NIC-NIELIT Scientist-B 2020
October 4, 2023October 4, 2023
Ethernet
| Question 9 |
Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 μs. The minimum frame size is:
| 94 | |
| 416 | |
| 464 | |
| 512 |
Question 9 Explanation:
Given RTT = 46.4 μs, B.w. = 10 Mbps
Round trip propagation delay is RTT = 2*Tp
Minimum frame size of Ethernet can be found by using formula Tt = 2*Tp
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*Tp
Minimum frame size of Ethernet can be found by using formula Tt = 2*Tp
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 9 Explanation:
Given RTT = 46.4 μs, B.w. = 10 Mbps
Round trip propagation delay is RTT = 2*Tp
Minimum frame size of Ethernet can be found by using formula Tt = 2*Tp
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*Tp
Minimum frame size of Ethernet can be found by using formula Tt = 2*Tp
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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