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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

please explain

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Best answer
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Answer : B

Explanation: Efficiency = transmission time/(transmission time + polling delay time)
Tt =1000 bytes/10Mbps =800μs.
Polling delay is = 80 μs
Efficiency=800/(800+80)= 10/11
Maximum throughput is =(10/11) * 10 Mbps= 100/11 Mbps

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shouldn't the bandwidth be divided among 10 nodes so that each gets 1Mbps bandwidth?
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