>OK, I'm still a little lost so you'll need to help me with this...
>
>I understand that the Latency Accurate is the Average of all calls and I also get that the standard deviation stuff you put in (I mean, I know what STD is). While I can see the usefulness to accuracy of getting the average (and even touched on that topic in my blog), here's my questions...
>
>What does the standard deviation tell us that is useful in diagnosing the execution speed of the program? I mean, I could see using something like this to throw out the elements that have a deviation that is much greater than the norm and then reaveraging what is left. This would increase the accuracy of our average by getting rid of any wild swings (out of bounds values). But I don't see how your use of it applies... I'm sure it does, I just don't understand it.
>
>Perhaps my first question is related to my being unable to comprehend the following...
>
>?"Then the precision is 1/" + TRANSFORM( ROUND(lnTicksPerSecond/SDev,0))
>
>...what are you showing here? I mean I can see that you are taking the TicksPerSecond and dividing it by the STD, but I don't see how that relates to precision.
>

Hi Craig,
for simplicity we suppose that lnTicksPerSecond = 999999, this is the accurate.
Well, to measure an interval of time that represents
the execution of a code, means measure a difference
of the counter that produces 999999 ticks/s.
If the events of Start and Stop were immediate,
the average error of measure it is of + / - 1 ticks,
while the error max it is of + / - 2 tickses.

There is instead a systematic error of some tickses,
owed to the delay with which Start finishes and Stop begins.

Being a systematic error, it is possible to compensate the result
removing it from the calculation.

But a casual error exists (noise) due to the fact
how other Thread or processes use the cpu.

This reduces the precision of the accurate number 999999.

For instance, if the error standard it is 10,
would every measure be of the type 99999? ( on average sense ),
where the last figure has not meant,
then the precision ( on average ) is 99999=999999/10.

Increasing the number of measures this error,
it can be reduced in average.

For example:
it considers your test of the loop of 100000 increases.
You repeat the measure 10 times:
* in my notebook in battery mode and VFP8
12.9ms
17.0ms
21.2ms
84.8ms
27.7ms
21.8ms
54.0ms
13.5ms
....