similar to: optim, constrOptim: setting some parameters equal to each other

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

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

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 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

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:

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)

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

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 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 ...

Hi everybody, please, could you give me help ? I scanned the help archives and didn't found hints... I want to solve a large sparse linear system subjected to an inequality constrains (all solutions positive) and an equality constrain (all solutions sum to 1), thus I tried to fool constrOptim using: x[1] + 0 + ... + 0 >= 0 ... 0 + 0 + ... + x[n] >= 0 x[1] + x[2] + ... +

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))

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/

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

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 <-

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

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 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

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 %*%

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)