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Suppose the r...

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Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 $μ$ s. The minimum frame size is:

A

464

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B

512

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C

416

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D

94

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Solution

The correct option is A 464
$Round trip propagation delay is 2 * T_{p}$ minimum frame size of ethernet can be found by using formula
$Frame size = (2 T_{p}) * (B a n d w i d t h)$
$= 46.4 m s * 10 M b p s$
$= 464 k b i t s$

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