简单线性规划

library(Rglpk)
obj<-c(2,4,3)
mat<-matrix(c(3,2,1,4,1,3,2,1,2),nrow=3)
dir<-c("<=","<=","<=")
rhs<-c(60,40,80)
Rglpk_solve_LP(obj,mat,dir,rhs,max=T)
Rglpk_solve_LP(obj,mat,dir,rhs,max=T)

$optimum  #为目标函数的最大值
[1] 90

$solution  #表示决策变量的最优解
[1]  0  0 30

$status # 表示最优解已经找到，非0时失败
[1] 0

$solution_dual
[1] -2.5 -2.0  0.0

$auxiliary
$auxiliary$primal
[1] 60 30 60

$auxiliary$dual
[1] 1.5 0.0 0.0

obj<-c(3,1,3)
mat<-matrix(c(-1,0,1,2,4,-3,1,-3,2),nrow=3)
dir<-rep("<=",3)
rhs<-c(4,2,3)
types<-c("I","C","I")
Rglpk_solve_LP(obj,mat,dir,rhs,types=types,max=TRUE)

$optimum
[1] 26.75

$solution
[1] 5.00 2.75 3.00

$status
[1] 0

$solution_dual
[1] NA

$auxiliary
$auxiliary$primal
[1] 3.50 2.00 2.75

$auxiliary$dual
[1] NA

2.2 专题:lpSolve包和运输问题

install.packages("lpSolve")
library(lpSolve)
costs <- matrix(c(4,2,8,12,10,5,4,3,11,11,9,6),nrow=3) #运费矩阵
row.signs <- rep("=", 3) #各家造纸厂的产量恰好可以售完，故都取等号
row.rhs <- c(16,10,22) #销量约束值
col.signs <- rep("=", 4) #各家客户需求量恰好可以满足，故都取等号
col.rhs <- c(8,14,12,14) #需求约束值
res <- lp.transport(costs,"min",row.signs,row.rhs,col.signs,col.rhs)
res #输出最小运费
res$solution
Success: the objective function is 244

 res$solution
      [,1] [,2] [,3] [,4]
[1,]    4    0   12    0
[2,]    4    0    0    6
[3,]    0   14    0    8

costs <- matrix (c(3,11,6,12,2,7,3,5,1,4,9,5),nrow=3); ##运费矩阵
row.signs <- rep ("<=", 3) #总产量大于总销量，故各家销量小于等于产量
row.rhs <- c(8,5,9) #销量约束值
col.signs <- rep ("=", 4) #各家客户需求量都可以满足，故都取等号
col.rhs <- c(4,3,5,6) #需求约束值
res <- lp.transport(costs,"min",row.signs,row.rhs,col.signs,col.rhs)
res #输出最小运费
res$solution
Success: the objective function is 49 
 res$solution
      [,1] [,2] [,3] [,4]
[1,]    4    0    0    4
[2,]    0    3    0    0
[3,]    0    0    5    2

2.3 专题： lpSolve包和指派问题（成本计算）

library(lpSolve)
x=matrix(c(4,7,6,6,6,8,9,9,7,9,7,17,12,14,12,15,14,
           8,6,10,12,10,7,10,6),ncol=5)
x
lp.assign(x)
lp.assign(x)$solution
> x
[,1] [,2] [,3] [,4] [,5]
[1,]    4    8    7   15   12
[2,]    7    9   17   14   10
[3,]    6    9   12    8    7
[4,]    6    7   14    6   10
[5,]    6    9   12   10    6
> lp.assign(x)
Success: the objective function is 34 
> lp.assign(x)$solution
      [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    1    0    0
[2,]    0    1    0    0    0
[3,]    1    0    0    0    0
[4,]    0    0    0    1    0
[5,]    0    0    0    0    1
library(goalprog)
coefficients=matrix(c(5,1,4,6,10,-2,4,8),4)
targets=c(60,0,36,48)
achievements=data.frame(objective=1:4,priority=c(1,2,3,4),
                        p=c(1,0,1,0),n=c(0,1,0,1))
soln=llgp(coefficients, targets, achievements)
soln$converged
soln$out

coefficients=matrix(c(50,1,0,2,1,0,30,0,1,1,3,0),6)
coefficients
targets<-c(4600,50,80,120,150,20)
achievements<-data.frame(objective=1:6,priority=c(1,2,3,4,4,5),
p=c(1,0,0,4,1,1),n=c(0,1,1,0,0,0))
achievements
soln<-llgp(coefficients,targets,achievements)
soln
soln$out