Question:medium
Consider a new TCP connection between a sender and a receiver. The receiver advertised window is constant at 48 KB, the maximum segment size (MSS) is 2 KB, and the slow start threshold for TCP congestion control is 16 KB. Assume that there are no timeouts or duplicate acknowledgements. The number of rounds of transmission required for the congestion control algorithm of the TCP connection to reach the congestion avoidance phase is ______.
Note: \(1\text{K}=2^{10}\)
Consider a new TCP connection between a sender and a receiver. The receiver advertised window is constant at 48 KB, the maximum segment size (MSS) is 2 KB, and the slow start threshold for TCP congestion control is 16 KB. Assume that there are no timeouts or duplicate acknowledgements. The number of rounds of transmission required for the congestion control algorithm of the TCP connection to reach the congestion avoidance phase is ______.
Note: \(1\text{K}=2^{10}\)
Note: \(1\text{K}=2^{10}\)
Show Hint
In slow start, cwnd doubles every round starting from 1 MSS. Find how many doublings it takes for cwnd to reach the slow start threshold of 16 KB (8 MSS).
Updated On: Jul 22, 2026
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Correct Answer: 4
Solution and Explanation
Step 1: Write cwnd as a formula.
In slow start, after $n$ rounds the window size is $2^{n-1}$ MSS (since it starts at $1$ MSS in round $1$ and doubles each round after that).
Step 2: Turn ssthresh into MSS units.
MSS is $2$ KB, so ssthresh of $16$ KB equals $\dfrac{16}{2}=8$ MSS.
Step 3: Solve for n.
We need the smallest $n$ such that $2^{n-1} \ge 8$. Since $8=2^{3}$, we need $n-1=3$, so $n=4$.
Step 4: Confirm with the numbers.
$n=4$ gives $2^{3}=8$ MSS $=16$ KB, matching ssthresh exactly. The advertised window of $48$ KB ($24$ MSS) is never a limiting factor since the window hits ssthresh well before that size.
Step 5: State the switch point.
Once cwnd equals ssthresh at round $4$, the connection leaves slow start and enters congestion avoidance from the next round onward.
\[ \boxed{4} \]
In slow start, after $n$ rounds the window size is $2^{n-1}$ MSS (since it starts at $1$ MSS in round $1$ and doubles each round after that).
Step 2: Turn ssthresh into MSS units.
MSS is $2$ KB, so ssthresh of $16$ KB equals $\dfrac{16}{2}=8$ MSS.
Step 3: Solve for n.
We need the smallest $n$ such that $2^{n-1} \ge 8$. Since $8=2^{3}$, we need $n-1=3$, so $n=4$.
Step 4: Confirm with the numbers.
$n=4$ gives $2^{3}=8$ MSS $=16$ KB, matching ssthresh exactly. The advertised window of $48$ KB ($24$ MSS) is never a limiting factor since the window hits ssthresh well before that size.
Step 5: State the switch point.
Once cwnd equals ssthresh at round $4$, the connection leaves slow start and enters congestion avoidance from the next round onward.
\[ \boxed{4} \]
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