GATE 2026 CS (CS1) – Question 44
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?
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Correct answer: (B) $10 \le t < 11$
Explanation
TCP starts in the slow-start phase with congestion window $CWND = 1$ segment. Since $CWND < \text{ssthresh} = 10000$, $CWND$ doubles every round-trip time ($RTT = 1\text{ ms}$).
Let us trace the segments transmitted in each RTT interval $[k - 1, k)$ ms for $k = 1, 2, \dots$:
- $k = 1$ ($t \in [0, 1)$): $CWND = 1$. Transmits segment 1. Cumulative segments = 1.
- $k = 2$ ($t \in [1, 2)$): $CWND = 2$. Transmits segments 2, 3. Cumulative = 3.
- $k = 3$ ($t \in [2, 3)$): $CWND = 4$. Transmits segments 4 to 7. Cumulative = 7.
- In general, during RTT $k$, $CWND = 2^{k-1}$. The cumulative segments sent up to the end of RTT $k$ is:
$$\sum_{i=1}^k 2^{i-1} = 2^k - 1$$
Evaluating cumulative totals:
- After $k = 10$ RTTs ($t = 10\text{ ms}$): cumulative sent = $2^{10} - 1 = 1023$ segments.
- After $k = 11$ RTTs ($t = 11\text{ ms}$): cumulative sent = $2^{11} - 1 = 2047$ segments.
The segments sent during RTT 11 (time interval $10 \le t < 11\text{ ms}$) are segments 1024 through 2047.
Since segment 2000 falls within this batch, its transmission begins in the interval $10 \le t < 11\text{ ms}$.
Therefore, option (B) is correct.