similar to: Relative constraint using constrOptim?

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 %*%

I'm using the function constrOptim together with the "SANN" method and my objective function (f) has two parameters. One of the parameters needs be into (2^(-10), 2^4) range and the other into (2^(-2), 2^12) range. How can I do it using constrOptim?? Thank you André Rossi Alertas do Yahoo! Mail em seu celular. Saiba mais em http://br.mobile.yahoo.com/mailalertas/

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

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

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)

Hello All, I am stuck in the following problem. Cexpt=c(0,25,50,100,150,300,250,125,40) t=c(0,0.2,0.4,0.6,1,4,8,12,24) theta0= vector of 6 parms (My initial parameter) A=Constraint matrix (hopefully 6*6) B= Constraint vector of length 6) Cfit=function(t,theta){ J(t)=function(theta,t) Cfit=function(J(t),constant) return(Cfit) } loss=function(theta,t,Cexpt) {

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)

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

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

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

Hello, How do I implement a cardinality constraint with constrOptim? I want to minimize (least square) a%*%x = 4 subject to x1<2 x2<1 x3<4 count(x1, x2, x3)= 2 (cardinality constraint) Is there a way to specify binary integer variables with constrOptim? Here's my code so far: a <-matrix(1:3,1,3) fr <- function(x) { (a%*%x-4)^2 }

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')

Hi, Is there any way to get the hessian matrix from the "constrOptim" function without supplying gradient function? Thanks. --------------------------------- Bring words and photos together (easily) with [[alternative HTML version deleted]]

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

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

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 helpers, I am using the function "constrOptim" to estimate a model with ML with an inequality constraint using the option method='Nelder-Mead'. When I specify the option: hessian = TRUE I obtain the response: Error in f(theta, ...) : unused argument(s) (hessian = TRUE) I guess the function "constrOptim" does not allow this argument which, on the other hand, is

Hi, i need to minimize a quadratic function with boundary condidtions and one equality condition. In order to do that i converted the equality constraint into 2 inequality constaints and passed everything cia constrOptim, as the manual said: everything included in the ... will be passed to Optim that will pass it back to fn in case it does not need it. My code is the following: mat <-

Does anybody know to introduce EQUALITY constraints in 'constrOptim' function? BR, Shubha This e-mail may contain confidential and/or privileged i...{{dropped:13}}

Trying to use constrOptim to minimize the sum of squared deviations. I put the objective function in as: sum((x %*% Y - Z)^2) so i'm trying to get values for x to minimize the sum of the squared deviations between the product of x and Y and Z. Anyways i have no problem using this when x is a 3x1 test variable. it works great with the constraints and everything. when i actually use it on