Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (10^8…
2015
Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (10^8 bits per second) over a 1 km (kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?
Answer: D. 20000 — Key idea: the minimum frame must be long enough to detect collisions, so it must last at least the round‑trip propagation time. Convert the minimum frame size…
- A.
8000
- B.
10000
- C.
16000
- D.
20000
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Correct answer: D
Key idea: the minimum frame must be long enough to detect collisions, so it must last at least the round‑trip propagation time.
Convert the minimum frame size to bits: 1250 bytes × 8 = 10,000 bits.
Use the CSMA/CD requirement: min_bits = 2 × bitrate × propagation_delay. So propagation_delay = min_bits / (2 × bitrate) = 10,000 / (2 × 100,000,000) = 5×10^−5 s.
Compute signal speed: speed = distance / delay = 1 km / 5×10^−5 s = 20,000 km/s.
Therefore the correct signal speed in the cable is 20,000 km/s.
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