Displaying 6 results from an estimated 6 matches for "nlregmod2".
2023 Aug 20
2
Issues when trying to fit a nonlinear regression model
Dear Bert,
Thank you so much for your kind and valuable feedback. I tried finding the
starting values using the approach you mentioned, then did the following to
fit the nonlinear regression model:
nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x),
start =
list(theta1 = 0.37,
theta2 = exp(-1.8),
theta3 = -0.05538), data=mod14data2_random)
However, I got this error:
Error in nls(y ~ theta1 - theta2 * exp(-theta3...
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
...2023 at 11:50?AM Paul Bernal <paulbernal07 at gmail.com> wrote:
> Dear Bert,
>
> Thank you so much for your kind and valuable feedback. I tried finding the
> starting values using the approach you mentioned, then did the following to
> fit the nonlinear regression model:
> nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x),
> start =
> list(theta1 = 0.37,
> theta2 = exp(-1.8),
> theta3 = -0.05538), data=mod14data2_random)
> However, I got this error:
> Error in nls(y ~...
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
...;paulbernal07 at gmail.com>
> wrote:
>
>> Dear Bert,
>>
>> Thank you so much for your kind and valuable feedback. I tried finding
>> the starting values using the approach you mentioned, then did the
>> following to fit the nonlinear regression model:
>> nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x),
>> start =
>> list(theta1 = 0.37,
>> theta2 = exp(-1.8),
>> theta3 = -0.05538), data=mod14data2_random)
>> However, I got this error:
&...
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
...gt;> wrote:
>>
>>> Dear Bert,
>>>
>>> Thank you so much for your kind and valuable feedback. I tried finding
>>> the starting values using the approach you mentioned, then did the
>>> following to fit the nonlinear regression model:
>>> nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x),
>>> start =
>>> list(theta1 = 0.37,
>>> theta2 = exp(-1.8),
>>> theta3 = -0.05538), data=mod14data2_random)
>>> However,...
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
I got starting values as follows:
Noting that the minimum data value is .38, I fit the linear model log(y -
.37) ~ x to get intercept = -1.8 and slope = -.055. So I used .37,
exp(-1.8) and -.055 as the starting values for theta0, theta1, and theta2
in the nonlinear model. This converged without problems.
Cheers,
Bert
On Sun, Aug 20, 2023 at 10:15?AM Paul Bernal <paulbernal07 at
2023 Aug 20
1
Determining Starting Values for Model Parameters in Nonlinear Regression
The cautions people have given about starting values are worth heeding. That nlxb() does well in many cases is useful,
but not foolproof. And John Fox has shown that the problem can be tackled very simply too.
Best, JN
On 2023-08-19 18:42, Paul Bernal wrote:
> Thank you so much Dr. Nash, I truly appreciate your kind and valuable contribution.
>
> Cheers,
> Paul
>
> El El