# [R] R_using non linear regression with constraints

J C Nash profjcnash at gmail.com
Sun Jun 18 20:52:16 CEST 2017

```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.

Notes:
1) nlsr has been flagged with a check error by CRAN (though it
is in the vignette, and also mentions pandoc a couple of times).
I'm working to purge the "bug", and found one on our part, but
not necessarily all the issues.
2) I used nlxb that requires an expression for the model. nlsLM
can use a function because it is using derivative approximations,
while nlxb actually gets a symbolic or automatic derivative if
it can, else squawks.

JN

#  Here's the script #
#
# Manoranjan Muthusamy <ranjanmano167 at gmail.com>
#

library(minpack.lm)
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), upper = c(3000,1))
summary(myfit)
library(nlsr)
r <- 2
myfitj=nlxb(y~a*b*(1-exp(-b*r*x)),data=mydata,start=list(a=2000,b=0.05), trace=TRUE)
summary(myfitj)
print(myfitj)

myfitj2<-nlxb(y~ab*(1-exp(-b*r*x)),data=mydata,start=list(ab=2000*0.05,b=0.05), trace=TRUE)
summary(myfitj2)
print(myfitj2)

myfitj2b<-nlxb(y~ab*(1-exp(-b*r*x)),data=mydata,start=list(ab=2000*0.05,b=0.05),
trace=TRUE, upper=c(1000, Inf))
summary(myfitj2b)
print(myfitj2b)
# End of script #

On 2017-06-18 01:29 PM, Bert Gunter wrote:
> 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 problematic.  As usual, if
> either John Nash or Ravi Varadhan comment, heed what they suggest. I'm
> pretty ignorant.
>
> Cheers,
> Bert
>
>
>
>
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Sun, Jun 18, 2017 at 9:43 AM, David Winsemius <dwinsemius at comcast.net> wrote:
>>
>>> 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))
>>>   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), upper = c(3000,1))
>>>
>>> It works. But now I would like to introduce a constraint which is a*b<1000.
>>
>> At the moment your coefficients do satisfy that constraint so that dataset is not suitable for testing. A slight modification of the objective function to include the logical constraint as an additional factor does not "break" that particular solution.:
>>
>> myfun2=function(a,b,r,t){
>>      prd=a*b*(1-exp(-b*r*t))*(a*b<1000)
>>      return(prd)}
>>
>>
>> myfit=nlsLM(y~myfun2(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05),
>>              lower = c(1000,0), upper = c(3000,1))
>>
>> #------------------
>> myfit
>> Nonlinear regression model
>>    model: y ~ myfun2(a, b, r = 2, t = x)
>>     data: mydata
>>          a         b
>> 3.000e+03 2.288e-02
>>   residual sum-of-squares: 38.02
>>
>> Number of iterations to convergence: 8
>> Achieved convergence tolerance: 1.49e-08
>> #--
>>
>> prod(coef(myfit))
>> #[1] 68.64909  Same as original result.
>>
>> How nlsLM will handle more difficult problems is not something I have experience with, but obviously one would need to keep the starting values within the feasible domain. However, if your goal was to also remove the upper and lower constraints on a and b, This problem would not be suitably solved by the a*b product without relaxation of the default maxiter:
>>
>>> myfit=nlsLM(y~myfun2(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05),
>> +             lower = c(0,0), upper = c(9000,1))
>>> prod(coef(myfit))
>> [1] 110.4382
>>> coef(myfit)
>>             a            b
>> 9.000000e+03 1.227091e-02
>>
>>>   myfit=nlsLM(y~myfun2(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05),
>> +              lower = c(0,0), upper = c(10^6,1))
>> Warning message:
>> In nls.lm(par = start, fn = FCT, jac = jac, control = control, lower = lower,  :
>>    lmdif: info = -1. Number of iterations has reached `maxiter' == 50.
>>
>> #---------
>>   myfit=nlsLM(y~myfun2(a,b,r=2,t=x),data=mydata,start=list(a=2000,b=0.05),
>>               lower = c(0,0), upper = c(10^6,1), control=list(maxiter=100))
>>   prod(coef(myfit))
>>
>>   coef(myfit)
>> #============
>>
>>
>>>   prod(coef(myfit))
>> [1] 780.6732  Significantly different than the solution at default maxiter of 50.
>>>
>>>   coef(myfit)
>>             a            b
>> 5.319664e+05 1.467524e-03
>>>
>>>
>>
>>
>> --
>> David.
>>
>>
>>> I had a look at the option available in nlsLM to set constraint via
>>> nls.lm.control. But it's not much of help. can somebody help me here or
>>> suggest a different method to to this?
>>>
>>>        [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
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>>> and provide commented, minimal, self-contained, reproducible code.
>>
>> David Winsemius
>> Alameda, CA, USA
>>
>> ______________________________________________
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