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
(A) 94
(B) 416
(C) 464
(D) 512
Answer: (D)
Explanation:
Td > 2 × Tp + Td( for jam signal)
Td > Rtt + ( Length of jam signal / Bandwidth )
Td > 46.4μs + ( 48 bit / 10 × 106 bits/sec)
Td > 46.4 × 10-6 sec + (4.8 × 10-6 sec)
Td > 51.2 × 10-6 sec
( Frame Size / Bandwidth) > 51.2 × 10-6 sec
Frame Size > 51.2 × 10-6 sec × Bandwidth
Frame Size > 51.2 × 10-6 sec × 10 × 106 bits/second
Frame Size > 512 bits
So , minimum frame size is 512 bits.
NOTE :-
Transmission delay = Td
Propagation delay = Tp
Round Trip Time = Rtt = 2 × Tp
Td = ( Frame Size / Bandwidth)