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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
in Computer Networks by Boss (16.3k points)
edited by | 2.8k views

2 Answers

+54 votes
Best answer
Propagation time is not given so that's negligible here.

efficiency $=\dfrac{\text{transmission time}}{\text{(transmission time + polling time)}}$

Tx$=\dfrac{1000\ bytes}{10\ Mbps} =800\mu s.$

Delay because of polling is $=80\mu s$

Efficiency of channel , $e=\dfrac{transmission-delay}{total-delay}$


Maximum throughput is $=\dfrac{10}{11}\times 10\ Mbps=\dfrac{100}{11} Mbps$

Correct Answer: $B$
by Boss (43.8k points)
edited by
Here capacity is givien. Not bandwidths howerver we can think 10Mbps as a Bandwidth by seeing its unit i.e Mbps.

and secondly how u take 80 as 80micro sec.
But why we have take polling delay just 80us suppose we start polling from first node and if the 10th node wants to use the medium then it would be 80*9=720us
we can find efficiency in this way also.

Efficiency $η = \frac {useful\ data\ sent\ in\ 880μs}{Total\ data\ that\ could\ be\ sent\ in\ 880μs}$

$η = \frac {1000*8}{880*10^{-6}*10*10^6}$

the number of nodes = 10

has no significance here ?

why ?


actually polling method is used to decide the station which is going to transmit next.....and

in this we use a algorithm to find suitable station and then allocate Time equal to Tt+Tp ...

once this time is over we ...again we use same for other station......

::no. of channel doesn't matter for polling...

+4 votes
The present best solution is using the concept of efficiency. I have tried solving the problem without using efficiency.

The time required to transmit 1000 bytes $= \frac{1000 \ bytes}{10 \times 10^6 \ bps}  = \frac{8 \times 10^3 \ bits}{10^7 \ bps} = 8 \times 10^{-4} \ s = 800 \ \mu s $

$ \therefore \ $ The total time required to transmit a 1000 byte data $= (800 + 80) \ \mu s  = 880 \ \mu s$

Now, in $ 880 \ \mu s $,  we can send $8000$ bits.

$ \implies $ in $ 1 \ s $, we can send $ \frac{8000}{880 \times 10^{-6}} $ bits $ = \frac{8000}{880} $ Mb $ = \frac{100}{11} $ Mb
by Junior (915 points)

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