R

时间:2016-12-09 13:39:44

标签: r linear-programming simplex-algorithm

我有一个线性编程问题,我试图在R中解决。我使用过lpSolve包。 lpSolve默认使用原始单纯形算法来获得解。如果我想将算法更改为双单纯形怎么办?结果在两种算法之间差异很大。是否有任何其他软件包可以使用双重单纯形算法帮助解决下面的问题。

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

R中使用lpSolve的Primal Simplex如下:

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

使用Lingo软件和SAS的Dual Simplex如下:

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

两种算法的目标函数相同2.14839

1 个答案:

答案 0 :(得分:1)

使用lpSolveAPI,您可以微调您的求解器:

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

有关详细信息,请参阅?lp.control.options。但是,我无法重现LINGO / SAS给出的解决方案。