###### NIC-NIELIT Scientist-B 2020

October 4, 2023October 4, 2023

# Computer-Networks

Question 45 |

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 45 Explanation:

Given RTT = 46.4 μs, B.w. = 10 Mbps

Round trip propagation delay is RTT = 2*T

Minimum frame size of Ethernet can be found by using formula T

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

Minimum frame size of Ethernet can be found by using formula T

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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