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我正在尝试使用 MIP 使用纸浆包将岛屿连接在一起或连接到终端。所需的解决方案是找到系统中的最小距离。

下面的 MIP 代码结果显示，最小距离是将所有岛屿连接到终端，尽管岛屿与终端之间的距离明显很高。代码应该将一些岛屿连接在一起。

我究竟做错了什么 ？感谢您的支持。

import itertools
import pulp

islands = {0: [(0, 0)], 1: [(0, 7)], 2: [(3, 3)]}  ## islands id = 0,1,2 and their locations
terminal = {3: [(1,1)]}  ## the terminal id = 3 and its location 
distances = {(0,1): 1.0, (0,2): 2.0, (1,2): 3.0, (0,3): 33.0, (1,3): 34.0, (2,3): 23.0} ## distances of connecting islands together and islands to terminal

islandsPair = [(m, n) for m in islands for n in islands if m < n] ## islands pair
terminalPair = [(b, c) for b in islands for c in terminal] ## island-terminal pair

x = pulp.LpVariable.dicts("x", distances.keys(), lowBound=0, upBound=1)
p = pulp.LpVariable.dicts("p", islandsPair, lowBound=0, upBound=1)
l = pulp.LpVariable.dicts("l", terminalPair, lowBound=0, upBound=1)

mod = pulp.LpProblem("Islands", pulp.LpMinimize)

# Objective
mod += sum([distances[k] * x[k] for k in distances])


## constraint that there has to be at least 3 connections in the system:
for island in range(len(islands)):
        mod += sum(p[(m, n)] for m in islands for n in islands if m < n) + sum(l[(b, c)] for b in islands for c in terminal)  >= 3 



# Solve and output
mod.solve()

## printing the solutions: 

print pulp.LpStatus[mod.status]

print '0,3',l[(0, 3)].value()
print '1,3',l[(1, 3)].value()
print '2,3',l[(2, 3)].value()

print '0,1',p[(0, 1)].value()
print '0,2',p[(0, 2)].value()
print '1,2',p[(1, 2)].value()
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1 回答 1

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您的代码有几个问题：

  • 您用于for island in range(len(islands))添加使用 3 个边的约束，但从不使用 island，因此不确定为什么要执行该循环而不是仅添加约束。
  • 你在 x 和 p, l 之间没有关系，所以你只是最小化与 x 的关系，x 值都是 0，因为它没有任何约束。
  • 当你已经有 x 来建模是否使用边缘时，我根本看不到 p 和 l 的需要。

总而言之，这就是我要做的：

import itertools
import pulp

distances = {(0,1): 1.0, (0,2): 2.0, (1,2): 3.0, (0,3): 33.0, (1,3): 34.0, (2,3): 23.0} ## distances of connecting islands together and islands to terminal

x = pulp.LpVariable.dicts("x", distances.keys(), lowBound=0, upBound=1)

mod = pulp.LpProblem("Islands", pulp.LpMinimize)

y = pulp.LpVariable("y", lowBound=0, upBound = sum(distances.values()))
y = sum([distances[k] * x[k] for k in distances])
# Objective
mod += y

## constraint that there has to be at least 3 connections in the system:
mod += sum(x[edge] for edge in distances)  >= 3 

# Solve and output
mod.solve()

## printing the solutions: 

print (pulp.LpStatus[mod.status], y.value())

for edge in distances:
    print (edge, x[edge].value())
于 2018-11-26T15:19:16.477 回答