Consider a CSMA/CD network that transmits data at 100 Mbps over a 1 km cable with no repeater. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?
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Correct answer: D. 20000
Explanation
Minimum frame = 2 × Tp × bandwidth, so 1250 × 8 = 10,000 bits = 2 × Tp × 10⁸, giving Tp = 50 μs. Speed = 1 km / 50 μs = 20,000 km/s.More Data Communication and Computer Networks questions
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