A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal…
2003
A 2 km long broadcast LAN has 107 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2 × 108 m/s. What is the minimum packet size that can be used on this network?
- A.
50 bytes
- B.
100 bytes
- C.
200 bytes
- D.
None of these
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Correct answer: D
Key idea: To detect collisions in CSMA/CD, a station must still be transmitting when a collision could return. Therefore the frame transmission time must be at least the round-trip propagation delay.
Step 1 — One-way propagation delay: delay = distance / speed = 2000 m / (2 × 10^8 m/s) = 10 μs.
Step 2 — Round-trip propagation delay: 2 × 10 μs = 20 μs.
Step 3 — Minimum bits required: bits = bandwidth × round-trip time = 10^7 bps × 20 μs = 200 bits.
Step 4 — Convert to bytes: 200 bits ÷ 8 = 25 bytes.
Conclusion: The minimum packet size required by CSMA/CD on this link is 200 bits = 25 bytes. None of the provided numerical choices (50 bytes, 100 bytes, 200 bytes) equals 25 bytes, so the correct answer is "None of these."
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