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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
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A broadcast channel has 10 nodes and total capacity of 10 Mbps. It use...
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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A broadcast channel has 10 nodes and total capacity of 10 Mbps. It use...
Calculation of Maximum Throughput:

To calculate the maximum throughput of the broadcast channel, we need to consider the transmission time, polling delay, and the number of nodes.

Transmission Time:
Given that the channel has a total capacity of 10 Mbps, we can calculate the transmission time for 1000 bytes as follows:

Transmission Time = (Number of bits / Channel Capacity)
= (1000 bytes * 8 bits/byte) / 10 Mbps
= 8000 bits / 10 Mbps
= 0.8 ms

Polling Delay:
After each node finishes its transmission, there is a polling delay of 80 μs to poll the next node.

Total Time:
The total time taken for a single cycle of polling for all nodes can be calculated as the sum of the transmission time and the polling delay.

Total Time = Transmission Time + Polling Delay
= 0.8 ms + 80 μs
= 0.8 ms + 0.08 ms
= 0.88 ms

Maximum Throughput:
The maximum throughput of the broadcast channel can be calculated by considering the time taken for a single cycle of polling and the number of nodes.

Maximum Throughput = (Number of nodes * Transmission Time) / Total Time
= (10 nodes * 0.8 ms) / 0.88 ms
= 8 ms / 0.88 ms
= 9.0909 Mbps

Therefore, the maximum throughput of the broadcast channel is approximately 9.0909 Mbps, which can be rounded off to 9 Mbps.

Answer:
The maximum throughput of the broadcast channel is 9 Mbps.
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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
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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 for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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 covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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