Auto-suggest helps you quickly narrow down your search results by suggesting possible matches as you type.

- JMP User Community
- :
- Discussions
- :
- Specialized Modeling->Fit Curve Growth Rates

- Subscribe to RSS Feed
- Mark Topic as New
- Mark Topic as Read
- Float this Topic for Current User
- Bookmark
- Subscribe
- Printer Friendly Page

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Specialized Modeling->Fit Curve Growth Rates

Apr 22, 2019 10:22 AM
(7064 views)

Hello,

I am looking for a resource to explain how the growth rate parameters are calculated from the tangent at the inflection points, specifically for 3P and Weibull models. I need the ability to convert the growth rates back to biologically relevant numbers.

For example: If my Weibull growth rate is ~8, how do I convert this back to meters per day in reference to plant growth.

The growth rate should be a number closer to 0.06 meters per day.

Any help would be greatly appreciated.

-Steve

9 REPLIES 9

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

Created:
Apr 22, 2019 11:34 AM
| Last Modified: Apr 22, 2019 11:38 AM
(7043 views)
| Posted in reply to message from andersst91 04-22-2019

I am not sure what you mean by '*how the growth rate parameters are calculated*.' The parameters are estimated using variety of numerical methods depending on the platform default options or ones that you select. You can find the function of the curves by selecting **Help** > **Books** > **Predictive and Specialized Modeling** > **Fit Curve** > **Statistical Details**. For example, here you will find that the parameter called '**growth rate**' by JMP is the power applied to ratio of **X** to the '**inflection point**' in the case of the Weibull model:

Here is the function for the 3P model you used:

I feel that I have missed your point, though. If you mean the instantaneous growth rate (change in Y with respect to change in X), then that would be the first derivative of this function with respect to the domain X. Because these functions are non-linear, the real growth rate will not be a constant.

Also, is time your X variable?

Learn it once, use it forever!

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

@markbailey Thank you for the response.

Yes, I was trying to find the proper way to phrase my question.

I can now rephrase and ask how can the "growth rate" parameter be explained in biological context/importance?

Form your explination, it is not a growth rate in the sense of a slope? Saying that the growth rate unit of measure is meters/days would be incorrect?

My X variable is a time in days and my Y variable is the height in meters.

I would like to be able to define the maximum growth rate (meters/day) at the inflection point.

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

Created:
Apr 22, 2019 12:18 PM
| Last Modified: Apr 22, 2019 12:21 PM
(7037 views)
| Posted in reply to message from andersst91 04-22-2019

No problem - I understand. It doesn't help that JMP developers had to come up with meaningful names for the function parameters when there are so many interpretations in different knowledge domains.

OK, you fit a model Y = f(X). We can use the estimated parameter values after differentiating the function: dY = f'(X). Here is the derivative for the Weibull model:

And here is the derivative for the Exponential 3P:

You can substitute the parameter estimates for the parameters and solve for 0 or plot these funtions versus X to see the maximum. (I don't think that the Exponential 3P has a maximum, though.)

Learn it once, use it forever!

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

I think that if you replace the parameters in the derivative with the estimates from your Weibull fit in this request to solve the equation, you will get the X value (time) when it is maximum:

You can then substitute X into the derivative to get the corresponding maximum growth rate (change in y versus change in X).

An easier approach might be to create a new column (Rate) with a formula for the derivative with the estimates instead of the parameters and profile it. You can use the default desirability function (maximize) in the Prediction Profiler to find the corresponding rate and time.

Learn it once, use it forever!

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

Great thank you for the excellent explanation. This has helped a lot!

I have one final question, would the "growth rate" parameter of the JMP output have a unit of measure? Does it relate to the X or Y variable in terms of units (e.g. m/days) if I were to present it in a figure or table?

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

I don't think so, but someone else might know otherwise.

I think that since Y = Weibull( X ), the parameter **a** (asymptotic plant height) has the unit *meter* and the parameter **c** (inflection point in time) has the unit *day*. It seem like the parameter **b** does not have any units. It is just a power, like a square or cube root.

Learn it once, use it forever!

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

Thank you again for your help.

I have a final question. Is there a publication that originally presents this form of the Weibull function?

I have seen several variants of the function and would like to reference the literature in which JMP used to develop their equation.

-Thanks

Highlighted
##

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

There is much literature about this subject. I looked it up for you. See **Help** > **Books** > **Reliability and Survival** > **References** chapter:

*Statistical Methods for Reliability Data*. New York: John Wiley & Sons.

Learn it once, use it forever!

Highlighted
##

For the 3P exponential growth model, if X has units of days, then the growth rate, c, would have units of (1/day). You can think of it as the log linear rate per day. (i.e. Y=a+b*exp(c*x), so ln(Y-a) = ln(b) + c*x)

- Mark as New
- Bookmark
- Subscribe
- Mute
- Subscribe to RSS Feed
- Get Direct Link
- Email to a Friend
- Report Inappropriate Content

Re: Specialized Modeling->Fit Curve Growth Rates

Article Labels

There are no labels assigned to this post.