LpSolve R条件约束

Question

我正在尝试回答以下ILP，其目标是最大限度地增加手术患者的类型，而最多只能手术两种不同的类型。

max 310x1 + 400x2 + 500x3 + 500x4 + 650x5 + 800x6 + 850x7
subject to
1.8x1 + 2.8x2 + 3.0x3 + 3.6x4 + 3.8x5 + 4.6x6 + 5.0x7 <= 25
250x1 + 300x2 + 500x3 + 400x4 + 550x5 + 800x6 + 750x7 >= 4000 
xj <= dj
d1 + d2 + d3 + d4 + d5 + d6 + d7 <= 2
xj >= 0 and integer

为了在R包lpSolve中编写这段代码，我有以下代码：

# Set coefficients of the objective function
f.obj <- c(310, 400, 500, 500, 650, 800, 850, 0, 0, 0, 0, 0, 0, 0)

# Set matrix corresponding to coefficients of constraints by rows
f.con <- matrix(c(1.8, 2.8, 3, 3.6, 3.8, 4.6, 5, 0, 0, 0, 0, 0, 0, 0,
                  250, 300, 500, 400, 550, 800, 750, 0, 0, 0, 0, 0, 0, 0,
                  0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1,
                  1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0,
                  0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0,
                  0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0,
                  0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0,
                  0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0,
                  0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0,
                  0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1), nrow = 10, byrow = TRUE)

# Set unequality/equality signs
f.dir <- c("<=",
           "<=",
           "<=",
           "<=",
           "<=",
           "<=",
           "<=",
           "<=",
           "<=",
           "<=")

# Set right hand side coefficients
f.rhs <- c(25, 4000, 2,0, 0, 0, 0, 0, 0,0)

# Final value (z)
lp("max", f.obj, f.con, f.dir, f.rhs, int.vec = 1:7, binary.vec = 8:14)

# Variables final values
lp("max", f.obj, f.con, f.dir, f.rhs, int.vec = 1:7, binary.vec = 8:14)$solution

但是，现在x不会超过1，因为d是二进制的。

有人知道我如何正确地编写这些约束吗？

Answer 1

你有7种不同的患者类型，从x1到x7，x是整数。您最多可以选择2个x为非零。为此，可以为每个x向b7添加二进制变量b1，并为每个x添加两个约束。

x >= -U + U*b
x <= U*b

其中U是最大x值的某个上限。

library(lpSolve)

# Set coefficients of the objective function
f.obj <- c(310, 400, 500, 500, 650, 800, 850, 0, 0, 0, 0, 0, 0, 0, 0)

U=999

# Set matrix corresponding to coefficients of constraints by rows
f.con <- matrix(c(1.8, 2.8, 3, 3.6, 3.8, 4.6, 5, 0, 0, 0, 0, 0, 0, 0, 0,
                  250, 300, 500, 400, 550, 800, 750, 0, 0, 0, 0, 0, 0, 0, 0,
                  0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0,
                  1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, 0, 0, 0, U,
                  1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, 0, 0, 0, 0,
                  0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, 0, 0, U,
                  0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, 0, 0, 0,
                  0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, 0, U,
                  0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, 0, 0,
                  0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, U,
                  0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0, 0,
                  0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, U,
                  0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0, 0,
                  0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, U,
                  0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0, 0,
                  0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, U,
                  0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -U, 0), nrow = 17, byrow = TRUE)

# Set unequality/equality signs
f.dir <- c("<=","<=","<=",rep(c(">=","<="),7))

# Set right hand side coefficients
f.rhs <- c(25, 4000, 2, rep(0,14))

# Final value (z)
res=lp("max", f.obj, f.con, f.dir, f.rhs, int.vec = 1:7, binary.vec = 8:14)

结果

> res$objval
[1] 4260

> res$solution
 [1] 11.000000  0.000000  0.000000  0.000000  0.000000  0.000000  1.000000  1.000000  0.000000  0.000000
[11]  0.000000  0.000000  0.000000  1.000000  0.998999

因此选择了第1和第7个患者类型，其中x1患者11例，x7患者1例。我们可以检查约束条件

> sum(c(1.8, 2.8, 3, 3.6, 3.8, 4.6, 5)*c(11,0,0,0,0,0,1))
[1] 24.8
> sum(c(250, 300, 500, 400, 550, 800, 750)*c(11,0,0,0,0,0,1))
[1] 3500