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I'm trying to solve the following optimization problem with R lpSolve package:

1.007825*x1 +12.000000*x2 +15.99492*x3 +14.00307*x4 +31.97207*x5 +30.97376*x6 >= 10
1.007825*x1 +12.000000*x2 +15.99492*x3 +14.00307*x4 +31.97207*x5 +30.97376*x6 <= 15
       1*x1        - 2*x2       + 0*x3        -1*x4        +0*x5        -3*x6 <=  2
         xi >= 0, where i = [1,2,3,4,5,6]

My objective function is:

1.007825*x1 +12.000000*x2 +15.99492*x3 +14.00307*x4 +31.97207*x5 +30.97376*x6

I create the matrix of constraints in the standard form (A):

           [,1] [,2]      [,3]      [,4]      [,5]      [,6]
 [1,] -1.007825  -12 -15.99492 -14.00307 -31.97207 -30.97376
 [2,]  1.007825   12  15.99492  14.00307  31.97207  30.97376
 [3,]  1.000000   -2   0.00000  -1.00000   0.00000  -3.00000
 [4,] -1.000000    0   0.00000   0.00000   0.00000   0.00000
 [5,]  0.000000   -1   0.00000   0.00000   0.00000   0.00000
 [6,]  0.000000    0  -1.00000   0.00000   0.00000   0.00000
 [7,]  0.000000    0   0.00000  -1.00000   0.00000   0.00000
 [8,]  0.000000    0   0.00000   0.00000  -1.00000   0.00000
 [9,]  0.000000    0   0.00000   0.00000   0.00000  -1.00000

I create the matrix of borders (b):

      [,1]
 [1,]  -10
 [2,]   15
 [3,]    2
 [4,]    0
 [5,]    0
 [6,]    0
 [7,]    0
 [8,]    0
 [9,]    0

and the objective function (f):

f = c(1.007825, 12.000000, 15.99492, 14.00307, 31.97207, 30.97376)

When I put it into the code:

out = lp("min",f,A,rep("<=",9),b,all.int=TRUE),

I get the solution c(0,0,0,1,0,0), although I know that the solution is c(0,1,0,0,0,0). If I change the right border (instead of 15 I make 13), everything works. What could be the problem?

Kirill
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1 Answers1

2

With lpSolveAPI, the problem solves fine:

library(lpSolveAPI)
lprec <- make.lp(0, ncol=6)
set.type(lprec, columns=seq(1,6), type="integer")
set.objfn(lprec, obj=c(1.007825, 12, 15.99492, 14.00307, 31.97207, 30.97376))

add.constraint(lprec, xt=c(1.007825, 12, 15.99492, 14.0030, 31.97207, 30.97376), type=">=", rhs=10)
add.constraint(lprec, xt=c(1.007825, 12, 15.99492, 14.0030, 31.97207, 30.97376), type="<=", rhs=15)
add.constraint(lprec, xt=c(1, -2, 0, -1, 0, -3), type="<=", rhs=2)

solve(lprec)
get.variables(lprec)

returns 

[1] 0 1 0 0 0 0
Karsten W.
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  • Thank you. I'll check it and let you know. – Kirill Nov 22 '16 at 12:30
  • Yep, it works. One more question: do you know if all MILP produce approximate integers; I mean that if a number is close enough, then it is considered as an integer. – Kirill Nov 23 '16 at 08:06
  • See [LP relaxation](https://en.wikipedia.org/wiki/Linear_programming_relaxation#Branch_and_bound_for_exact_solutions) for a description of an algorithm that computes exact solutions. – Karsten W. Nov 23 '16 at 19:12