suppose the round trip propagation delay for a 10 mbps ethernet having 48-bit jamming signal is 42.6 ms. what is the minimum frame size
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"if we keep the question in mind then, the minimal value of the frame size is 464.
Here transmission speed will be 10 Mbps
The total round-trip propagation delay will be 46.4 ms
Hence, the frame size = transmission speed * round-trip propagation speed
Frame size = 46.4 * (10^-3) * 10 * (10^6)
Frame size = 464 Kbits
"
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