A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1…
2005
A network with CSMA/CD protocol in the MAC layer is running at 1 Gbps over a 1 km cable with no repeaters. The signal speed in the cable is 2 x 108 m/sec. The minimum frame size for this network should be
- A.
10000 bits
- B.
10000 bytes
- C.
5000 bits
- D.
5000 bytes
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Correct answer: A
Answer: 10000 bits.
Compute one-way propagation delay: distance / speed = 1000 m / (2 x 10^8 m/s) = 5 x 10^-6 s = 5 µs.
Round-trip propagation delay = 2 × one-way = 10 µs.
Minimum transmission time must be at least the round-trip delay so collisions can be detected.
Minimum frame length in bits = bitrate × round-trip time = 1 x 10^9 bps × 10 x 10^-6 s = 10000 bits.
Conversion to bytes: 10000 bits = 10000 / 8 = 1250 bytes.
Therefore, 10000 bits (1250 bytes) is the minimum frame size. Values like 5000 bits use only the one-way delay and are too short; answers given in bytes that do not equal 1250 bytes are incorrect.
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