nls problem with R
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Date: Thu, 5 May 2011 01:20:33 -0700 From: sterlesser at hotmail.com To: r-help at r-project.org Subject: Re: [R] nls problem with R ID1 ID2 t V(t) 1 1 0 6.053078443 2 1 0.3403 5.56937391 3 1 0.4181 5.45484486 4 1 0.4986 5.193124598 5 1 0.7451 4.31386722 6 1 1.0069 3.645422269 7 1 1.5535 3.587710965 8 1 1.8049 3.740362689 9 1 2.4979 3.699837726 10 1 6.4903 2.908485019 11 1 13.5049 1.888179494 12 1 27.5049 1.176091259 13 1 41.5049 1.176091259 The model (1) V(t)=V0[1-epi+ epi*exp(-c(t-t0))]
A=Vo, B-Vo*epi, C=exp(-c*t0) V(t)=A-B+B*C*exp(-ct) or further, D=A-B, F=B*C, V(t)=D+F*exp(-ct) this model only really has 3 attriubtes: initial value, final value, and decay constant yet you ask for 4 parameters. There is no way to get a unique answer. For some reason this same form comes up a lot here, I think this is about third time I've sene this in last few weeks. I guess when fishing or shopping for forms to fit, it is tempting to throw a bunch of parameteres into your model but this can create intractable ambiguities. Indeed, if I just remove t0 and use your first 8 points I get this ( random starting values, but convewrged easily you still need to plot etc) [1] "1?? v= 8.77181162126362? epi= 0.672516376478598? cl= 1.90973175223917 t0= 0 .643481321167201"
summary(nls2)
Formula: V2 ~ v0 * (1 - epi + epi * exp(-cl * (T2))) Parameters: ??? Estimate Std. Error t value Pr(>|t|) v0??? 6.2901???? 0.3384? 18.585? 8.3e-06 *** epi?? 0.5430???? 0.1373?? 3.955?? 0.0108 * cl??? 0.9684???? 0.5491?? 1.763?? 0.1381 --- Signif. codes:? 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.3579 on 5 degrees of freedom Number of iterations to convergence: 11 Achieved convergence tolerance: 4.057e-06
(2) V(t)=V0{A*exp[-lambda1(t-t0)]+(1-A)*exp[-lambda2(t-t0)]}
in formula (2) lambda1=0.5*{(c+delta)+[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
lambda2=0.5*{(c+delta)-[(c-delta)^2+4*(1-epi)*c*delta]^0.5}
A=(epi*c-lambda2)/(lambda1-lambda2)
The regression rule :
for formula (1):(t<=2,that is) first 8 rows are used for non-linear
regression
epi,c,t0,V0 parameters are obtained
for formula (2):all 13 rows of results are used for non-linear regression
lambda1,lambda2,A (with these parameters, delta can be calculated from them)
Thanks for help
Ster Lesser
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