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I'am doing solve the linear program by lpSolveAPI and I want to substitute object function's variable for subject to's variable.

for example, i mimic orignal my lp like this

x5 <- x1 + x2
x6 <- x2 + x3 + x4

object funtion 
min x5 + x6 

subject to
x1 + x2                <=  2
         x3 +  x4      <=  5
      x2    +  x4      <=  3
x1, x2, x3, x4 = binary
x5, x6 = integer

to do this by lpSolveAPI, I try it like below


lprec <- make.lp(3, 4) # make.lp
var_lp <- matrix(c( rep(1, 12)), nrow = 3) # variable as var_lp

set.column(lprec, 1, var_lp[1,1], indices = 1) # s.t.
set.column(lprec, 2, c(var_lp[1,2], var_lp[3,2], indices = c(1,3))
set.column(lprec, 3, var_lp[2,3], indices = 2)
set.column(lprec, 4, c(var_lp[2,4], var_lp[3,4], indices = c(2,3))

i don't know any more how to express x5, x6 at lpSolveAPI

Thank you for your leading. i hope someone give me a answer about that

*plus, my orignal data has about 5,000 constraint. is it possible to analyze by lpSolveAPI??

MINGYU LEE
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1 Answers1

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The set.type function allows you to specify integer domains

set.type(lprec, 5, type="integer")

You would need to increase your definition of the problem, though:

lprec <- make.lp(5,6)

Edit: Code fragment for this problem

library(lpSolveAPI)
lprec <- make.lp(5,6)
set.objfn(lprec, obj=c(1,1), indices=c(5,6)) # sense defaults to "min"
set.row(lprec, 1, xt=c(1,1), indices=c(1,2))
set.row(lprec, 2, xt=c(1,1), indices=c(3,4))
set.row(lprec, 3, xt=c(1,1), indices=c(2,4))
set.row(lprec, 4, xt=c(1,1,-1), indices=c(1,2,5))
set.row(lprec, 5, xt=c(1,1,1,-1), indices=c(2,3,4,6))
set.type(lprec, 1, type="binary")
set.type(lprec, 2, type="binary")
set.type(lprec, 3, type="binary")
set.type(lprec, 4, type="binary")
set.type(lprec, 5, type="integer")
set.type(lprec, 6, type="integer")
set.constr.type(lprec, types=c(rep("<=",3), rep("=",2)))
set.rhs(lprec, b=c(2,5,3,0,0))

You can use print or write.lp to check that the specified program is the program you want to solve. With solve you calculate the solution which would be in this all zero:

solve(lprec)
get.variables(lprec)

(I believe that implicitly, lpSolve assumes non-negativity of all variables.)

Karsten W.
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