Question

Suppose the round trip propagation delay for a 10 Mbps Ethernet having a 48-bit jamming signal is 46.4 ms. What is the minimum frame size?

Answer 1

Transmission Speed = 10Mbps.

Round trip propagation delay = 46.4 ms

The minimum frame size = (Round Trip Propagation Delay) * (Transmission Speed) = 10*(10^6)*46.4*(10^-3) = 464 * 10^3 = 464 Kbit