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A network using csma/cd has a bandwidth of 10 mbps. If the maximum propagation time including the device delays is 25.6 s, what is the minimum size of the frame? |
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Answer» Explanation: The minimum frame transmission time is Tfr = 2 × TP = 51.2 μs. This means, in the WORST CASE, a station needs to transmit for a period of 51.2 μs to detect the collision.The minimum size of the frame is 10 Mbps × 51.2 μs = 512 bits or 64 bytes. This is actually the minimum size of the frame for Standard ETHERNET... |
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