我要用纸浆来解决一个整数线性优化问题。我解决了这个问题,得到的优化值等于42。但当我编写更通用的代码时,比如在循环中声明变量、在循环中定义约束以及使用lpSum函数定义优化,我没有得到解决方案。 我想我的问题是定义下一个约束。在
for a in itemset_dict.keys():
for b in itemset_dict[a][0]:
my_lp_program +=b >= a, "2Constraint"
我收到了下一个警告:
^{pr2}$谢谢。在
from pulp import *
# defining list of products
products = ['cola','peanuts', 'cheese', 'beer']
itemsets = ['x1','x2', 'x3']
#disctionary of the costs of each of the products is created
costs = {'cola' : 5, 'peanuts' : 3, 'cheese' : 1, 'beer' : 4 }
# dictionary of frequent itemsets
itemset_dict = { "x1" : (("cola", "peanuts"),10),
"x2" : (("peanuts","cheese"),20),
"x3" : (("peanuts","beer"),30)
}
products_var=LpVariable.dicts("Products", products, 0)
itemsets_var=LpVariable.dicts("Itemsets", itemsets, 0)
# defining itemsets variables
'''
for x in itemsets:
x = LpVariable('x', lowBound=0, upBound=1, cat='Binary')
'''
#x = LpVariable.dicts("x", itemsets, lowBound=0, upBound=1, cat='Binary')
#option2
i1=LpVariable.dict("itemsets", itemsets, lowBound=0, upBound=1, cat='Binary')
#print(i1);
#print(i1['x1'])
#print(type(i1['x1']))
'''
x1 = LpVariable('x1', lowBound=0, upBound=1, cat='Binary')
x2 = LpVariable('x2', lowBound=0, upBound=1, cat='Binary')
x3 = LpVariable('x3', lowBound=0, upBound=1, cat='Binary')
'''
# defining products variables
'''
for p in products:
p = LpVariable(p, lowBound=0, upBound=1, cat='Binary')
'''
#p = [LpVariable.dicts("p", i, lowBound=0, upBound=1, cat='Binary') for i in products]
'''
option1
p=[LpVariable(i, lowBound=0, upBound=1, cat='Binary') for i in products]
print(p[2])
print(type(p[2]))
'''
#option2
p1=LpVariable.dict("products", products, lowBound=0, upBound=1, cat='Binary')
#print(p1);
#print(p1['cola'])
#print(type(p1['cola']))
'''
cola = LpVariable('cola', lowBound=0, upBound=1, cat='Binary')
peanuts = LpVariable('peanuts', lowBound=0, upBound=1, cat='Binary')
cheese = LpVariable('cheese', lowBound=0, upBound=1, cat='Binary')
beer = LpVariable('beer', lowBound=0, upBound=1, cat='Binary')
'''
my_lp_program = LpProblem('My LP Problem', LpMaximize)
#my_lp_program += 10*x1+20*x2+30*x3-5*p1-3*p2-1*p3-4*p4 , "Maximization"
#my_lp_program += 10*x1+20*x2+30*x3 - lpSum([costs[i] * p1[i] for i in products]) , "Maximization"
my_lp_program += lpSum([itemset_dict[i][1] * i1[i] for i in itemsets]) - lpSum([costs[i] * p1[i] for i in products]) , "Maximization"
#my_lp_program += lpSum([itemset_dict[i][1] * itemsets_var[i] for i in itemsets]) - lpSum([costs[i] * products_var[i] for i in products]) , "Maximization"
#my_lp_program +=cola+peanuts+cheese+beer<=3, "1Constained"
my_lp_program +=lpSum([p1[i] for i in products]) <= 3, "1Constaint"
'''
my_lp_program +=cola>=x1, "2Constained"
my_lp_program +=peanuts>=x1, "3Constained"
my_lp_program +=peanuts>=x2, "4Constained"
my_lp_program +=cheese>=x2, "5Constained"
my_lp_program +=peanuts>=x3, "6Constained"
my_lp_program +=beer>=x3, "7Constained"
'''
for a in itemset_dict.keys():
for b in itemset_dict[a][0]:
my_lp_program +=b >= a, "2Constraint"
my_lp_program.writeLP("CheckLpProgram.lp")
my_lp_program.solve()
print("Status:", LpStatus[my_lp_program.status])
print("Total Optimum=", value(my_lp_program.objective))
for v in my_lp_program.variables():
print(v.name, "=", v.varValue)
我在这里重写了您的代码,提供了一个有效的解决方案。请看一下这些注释,它们给出了一些关于调试此类模型的有用信息的提示。在
输出:
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