ACCEPTED SOLUTION
Accepted Solutions
Without seeing the sheet, a distribution function that is often used to mimic many kinds of distributions is the Beta distribution.
In a more elaborate form than what Mathcad provides, it is:
Hope this helps.
Success!
Luc
Ah, with this kind of shape you will want to use the lognormal distribution. Why? Distribution Fitting - Chemical Yield Analysis
Use the dlnorm function. Here's an attempt:
You mention that you need asymptotes for the point t(i) and v(k). It would help if they're not exactly zero, but slightly positive.
Do you have more points along the curve?
Success!
Luc
Here I played with dlnorm. values changed manually as genfit was not able to find a solution.
> Levenberg Marquardt was already available in Mathcad 6, if I remember well.
Yes but in MC15 I have the option for an Optimzed LM as well:
When I chose it, genfit provides the solution a had shown.
I have no idea why g_ is labeled as "constant" when defining that function and I have no idea why genfit works even though g_ is labeled as "variable" here. I would have expected the undefined error you describe.
> Mathcad complains that there must be more data points than variables to solve for
Yes, thats the error in the first call of "genfit" in my sheet (I made a comment about this in the file) and that also was the motivation for adding some (more or less) arbitrary additional points.
The whole thing is VERY sensible, not only concerning the guess values but also concerning the coordinates of those additionally added points.
With datapoints we mean what you call benchmarks. Points you want to fit a certain function to.
With only start, peak and end its hard to fit a suitable bell curve. You can also fit a perfect paraobola to run through those 3 points.
As you see I had added a few other points to your three before I tried with your function (slightly modified by adding a parameter for vertical offset.
You could also use a solveblock with minerr (or even minimize) to get the "optimal" parameters mimizing the sum of squares of the errors. (least squares fit).
You can also see the equation using the symbolic evaluation.
The region with V:V= ist mandatory to get a meaningful symbolic result :
> The benchmarks are the only spots where I needed a large degree of accuracy. The left and right slopes can generally be anything as long as they go up and then down smoothly.
Thats not much you damand for 😉
Then simple polynomials (chemical processes usually don't follow polynomials) would have done the job as well!?
Here polynomials of third and sixth order. The provide not just an approximation but a perfect fit at your benchmarks and the polynomial of sixth order provides a slope of zero at all three benchmarks. You never answered the question as to if this would be important to you.
With just these three points, and a desire for smooth transition, this is a viable solution:
Luc
