[R] Semi continous variable- define bounds using lpsolve

[pragathichi]

thank a lot it works

Hans W. Borchers wrote:
> 
> 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
>> 
> 
> 

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