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I have a linear programming problem that I'm trying to solve in R. I have used lpSolve package. lpSolve by default uses primal simplex algorithm to obtain solution. What if I want to change the algorithm to dual simplex ? The results vary widely between the two algorithms. Are there any other packages that would help to solve the problem below using dual simplex algorithm.

library("lpSolve")

f.obj <- c(rep(1,12),rep(0,4))
f.cons <- matrix(c(1,-1,0,0,0,0,0,0,0,0,0,0,1,-1,0,0,
                   0,0,1,-1,0,0,0,0,0,0,0,0,1,0,-1,0,
                   0,0,0,0,1,-1,0,0,0,0,0,0,1,0,0,-1,
                   0,0,0,0,0,0,1,-1,0,0,0,0,0,1,-1,0,
                   0,0,0,0,0,0,0,0,1,-1,0,0,0,1,0,-1,
                   0,0,0,0,0,0,0,0,0,0,1,-1,0,0,1,-1),nrow=6,byrow=T)

f.dir <- rep("=",6)

f.rhs <- c(-1.0986,1.6094,-1.0986,1.94591,1.3863,-1.7917)

g <- lp ("min", f.obj, f.cons, f.dir, f.rhs,compute.sens=TRUE)
g$solution

Primal Simplex using lpSolve in R is as follows:

0 0 0 0 0 0.91630 0.0 0.76209 0.47 0 0 0 1.60940 2.70800 0 1.79170

Dual Simplex using Lingo software and SAS is as follows:

0 0.76214 0 0 1.23214 0 0 0 0.15415 0 0 0 0.8473 1.9459 0 1.7918

The objective function is same for both the algorithms is 2.14839

forecaster
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    If you use `lpSolveAPI`, you can control the simplex type using e.g. `lp.control(lprec, simplextype="dual")` – Karsten W. Dec 09 '16 at 17:26
  • @KarstenW. excellent it worked, thank you. Can you please put your comment as answer so that I can accept it ? – forecaster Dec 09 '16 at 20:24

1 Answers1

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With lpSolveAPI, you can finetune your solver:

lprec <- make.lp(0, ncol=16) 
set.objfn(lprec, obj=c(rep(1,12), rep(0,4)))

add.constraint(lprec, xt=c(1,-1,1,-1), indices=c(1, 2, 13, 14), type="=", rhs=-1.0986)
add.constraint(lprec, xt=c(1,-1,1,-1), indices=c(3, 4, 13, 15), type="=", rhs=1.6094)
add.constraint(lprec, xt=c(1,-1,1,-1), indices=c(5, 6, 13, 16), type="=", rhs=-1.0986)
add.constraint(lprec, xt=c(1,-1,1,-1), indices=c(7, 8, 14, 15), type="=", rhs=1.94591)
add.constraint(lprec, xt=c(1,-1,1,-1), indices=c(9, 10, 14, 16), type="=", rhs=1.3863)
add.constraint(lprec, xt=c(1,-1,1,-1), indices=c(11, 12, 15, 16), type="=", rhs=-1.7917)

lp.control(lprec, simplextype="dual", pivoting="dantzig", verbose="detailed")
solve(lprec)
get.variables(lprec)
#  [1] 0.00000 0.00000 0.76209 0.00000 0.00000 0.15421 0.00000 0.00000 1.23209
# [10] 0.00000 0.00000 0.00000 0.84731 1.94591 0.00000 1.79170

See ?lp.control.options for more details. However, I could not reproduce the solution given by LINGO/SAS.

Karsten W.
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