The network using CSMA/CD has a bandwidth of 20 Mbps. If the maximum one-way…
2025
The network using CSMA/CD has a bandwidth of 20 Mbps. If the maximum one-way propagation time is 25 μs, what is the minimum frame size?
Answer: B. 1000 bits — CONCEPT In CSMA/CD, a sender must still be transmitting when a worst-case collision propagates back. Therefore, the frame transmission time must be at least…
- A.
500 bits
- B.
1000 bits
- C.
1500 bits
- D.
2000 bits
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Correct answer: B
CONCEPT
In CSMA/CD, a sender must still be transmitting when a worst-case collision propagates back. Therefore, the frame transmission time must be at least twice the maximum one-way propagation time.
APPLICATION
Find the round-trip propagation time: 2 × 25 μs = 50 μs.
Convert the bandwidth: 20 Mbps = 20,000,000 bits/s.
Compute the bits sent during the round-trip interval: 20,000,000 × 50 × 10−6 = 1000 bits.
CROSS-CHECK
A 1000-bit frame at 20 Mbps takes 1000 ÷ 20,000,000 s = 50 μs to transmit, exactly matching the required round-trip interval.
Result: The minimum frame size is 1000 bits (125 bytes).
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