NICNIELIT ScientistB 2020
October 4, 2023October 4, 2023
ComputerNetworks
Question 45

Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48bit jamming signal is 46.4 μs. The minimum frame size is:
94


416


464


512

Question 45 Explanation:
Given RTT = 46.4 μs, B.w. = 10 Mbps
Round trip propagation delay is RTT = 2*T_{p}
Minimum frame size of Ethernet can be found by using formula T_{t} = 2*T_{p}
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L = 464 Kbits
It has nothing to do with jamming signal.
Round trip propagation delay is RTT = 2*T_{p}
Minimum frame size of Ethernet can be found by using formula T_{t} = 2*T_{p}
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L = 464 Kbits
It has nothing to do with jamming signal.
Correct Answer: C
Question 45 Explanation:
Given RTT = 46.4 μs, B.w. = 10 Mbps
Round trip propagation delay is RTT = 2*T_{p}
Minimum frame size of Ethernet can be found by using formula T_{t} = 2*T_{p}
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L = 464 Kbits
It has nothing to do with jamming signal.
Round trip propagation delay is RTT = 2*T_{p}
Minimum frame size of Ethernet can be found by using formula T_{t} = 2*T_{p}
Let L is minimum frame size. Then L / 10Mbps = 46.4 μs
L = 464 Kbits
It has nothing to do with jamming signal.
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