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I am using a formula from a paper to find the throughput upper limit across a computer network however my calculation gives a different answer to what is given in the paper. Below are the parameters and the formula used:

Parameters for 802.11B

parameters for 802.11b

Throughput upper limit formula

Throughput upper limit formula

The answer stated in the paper when LData is 1000 is 11.49, however the answer i continue to get is 10.58.

Could anyone shed any light on how to get the correct answer?

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    I think I am getting the answer 10.58. I'll write the answer for you.2012-04-19

1 Answers 1

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I'll write out the parameters set of 802.11B separately for convenience

$\begin{array}{|c|c|}\hline T_P & 144 \mu s\\ T_{PHY} & 48 \mu s \\ \tau & 1 \mu s \\ T_{DIFS} & 50 \mu s\\ T_{SIFS} & 10 \mu s\\ CW_{min}&31\\ T_{slot} & 10 \mu s\\ \hline \end{array}$

Along with this we also have $L_{DATA}=1000$. So, let's compute the quantity $TUL$:

  1. Numerator is easy to compute: $8L_{DATA}=8000$.

  2. Denominator is the following expression:

\begin{align*} &2T_P +2 T_{PHY}+2\tau+T_{DIFS}+T_{SIFS}+\frac{CW_{min} T_{Slot}}{2}\\&=2\cdot 144+2 \cdot 48+2 \cdot 1+50+10+\frac{31\cdot 20}{2}\\&=288+96+2+50+10+310\\&=756 \end{align*}

So, the final thing is $TUL=\frac {8000}{756}=10.5820106\cdots$

which means you're right or the recipe we are working with is wrong!

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    Me neither, can have the accept for helping as you have.2012-04-19