similar to: Minimise a parameter of a given function f, with f > 0

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

Hello all, I have an optimization routine that is giving me good results, but the results are not in the nice "model" format like "lm". How can I get optim results into a model so that I can use the clever 'fitted', 'residuals', and 'summary' functions? Using optim is the only way that I was able to make a model that 1) sums the betas to 1, 2)

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

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

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:

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

Hello, and apologies for the upcoming naive questions. I am a biologist who is trying to teach himself the appropriate areas of math and stats. I welcome pointers to suggested background reading just as much as I do direct answers to my question. Let's say I have a function F() that takes variables (a,b,c,a1,b1,c1) and returns x, and I want to find the values of these variables that result in

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 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 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 ...

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, 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

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, 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 <-

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))

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

Dear R helpers I have a question regarding the constrainOptim. I'm coding the nested logit and would like to set a bound of rho to (0,1] as an extreme value distribution where rho = exp(lambda)/1+exp(lambda) I wonder if I can do that directly in optim (say rho > 0 & <= 1) or need to use constrainOptim I read the help but still don't know how to set ui and ci Thanks, June

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