nls convergence trouble
# 1. Supplying the derivatives results in convergence:
lgmg <- function(a, b, C0, m, V) {
+ e <- expression((V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * + C0 * b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * + a * b + (b * m * a)^2))/(2 * b * m)) + val <- eval(e) + attr(val, "gradient") <- cbind(a = eval(D(e, "a")), b = eval(D(e, "b"))) + val + }
nls(Qe ~ lgmg(a, b, C0, m, V), bois.DATA, start = c(a = 300, b = 1))
Nonlinear regression model
model: Qe ~ lgmg(a, b, C0, m, V)
data: bois.DATA
a b
337.74912 0.03864
residual sum-of-squares: 15473
Number of iterations to convergence: 9
Achieved convergence tolerance: 3.16e-06
# 2. As mentioned before squaring both sides results in convergence:
# even without derivatives
# though admittedly that is a slightly different objective.
lgm <- function(a, b, C0, m, V) {
+ (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * + C0 * b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * + a * b + (b * m * a)^2))/(2 * b * m) + }
nls(Qe^2 ~ lgm(a, b, C0, m, V)^2, bois.DATA, start = c(a = 300, b = 1))
Nonlinear regression model
model: Qe^2 ~ lgm(a, b, C0, m, V)^2
data: bois.DATA
a b
225.6474 0.3568
residual sum-of-squares: 9.98e+10
Number of iterations to convergence: 16
Achieved convergence tolerance: 6.096e-06
# 3. Also using the reciprocal it converges # without derivatives nls(Qe ~ lgm(a, 1/b, C0, m, V), bois.DATA, start = c(a = 300, b = 1))
Nonlinear regression model
model: Qe ~ lgm(a, 1/b, C0, m, V)
data: bois.DATA
a b
337.75 25.88
residual sum-of-squares: 15473
Number of iterations to convergence: 12
Achieved convergence tolerance: 1.722e-06
transform(as.list(coef(.Last.value)), b = 1/b)
a b 1 337.7492 0.03863738
On Wed, Sep 3, 2008 at 10:36 AM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
Try squaring both sides of the formula. On Wed, Sep 3, 2008 at 10:01 AM, Benoit Boulinguiez <benoit.boulinguiez at ensc-rennes.fr> wrote:
Hi,
Parameters assessment in R with nls doesn't work, though it works fine with
MS Excel with the internal solver :(
I use nls in R to determine two parameters (a,b) from experimental data.
m V C0 Ce Qe
1 0.0911 0.0021740 3987.581 27.11637 94.51206
2 0.0911 0.0021740 3987.581 27.41915 94.50484
3 0.0911 0.0021740 3987.581 27.89362 94.49352
4 0.0906 0.0021740 5981.370 82.98477 189.37739
5 0.0906 0.0021740 5981.370 84.46435 189.34188
6 0.0906 0.0021740 5981.370 85.33213 189.32106
7 0.0911 0.0021740 7975.161 192.54276 233.30310
8 0.0911 0.0021740 7975.161 196.52891 233.20797
9 0.0911 0.0021740 7975.161 203.07467 233.05176
10 0.0906 0.0021872 9968.951 357.49157 328.29824
11 0.0906 0.0021872 9968.951 368.47609 328.03306
12 0.0906 0.0021872 9968.951 379.18904 327.77444
13 0.0904 0.0021740 13956.532 1382.61955 350.33391
14 0.0904 0.0021740 13956.532 1389.64915 350.16485
15 0.0904 0.0021740 13956.532 1411.87726 349.63030
16 0.0902 0.0021740 15950.322 2592.90486 367.38460
17 0.0902 0.0021740 15950.322 2606.34599 367.06064
18 0.0902 0.0021740 15950.322 2639.54301 366.26053
19 0.0906 0.0021872 17835.817 3894.12224 336.57036
20 0.0906 0.0021872 17835.817 3950.35273 335.21289
21 0.0906 0.0021872 17835.817 3972.29367 334.68320
the model "LgmAltformula" is
Qe ~ (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * C0 *
b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a *
b + (b * m * a)^2)^(1/2))/(2 * b * m)
the command in R is
nls(formula=LgmAltFormula,data=bois.DATA,start=list(a=300,b=0.01),trace=TRUE
,control=nls.control(minFactor=0.000000009))
R has difficulties to converge and stops after the maximum of iterations
64650.47 : 2.945876e+02 3.837609e+08
64650.45 : 2.945876e+02 4.022722e+09
64650.45 : 2.945876e+02 1.695669e+09
64650.45 : 2.945876e+02 5.103971e+08
64650.44 : 2.945876e+02 8.497431e+08
64650.41 : 2.945876e+02 1.515243e+09
64650.36 : 2.945877e+02 5.482744e+09
64650.36 : 2.945877e+02 2.152294e+09
64650.36 : 2.945877e+02 7.953167e+08
64650.35 : 2.945877e+02 7.625555e+07
Erreur dans nls(formula = LgmAltFormula, data = bois.DATA, start = list(a =
300, :
le nombre d'it?rations a d?pass? le maximum de 50
The parameters "a" and "b" are estimated to be 364 and 0.0126 with Excel
with the same data set.
I tried with the algorithm="port" with under and upper limits. One of the
parameter reaches the limit and the regression stops.
How can I succeed with R to make this regression?
Regards/Cordialement
-------------
Benoit Boulinguiez
Ph.D
Ecole de Chimie de Rennes (ENSCR) Bureau 1.20
Equipe CIP UMR CNRS 6226 "Sciences Chimiques de Rennes"
Campus de Beaulieu, 263 Avenue du G?n?ral Leclerc
35700 Rennes, France
Tel 33 (0)2 23 23 80 83
Fax 33 (0)2 23 23 81 20
http://www.ensc-rennes.fr/
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