Displaying 10 results from an estimated 10 matches for "maxfeval".
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maxeval
2012 Nov 05
1
Error message in nmkb()
Hallo together,
I am trying to use the nmkb() optimizer and I have problems using the
function, as it causes the following error message
Fehler (error)* in while (nf < maxfeval & restarts < restarts.max & dist >
ftol & :
Fehlender Wert (missing value)* , wo (where)* TRUE/FALSE n?tig ist (is
required)*
*translation
Do I need to adjust the control ?
my.optpar3<-nmkb(par=par.start,fn=my.loglike.normal,lower=constLo,upper=999,
control=list(restarts...
2025 Mar 27
1
Problem with minimization that I failed to understand
...2, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04,
0.09, 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02),
fn = Fn,
A = matrix(c(rep(0, 20), -1), nrow = 1), b = -2.05/100, Aeq =
matrix(c(rep(1, 20), 1), nrow = 1), beq = 1,
lb = rep(0.01, 21),
tol = 1e-16, maxfeval = 10000000, maxiter = 5000000)
Fn(q1$par) < Fn(c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09,
0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02))
#FALSE
On Fri, 28 Mar 2025 at 00:58, Rui Barradas <ruipbarradas at sapo.pt> wrote:
> ?s 18:35 de 27/03/2025, D...
2025 Mar 27
1
Problem with minimization that I failed to understand
...0.07, 0.04,
> 0.09, 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02),
> fn = Fn,
> A = matrix(c(rep(0, 20), -1), nrow = 1), b = -2.05/100, Aeq =
> matrix(c(rep(1, 20), 1), nrow = 1), beq = 1,
> lb = rep(0.01, 21),
> tol = 1e-16, maxfeval = 10000000, maxiter = 5000000)
>
>
> However with above code, I got sub-optimal value in terms of minimization
> of the objective function:
>
> q1$value
> #0.1632184
> Fn(c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09, 0.08, 0.02,
> 0.02, 0.03, 0.06, 0.02, 0, 0...
2025 Mar 27
1
Problem with minimization that I failed to understand
...0.07, 0.04,
> 0.09, 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02),
> fn = Fn,
> A = matrix(c(rep(0, 20), -1), nrow = 1), b = -2.05/100, Aeq =
> matrix(c(rep(1, 20), 1), nrow = 1), beq = 1,
> lb = rep(0.01, 21),
> tol = 1e-16, maxfeval = 10000000, maxiter = 5000000)
>
> Fn(q1$par) < Fn(c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09,
> 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02))
> #FALSE
>
>
> On Fri, 28 Mar 2025 at 00:58, Rui Barradas <ruipbarradas at sapo.pt> wro...
2025 Mar 27
2
Problem with minimization that I failed to understand
...2, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04,
0.09, 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02),
fn = Fn,
A = matrix(c(rep(0, 20), -1), nrow = 1), b = -2.05/100, Aeq =
matrix(c(rep(1, 20), 1), nrow = 1), beq = 1,
lb = rep(0.01, 21),
tol = 1e-16, maxfeval = 10000000, maxiter = 5000000)
However with above code, I got sub-optimal value in terms of minimization
of the objective function:
q1$value
#0.1632184
Fn(c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09, 0.08, 0.02,
0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02))
#0.1586207
Cou...
2025 Mar 28
3
Problem with minimization that I failed to understand
....02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02,
> 0.02),
> > fn = Fn,
> > A = matrix(c(rep(0, 20), -1), nrow = 1), b = -2.05/100, Aeq =
> > matrix(c(rep(1, 20), 1), nrow = 1), beq = 1,
> > lb = rep(0.01, 21),
> > tol = 1e-16, maxfeval = 10000000, maxiter = 5000000)
> >
> >
> > However with above code, I got sub-optimal value in terms of minimization
> > of the objective function:
> >
> > q1$value
> > #0.1632184
> > Fn(c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09, 0.08, 0....
2025 Mar 28
1
Problem with minimization that I failed to understand
...0.07, 0.04,
> 0.09, 0.08, 0.02, 0.02, 0.03, 0.06, 0.02, 0, 0.07, 0.05, 0.02, 0.02, 0.02),
> fn = Fn,
> A = matrix(c(rep(0, 20), -1), nrow = 1), b = -2.05/100, Aeq =
> matrix(c(rep(1, 20), 1), nrow = 1), beq = 1,
> lb = rep(0.01, 21),
> tol = 1e-16, maxfeval = 10000000, maxiter = 5000000)
>
>
> However with above code, I got sub-optimal value in terms of minimization
> of the objective function:
>
> q1$value
> #0.1632184
> Fn(c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0.07, 0.04, 0.09, 0.08, 0.02,
> 0.02, 0.03, 0.06, 0.02, 0, 0...
2012 Sep 16
1
trying to obtain same nls parameters as in example
Dear R-users;
I'm working with a a dataset that was previously used to fit a
nonlinear model of the form:
Y ~ a * (1 + b * log(1 - c * X^d))
The parameters published elsewhere are:
a = 1.758863, b = .217217, c = .99031, and d = .054589
However, there is no way I can replicate this result. I've tried
several options (including SAS) w/o success.
The data is:
X <-
2025 Mar 28
1
Problem with minimization that I failed to understand
...0.02, 0, 0.07, 0.05, 0.02, 0.02,
>> 0.02),
>>> fn = Fn,
>>> A = matrix(c(rep(0, 20), -1), nrow = 1), b = -2.05/100, Aeq =
>>> matrix(c(rep(1, 20), 1), nrow = 1), beq = 1,
>>> lb = rep(0.01, 21),
>>> tol = 1e-16, maxfeval = 10000000, maxiter = 5000000)
>>>
>>>
>>> However with above code, I got sub-optimal value in terms of minimization
>>> of the objective function:
>>>
>>> q1$value
>>> #0.1632184
>>> Fn(c(0.12, 0.04, 0.07, 0.03, 0.06, 0.07, 0....
2011 Mar 15
1
Problem with nls.lm function of minpack.lm package.
Dear R useRs,
I have a problem with nls.lm function of minpackl.lm package.
I need to fit the Van Genuchten Model to a set of data of Theta and hydraulic conductivity with nls.lm function of minpack.lm package.
For the first fit, the parameter estimates keep changing even after 1000 iterations (Th)
and
I have a following error message for fit of hydraulic conductivity (k);
Reason for