Displaying 6 results from an estimated 6 matches for "r_use".
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2017 Jun 18
0
R_using non linear regression with constraints
> On Jun 18, 2017, at 6:24 AM, Manoranjan Muthusamy <ranjanmano167 at gmail.com> wrote:
>
> I am using nlsLM {minpack.lm} to find the values of parameters a and b of
> function myfun which give the best fit for the data set, mydata.
>
> mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45))
>
> myfun=function(a,b,r,t){
> prd=a*b*(1-exp(-b*r*t))
>
2017 Jun 18
2
R_using non linear regression with constraints
I am using nlsLM {minpack.lm} to find the values of parameters a and b of
function myfun which give the best fit for the data set, mydata.
mydata=data.frame(x=c(0,5,9,13,17,20),y = c(0,11,20,29,38,45))
myfun=function(a,b,r,t){
prd=a*b*(1-exp(-b*r*t))
return(prd)}
and using nlsLM
myfit=nlsLM(y~myfun(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05),
lower = c(1000,0),
2017 Jun 18
0
R_using non linear regression with constraints
I ran the following script. I satisfied the constraint by
making a*b a single parameter, which isn't always possible.
I also ran nlxb() from nlsr package, and this gives singular
values of the Jacobian. In the unconstrained case, the svs are
pretty awful, and I wouldn't trust the results as a model, though
the minimum is probably OK. The constrained result has a much
larger sum of squares.
2017 Jun 18
0
R_using non linear regression with constraints
I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have
a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is
the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful,
but it takes some time to sort them all out and
2017 Jun 18
3
R_using non linear regression with constraints
https://cran.r-project.org/web/views/Optimization.html
(Cran's optimization task view -- as always, you should search before posting)
In general, nonlinear optimization with nonlinear constraints is hard,
and the strategy used here (multiplying by a*b < 1000) may not work --
it introduces a discontinuity into the objective function, so
gradient based methods may in particular be
2017 Jun 18
3
R_using non linear regression with constraints
I am not as expert as John, but I thought it worth pointing out that the
variable substitution technique gives up one set of constraints for
another (b=0 in this case). I also find that plots help me see what is
going on, so here is my reproducible example (note inclusion of library
calls for completeness). Note that NONE of the optimizers mentioned so far
appear to be finding the true best