[R] R_using non linear regression with constraints

Jeff Newmiller jdnewmil at dcn.davis.ca.us
Mon Jun 19 00:23:34 CEST 2017


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 fit. The fact that myfun() yields 0 
always if t=0 and that condition is within the data given seems likely to 
be part of the problem. I don't know how to resolve this... perhaps John 
will look at it again.

David: Your thinking makes fine sense if you are using a Monte Carlo or 
brute force solution, but the fact that it creates a discontinuity in 
the objective function will confuse any optimizer that uses analytic or 
numerically estimated slopes.

##----------begin
library(minpack.lm)
library(ggplot2)

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 ) {
   a * b * ( 1 - exp( -b * r * t ) )
}

objdta <- expand.grid( a = seq( 1000, 3000, by=20 )
                      , b = seq( -0.01, 1, 0.01 )
                      , rowidx = seq.int( nrow( mydata ) )
                      )
objdta[ , c( "y", "t" ) ] <- mydata[ objdta$rowidx
                                    , c( "y", "x" ) ]
objdta$tf <- factor( objdta$t )
objdta$myfun <- with( objdta
                     , myfun( a = a, b = b, r = 2, t = t )
                     )
objdtass <- aggregate( ( objdta$myfun - objdta$y )^2
                      , objdta[ , c( "a", "b" ) ]
                      , FUN = function( x )
                               sum( x, na.rm=TRUE )
                      )
objdtassmin <- objdtass[ which.min( objdtass$x ), ]

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 )
               )
a <- as.vector( coef( myfit )[ "a" ] )
b <- as.vector( coef( myfit )[ "b" ] )

brks <- c( 500, 1e7, 2e7, 3e7, 4e7 )
ggplot( objdtass, aes( x=a, y=b, z = x, fill=x ) ) +
     geom_tile() +
     geom_contour( breaks= brks ) +
     geom_point( x=a, y=b, colour="red" ) +
     geom_point( x=objdtassmin$a
               , y=objdtassmin$b
               , colour="green" ) +
     scale_fill_continuous( name="SumSq", breaks = brks )
# Green point is brute-force solution
# Red point is optimizer solution for myfun

##############

myfun2 <- function( a, log1ab, r, t ) {
   ab <- 1000 - exp( log1ab )
   ab * ( 1 - exp( -ab/a * r * t ) )
}

objdta$log1ab <- with( objdta, log( 1000 - a * b ) )
objdta$myfun2 <- with( objdta
                      , myfun2( a = a
                              , log1ab = log1ab
                              , r = 2
                              , t = t
                              )
                      )
objdtass2 <- aggregate( ( objdta$myfun2 - objdta$y )^2
                       , objdta[ , c( "a", "b" ) ]
                       , FUN = function( x )
                                if ( all( is.na( x ) ) ) NA
                                else sum( x, na.rm=TRUE )
                       )
objdtass2min <- objdtass2[ which.min( objdtass2$x ), ]

myfit2 <- nlsLM( y ~ myfun2( a, log1ab, r = 2, t = x )
                , data = mydata
                , start = list( a = 2000
                              , log1ab = 4.60517
                              )
                , lower = c( 1000, 0 )
                , upper = c( 3000, 8.0063 )
                )
a2 <- as.vector( coef( myfit2 )[ "a" ] )
b2 <- ( 1000
       - exp( as.vector( coef( myfit2 )[ "log1ab" ] ) )
       ) / a

brks <- c( 500, 1e6, 2e6, 3e6, 4e6 )
ggplot( objdtass2, aes( x=a, y=b, z = x, fill=x ) ) +
     geom_tile() +
     geom_contour( breaks = brks ) +
     geom_point( x=a2, y=b2, colour="red" ) +
     geom_point( x=objdtass2min$a
               , y=objdtass2min$b
               , colour="green" ) +
     scale_fill_continuous( name="SumSq", breaks = brks )
# Green point is brute-force solution
# Red point is optimizer solution for myfun2

##----------end

On Sun, 18 Jun 2017, J C Nash wrote:

> 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]]
>>>> 
>>>> ______________________________________________
>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>> PLEASE do read the posting guide 
>>>> http://www.R-project.org/posting-guide.html
>>>> and provide commented, minimal, self-contained, reproducible code.
>>> 
>>> David Winsemius
>>> Alameda, CA, USA
>>> 
>>> ______________________________________________
>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide 
>>> http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>> 
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide 
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>> 
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

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