MacLeamy curve fit

[ATG1964]

I'd like to find an equation that best fits this curve... Seems simple enough right? Y axis = effort in hours/day, X axis is time in days. Note that the effort peaks early and then decays to a lower limit. The area under the curve should be the total effort in hours for the project.

This seams so simple - but I'm crying uncle and need some thoughts to further this exercise.

Thank you - Andy

 

[Cubist]

You say "best fits". Do you need the function to go exactly through all the points of your supplied data? Or, perhaps, do you need a function that would be easy to manipulate (a simple function) that is fairly close to the supplied points. For the latter, you could try something like this...
[math]\dfrac{c_1}{1+(x-c_2)^2}[/math]

 

[ATG1964]

You say "best fits". Do you need the function to go exactly through all the points of your supplied data? Or, perhaps, do you need a function that would be easy to manipulate (a simple function) that is fairly close to the supplied points. For the latter, you could try something like this...
[math]\dfrac{c_1}{1+(x-c_2)^2}[/math]

The latter... Love it... c1=amplitude, c2-phase shift... Thank you!

 

[ATG1964]

The latter... Love it... c1=amplitude, c2-phase shift... Thank you!

does this function have a name?

 

[Cubist]

The latter... Love it... c1=amplitude, c2-phase shift... Thank you!

You're welcome

does this function have a name?

I'm not aware of a name. (This doesn't mean that there isn't a name for it, there might well be)

I came up with this expression because I thought that your graph seemed symmetrical about a peak and this made me think of a parabola: -x^2. However, x^2 heads off to infinity as x grows in either direction. To change this behaviour, and to have the value approach zero as x becomes large, then we could just take the reciprocal: 1/x^2. However. this would then cause values near the centre of the curve, around x=0, to become large without limit. To stop this, we just need to add 1 to the denominator 1/(1 + x^2).

Of course, you'll need to scale the width of the peak. And it's also possible to move the horizontal asymptote upwards, since your graph doesn't seem to be heading down towards 0. A lower value of around 3 might be better. Can you think of how to improve the given expression to incorporate these features?