Those are in fact the coefficients for p and q they are estimating, though their M is different. Who knows what they did with that.
The data source was: Title: Diffusion models of mobile telephony in Greece Source: Telecommunications policy [0308-5961] Michalakelis yr:2008 vol:32 iss:3-4 pg:234 Gabor Grothendieck wrote: > > On Wed, Jun 15, 2011 at 6:05 PM, Daniel Malter <dan...@umd.edu> > wrote: >> There may be two issues here. The first might be that, if I understand >> the >> Bass model correctly, the formula you are trying to estimate is the >> adoption >> in a given time period. What you supply as data, however, is the >> cumulative >> adoption by that time period. >> >> The second issue might be that the linear algorithm may fail and that it >> may >> be preferable to use Newton-Raphson (the standard) as this may provide >> better values in the iterations. >> >> If you do both, i.e., you do NLS on period adoption and use >> Newton-Raphson, >> you get an estimate. Though, I am of course not sure whether that is >> "correct" in the sense that it is what you would expect to find. >> >> >> adoption <- >> c(167000,273000,531000,938000,2056452,3894103,5932090,7963742,9314687,10469060,11393302,11976340) >> time <- seq(from = 1,to = 12, by = 1) >> >> adoption2<-c(0,adoption[1:(length(adoption)-1)]) >> S<-(adoption-adoption2)/max(adoption) >> >> ## Models >> Bass.Model <- S ~ M*((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) * >> exp(-(p + q) * time) + 1)^2) >> ## Starting Parameters >> Bass.Params <- list(p = 0.1, q = 0.1, M=1) >> ## Model fitting >> Bass.Fit <- nls(formula = Bass.Model, start = Bass.Params) >> summary(Bass.Fit) >> > > If your hypothesis regarding the cumulative vs. adoptions is correct > then it may be that poster wants this: > >> S <- diff(adoption) >> time <- seq_along(S) >> Bass2 <- S ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q / p) * > + exp(-(p + q) * time) + 1)^2) >> nls(formula = Bass2, start = c(p = 0.03, q = 0.4, m = max(S))) > Nonlinear regression model > model: S ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q/p) * > exp(-(p + q) * time) + 1)^2) > data: parent.frame() > p q m > 8.65635536465e-03 6.52817192695e-01 1.23485254536e+07 > residual sum-of-squares: 321990186229 > > Number of iterations to convergence: 16 > Achieved convergence tolerance: 8.10600476229e-06 > >> # or equivalently in terms of "plinear" where S, time and Bass2 are >> # as written just above >> m <- 1 # set m to 1 since we are using .lin instead >> nls(formula = Bass2, start = c(p = 0.03, q = 0.4), alg = "plinear") > Nonlinear regression model > model: S ~ m * ((p + q)^2/p) * (exp(-(p + q) * time)/((q/p) * > exp(-(p + q) * time) + 1)^2) > data: parent.frame() > p q .lin > 8.65637919209e-03 6.52816636341e-01 1.23485299874e+07 > residual sum-of-squares: 321990186247 > > Number of iterations to convergence: 9 > Achieved convergence tolerance: 5.6090474901e-06 > > > > > -- > Statistics & Software Consulting > GKX Group, GKX Associates Inc. > tel: 1-877-GKX-GROUP > email: ggrothendieck at gmail.com > > ______________________________________________ > R-help@r-project.org mailing list > 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. > -- View this message in context: http://r.789695.n4.nabble.com/Problems-with-nls-tp3600409p3603030.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list 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.
