GATE IT 2007 | Question: 64

Question

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

Comments

Option B is correct.

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Shouldn’t the propogation time be given though? Atleast they should mention it in the question.

Answer 1

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: $B$

Comments

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.

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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

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we can find efficiency in this way also.

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

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

   $=\frac{10}{11}$

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the number of nodes = 10

has no significance here ?

why ?

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@ A_i_$_h

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...

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capacity = propogation delay * bandwidth (for half duplex), so how can we take it as bandwith, i think the question is a little wrong as we cannot extract propogation delay or the capacity out

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I also have the same doubt

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tanmay patel Arghya Jana

capacity also difine BW. for avoid confusion look at Unit of the capacity..

Bw(unit) = bps 

capacity(unit) = Tp*Bw = s * bps = bits

Answer 2

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

Comments

Nice

Answer 3

It is not in GATE as per new GATE syllabus.

Comments

Brother, I posted a comment on your youtube channel yesterday. I need a bit of your help. I'm also preparing for GATE 2021.

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@shashankrustagi Please provide gate full details syllabus.. (new) 

Answer 4

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= $\frac{100}{11}Mbps$