search for: rglpk_solve_lp

Hi list, in the Rglpk_solve_LP function (::Rglpk), on line 26, the function calls a function as.glp_bounds() that i cannot access. i'm trying to alter the Rglpk_solve_LP function to add a line to retrieve column/row dual values. everytime i change the slightest line of code inside Rglpk_solve_LP() [to even add a print] i...

...negative integers ## x_2 is a non-negative real number obj <- c(3, 1, 3) mat <- matrix(c(-1, 0, 1, 2, 4, -3, 1, -3, 2), nrow = 3) dir <- c("<=", "<=", "<=") rhs <- c(4, 2, 3) types <- c("I", "C", "I") max <- TRUE Rglpk_solve_LP(obj, mat, dir, rhs, types, max) ## Same as before but with bounds replaced by ## -Inf < x_1 <= 4 ## 0 <= x_2 <= 100 ## 2 <= x_3 < Inf bounds <- list(lower = list(ind = c(1L, 3L), val = c(-Inf, 2)), upper = list(ind = c(1L, 2L), val = c(4, 100))) Rglpk_solve_LP(obj, mat, dir...

...c(0,1,0,0,1,0,1,-2.057855981,0) col3 <-c(0,0,1,0,0,1,1,-1.885177032,0) col4 <-c(-1,-1,-1,1,1,1,0,0,1) mat <- cbind(col1, col2, col3, col4) dir <- c(rep("<=", 3), rep(">=", 3), rep("==", 2), ">=") rhs <- c(rep(0, 7), 1, 0) sol <- Rglpk_solve_LP(obj, mat, dir, rhs, types = NULL, max = FALSE, bounds = c(-100,100), verbose = TRUE) The R output says there is no feasible solution, but e.g. (-2.3756786, 0.3297676, 2.0459110, 2.3756786) is feasible. The output is "GLPK Simplex Optimizer, v4.42 9 rows, 4 columns, 19 non-zeros 0:...

..., rep(-1,XROWS)), cbind(diag(XROWS), rep(1,XROWS)), cbind(t(X), rep(0,XCOLS)), cbind(t(rep(0,XROWS)), 1)), nrow=2*XROWS+XCOLS+1) dir <- c(rep("<=", XROWS), rep(">=", XROWS), rep("==", XCOLS), ">=") rhs <- c(rep(0, 2*XROWS), e_2, 0) sol <- Rglpk_solve_LP(obj, mat, dir, rhs, types = NULL, max = FALSE, bounds = c(-5,5), verbose = TRUE) ........................................................................... The output is " GLPK Simplex Optimizer, v4.42 203 rows, 101 columns, 601 non-zeros 0: obj = 0.000000000e+000 infeas = 1.000e+0...

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