But of course, it is always possible to emulate a semi-continuous variable by introducing a binary variable and use some "big-M" trick. That is, with a new binary variable b we add the following two conditions:
x3 - 3.6 * b >= 0 and x3 - 10 * b <= 0 # Big-M trick, here M >= 10 (If b = 0, then x3 = 0, and if b = 1, then x3 >= 3.6 !) As I do not trust 'lpSolve' too much anymore I used package 'Rglpk' with the following code: #-- snip --- library(Rglpk) obj<- c(5, 9, 7.15, 0.1, 0) mat <- matrix(c(1,1,1,1,0, 1,0,0,1,0, 0,0,1,0,-3.6, 0,0,1,0,-10, 0,0,0,0,1), byrow=TRUE, ncol=5) dir <- c("==", "<=", ">=", "<=", "<=") rhs <- c(9, 6.55, 0, 0, 1) types <- c("C", "C", "C", "C", "I") max <- FALSE Rglpk_solve_LP(obj, mat, dir, rhs, types, max = max) # $optimum # [1] 22.705 # # $solution # [1] 0.00 2.45 0.00 6.55 0.00 # # $status # [1] 0 #-- snap --- Semi-continuous variables are sometimes preferred as with a good implementation the solution is reached much faster (that's why I suggested them), but they can always be modelled with binary variables. Hans Werner pragathichi wrote: > > 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 message in context: http://www.nabble.com/Semi-continous-variable--define-bounds-using-lpsolve-tp25530668p25530929.html Sent from the R help mailing list archive at Nabble.com. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.