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python - 如何修复不满足 biner 约束的 PuLP VRP 模型

转载 作者:行者123 更新时间:2023-12-01 08:05:45 25 4
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我使用 python PuLP 创建了一个 VRP(车辆路径问题模型),但它找不到满足所有约束的最佳解决方案。

使用此 xls 文件:https://drive.google.com/file/d/1s7rOQCULynGxQk8_IMlvHl286d4WfdPt/view?usp=sharing

import pulp, pandas, itertools
import numpy as np

xls =pandas.ExcelFile('data node VRP 2.xls')
weight = pandas.read_excel(xls,'Sheet1')
sheet2 = pandas.read_excel(xls, 'Sheet2')
matrixjarak = pandas.read_excel(xls, 'matrixjarak')
#weight=sheet1.as_matrix()
vehicle=sheet2.as_matrix() #vehicle
matrixjarak=matrixjarak.as_matrix()

model = pulp.LpProblem("VRP Problem", pulp.LpMinimize)

d = weight['demand']
c = matrixjarak
J = np.arange(len(c)-5) #create array 0..
p = np.arange(len(vehicle))
C = vehicle

x = pulp.LpVariable.dicts("nodes to nodes",
((r,i,j) for i in J for j in J for r in p),
lowBound=0,
cat='Biner')

model += (
pulp.lpSum([
c[i][j]*x[(r,i,j)]
for i in J for j in J for r in p if i != j])
)
#1 in out always 1
for i in range(1,len(J)-1):
model += pulp.lpSum([x[r,i,j] for j in range(1,len(J)-1) for r in p if i != j]) == 1
#model += pulp.lpSum([x[r,i,j] for j in range(1,len(J)-1) for r in p if i != j]) == 1

for j in range(1,len(J)-1):
model += pulp.lpSum([x[r,i,j] for i in range(1,len(J)-1) for r in p if j != i]) == 1
#model += pulp.lpSum([x[r,i,j] for j in range(1,len(J)-1) for r in p if i != j]) == 1

#2 capacity
for r in p:
model += pulp.lpSum([d[i]*x[r,i,j] for i in J for j in J if i != j]) <= 70 #l[v]


#3 go from depot
for r in p:
model += pulp.lpSum([x[r,0,j] for j in J for r in p]) == 1

#4 back to depot
for r in p:
model += pulp.lpSum([x[r,i,0] for j in J for r in p]) == 1

#5
for r in p:
for h in J:
model += pulp.lpSum([x[r,i,h] for i in J if i != h]) - pulp.lpSum([x[r,h,j] for j in J if h != j]) == 0


model.solve()
pulp.LpStatus[model.status]
for var in x:
var_value = x[var].varValue
print("nodes", var[1]," move to nodes ",var[2],"with vehicle ",var[0],"adalah", var_value)

print("cost optimal",pulp.value(model.objective))


我预计 x[r,i,j](决策变量)将得到 0 和 1 输出。但它会导致十进制输出:

nodes 0  go to nodes  0  with vehicle 0 are 0.875
nodes 0 go to nodes 6 with vehicle 0 are 0.125
nodes 1 go to nodes 2 with vehicle 1 are 1.0
nodes 2 go to nodes 1 with vehicle 1 are 1.0
nodes 3 go to nodes 6 with vehicle 1 are 0.23333333
nodes 3 go to nodes 6 with vehicle 2 are 0.76666667
nodes 4 go to nodes 5 with vehicle 2 are 1.0
nodes 5 go to nodes 4 with vehicle 2 are 1.0
nodes 6 go to nodes 0 with vehicle 0 are 0.125
nodes 6 go to nodes 3 with vehicle 1 are 0.23333333
nodes 6 go to nodes 3 with vehicle 2 are 0.76666667

cost optimal adalah 2.8

有什么线索可以解决这个问题吗?

最佳答案

类pulp.LpVariable定义为:

pulp.LpVariable(name, lowBound=None, upBound=None, cat='Continuous', e=None)

with:

cat – The category this variable is in, Integer, Binary or Continuous(default)

此外src/pulp/constants.py定义:

LpCategories = {LpContinuous: "Continuous", LpInteger: "Integer",
LpBinary: "Binary"}

含义:

  • 您要求cat='Biner'
  • PuLP 想要被要求 cat='Binary'
    • 否则将生成一个连续变量,从而导致您观察到的结果
    • 查看有关 that selection 的原始资料
      • (我想我会在该例程中引入更“积极”的检查)

关于python - 如何修复不满足 biner 约束的 PuLP VRP 模型,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55543535/

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