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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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Best answer
74 votes
74 votes
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 , $e=\dfrac{\text{transmission delay}}{\text{total delay}}$

$=\dfrac{800}{800+80}=\dfrac{10}{11}$

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

Correct Answer: $B$
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20 votes
20 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
4 votes
4 votes
It is not in GATE as per new GATE syllabus.
2 votes
2 votes
Bandwidth given=10 Mbps

Transmission time = $\frac{L}{BW}$= $\frac{1000 B}{10 Mbps}$=$800 \mu s$

total time taken to transfer  1000 byte packet= 800$\mu s$+80$\mu s$(polling delay)=880$\mu s$

You can send at max 1000B of data in 880$\mu$s.

for throughput, we have to find how much data we can send in 1 second= $\frac{1000B}{880\mu }$=9.09Mb=$\frac{100}{11}$Mb

hence , throughput= $\frac{100}{11}Mbps$
Answer:

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