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. >--------------------------------------------------------------------------- Jeff Newmiller The ..... ..... Go Live... DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/Batteries O.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k

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 make sense of the problem as a whole. Right now I'm getting ready to go to UseR!, so probably won't have time to spend sorting things out, though I'll have a go if I get time. David: Did you get a crash with the Mac version of nlsr? (You have the latest version I uploaded to CRAN.) I don't have Mac, just Linux and a very ancient Win XP. I suspect the latter is too old to take seriously. I use Win-Builder for package checks. I don't think there's anything seriously wrong with nlsr functions, but there is more to be done to the vignettes and better manual pages are always desirable. It is the vignettes that are popping up warnings on CRAN checks, and I'll see if I can fix those. Cheers, JN On 2017-06-18 06:23 PM, Jeff Newmiller wrote:> 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. >> > > --------------------------------------------------------------------------- > Jeff Newmiller The ..... ..... Go Live... > DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... > Live: OO#.. Dead: OO#.. Playing > Research Engineer (Solar/Batteries O.O#. #.O#. with > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k > ---------------------------------------------------------------------------

> On Jun 18, 2017, at 3:47 PM, J C Nash <profjcnash at gmail.com> wrote: > > 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 make sense of the problem as a whole. Right now I'm getting ready > to go to UseR!, so probably won't have time to spend sorting things out, though I'll have a go if I get time. > > David: Did you get a crash with the Mac version of nlsr? (You have the latest version I uploaded to CRAN.)No. Everything proceeded without error.> summary(myfitj)$residuals [1] 0.000000 0.229245 0.189927 0.130466 0.050909 -0.271920 attr(,"gradient") a b [1,] 0.0000e+00 0 [2,] 7.9014e-07 79784 [3,] 1.4207e-06 143370 [4,] 2.0498e-06 206741 [5,] 2.6774e-06 269898 [6,] 3.1473e-06 317126 $sigma [1] 0.21341 $df [1] 2 4 $cov.unscaled a b a NA NA b NA NA $param Estimate Std. Error t value Pr(>|t|) a 1.4212e+07 NA NA NA b 2.8129e-04 NA NA NA $resname [1] "myfitj" $ssquares [1] 0.18218 $nobs [1] 6 $ct [1] " " " " $mt [1] " " " " $Sd [1] 4.9301e+05 2.7070e-09 $gr [,1] a -1.1226e-09 b -2.0297e-01 $jeval [1] 525 $feval [1] 656 attr(,"pkgname") [1] "nlsr" attr(,"class") [1] "summary.nlsr"> print(myfitj)nlsr object: x residual sumsquares = 0.18218 on 6 observations after 525 Jacobian and 656 function evaluations name coeff SE tstat pval gradient JSingval a 14211701 NA NA NA -1.123e-09 493005 b 0.000281292 NA NA NA -0.203 2.707e-09 And (after considerable "verbosity":> summary(myfitj2)$residuals [1] 0.000000 0.195004 0.149715 0.103277 0.055691 -0.230752 attr(,"gradient") ab b [1,] 0.00000000 0 [2,] 0.00016034 698106 [3,] 0.00028859 1256430 [4,] 0.00041682 1814611 [5,] 0.00054504 2372649 [6,] 0.00064119 2791083 $sigma [1] 0.1785 $df [1] 2 4 $cov.unscaled ab b ab NA NA b NA NA $param Estimate Std. Error t value Pr(>|t|) ab 6.9822e+04 NA NA NA b 1.6035e-05 NA NA NA $resname [1] "myfitj2" $ssquares [1] 0.12746 $nobs [1] 6 $ct [1] " " " " $mt [1] " " " " $Sd [1] 4.3293e+06 6.2886e-08 $gr [,1] ab -8.2085e-08 b -2.6333e+02 $jeval [1] 736 $feval [1] 1024 attr(,"pkgname") [1] "nlsr" attr(,"class") [1] "summary.nlsr"> print(myfitj2)nlsr object: x residual sumsquares = 0.12746 on 6 observations after 736 Jacobian and 1024 function evaluations name coeff SE tstat pval gradient JSingval ab 69821.8 NA NA NA -8.208e-08 4329320 b 1.6035e-05 NA NA NA -263.3 6.289e-08 The estimates were the same order of magnitude but off by a factor of 3 or so. I wondered whether some sort of scaling would have improved the estimates. I did try plotting the geometric picture I imagined using `persp`, but my R-fu was weak. Best; David.> I don't have Mac, just Linux and a very ancient Win XP. I suspect the latter is too old to take seriously. I use Win-Builder for package checks. I don't think there's anything seriously wrong with nlsr functions, but there is > more to be done to the vignettes and better manual pages are always desirable. It is the vignettes that are > popping up warnings on CRAN checks, and I'll see if I can fix those. > > Cheers, JN > > > On 2017-06-18 06:23 PM, Jeff Newmiller wrote: >> 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. >>> >> --------------------------------------------------------------------------- >> Jeff Newmiller The ..... ..... Go Live... >> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live Go... >> Live: OO#.. Dead: OO#.. Playing >> Research Engineer (Solar/Batteries O.O#. #.O#. with >> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k >> ---------------------------------------------------------------------------David Winsemius Alameda, CA, USA

I took another look at the problem. Essentially there isn't enough information in the data to estimate the parameters. What there is doesn't have the right shape for the model. You can see this by plotting the data -- essentially a straight line. Get the slope as follows: rr <- mean(mydata$y/mydata$x, na.rm=TRUE) rr mydata$yx <- mydata$y - rr*mydata$x with(mydata, plot(x, yx)) Shows that the curve is bending the wrong way for the model. More data is needed, especially at larger values of x. JN On 2017-06-18 06:23 PM, Jeff Newmiller wrote:> I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one > set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my > reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far > appear to be finding the true best fit. The fact that myfun() yields 0 always if t=0 and that condition is within the > data given seems likely to be part of the problem. I don't know how to resolve this... perhaps John will look at it again. > > David: Your thinking makes fine sense if you are using a Monte Carlo or brute force solution, but the fact that it > creates a discontinuity in the objective function will confuse any optimizer that uses analytic or numerically estimated > slopes. > > ##----------begin > library(minpack.lm) > library(ggplot2) > > mydata <- data.frame( x = c( 0, 5, 9, 13, 17, 20 ) > , y = c( 0, 11, 20, 29, 38, 45 ) > ) > > myfun <- function( a, b, r, t ) { > a * b * ( 1 - exp( -b * r * t ) ) > } > > objdta <- expand.grid( a = seq( 1000, 3000, by=20 ) > , b = seq( -0.01, 1, 0.01 ) > , rowidx = seq.int( nrow( mydata ) ) > ) > objdta[ , c( "y", "t" ) ] <- mydata[ objdta$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. >> > > --------------------------------------------------------------------------- > Jeff Newmiller The ..... ..... 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