GATE CS 2026 Set 2 Q48 — Transport Layer: Flow Control, Congestion Control, UDP, TCP, Sockets

To find the number of transmission rounds required to reach the congestion avoidance phase, we trace the growth of the TCP congestion window ($cwnd$) starting from the initial state:

• Given parameters:
  • Maximum Segment Size ($ext{MSS}$) = $2 ext{ KB}$
  • Slow start threshold ($ext{ssthresh}$) = $16 ext{ KB}$
  • Receiver advertised window ($ext{rwnd}$) = $48 ext{ KB}$
• Initial State: The connection starts in the Slow Start phase with an initial congestion window of $1 ext{ MSS}$: $$ext{cwnd}_1 = 1 imes ext{MSS} = 2 ext{ KB}$$

We trace the transmission rounds step-by-step:

1. Round 1: $ext{cwnd} = 2 ext{ KB}$. Since $ext{cwnd} < ext{ssthresh}$ ($2 ext{ KB} < 16 ext{ KB}$), the sender is in the slow start phase. After successful transmission and acknowledgment of all segments, $ext{cwnd}$ doubles: $$ext{cwnd} = 2 imes 2 ext{ KB} = 4 ext{ KB}$$
2. Round 2: $ext{cwnd} = 4 ext{ KB}$. Since $ext{cwnd} < ext{ssthresh}$ ($4 ext{ KB} < 16 ext{ KB}$), the sender remains in slow start. After successful ACKs, $ext{cwnd}$ doubles: $$ext{cwnd} = 2 imes 4 ext{ KB} = 8 ext{ KB}$$
3. Round 3: $ext{cwnd} = 8 ext{ KB}$. Since $ext{cwnd} < ext{ssthresh}$ ($8 ext{ KB} < 16 ext{ KB}$), the sender remains in slow start. After successful ACKs, $ext{cwnd}$ doubles: $$ext{cwnd} = 2 imes 8 ext{ KB} = 16 ext{ KB}$$
4. Round 4: $ext{cwnd} = 16 ext{ KB}$. Now, the congestion window has reached the slow start threshold ($ext{cwnd} = ext{ssthresh} = 16 ext{ KB}$). At this point, the TCP congestion control algorithm transitions to the Congestion Avoidance phase.

Thus, the $4^{ ext{th}}$ round of transmission is the round in which the congestion avoidance phase is reached and active.