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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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## 1 Answer

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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

by Boss (45.1k points)
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0
what is the use of number of nodes???
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ok ..up to effeciency..i m not getting throughput part
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They are  asking maximum throughput of the broadcast channel .I have not seen any use of node here .at-east in deriving the answer...i just used the formula and got the answer
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But it is given in question
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why u multiplied by 10
+1
What part you are not understanding

Packet transmission time = Packet size / Bit rate   (formula you can see any test book for it)

I have put the all values given in question in formula and got the value 1000

Polling delay is given in question =80

then i have written the efficiency formula and put the all derived value in the formula and got hte efficiency

after that i put all the values in throughput formula which is

Throughput = efficiency * Bandwidth

Now tell me which part you don't understand
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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

+1

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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