Suppose the round trip propagation delay for a 10 Mbps Ethernet

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

1. 94
2. 416
3. 464
4. 512

Correct Option: D

We are given with Jamming Signal = 48 bit
Round trip propagation delay = 46.4 ps
Now, To get actual round propagation time, subtract transmission time of 48 bit We get actual round propagation time = 51.2 µs.
Now, the frame length must be such that to its transmission time must be more than 51.2 µs, so minimum frame length = 51.2 * 10^–6 × 10 × 10^–6 = 512 bits.