Question 65

Given parameters:

• $\text{RTT} = 6\text{ ms}$
• $\text{MSS} = 2\text{ KB}$
• $\text{ssthresh} = 32\text{ KB}$
• $\text{rwnd} = 50\text{ KB}$
• Time interval: $\Delta t = 60\text{ ms} \implies \frac{60\text{ ms}}{6\text{ ms}} = 10\text{ RTTs}$

Trace of $\text{cwnd}$ across 10 RTT rounds:

• Initially at $t = 0$: $\text{cwnd} = 1\text{ MSS} = 2\text{ KB}$. (Slow Start)
• Round 1 ($t = 6\text{ ms}$): cwnd doubles $\implies 4\text{ KB}$
• Round 2 ($t = 12\text{ ms}$): cwnd doubles $\implies 8\text{ KB}$
• Round 3 ($t = 18\text{ ms}$): cwnd doubles $\implies 16\text{ KB}$
• Round 4 ($t = 24\text{ ms}$): cwnd doubles $\implies 32\text{ KB}$
  • Here $\text{cwnd} = \text{ssthresh} = 32\text{ KB}$. TCP transitions from Slow Start to Congestion Avoidance (additive increase of $1\text{ MSS} = 2\text{ KB}$ per RTT).
• Round 5 ($t = 30\text{ ms}$): $\text{cwnd} = 32 + 2 = 34\text{ KB}$
• Round 6 ($t = 36\text{ ms}$): $\text{cwnd} = 34 + 2 = 36\text{ KB}$
• Round 7 ($t = 42\text{ ms}$): $\text{cwnd} = 36 + 2 = 38\text{ KB}$
• Round 8 ($t = 48\text{ ms}$): $\text{cwnd} = 38 + 2 = 40\text{ KB}$
• Round 9 ($t = 54\text{ ms}$): $\text{cwnd} = 40 + 2 = 42\text{ KB}$
• Round 10 ($t = 60\text{ ms}$): $\text{cwnd} = 42 + 2 = 44\text{ KB}$

At $t + 60\text{ ms}$, $\text{cwnd} = 44\text{ KB}$, which is strictly below $\text{rwnd} = 50\text{ KB}$. Therefore, the congestion window size is 44 KB.