R help - Sep 2009 - Semi continous variable- define bounds using lpsolve

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-tp25530668p25530668.html
Sent from the R help mailing list archive at Nabble.com.

I played around a bit with the original 'lp-solve' program --- i.e., not
the
R package but the program to be downloaded from Sourceforge ---, at least
version 5.5.0.15 through its IDE.  I was not even able to reproduce the
example on semi-continuous variables in the reference documentation at
<http://lpsolve.sourceforge.net/5.5/>.

It appears as if a bug has been introduced in newer versions of
'lp-solve'
that prevent these semi-continuous variables from being handled correctly.
So maybe it's not the fault of the R package 'lpsolve' and can only
be
amended by the 'lp-solve' team itself.

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-tp25530668p25530901.html
Sent from the R help mailing list archive at Nabble.com.

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.