The minimum frame size required for a CSMA/CD based computer network running…
2008
The minimum frame size required for a CSMA/CD based computer network running at 1 Gbps on a 200m cable with a link speed of 2 × 108m/s is
- A.
125 bytes
- B.
250 bytes
- C.
500 bytes
- D.
None of these
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Show answer & explanation
Correct answer: B
Key idea: for CSMA/CD the frame transmission time must be at least the round-trip propagation delay so a transmitting station can detect collisions.
One-way propagation delay = cable length ÷ propagation speed = 200 m ÷ (2 × 10^8 m/s) = 1 × 10^-6 s (1 µs).
Round-trip propagation delay = 2 × 1 µs = 2 µs.
Required transmission time ≥ 2 µs. At 1 Gbps (1 × 10^9 bits/s), bits transmitted in 2 µs = 1 × 10^9 × 2 × 10^-6 = 2,000 bits.
Convert bits to bytes: 2,000 bits ÷ 8 = 250 bytes. Therefore the minimum frame size required is 250 bytes.
Answer: 250 bytes.
Note: Frames are measured in whole bytes; in this case the calculation yields an exact integer byte value (250 bytes).
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