Displaying 4 results from an estimated 4 matches for "log1ab".
2017 Jun 18
3
R_using non linear regression with constraints
...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...
2017 Jun 18
0
R_using non linear regression with constraints
..., 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
&...
2017 Jun 18
0
R_using non linear regression with constraints
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.
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