###### Computer-Networks

October 9, 2023###### Computer-Networks

October 9, 2023# Ethernet

Question 1 |

A 2 km long broadcast LAN has 10^{7} bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2×10^{8} m/s. What is the minimum packet size that can be used on this network?

50 bytes | |

100 bytes | |

200 bytes | |

None of the above |

Question 1 Explanation:

Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. T

d= 2 km = 2 x 10

T

Let L bits be minimum size of frame, then T

Now, T

L/10

_{t}= 2T_{p}d= 2 km = 2 x 10

^{3}m, v = 2 x 10^{8}m/s, B= 10^{7}T

_{p}= d / v = 2 x 10^{3}/(2 x 10^{8}) seconds = 10^{-5}secondsLet L bits be minimum size of frame, then T

_{t}= t L / B = L / 10^{7}secondsNow, T

_{t}= 2T_{p}L/10

^{7}= 2 x 10^{-5}= 200 bits = (200 / 8) bytes = 25 bytesCorrect Answer: D

Question 1 Explanation:

Minimum packet size for a CSMA/CD LAN is the frame which cover whole RTT(round trip time). i.e. T

d= 2 km = 2 x 10

T

Let L bits be minimum size of frame, then T

Now, T

L/10

_{t}= 2T_{p}d= 2 km = 2 x 10

^{3}m, v = 2 x 10^{8}m/s, B= 10^{7}T

_{p}= d / v = 2 x 10^{3}/(2 x 10^{8}) seconds = 10^{-5}secondsLet L bits be minimum size of frame, then T

_{t}= t L / B = L / 10^{7}secondsNow, T

_{t}= 2T_{p}L/10

^{7}= 2 x 10^{-5}= 200 bits = (200 / 8) bytes = 25 bytes Subscribe

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