# Throughput upper limit formula, not giving expected result

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

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 get with all these: $\dfrac{8000}{159.5}=50.1567398$? So, do I have to do some conversions or just plug in the parameters in the formula? – user21436 Apr 19 '12 at 8:36
did you use the 802.11b parameters? there are parameters for both 802.11a and 802.11b in the table. – Darren Burgess Apr 19 '12 at 8:38
I used the first set of parameters... – user21436 Apr 19 '12 at 8:39
sorry it was the second set, i should have been specific. – Darren Burgess Apr 19 '12 at 8:39
I think I am getting the answer 10.58. I'll write the answer for you. – user21436 Apr 19 '12 at 8:41

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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That is exactly how i worked it out, though I'm still confused how the author of the paper has come to get 11.49. – Darren Burgess Apr 19 '12 at 8:57
What's his answer with parameter set 1? – user21436 Apr 19 '12 at 9:00
The first set of parameters gives the result 50.2 which is what you first calculated. So assuming you used to the same calculation for both sets of parameters this should work. – Darren Burgess Apr 19 '12 at 9:04
So, then, he has a typo in his paper or something then? – user21436 Apr 19 '12 at 9:05
I would hope not, it was published in the IEEE communications letters. – Darren Burgess Apr 19 '12 at 9:07