search for: myfitj

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2017 Jun 18
0
R_using non linear regression with constraints
...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(...
2017 Jun 18
0
R_using non linear regression with constraints
...xp(-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) >> &...
2017 Jun 18
3
R_using non linear regression with constraints
...un=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<...
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