LPR条件约束的求解

我试图回答以下ILP问题，其目标是最大限度地提高手术患者的类型，而最多只能手术两种不同类型的患者

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

最大值310x1+400x2+500x3+500x4+650x5+800x6+850x7
从属于
1.8x1+2.8x2+3.0x3+3.6x4+3.8x5+4.6x6+5.0x7=4000
xj您有7种不同的患者类型，x1到x7，x是整数。您最多可以选择2个x为非零。可以通过为每个x添加二进制变量b1到b7，并为每个x添加两个约束来实现这一点

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

因此，选择了第一和第七种患者类型，其中11种为x1，1种为x7。
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

x代表什么？x是不同类型的病人谢谢你的帮助！不幸的是，这不起作用，因为x每个组可以有更多的患者，只是总共不超过2个不同的患者组。@Sam，那么我不明白，您需要对x进行一些额外的限制，每个类型最多有多少患者？否则，您只能选择一种患者类型，并使x等于无穷大以获得最大利润。否，因为前两个约束条件阻止了这一点。所有患者最多有25个手术小时和4000个护理分钟。@Sam看到了新的答案。非常感谢！你真是帮了大忙！
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)

> 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

> 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