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: 1K=$2^{10}$

Given parameters:

• $\text{MSS} = 2\text{ KB}$

• $\text{ssthresh} = 16\text{ KB} = 8\text{ MSS}$

• Receiver advertised window ($\text{rwnd}$) = $48\text{ KB}$

Initially, the TCP sender starts in the Slow Start phase with $\text{CWND} = 1\text{ MSS} = 2\text{ KB}$.

In the slow start phase, the congestion window doubles every round of transmission (assuming all ACKs are received successfully):

• Round 1: Transmit $\text{CWND} = 1\text{ MSS} = 2\text{ KB}$. Upon receiving ACK, $\text{CWND}$ increases to $2\text{ MSS} = 4\text{ KB}$.

• Round 2: Transmit $\text{CWND} = 2\text{ MSS} = 4\text{ KB}$. Upon receiving ACKs, $\text{CWND}$ increases to $4\text{ MSS} = 8\text{ KB}$.

• Round 3: Transmit $\text{CWND} = 4\text{ MSS} = 8\text{ KB}$. Upon receiving ACKs, $\text{CWND}$ increases to $8\text{ MSS} = 16\text{ KB}$.

• Round 4: $\text{CWND} = 16\text{ KB} = \text{ssthresh}$. At this point, $\text{CWND} \ge \text{ssthresh}$, so the connection reaches and enters the Congestion Avoidance phase during this round of transmission.

Therefore, the number of transmission rounds required to reach the congestion avoidance phase is 4.