search for: dmat

...nction doesn't lead the minimal "value" calculated by solve.QP I would be very happy, if anyone could help and tell me, where's my mistake. Thank you very much. Fabian My R-code also containing the sampel of quadprog starts here: #sample from quadprog package library(quadprog) Dmat <- matrix(0,3,3) diag(Dmat) <- 1 dvec <- c(0,5,0) Amat <- matrix(c(-4,-3,0,2,1,0,0,-2,1),3,3) bvec <- c(-8,2,0) erg<-solve.QP(Dmat,dvec,Amat,bvec=bvec) print(erg) erg<-erg$solution -dvec%*%erg+.5*erg^T%*%Dmat%*%erg # my "non working" sampl...

...ets from the distance matrix in a simpler manner than my example code below? ############## # ex_dist.R # example for # manipulating # distance matrices #################### set.seed<-12345 a<-sample(20:40, 10) b<-sample(80:100, 10) c<-sample(0:40, 10) dat<-data.frame(a,b,c) dat dmat<-dist(dat, method="euclidean") dmat dmat[1:6] #vector that stores the distance matrix runs descending down columns, left to right #in a 10-element distance matrix, column lengths are 9,8,7,6....1 #get comparisons of rows 1:4 (from dat) ONLY #top-left matrix will consist of top 3 of...

...constrains including not only the variables P1, P2, P3, P4, P5, P6, but also the variables X1, X2,X3,X4,X5,X6,X7,X8,X9. As the set of variables X's are not affecting the objective function, I assume that I have to enter them as zero's in the vector "dvec" and the matrix "Dmat". My program states as: mat<-matrix(0,15,15) diag(Dmat)<-c(10,5,8,10,20,10,0,0,0,0,0,0,0,0,0) dvec<- c(-200,-100,-160,-50,-50,-50,0,0,0,0,0,0,0,0,0) Amat<- matrix(c(-1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,-1,0,1, 0,0,0,0,0,0,0,0,0,0,0,0,0,-1,1,0,0,0,0,...

...orized" argument in solve.QP (part of the quadprog library) to work as expected. The helpfile states that when the factorized argument is set to TRUE, then the function requires the inverse of a square-root factor of the Hessian instead of the Hessian itself. That is, when factorized=TRUE, the Dmat argument should be a matrix R^(-1), such that the Hessian of the objective function is t(R) %*% R. I modified the example in the helpfile slightly to test this out: R = matrix(rnorm(9),3,3) R.inv = solve(R) Dmat = t(R) %*% R dvec = c(0,5,0) Amat = matrix(c(-4,-3,...

...= bvec and [bLo=0 <= w #<=1=bUp] to Amat <- rbind(-1, 1, -mu, mu) dim(bLo) <- c(n,1) dim(bUp) <- c(n,1) bvec <- rbind(-1, 1, mu.target, Inf, bLo, -bUp) zMat <- matrix(rep(0,2*n*n),ncol=n, nrow=n*2) zMat[,1] <- c(rep(1,n), rep(-1,n)) Amat <- t(rbind(Amat, zMat)) #So I set Dmat=Cov and set dvec=0 Dmat=Cov dvec=rep(0, nrow(Amat)) #The first two rows of Amat should be equality constraints (so weights sum to 1) meq <- 2 sol <- solve.QP(Dmat=Dmat, dvec=dvec, Amat=Amat, bvec=bvec, meq) sol

...; is a Nx1 vector of the expected returns and \mu is a number, the postualted return of the investor I've done the following code but it doesn't make what I want it to do. The weights aren't all positive and the \mu isn't reached. What's wrong with my code? Require(quadprog) Dmat<-diag(1,7,7) # muss als quadratische Matrix eingegeben werden Dmat dvec<-matrix(0,7,1) # muss als Spaltenvektor eingegeben werden dvec mu<-0 # (in Mio. €) bvec<-c(1,mu,matrix(0,7,1)) # muss als Spaltenvektor eingegeben werden bvec mu_r<-c(19.7,33.0,0.0,49.7, 82.5, 39.0,11.8) Amat&lt...

Hello: Does anyone know if there exists a package that handles methods for [ for dist objects? I would like to access a dist object using matrix notation e.g. dMat = dist(x) dMat[i,j] Thanks in advance to anyone who can point me in the right direction. [[alternative HTML version deleted]]

...traints by creating pairs of positive and negative values within the constraint matrix ( http://tolstoy.newcastle.edu.au/R/help/05/10/14251.html), but when I pass this expanded constraint matrix to solve.QP it complains that Amat and dvec are incompatible. How should I expand dvec (and consequently Dmat) to accomodate the larger Amat? Moreover, I am unclear how to apply the meq equality constraint across more than one cell (i.e. rows summing to one) although I have attempted a guess below. Any help warmly received. Selwyn. ################ #examples below ################ library(quadprog) #pai...

Hello. I'm trying to solve a quadratic programming problem of the form min ||Hx - y||^2 s.t. x >= 0 and x <= t using solve.QP in the quadprog package but I'm having problems with Dmat not being positive definite, which is kinda okay since I expect it to be numerically semi-definite in most cases. As far as I'm aware the problem arises because the Goldfarb and Idnani method first solves the unconstrained problem requiring a positive definite matrix. Are there any (fast) packa...

...I have been reading the documentation and it seems like all the examples use equations instead of vector manipulation. All of my parameters are vectors and matrices and they can be quite large. Here is my problem: X<-([Cf]+[H])%*%[A] Y<-([Cf]+[H]-[R])%*%[B]I want to find H that minimizes Y%*%Dmat%*%t(Y) for a given value of X%*%Dmat%*%t(X) Cf, R, A, Dmat and B are matrices of constants. The values for H sohould be between 0 and 1. Is it possible to use Rsolnp to find the vector H even though the input functions will all return other vectors? -- View this message in context: http://r....

...( function is in the quadprog package ) and the code is below. ( I'm not even sure there if there is a reasonable solution because I made the problem up ). But, when I try to use solve.QP to solve it, I get the error that D in the quadratic function is not positive definite. This is because Dmat is zero because I don't have a quadratic term in my objective function. So, I was wondering if it was possible to use solve.QP when there isn't a quadratic term in the objective function. I imagine that there are other functions in R that can be used but I would like to use solve.QP beca...

...*/ break; case DLL_PROCESS_DETACH: /* clean-up code here */ break; } return(TRUE); } #endif void nlme_two_comp_zero_CI_lag (long int *norow, double *maxoorder, double *OMAT, long int *ndrow, double *maxdorder, double *DMAT, long int *logparam, long int *logresp, double *Resp) { long int i, j, No = *norow, Nd = *ndrow, LogParam = *logparam, LogResp = *logresp; double Tdiff, a, b, k21, origReset, counterDose, id, T1, constReseti, constResetj, MaxOORDER = *maxoorder, MaxDORDER = *maxdorder, *OTime, *ITI...

...talling R I downloaded and installed the contributed quadprog library. I ran this simple solver example using Python+rpy which works (well, except for some R error I haven't figured out yet) in the old installation: import numpy import rpy rpy.r("library(quadprog)") Dmat = numpy.identity(3, numpy.float_) print Dmat rpy.r.assign("Dmat", Dmat) dvec = numpy.array([0,5,0], numpy.float_) print dvec rpy.r.assign("dvec", dvec) Amat = numpy.array([[-4,-3,0], [2,1,0], [0,-2,1]], numpy.float_) print Amat rpy.r.assign(&quo...

...code example: ######################################## library(mvtnorm) c<-data.frame(rnorm(1000,5,sd=1),rnorm(1000,6,sd=1)) c2<-data.frame(rnorm(1000,10,sd=2),rnorm(1000,7,sd=1)) xx=seq(0,20,0.1) yy=seq(0,20,0.1) xmult=cbind(rep(yy,201),rep(xx,each=201)) dens=dmvnorm(xmult,mean(c),cov(c)) dmat=matrix(dens,ncol=length(yy),nrow=length(xx),byrow=F) dens2=dmvnorm(xmult,mean(c2),cov(c2)) dmat2=matrix(dens2,ncol=length(yy),nrow=length(xx),byrow=F) contour(xx,yy,dmat,lwd=2) contour(xx,yy,dmat2,lwd=2,add=T) ############################################## Is their an easy way to do this (maybe w...

Hi, I have a least square fitting problem with linear inequality constraints. pcls seems capable of solving it so I tried it, unfortunately, it is stuck with the following error: > M <- list() > M$y = Dmat[,1] > M$X = Cmat > M$Ain = as.matrix(Amat) > M$bin = rep(0, dim(Amat)[1]) > M$p=qr.solve(as.matrix(Cmat), Dmat[,1]) > M$w = rep(1, length(M$y)) > M$C = matrix(0,0,0) > p<-pcls(M) Error in t(qr.qty(qra, t(M$X))[(j + 1):k, ]) : error in evaluating the argument 'x' in...

...7565 -0.33015962 0.09136359 -5.426254 -0.8201206 [5,] -0.68977432 0.01977374 0.61772506 3.751978 0.4348802 %Matlab Matrix PA9900 =[11/24 10/53 0/1 0/1 29/43 ;1/24 27/53 0/1 0/1 13/43 ;14/24 178/53 146/244 17/23 15/43 ;2/24 4/53 0/1 2/23 2/43 ;4/24 58/53 26/244 0/1 5/43] %MatLab-syntax [wmat,dmat]=eig(mat) %MatLab-output wmat = -0.2250 0.4330 -0.4998 -0.1795 -0.1854 -0.1083 0.1771 0.1599 -0.0614 -0.0583 -0.9403 -0.7191 -0.8457 0.9617 0.9708 -0.0327 -0.0752 -0.0967 -0.1750 -0.1160 -0.2289 -0.5083 0.0058 0.0928 0.0802 dmat =...

Dear list members, I am trying to get information on how to fit a linear regression with constrained parameters. Specifically, I have 8 predictors , their coeffiecients should all be non-negative and add up to 1. I understand it is a quadratic programming problem but I have no experience in the subject. I searched the archives but the results were inconclusive. Could someone provide suggestions

I have a matrix y: > dimnames(y) $x93 [1] "1" "2" $x94 [1] "0" "1" "2" .................. so on (there are other dimensions as well) I need to access a particular dimension, but a random mechanism tells me which dimension it would. So, sometimes I might need to access dimnames(y)$x93, some other time it would be dimnames(y)$x94.. and so

Hello everyone, I have 3 variables Y, X1 and X2. Each variables lies between 0 and 1. I want to do a constrained regression such that a>0 and (1-a) >0 for the model: Y = a*X1 + (1-a)*X2 I tried the help on the constrained regression in R but I concede that it was not helpful. Any help is greatly appreciated -- Thanks, Jim. [[alternative HTML version deleted]]

Hi there, I have a major problem (major for me that is) with solve.QP and I'm new at this. You see, to solve my quadratic program I need to have the lagrange multipliers after each iteration. Solve.QP gives me the solution, the unconstrained solution aswell as the optimal value. Does anybody have an idea for how I could extract the multipliers? Thanx, Serge "Beatus qui prodest quibus