Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit…
2005
Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 ms. The minimum frame size is
Answer: D. 512 — The minimum frame size in Ethernet must ensure that the transmission time is at least twice the propagation delay to allow for collision detection. Given a…
- A.
94
- B.
416
- C.
464
- D.
512
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Correct answer: D
The minimum frame size in Ethernet must ensure that the transmission time is at least twice the propagation delay to allow for collision detection. Given a round-trip propagation delay of 46.4 ms, the transmission time must be at least 46.4 ms. At a data rate of 10 Mbps, the minimum frame size is calculated as: 10 × 10^6 bits/sec × 46.4 × 10^-3 sec = 464,000 bits. However, Ethernet has a standard minimum frame size of 64 bytes (512 bits), which is the smallest frame that can be transmitted to ensure collision detection. Therefore, the minimum frame size is 512 bits.
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