Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that…
2004
Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that needs to transmit seizes the token. Then it sends a frame of 1000 bytes, removes the frame after it has circulated all around the ring, and finally releases the token. This process is repeated for every frame. Assuming that only a single host wishes to transmit, the effective data rate is
- A.
1Mbps
- B.
2Mbps
- C.
5Mbps
- D.
6Mbps
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Correct answer: C
Answer: 5 Mbps
Reasoning and calculation:
Frame size = 1000 bytes = 8000 bits.
Transmission time at 10 Mbps = 8000 bits / 10,000,000 bps = 800 µs.
Ring latency = 400 µs. After finishing transmission, the last bit returns after an additional 400 µs, so the sender removes the frame at time 800 + 400 = 1200 µs from the start.
After removing the frame the sender releases the token. The token must travel the ring and return before the sender can seize it again, adding another 400 µs.
Total cycle time per frame = transmission time + frame-circulation latency + token-return latency = 800 + 400 + 400 = 1600 µs.
Effective data rate = 8000 bits / 1600 µs = 8000 / 0.0016 s = 5,000,000 bps = 5 Mbps.
Key takeaway: include both the frame circulation time (to remove the frame) and the token-return time when computing the cycle for repeated single-host transmissions.