similar to: lpSolve

Dear R users, I would like to optimize a linear approximation of a quadratic function using lpSolve. My code runs without any error or warning message but the constraints that I set don't seem to work properly. Nevertheless, I am certain that my code is somewhere wrong. I would like to solve the following problem: max 2x-x^2+y subject to 2x^2 + 3y^2 <= 6 2>= x,y >= 0 I would

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 all, I'm working on a very complex linear optimization problem using the lp.transport function in lpSolve. My PC has 10 cores, but by default R uses only one of them. Is there a straightforward way to make lp.transport use all cores available? I had a look at "High-performance and parallel computing in R" ( http://cran.r-project.org/web/views/HighPerformanceComputing.html),

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

Hey, I have stacked a couple of garchFit objects in a list with names $fit1, $fit2, ..., $fiti assigning objects names using a loop, i.e. after running the loop modelStack = list($fit1, $fit2,...,$fiti). Thus the following apply; a = modelStack$fit2, then a is the second garchFit object of formal class 'fGarch' with 11 slots, @call, @formula... etc. I then want to extract information in

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

Hi, Using the package lpSolve API, I need to build a 2000*100000 constraint matrix. I wonder which method is faster: (a) model = make.lp(0,0) add.constraint(model, ...) or (b) model = make.lp(2000,100000) set.constraint(model,...) Thanks ------------ KC

Hi. I've tried to execute &#34;example(lp.assign)&#34; in lpSolve package. However, this solution seemed not to return correct answer, all matrix elements were zeros. Could you give me any &#34;solution&#34; about this? Regards.

Hi. In the help page for "lp" in package lpSolve, regarding "const.dir" it says: const.dir: Vector of character strings giving the direction of the constraint: each value should be one of "<," "<=," "=," "==," ">," or ">=". (In each pair the two values are identical.) I am having trouble understanding

Hi Bert, I won't post any more messages on this thread as problem has shifted from Optimization in R to Graph Algorithms. Rest fine Khris. On Aug 2, 2012, at 9:13 PM, Bert Gunter [via R] wrote: > This discussion needs to be taken off (this) list, as it appears to > have nothing to do with R. > > -- Bert > > On Thu, Aug 2, 2012 at 2:27 AM, khris <[hidden email]>

Dear all, I am a beginner with lpSolve package (and not an expert in the others). I can not understand why I am doing wrong, and I would be very grateful if anyone could please help me on this. I am trying to optimize ("min") the sum of columns/variables, constrained to >=1. Each column/variables has its weight - given by values at f.obj. The matrix for numeric constraint

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

Hello R users, I have two quick questions while using "lpSolve" package for linear programming. (1) the result contains both characters and numbers, e.g., Success: the objective function is 40.5, but I only need the number, can I only store the number? (2) How to set boundaries for variables? e.g., all variable are positive. Thanks a lot! Xiaoxi

Under freebsd 6.2-p7 i386, R 2.5.1,gcc-4.2.2 I'm unable to compile package lpSolve because: hpbsd# R CMD INSTALL lpSolve_5.5.8.tar.gz * Installing to library '/usr/local/lib/R/library' * Installing *source* package 'lpSolve' ... ** libs cc -std=gnu99 -I/usr/local/lib/R/include -I/usr/local/lib/R/include -I . -DINTEGERTIME -DPARSER_LP -DBUILDING_FOR_R -DYY_NEVER_INTERACTIVE

Hi, I am working with lpSolve "R" package by Sam Buttrey, buttrey at nps.navy.mil , which is interface to lp_solve linear/integer programming system. You can find information about lp_solve at http://groups.yahoo.com/group/lp_solve/ (free registration required). lpSolve (R package) supports linear and integer programming but it does not support semi-continuous variables

Dear Friends, I am trying to install a few packages in R and am receiving error messages. Since the error messages are different, I am posting them separately. The second error is with the installation of lpSolve. The core error message is: ======================================================================== In file included from /usr/include/bits/posix1_lim.h:153, from

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. ..+

Hello all, I would like to use the lp function (lpSolve package) on a linux system. Using sample code from the lp function help, I can solve the linear program with no errors or problems. However, when I copy the exact same code to R in linux (after loading the lpSolve package), I get the following error: *** caught segfault *** address 0x3, cause 'memory not mapped' Traceback: 1:

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 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: