Question

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 ___________. (answer in integer)
Note: \(1\mathrm{K}=2^{10}\)

Hint:

In slow start, cwnd starts at 1 MSS and doubles every round (1, 2, 4, 8 MSS or 2, 4, 8, 16 KB). Congestion avoidance begins once cwnd equals ssthresh (16 KB = 8 MSS). Count the rounds needed to reach that value; the 48 KB advertised window is not the limiting constraint here.

Answer 1

Correct Answer: 4

This question is about TCP slow start and how many rounds of transmission are needed before the sender's congestion window (cwnd) grows large enough to trigger the switch from slow start to congestion avoidance.

Given data:

• Receiver advertised window = 48 KB
• Maximum Segment Size (MSS) = 2 KB
• Slow start threshold (ssthresh) = 16 KB
• No timeouts or duplicate acknowledgements occur

Step 1: Recall the slow start rule.

In slow start, the sender begins with \( cwnd = 1 \times MSS \). After every round trip in which all segments of the current window are acknowledged, cwnd is doubled (exponential growth), as long as \( cwnd < ssthresh \). Once \( cwnd \) reaches or exceeds ssthresh, the connection moves into the congestion avoidance phase, where growth becomes linear (additive increase).

Step 2: Track cwnd round by round in KB.

\[ \text{Round 1: } cwnd = 1 \times MSS = 2\ KB \]\[ \text{Round 2: } cwnd = 2 \times MSS = 4\ KB \]\[ \text{Round 3: } cwnd = 4 \times MSS = 8\ KB \]\[ \text{Round 4: } cwnd = 8 \times MSS = 16\ KB \]

Step 3: Compare cwnd with ssthresh at each round.

After round 1, cwnd = 2 KB, still less than ssthresh (16 KB), so slow start continues.
After round 2, cwnd = 4 KB, still less than ssthresh, slow start continues.
After round 3, cwnd = 8 KB, still less than ssthresh, slow start continues.
In round 4, cwnd becomes 16 KB, which is exactly equal to ssthresh. Once \( cwnd = ssthresh \), the TCP congestion control algorithm switches to the congestion avoidance phase.

Step 4: Note the role of the advertised window.

The receiver's advertised window of 48 KB is much larger than 16 KB, so it never restricts cwnd during this slow start process; it only matters later in congestion avoidance or when cwnd would otherwise exceed it.

Step 5: Count the rounds.

The doubling of cwnd took place over exactly 4 rounds of transmission (rounds 1 through 4) for cwnd to reach the ssthresh value of 16 KB, after which the algorithm enters congestion avoidance.

\[ \boxed{\text{Number of rounds required} = 4} \]