similar to: semi-continuous variables in lpSolve package

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

Get articles from http://www.jstatsoft.org Firth Bradley-Terry Models in R Volume 12, Issue 01 Sturtz, Ligges, and Gelman R2WinBUGS: A Package for Running WinBUGS from R Volume 12, Issue 03 Mineo and Ruggieri A Software Tool for the Exponential Power Distribution: The normalp Package Volume 12, Issue 04 Ritz and Streibig Bioassay Analysis Using R Volume 12, Issue 05 Baddeley and Turner

Get articles from http://www.jstatsoft.org Firth Bradley-Terry Models in R Volume 12, Issue 01 Sturtz, Ligges, and Gelman R2WinBUGS: A Package for Running WinBUGS from R Volume 12, Issue 03 Mineo and Ruggieri A Software Tool for the Exponential Power Distribution: The normalp Package Volume 12, Issue 04 Ritz and Streibig Bioassay Analysis Using R Volume 12, Issue 05 Baddeley and Turner

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

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 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

Hey, Anyone who knows what the following error report when calling lp() in package lpSolve means? Error: status 3 JT -- View this message in context: http://n4.nabble.com/lpSolve-tp1599380p1599380.html Sent from the R help mailing list archive at Nabble.com.

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

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

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

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.

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. 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

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

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-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,

Hi R People: If you have a linear programming problem in which some of the constraints have the "<=", some have ">=" and some have "=", all in the same problem, should the solver work? I'm having trouble with that. Any help much appreciated! Sincerely, Erin Hodgess mailto: hodgess at gator.uhd.eud whoops! mailto: hodgess at gator.uhd.edu

Hi. This feels like a bug to me, or at least an undocumented feature, but I thought I'd see what people here thought of it. Consider a POSIXlt object like this one: > a <- as.POSIXlt ("2011-01-23 12:45:45") > class (a) [1] "POSIXlt" "POSIXt" Fine. Now, if I do some arithmetic on that object, the result is converted to POSIXct. > class (a

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]>