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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:

  1. $1$ Mbps
  2. $100/11$ Mbps
  3. $10$ Mbps
  4. $100$ Mbps
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Here in this question it is not given whether polling is used for first node or first node start transmission without polling as polling definition given above is here is a polling delay of 8080 μs to poll the next node 

if we start polling from first node then question has been already answered-- 

but in case it starts from first node then total channel efficiency = 10*transmission time/(10*transmission time + 9*polling time) = 100/109

throughput = eff *  B = 10*(100/109) ~ 100/11

please let me know if i am wrong somewhere

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I would approach it this way...

Data we have transfer is 1000 bytes or 8000 bits.
Capacity or bandwidth  = 10Mbps
Wasted time in poling = 80ms

using bandwidth to find the time that I would need for 8000 bits.

1sec = 10 * 10^6 bits
10^6ms = 10 * 10^6 bits

(1/10)ms = 1 bit (time)
(8/10)ms = 1 byte (time)
[ (8*1000)/10 ] ms = 1000bytes ( time )
so to xfr 1000 bytes we take 800ms purely

but wait there some time 80ms polling
therefore 80+800 = 880



No we have to find answer in Mbps
880ms = 8000 bits
880 * 10^6 sec = 8000 * 10^6 Mb
Hence 8000/880 = (100/11)Mbps.


Please correct if approach is wrong..


 

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