Computer-Networks
August 30, 2024Computer-Networks
August 30, 2024Computer-Networks
| Question 786 |
Suppose a network using CSMA/CD has a bandwidth of 10 Mbps. If the maximum propagation time is 25 microsec, that what will be the minimum frame size?
| 500 bits | |
| 50 bits | |
| 500 bytes
| |
| 4*1011 bits
|
Question 786 Explanation:
→ The minimum frame transmission time is Tfr = 2 × Tp = 50 μs.
→ This means, in the worst case, a station needs to transmit for a period of 50 μs to detect the collision.
→ The minimum size of the frame is 10 Mbps × 50 μs = 500 bits or 62.5 bytes.
→ This is actually the minimum size of the frame for Standard Ethernet.
→ This means, in the worst case, a station needs to transmit for a period of 50 μs to detect the collision.
→ The minimum size of the frame is 10 Mbps × 50 μs = 500 bits or 62.5 bytes.
→ This is actually the minimum size of the frame for Standard Ethernet.
Correct Answer: A
Question 786 Explanation:
→ The minimum frame transmission time is Tfr = 2 × Tp = 50 μs.
→ This means, in the worst case, a station needs to transmit for a period of 50 μs to detect the collision.
→ The minimum size of the frame is 10 Mbps × 50 μs = 500 bits or 62.5 bytes.
→ This is actually the minimum size of the frame for Standard Ethernet.
→ This means, in the worst case, a station needs to transmit for a period of 50 μs to detect the collision.
→ The minimum size of the frame is 10 Mbps × 50 μs = 500 bits or 62.5 bytes.
→ This is actually the minimum size of the frame for Standard Ethernet.
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