我正在使用'lpSolveAPI'来解决R中的多个二进制线性规划问题。我如何得到它也返回所有变量组合以实现最大化问题。
我搜索了文档,但找不到这个命令。试图从包'lpSolve'切换,因为它不一致地崩溃R.
这是一个示例问题:
library(lpSolveAPI)
#matrix of constraints
A <- matrix(rbind(
c(1,1,0,0,0,0,0,0),
c(1,0,0,0,1,0,0,0),
c(0,0,1,0,0,1,0,0),
c(0,0,1,0,0,0,1,0),
c(0,0,1,0,0,0,0,1),
c(0,0,0,1,1,0,0,0),
c(0,0,0,0,0,0,1,1)), 7, 8)
#create an LP model with 7 constraints and 8 decision variables
num_con <- nrow(A)
num_points <- ncol(A)
lpmodel <- make.lp(num_con,num_points)
# all right hand
set.constr.type(lpmodel,rep("<=",num_con))
set.rhs(lpmodel, rep(1,num_con))
set.type(lpmodel,columns = c(1:num_points), "binary")
# maximize
lp.control(lpmodel,sense="max")
# add constraints
for (i in 1:num_points){
set.column(lpmodel,i,rep(1,length(which(A[,i]==1))),which(A[,i]==1))
}
set.objfn(lpmodel, rep(1,num_points))
solve(lpmodel)
get.variables(lpmodel)
返回:
"[1] 0 1 0 0 1 1 0 1"
我知道这个问题有6个可能的解决方案:
[1] 0 1 0 0 1 1 0 1
[2] 1 0 0 1 0 1 0 1
[3] 1 0 0 1 0 1 1 0
[4] 0 1 0 1 0 1 0 1
[5] 0 1 0 1 0 1 1 0
[6] 0 1 0 0 1 1 1 0
我如何得到它也让我回归所有这些?
发现这是骗子:Get multiple solutions for 0/1-Knapsack MILP with lpSolveAPI
以下是我用来解决此问题的代码:
# find first solution
status<-solve(lpmodel)
sols<-list() # create list for more solutions
obj0<-get.objective(lpmodel) # Find values of best solution (in this case four)
counter <- 0 #construct a counter so you wont get more than 100 solutions
# find more solutions
while(counter < 100) {
sol <- get.variables(lpmodel)
sols <- rbind(sols,sol)
add.constraint(lpmodel,2*sol-1,"<=", sum(sol)-1)
rc<-solve(lpmodel)
if (status!=0) break;
if (get.objective(lpmodel)<obj0) break;
counter <- counter + 1
}
sols
这找到所有六个解决方案:
> sols
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
sol 1 0 0 1 0 1 0 1
sol 1 0 0 1 0 1 1 0
sol 0 1 0 1 0 1 0 1
sol 0 1 0 1 0 1 1 0
sol 0 1 0 0 1 1 1 0
sol 0 1 0 0 1 1 0 1
对不起,每个人都是骗子。如果有人知道'lpSolveAPI'包中内置的另一种方法,我仍然很想知道。