In TCP congestion control, the transmission starts in the slow-start phase. During slow-start, the congestion window size ($\mathrm{cwnd}$) starts at 1 segment and doubles every Round-Trip Time (RTT) until it reaches the slow-start threshold ($\mathrm{ssthresh}$).

We are given the following parameters:

• Initial congestion window (${\mathrm{cwnd}}_1$) = 1 segment
• Slow-start threshold ($\mathrm{ssthresh}$) = 10,000 segments
• Round-Trip Time ($\mathrm{RTT}$) = 1 millisecond (ms)
�� The segments are numbered starting from 1, and no segments are lost.
• The sender always has data ready to send.

Let us analyze the window size and the cumulative number of segments sent in each RTT interval:

• Interval 1 (Time t = 0 to 1 ms):
The congestion window size is ${\mathrm{cwnd}}_1=2^0=1$ segment.
Segment sent in this interval: Segment 1.
Cumulative segments sent: 1.

• Interval 2 (Time t = 1 to 2 ms):
Upon receiving the ACK for the first segment at 1 ms, the window size doubles: ${\mathrm{cwnd}}_2=2^1=2$ segments.
Segments sent: Segments 2 to 3.
Cumulative segments sent: 1 + 2 = 3.

• Interval 3 (Time t = 2 to 3 ms):
The window size doubles: ${\mathrm{cwnd}}_3=2^2=4$ segments.
Segments sent: Segments 4 to 7.
Cumulative segments sent: 3 + 4 = 7.

• Interval 4 (Time t = 3 to 4 ms):
The window size is ${\mathrm{cwnd}}_4=2^3=8$ segments.
Segments sent: Segments 8 to 15.
Cumulative segments sent: 7 + 8 = 15.

• Interval 5 (Time t = 4 to 5 ms):
The window size is ${\mathrm{cwnd}}_5=2^4=16$ segments.
Segments sent: Segments 16 to 31.
Cumulative segments sent: 15 + 16 = 31.

• Interval 6 (Time t = 5 to 6 ms):
The window size is ${\mathrm{cwnd}}_6=2^5=32$ segments.
Segments sent: Segments 32 to 63.
Cumulative segments sent: 31 + 32 = 63.

• Interval 7 (Time t = 6 to 7 ms):
The window size is segments.
Segments sent: Segments 64 to 127.
Cumulative segments sent: 63 + 64 = 127.

• Interval 8 (Time t = 7 to 8 ms):
The window size is ${\mathrm{cwnd}}_8=2^7=128$ segments.
Segments sent: Segments 128 to 255.
Cumulative segments sent: 127 + 128 = 255.

• Interval 9 (Time t = 8 to 9 ms):
The window size is ${\mathrm{cwnd}}_9=2^8=256$ segments.
Segments sent: Segments 256 to 511.
Cumulative segments sent: 255 + 256 = 511.

• Interval 10 (Time t = 9 to 10 ms):
The window size is ${\mathrm{cwnd}}_{10}=2^9=512$ segments.
Segments sent: Segments 512 to 1023.
Cumulative segments sent: 511 + 512 = 1023.

• Interval 11 (Time t = 10 to 11 ms):
The window size is ${\mathrm{cwnd}}_{11}=2^{10}=1024$ segments.
Segments sent in this window range from segment 1024 to 2047.
Cumulative segments sent: 1023 + 1024 = 2047.

Since the threshold is 10,000 segments, the TCP sender is still in the slow-start phase because the window size of 1024 is well below the threshold of 10,000.

We are looking for the transmission start time $t$ of segment number 2000.
Since segment 2000 falls within the range [1024, 2047], it is transmitted during Interval 11, which spans from 10 ms to 11 ms.

Thus, the transmission of segment 2000 starts at a time $t$ such that:

$10\le t<11$