使用lpSolveAPI获得针对0/1-背包MILP的多种解决方案 [英] Get multiple solutions for 0/1-Knapsack MILP with lpSolveAPI

问题描述

可复制的示例:

我描述了一个简单的 0/1-背包问题，其中包含 R ，它应该返回2个解决方案:

I described a simple 0/1-Knapsack problem with lpSolveAPI in R, which should return 2 solutions:

library(lpSolveAPI)

lp_model= make.lp(0, 3)
set.objfn(lp_model, c(100, 100, 200))
add.constraint(lp_model, c(100,100,200), "<=", 350)
lp.control(lp_model, sense= "max")
set.type(lp_model, 1:3, "binary")

lp_model
solve(lp_model)
get.variables(lp_model)
get.objective(lp_model)
get.constr.value((lp_model))
get.total.iter(lp_model)
get.solutioncount(lp_model)

问题:

但是get.solutioncount(lp_model)显示只有找到的1解决方案:

But get.solutioncount(lp_model) shows that there's just 1 solution found:

> lp_model
Model name: 
           C1   C2   C3         
Maximize  100  100  200         
R1        100  100  200  <=  350
Kind      Std  Std  Std         
Type      Int  Int  Int         
Upper       1    1    1         
Lower       0    0    0         
> solve(lp_model)
[1] 0
> get.variables(lp_model)
[1] 1 0 1
> get.objective(lp_model)
[1] 300
> get.constr.value((lp_model))
[1] 350
> get.total.iter(lp_model)
[1] 6
> get.solutioncount(lp_model)
[1] 1

我希望有两种解决方案:1 0 1和0 1 1.

I would expect that there are 2 solutions: 1 0 1 and 0 1 1.

我尝试传递 lpSolve num.bin.solns参数. >与solve(lp_model, num.bin.solns=2)，但解决方案的数量仍为1.

I tried to pass the num.bin.solns argument of lpSolve with solve(lp_model, num.bin.solns=2), but the number of solutions remained 1.

问题:

如何获得两个正确的解决方案? 我更喜欢使用 lpSolveAPI ，因为该API非常好. 如果可能的话，我想避免直接使用 lpSolve .

How can I get the two correct solutions? I prefer using lpSolveAPI as the API is really nice. If possible I'd like to avoid to use lpSolve directly.

推荐答案

看起来像坏了.这是您特定模型的DIY方法:

Looks like that is broken. Here is a DIY approach for your specific model:

# first problem
rc<-solve(lp_model)
sols<-list()
obj0<-get.objective(lp_model)
# find more solutions
while(TRUE) {
   sol <- round(get.variables(lp_model))
   sols <- c(sols,list(sol))
   add.constraint(lp_model,2*sol-1,"<=", sum(sol)-1)
   rc<-solve(lp_model)
   if (rc!=0) break;
   if (get.objective(lp_model)<obj0-1e-6) break;
}
sols

这个想法是通过添加一个约束来切断当前的整数解.然后解决.当不再最佳时或目标开始恶化时停止. 此处是一些数学背景.

The idea is to cut off the current integer solution by adding a constraint. Then resolve. Stop when no longer optimal or when the objective starts to deteriorate. Here is some math background.

您现在应该看到:

> sols
[[1]]
[1] 1 0 1

[[2]]
[1] 0 1 1

更新

在下面的评论中，有人问为什么切口系数的形式为 2 * sol-1 .再次查看推导.这是一个反例:

Below in the comments it was asked why coefficients of the cut have the form 2*sol-1. Again have a look at the derivation. Here is a counter example:

           C1   C2        
Maximize    0   10        
R1          1    1  <=  10
Kind      Std  Std        
Type      Int  Int        
Upper       1    1        
Lower       0    0       

使用我的"切割将产生:

With "my" cuts this will yield:

> sols
[[1]]
[1] 0 1

[[2]]
[1] 1 1

在使用建议的错误"切割时，只会给出:

while using the suggested "wrong" cuts will give just:

> sols
[[1]]
[1] 0 1

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