Topic: Computer Networks
A host in one organisation is connected to another host in other organisation using 1 Gbps link with 25m sec RTT. Network administrator of this network tuned the packet size to 1000 bytes to transfer large files optimally over this network. In order to have channel utilisation > 96%, how big window size (in terms of packets) should be (assuming ACK packets have no impact)?
- A. 2000 packets
- B. 2400 packets
- C. 3000 packets
- D. 4000 packets
Answer: C. 3000 packets
Explanation: Bandwidth (B) = 1 Gbps = $10^9$ bps. Packet size (L) = 1000 bytes = 8000 bits. Transmission time ($T_t$) = L / B = $8000 / 10^9 = 8 \mu s$. RTT = $25 ms$. Channel utilization $U = \frac{W \times T_t}{RTT + T_t}$. We require $U > 0.96$. $0.96 = \frac{W \times 8 \times 10^{-6}}{25.008 \times 10^{-3}} \implies W \approx 3000$. Thus, a window size slightly over 3000 achieves > 96% utilization, making 3000 the boundary threshold representing optimal efficiency.