similar to: Binary Quadratic Opt

Hi Petr, The following is different line of thought which is posted in different form, maybe you have some wise input on it. "I need to find Efficient(tracktable) deterministic algorithm for Matching Weighted Graphs with bounded degree. Now we all know Graph matching is non-tractable but when degree of vertex has upper bound are there any tractable algorithm? Does this special case comes

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

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 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

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.

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

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 aware of three different R packages for linear programming: glpk, linprog, lpSolve. From what I can tell, if there are N variables and M constraints, all these solvers require the full NxM constraint matrix. Some linear solvers I know of (not in R) have a sparse matrix input format. Are there any linear solvers in R that have a sparse matrix input format? (including the

Apart from the fact that the statement "such that t1+t2+t3+t4=2970 (as it must)" is not correct, the LP can be implemented as follows: library(lpSolve) LHS <- rbind( c(0,0,0,0, 1, 0, 0,0), c(1,0,0,0,-1, 1, 0,0), c(0,1,0,0, 0,-1, 1,0), c(0,0,1,0, 0, 0,-1,1), cbind(-diag(4),diag(4)), c(0,0,0,0,0,1,0,0), c(0,0,0,0,0,0,1,0), c(0,0,0,0,0,0,0,1) ) RHS <-

HI All, I am unable to solve a optimization Problem Please Help Me out of this to solve. The Optimization problem is as follows :- My objective function is linear and one of the constraint is quadratic. Min z = 5 * X1 + 9* X2 + 7.15 *X3 + 2 * X4 subject to X1 + X2 + X3 +X4 = 9 X1 + X4 < = 6.55 X3(X3 - 3.5) >=0 X1,X2,X3,X4 >=0 Now the problem is how to solve this kind of

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:

Dear all I can not understand how to install the package lpsolve_1.1.9.zip I have read the FAQ and the help pages carefully, but it still not clear for me. I have tried the following (and obtained the respective error messages): >install.packages("c:/ProgramFiles/R/rw2011/library/lpSolve_1.1.9",destdir=" c:/ProgramFiles/R/rw2011/library/lpSolve") Mensagem de aviso:

I was playing around with a simple example using solve.qp ( 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

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 =

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

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

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:

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.