Question

我有以下简单的线性编程模型，我用R中的linprog包解决了这个问题：

install.packages("linprog")
library(linprog)

function_opt <- c(8, 13, 9, 8, 9, 11, 12, 10, 7, 8, 10, 9)
names(function_opt) <- c("F1A1","F1A2","F1A3","F1A4","F2A1","F2A2","F2A3","F2A4","F3A1","F3A2","F3A3","F3A4")
##Order: 3 factory capacities, 4 customer demands
cons_indep_term <- c(60, 70, 80, 75, 45, 40, 50)
names(cons_indep_term) <- c("F1","F2","F3","A1","A2","A3","A4")

r1 <- c(1,1,1,1,0,0,0,0,0,0,0,0)
r2 <- c(0,0,0,0,1,1,1,1,0,0,0,0)
r3 <- c(0,0,0,0,0,0,0,0,1,1,1,1)
r4 <- c(1,0,0,0,1,0,0,0,1,0,0,0)
r5 <- c(0,1,0,0,0,1,0,0,0,1,0,0)
r6 <- c(0,0,1,0,0,0,1,0,0,0,1,0)
r7 <- c(0,0,0,1,0,0,0,1,0,0,0,1)
cons_coef <- rbind(r1,r2,r3,r4,r5,r6,r7)

res <- solveLP(function_opt, cons_indep_term, cons_coef, maximum=FALSE, const.dir = c("<=","<=","<=",">=",">=",">=",">="))

print (res)

第六个约束条件要求FxA3变量之和必须至少为40.但是，解决方案如下：

Results of Linear Programming / Linear Optimization

Objective function (Minimum): 1355 

Iterations in phase 1: 6
Iterations in phase 2: 3
Solution
     opt
F1A1  10
F1A2   0
F1A3   0
F1A4  50
F2A1  30
F2A2   0
F2A3   0
F2A4   0
F3A1  35
F3A2  45
F3A3   0
F3A4   0

Basic Variables
     opt
F1A1  10
F1A4  50
F2A1  30
F3A1  35
F3A2  45
S F2  40
S A3  40

Constraints
   actual dir bvec free dual dual.reg
F1     60  <=   60    0    1       10
F2     30  <=   70   40    0       40
F3     80  <=   80    0    2       35
A1     75  >=   75    0    9       40
A2     45  >=   45    0   10       35
A3     80  >=   40   40    0       40
A4     50  >=   50    0    9       10

All Variables (including slack variables)
     opt cvec min.c max.c marg marg.reg
F1A1  10    8    -9     9   NA       NA
F1A2   0   13    99    77    4       10
F1A3   0    9    99    77   10       10
F1A4  50    8   -17     9   NA       NA
F2A1  30    9   -10    10   NA       NA
F2A2   0   11    99    77    1       30
F2A3   0   12    99    77   12       40
F2A4   0   10    99    77    1       30
F3A1  35    7    -8     9   NA       NA
F3A2  45    8   -18     9   NA       NA
F3A3   0   10    99    77   12       35
F3A4   0    9    99    77    2       35
S F1   0    0    -1   Inf    1       10
S F2  40    0    NA     1    0       NA
S F3   0    0    -2   Inf    2       35
S A1   0    0    -9   Inf    9       40
S A2   0    0   -10   Inf   10       35
S A3  40    0    NA    10    0       NA
S A4   0    0    -9   Inf    9       10

所有三个FxA3变量都设置为0，这意味着违反了第六个约束。问题是什么？我已经三次检查了一切，但仍然不知道。

Answer 1

这是......非常奇怪，我无法在您的代码中发现任何问题。由于这是一个如此简单的LP，你可能会考虑向软件包维护者提交一个bug。

话虽如此，您应该能够通过使用lpSolve包来解锁，该包具有几乎相同的界面（实际上，您所询问的linprog包使用了该包）：

library(lpSolve)
mod2 = lp(direction = "min",
          objective.in = function_opt,
          const.mat = cons_coef,
          const.dir = c("<=","<=","<=",">=",">=",">=",">="),
          const.rhs = cons_indep_term)
setNames(mod2$solution, names(function_opt))
# F1A1 F1A2 F1A3 F1A4 F2A1 F2A2 F2A3 F2A4 F3A1 F3A2 F3A3 F3A4 
#    0    0   40   20   40    0    0   30   35   45    0    0 
mod2$objval
# [1] 1785

在最佳解决方案中，我们有F1A3 = 40，F2A3 = 0和F3A3 = 0，因此符合第六个约束条件。

对于简单的线性编程模型，返回不可行的结果

1 个答案: