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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]]

Hi All, I'm trying to do the following constrained optimization example. Maximize x1*(1-x1) + x2*(1-x2) + x3*(1-x3) s.t. x1 + x2 + x3 = 1 x1 >= 0 and x1 <= 1 x2 >= 0 and x2 <= 1 x3 >= 0 and x3 <= 1 which are the constraints. I'm expecting the answer x1=x2=x3 = 1/3. I tried the "constrOptim" function in R and I'm running into some issues. I first start off

Hi, I apologize for the long message but the problem I encountered can't be stated in a few lines. I am having some problems with the function constrOptim. My goal is to maximize the likelihood of product of K multinomials, each with four catagories under linear constraints on the parameter values. I have found that the function does not work for many data configurations. #The likelihood

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 ...

Full_Name: Haobo Ren Version: 2.1.1 OS: Windows 2000 Submission from: (NULL) (192.11.226.116) When running constrOptim, there is error message Error: subscript out of bounds

Dear R users, I?m a graduate students and in my master thesis I must obtain the values of the parameters x_i which maximize this Multinomial log?likelihood function log(n!)-sum_{i=1]^4 log(n_i!)+sum_ {i=1}^4 n_i log(x_i) under the following constraints: a) sum_i x_i=1, x_i>=0, b) x_1<=x_2+x_3+x_4 c)x_2<=x_3+x_4 I have been using the ?ConstrOptim? R-function with the instructions

Hello All, I would like to optimize a (log-)likelihood function subject to a number of linear constraints between parameters. These constraints are equality constraints of the form A%*%theta=c, ie (1,1) %*% 0.8,0.2)^t = 1 meaning that these parameters should sum to one. Moreover, there are bounds on the individual parameters, in most cases that I am considering parameters are bound between zero

Hi list: This is my first time to post my question on the list. Thanks for your help. I am solving a quadratic programming using R. Here is my question: w = arg min 0.5*w'Mw - w'N s. t. sum(w) = 1; w>0 note: w is weight vector, each w_i must >=0, and the sum of w =1. Here is my R code: A <-matrix(c(2.26,1.26,1.12,1.12,2.27,1.13,1.12,1.13,2.2),3,3); B <-

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 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)

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 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

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

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))

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 %*%

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:

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 }

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)