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 

Answer: A. 10000 bits — 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 ×…

  1. A.

    10000 bits

  2. B.

    10000 bytes

  3. C.

    5000 bits

  4. D.

    5000 bytes

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Correct answer: A

Answer: 10000 bits.

  1. Compute one-way propagation delay: distance / speed = 1000 m / (2 x 10^8 m/s) = 5 x 10^-6 s = 5 µs.

  2. Round-trip propagation delay = 2 × one-way = 10 µs.

  3. Minimum transmission time must be at least the round-trip delay so collisions can be detected.

  4. 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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