In a stop-and-wait protocol, the sender sends a frame and waits for an ACK. If the round-trip time is 40 ms and the transmission time is 10 ms, what is the bandwidth-delay product?
Q1. In a stop-and-wait protocol, the sender sends a frame and waits for an ACK. If the round-trip time is 40 ms and the transmission time is 10 ms, what is the bandwidth-delay product?
Answer: 800 bits
Explanation: Bandwidth-delay product = bandwidth * round-trip time. Assuming bandwidth is 20 kbps, product = 20 kbps * 40 ms = 800 bits.