我正在尝试使用Python PuLP中的整数编程公式解决Bin装箱问题。该问题的模型如下:
我使用PuLP库编写了以下Python代码
from pulp import *
#knapsack problem
def knapsolve(bins, binweight, items, weight):
prob = LpProblem('BinPacking', LpMinimize)
y = [LpVariable("y{0}".format(i+1), cat="Binary") for i in range(bins)]
xs = [LpVariable("x{0}{1}".format(i+1, j+1), cat="Binary")
for i in range(items) for j in range(bins)]
#minimize objective
nbins = sum(y)
prob += nbins
print(nbins)
#constraints
prob += nbins >= 1
for i in range(items):
con1 = sum(xs[(i * bins) + j] for j in range(bins))
prob += con1 == 1
print(con1)
for k in range(bins):
x = xs[k*bins : (k+1)*bins]
con1 = sum([x1*y for x1, y in zip(x, weight)])
prob += con1 <= binweight[k]
print(con1)
exec('prob')
status = prob.solve()
print(LpStatus[status])
print("Objective value:", value(prob.objective))
print ('\nThe values of the variables : \n')
for v in prob.variables():
print(v.name, "=", v.varValue)
return
def knapsack():
#bins
bins = int(input ('Enter the upper bound on the number of bins:'))
print ('\nEnter {0} bins\' capacities one by one'.format(bins))
binweight = []
for i in range(0, bins):
print('Enter {0} bin capacity'.format(i+1))
binweight.append(int(input()))
for i in range(0, bins):
print('The capacity at {0} is {1}'.format(i, binweight[i]))
#items
items = int(input('Enter the number of items:'))
weight = []
print ('\nEnter {0} items weights one by one'.format(items))
for i in range(0, items):
print('Enter {0} item weight'.format(i+1))
weight.append(int(input()))
for i in range(0, items):
print('The weight at {0} is {1}'.format(i, weight[i]))
knapsolve(bins, binweight, items, weight)
return
knapsack()
以下是代码的示例运行:
Enter the upper bound on the number of bins:3
Enter 3 bins' capacities one by one
Enter 1 bin capacity
6
Enter 2 bin capacity
4
Enter 3 bin capacity
5
The capacity at 0 is 6
The capacity at 1 is 4
The capacity at 2 is 5
Enter the number of items:3
Enter 3 items weights one by one
Enter 1 item weight
5
Enter 2 item weight
1
Enter 3 item weight
2
The weight at 0 is 5
The weight at 1 is 1
The weight at 2 is 2
y1 + y2 + y3
x11 + x12 + x13
x21 + x22 + x23
x31 + x32 + x33
5*x11 + x12 + 2*x13
5*x21 + x22 + 2*x23
5*x31 + x32 + 2*x33
Optimal
Objective value: 1.0
The values of the variables :
x11 = 0.0
x12 = 1.0
x13 = 0.0
x21 = 0.0
x22 = 0.0
x23 = 1.0
x31 = 0.0
x32 = 1.0
x33 = 0.0
y1 = 0.0
y2 = 0.0
y3 = 1.0
输出不符合预期。如何正确指定上述约束以获得正确的输出?
答案 0 :(得分:1)
您可以在构建问题后将其写入文件来检查生成的LP / MIP模型:
...
prob.writeLP("binpacking")
status = prob.solve()
...
现在,如果你看一下binpacking文件:
\* BinPacking *\
Minimize
OBJ: y1 + y2 + y3
Subject To
_C1: y1 + y2 + y3 >= 1
_C2: x11 + x12 + x13 = 1
_C3: x21 + x22 + x23 = 1
_C4: x31 + x32 + x33 = 1
_C5: 5 x11 + x12 + 2 x13 <= 6
_C6: 5 x21 + x22 + 2 x23 <= 4
_C7: 5 x31 + x32 + 2 x33 <= 5
Binaries
x11
x12
x13
x21
x22
x23
x31
x32
x33
y1
y2
y3
End
容器容量的限制是不对的。它们的工作就好像使用了所有的箱子而没有为变量分配1。这是因为您在使用项目权重时会覆盖y
值。
你需要改变这样的约束:
for k in range(bins):
x = xs[k*bins : (k+1)*bins]
con1 = sum([x1*w for x1, w in zip(x, weight)])
prob += con1 <= binweight[k] * y[k]
print(con1)
现在他们将按如下方式建模:
_C5: 5 x11 + x12 + 2 x13 - 6 y1 <= 0
_C6: 5 x21 + x22 + 2 x23 - 4 y2 <= 0
_C7: 5 x31 + x32 + 2 x33 - 5 y3 <= 0
此外,项目约束的索引不正确。而不是x11 + x12 + x13 = 1
它应该是x11 + x21 + x31 = 1
你可以这样纠正:
for i in range(items):
con1 = sum(xs[(i + j*bins)] for j in range(bins))
prob += con1 == 1
print(con1)
约束将是:
_C2: x11 + x21 + x31 = 1
_C3: x12 + x22 + x32 = 1
_C4: x13 + x23 + x33 = 1