similar to: [Fwd: need some help]

How to define bounds for a semi continous variable in lp_solve. Min 5x1 +9x2 +7.15x3 +0.1x4 subject to x1+x2+x3+x4=6.7 x1+x4 <= 6.5 And x3 can be 0 or greater than 3.6 hence x3 is a semi continous variable how to define bounds as well as semicontinous function because using set.semicont and set. bound simantaneously doesn't seem to work.Thanks in advance for the help -- View this

Hi list, I want to find the total maximum resources I can spend given a set allocation proportion and some simple budget constraints. However, I get suboptimal results via lp and friends (i.e. lpSolve and simplex in the linprog and boot) . For example: library(lpSolve) proportions = c( 0.46, 0.28, 0.26) constraints = c( 352, 75, 171) lp(objective.in = proportions, const.mat =

Dear all, I'm looking for some material on data mining with R. I have something from Luis Torgo but I'd like to see something else. If anybody could help me I'll be thankful Adri??n

Gracias, Jorge. Y cual fue la solucion a la que llegaron? --JIV Sent from my phone. Please excuse my brevity and misspelling. On Aug 25, 2013, at 8:36 AM, Jorge Ayuso Rejas <jayusor@gmail.com> wrote: Esto lo hice yo en una práctica en la universidad, Definíamos un problema de optimización entera minimizando el error de redondeo y restringiendo a la suma de filas y columnas. El 23 de

Dear UseRs, I wrote following function in order to solve Data Envelopment Analysis. Reason for posting is that the function is slow when nrow(dat) is large. I wonder if other functions could substitute the for() loop in the code, such as mapply(). Can anybody help to rewrite the dea() function as efficiently as possible? The code is as follows:

Dear R-helpers, I am struggling with an optimization problem at the moment and decided to write the list looking for some help. I will use a very small example to explain what I would like to. Thanks in advance for your help. We would like to distribute resources from 4 warehouses to 3 destinations. The costs associated are as follows: Destination >From 1 2 3 Total

Dear all, I am trying to find the solution for the optimization problem focused on the finding minimum cost. I used the solution proposed by excel solver, but there is a restriction in the number of variables. My data consists of 300 rows represent cities and 6 columns represent the centres. It constitutes a cost matrix, where the cost are distances between each city and each of six centres. ..+

SImplify your call to lm using the "." argument instead of manipulating formulas. > strt <- lm(y1 ~ ., data = dat) and you do not need to explicitly specify the "1+" on the rhs for lm, so > frm2<-as.formula(paste(trg," ~ ", paste(xvars,collapse = "+"))) works fine, too. Anyway, doing this gives (but see end of output)" bst <-

Jorge, First of all, I really do think that questions such as these should be posted directly to R-help. More people will then see it with a greater chance of getting useful replies. I am subscribed to R-help so will see posts. I have cc'ed this reply to R-help so that you may get more answers than I can give you. I know absolutely nothing about transportation problems. For comments see

Dear all, I am interested in solving a MIP problem with binary outcomes and continuous variables, which ARE NOT RESTRICTED TO BE NEGATIVE. In particular, Max {z1,z2,z3,b1} z1 + z2 + z3 (s.t.) # 7 z1 + 0 z2 + 0 z3 + b1 <= 5 # 0 z1 + 8 z2 + 0 z3 - b1 <= 5 # 0 z1 + 0 z2 + 6 z3 + b1 <= 7 # z1, z2, z3 BINARY {0,1} # -5<= b1 <=5 (i.e. b1 <= 5; -b1 <= 5 ) Using

Dear R-users, i try to solve to following linear programm in R 0 * x_1 + 2/3 * x_2 + 1/3 * x_3 + 1/3 * x_4 = 0.3 x_1 + x_2 + x_3 + x_4 = 1 x_1, x_2, x_3, x_4 > 0, x_1, x_2, x_3, x_4 < 1 as you can see i have no objective function here besides that i use the following code. library(lpSolve) f.obj<-c(1,1,1,1) f.con<-matrix(c(0,2/3,1/3,1/3, 1,1,1,1,

Hello! I have a curious problem, which I cannot solve. With my code I solve thousands of small linear programs with the package lpSolve automatically. But R crashes sometimes (~always, but always on different linear programs) in a strange way. For illustration, I tried to prepare a simple example, which shows the nature of the problem. The function aaa (see below) declares some constants (only in

Dear R-List, (I am not sure whether this list is the right place for my question...) I have a dataframe df.cfa

Hi. I am trying to use the linear optimizer from package lpSolve in R 2.4.1 on Windows XP (Version 5.1). The problem I am trying to solve has 2843 variables (2841 integer, 2 continuous) and 8524 constraints, and I have 2 Gb of memory. After I load the input data into R, I have at most 1.5 Gb of memory available. If I start the lp with significantly less memory available (say 1 Gb), I get

Peter, As I understand your Q. You probably have data that is similar to each other like stock Prices for all RHS variable. In that case the difference between corr and cov is not significant; however, if your RHS contains totally dissimilar variables it matters a great deal. If x1 income, x2 job type, x3 Education level, etc..., then taking cov of these variables would not be desireable

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

Failed? What was the error message? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Aug 22, 2017 at 8:17 AM, Stephen O'hagan <SOhagan at manchester.ac.uk> wrote: > I'm trying to use boot.stepAIC for

The error is "the model fit failed in 50 bootstrap samples Error: non-character argument" Cheers, SOH. On 22/08/2017 17:52, Bert Gunter wrote: > Failed? What was the error message? > > Cheers, > > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka

Jorge, como lo comentó Jorge Ayuso se puede efectivamente ver tu problema como un problema de optimización. Aquí redacté un pequeña función seguramente mejorable que creo soluciona el problema. Utiliza la librería lpSolve y adopté las notaciones que aparecen en el ejemplo de la función lp. Un saludo. Olivier redondeo.matriz<-function(m) { require(lpSolve) n=dim(m)[1] M = m-floor(m)

Hi I'm having a problem with model.frame, encapsulated in this example: y1 <- matrix(c(3,1,0,1,0,1,1,0,0,0,1,0,0,0,1,1,0,1,1,1), nrow = 5, byrow = TRUE) y1 <- as.data.frame(y1) rownames(y1) <- paste("site", 1:5, sep = "") colnames(y1) <- paste("spp", 1:4, sep = "") y1 model.frame(~ y1) Error in model.frame(formula, rownames,