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GATE | Gate IT 2007 | Question 62

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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

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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Last Updated : 28 Jun, 2021
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