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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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2 Answers

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Best answer

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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4 Comments

Don't worry about the question you got the answer it enough you can't debate in 3 hours gate paper why you have given node in question if there is no use of it .
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can u clear me this line:

Throughput = efficiency * Bandwidth

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efficiency = use-full time / total time  ===> Tt / Tt +2 Tp

Divide the complete formula with Tt ===> Efficiency = 1/1+2a        consider Tp/Tt =a

Throughput also called as effective bandwidth or bandwidth utilization 

in CN throughput =no of bits we are able to send per sec if L is the no of bit then

Throughput =L / Tt + 2 Tp (total cycle time)

Tt + 2 Tp  is the total cycle time in which we are able to send L bits

Now multiply and divide L with Bandwidth(B) in formula  L*(B/B) / Tt + 2 Tp

Now L / B =Tt  So it will become   Tt*B / Tt + 2 Tp

Now divide the formula with Tt then you get 

(1/1+2a) * B  ===> efficiency * bandwidth   (I have given efficiency formula at the starting)

Throughput= efficiency * Bandwidth

if you dont understand it here write it down of a paper the you will understand .Try to do it by yourself

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

1 comment

It's asking for throughput of whole channel
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Answer:

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