A broadcast channel has 10 nodes and total capacity of 10 Mbps. It uses…
2007
A broadcast channel has 10 nodes and total capacity of 10 Mbps. It uses polling for medium access. Once a node finishes transmission, there is a polling delay of 80 μs to poll the next node. Whenever a node is polled, it is allowed to transmit a maximum of 1000 bytes. The maximum throughput of the broadcast channel is
- A.
1 Mbps
- B.
100/11 Mbps
- C.
10 Mbps
- D.
100 Mbps
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Correct answer: B
Answer: 100/11 Mbps (≈9.09 Mbps)
Reasoning:
Data per polled transmission: 1000 bytes = 8000 bits.
Transmission time at 10 Mbps: 8000 bits / 10,000,000 bps = 0.0008 s = 800 μs.
Add polling delay: 80 μs, so time per polled transmission = 800 μs + 80 μs = 880 μs.
For 10 nodes, cycle time = 10 × 880 μs = 8800 μs = 0.0088 s. Total data per cycle = 10 × 8000 bits = 80,000 bits.
Throughput = total bits per cycle / cycle time = 80,000 bits / 0.0088 s = 9,090,909.09 bps = 100/11 Mbps ≈ 9.09 Mbps.
Note: The raw channel capacity is 10 Mbps, but polling overhead reduces the effective maximum throughput to about 9.09 Mbps.
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