IP-Address
September 12, 2024Routing
September 12, 2024Ethernet
| Question 4 |
The minimum frame size required for a CSMA/CD based computer network running at 1 Gbps on a 200 m cable with a link speed of 2 × 108 m/s is
| 125 bytes | |
| 250 bytes | |
| 500 bytes | |
| None of these |
Question 4 Explanation:
For CSMA/CD protocol to work, the transmission time of the frame must be more than minimum value. This minimum value is the RTT.
So,
Tt ≥ 2 × Tp
L/B ≥ 2 × Tp
L ≥ 2 × Tp × B
L ≥ 2 × 200/(2×108) × 109
L ≥ 2000 bits
L ≥ 250 Bytes
So,
Tt ≥ 2 × Tp
L/B ≥ 2 × Tp
L ≥ 2 × Tp × B
L ≥ 2 × 200/(2×108) × 109
L ≥ 2000 bits
L ≥ 250 Bytes
Correct Answer: B
Question 4 Explanation:
For CSMA/CD protocol to work, the transmission time of the frame must be more than minimum value. This minimum value is the RTT.
So,
Tt ≥ 2 × Tp
L/B ≥ 2 × Tp
L ≥ 2 × Tp × B
L ≥ 2 × 200/(2×108) × 109
L ≥ 2000 bits
L ≥ 250 Bytes
So,
Tt ≥ 2 × Tp
L/B ≥ 2 × Tp
L ≥ 2 × Tp × B
L ≥ 2 × 200/(2×108) × 109
L ≥ 2000 bits
L ≥ 250 Bytes
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