similar to: constraint in constrOptim

Hi Everyone, I'm trying to minimize a function using constrOptim with the simulated annealing method SANN. If I understand constrOptim well, it basically passes most of its arguments to optim while somehow enforcing the constraints. My problem is, that since SANN does not need gradients, when using optim with SANN, the gr argument of optim is used to specify a function to create the next

Dear All, I am having trouble in using R function "constrOptim" to do constraint optimization. It seems that "constrOptim" calls function "optim" when it does the optimization, and "optim" allows us to set "method" to be "SANN" if we want to use simulated annealing. In "optim", the function allows us to set gradient to be

Hello. I have some problems, when I try to model an optimization problem with some constraints. The original problem cannot be solved analytically, so I have to use routines like "Simulated Annealing" or "Sequential Quadric Programming". But to see how all this works in R, I would like to start with some simple problem to get to know the basics: The Problem: min f(x1,x2)=

Hi Everyone, I'm trying to minimize a function using constrOptim with the simulated annealing method SANN. If I understand constrOptim well, it basically passes most of its arguments to optim while somehow enforcing the constraints. My problem is, that since SANN does not need gradients, when using optim with SANN, the gr argument of optim is used to specify a function to create the next

Dear All, I would like to minimise a nonlinear function subject to linear inequality constraints as part of an R program. I have been using the constrOptim function. I have tried all of the methods that come with Optim, but nothing finds the correct solution. If I use the correct solution as the vector of starting values, though, my program does output the correct solution and optimum - the

> optim(c(1,1),LL,method="SANN",control=list(fnscale=-1),trans=trans,times=times) $par [1] 17.422635 -1.606859 How could i constraint that the parameters should be both positive in my maximizing problem? I check constrOptim but here i could only constraint the variables trans and times and not my parameters? many thanks, regards Christian

Hello, I'm trying to run constrOptim. It returns to me an error about the fact that constraints arguments (ui and ci) are non compatibles: > optout= constrOptim(startparams, f=ImpulseSS, grad=grImpulse, ui=UI, ci=CI, data=gexp[k,], t=t) Error in ui %*% theta : non-conformable arguments I would like to point out that I can calculate that product in the command line: > UI %*%

Hi, I think that there something wrong with the 'constrOptim' max/minimization function because she doesn?t send extra arguments to 'optim' call. Fact: When I use optim in a f(x,theta)-like function, everything goes ok. But using constrOptim with the same function leads to error... Proof: Make a small change in the 'Rosenbrock Banana function' (taken from the Examples

I have a function myFunction(beta,x) where beta is a vector of coefficients and x is a data frame (think of it as a matrix). I want to optimize the function myFunction() by ONLY changing beta, i.e. x stays constant, with 4 constraints. I have the following code (with a separate source file for the function): rm(list=ls()) source('mySourceFile')

Dear List, I am using constrOptim to solve the following fr1 <- function(x) { b0 <- x[1] b1 <- x[2] ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 } As you can see, my objective function is ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 and I would like to solve for both b0 and b1. If I were to use optim then I would derive the gradient of the

Hello list reader, I am trying to form some constraints for an optimization I am working on. I think I have understand the use of the constraints in matrix form. I use them like: constr_mat<- -diag(2) constr_vec<- rep(-0.05,2) constr_mat<- rbind(constr_mat, diag(2)) constr_vec<- c(constr_vec, rep(-1, 2)) To get parameters in the interval [-1, 0.05]. (And this works so far) Now I

There is a contradiction between what the help page says and what constrOptim actually does with the constraints. The issue is what happens on the boundary. The help page says The feasible region is defined by ?ui %*% theta - ci >= 0?, but the R code for constrOptim reads if (any(ui %*% theta - ci <= 0)) stop("initial value not feasible") The following example

Dear List, I have the code below, where I am using the constrained optimisation package, 'constrOptim.nl' to find the values of two values, b0 and b1. I have no problems when I enter further variable information DIRECTLY into the functions, fn, and heq. In this instance I require fn to have -0.0075 appended to it, and in the case of heq, h[1] has -0.2. library(alabama)

Dear R wizards: I am playing (and struggling) with the example in the constrOptim function. simple example. let's say I want to constrain my variables to be within -1 and 1. I believe I want a whole lot of constraints where ci is -1 and ui is either -1 or 1. That is, I have 2*N constraints. Should the following work? N=10 x= rep(1:N) ci= rep(-1, 2*N) ui= c(rep(1, N), rep(-1, N))

Dear all, There appears to be a bug in how constrOptim handles ... arguments that are suppose to be passed to optim, according to the documentation. This means you can't get the hessian to be returned, for example (so this is a real problem, and not just a question of mistaken documentation). Looking at the code, it appears that a call to the user-defined f includes the ..., when the ...

How can I use a function with some additional input parameters in constrOptim? For example, something like fr <- function(x,a) { ## Rosenbrock Banana function x1 <- x[1] x2 <- x[2] a * (x2 - x1 * x1)^2 + (1 - x1)^2 } where the optimum is to be found w.r.t. x. Calling optim(c(-1.2,1), fr, NULL, a=100) works as expected, but I fail to provide the a=100 in the constrained case:

Hi, I'm having trouble using the constrOptim function to generate the 9-component vector argmin of the function ELfsds: ELfsds <- function(pvechat){ LG=0 for(i in 1:9){ LG=LG+log(pvechat[i]) } return(-LG) } with accompanying gradient function: gradfunc <- function(thetavec){ g=1/(9*thetavec) return(g) } The constraints on the optimization problem are: 1 - components of

Hello, I got a question with the R function constrOptim. >From the R help, it says that the return values of "constrOptim" are the same as "optim". For the return value "convergence" of the function "optim", the values should be 0, 1, 10, 51 and 52. See http://www.maths.lth.se/help/R/.R/library/stats/html/optim.html When I use constrOptim, I get

Hello at all, working with a dataset I try to optimize a non-linear function with constraint. test<-read.csv2("C:/Users/Herb/Desktop/Opti/NORM.csv") fkt<- function(x){ a<-c(0) s<-c(0) #Minimizing square error for(j in 1:107){ s<-(test[j,2] - (x[1] * test[j,3]) - (x[2] * test[j,4]) - (x[3]*test[j,5]) - (x[4]*test[j,6]) - (x[5]*test[j,7]))^2 a<- a+s} a<-as.double(a)

Hi, I am having trouble using constrOptim. My target is to do a portfolio optimization and there some constraints have to be fulfilled. 1) The weight of each share of the portfolio has to be greater than 0 2) The sum of these weights has to be 1 I am able to fulfill either the first or the second constraint but not both. One simple way would be to fulfill the first constraint by using optim as