In standard Ethernet with a transmission rate of 10 Mbps, assume the length of the medium is 2500 m and the size of a frame is 512 bytes. The propagation speed of a signal in the cable is 2 × 108 m/s. Calculate the transmission delay and propagation delay.
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Correct answer: B. 51.2 μs and 12.5 μs
Explanation
Propagation = 2500 / (2 × 108) = 12.5 μs. Transmission of the 512-bit minimum frame = 512 / 107 = 51.2 μs (the paper says bytes, but only the 512-bit value matches an option).More Data Communication and Computer Networks questions
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