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The round trip delay for a 100 Mbps Ethernet having 48 bit jamming signal is 64 * 10^-6 sec. What is the minimum frame size? 1. 400 bytes 2. 800 bytes 3. 600 bytes 4. 1400 bytes

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Best answer
9 9 votes
2 * propagation delay  + transmission time for jamming signal <= transmission time for frame
64 μs + 48/100 μs <= transmission time for frame
64.48 μs <= l/100 μs
l >= 6448 bits >= 806 bytes.
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6 6 votes

L>= 2* Tp *B

L-> Length of frame

Tp -> Propagation time = RTT/2 = 64 * 10- 6/2 = 32 * 10-6

B -> Bandwidth = 108 bps

So, L>= 2 * 32 * 10-6 * 108 

     L>= 6400 b

Converting in Bytes 6400/8= 800 B 

So option 2

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