使用 PuLP 进行整数线性优化

Question

我必须用纸浆解决整数线性优化问题。我解决了这个问题并得到了等于 42 的优化值。但是当我编写更通用的代码时，比如在循环内声明变量、在循环内定义约束以及使用 lpSum 函数定义优化，我没有解决方案。 我认为我的问题在于定义下一个约束。

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：

C:\Program Files\Python36\lib\site-packages\pulp\pulp.py:1353: UserWarning: Overwriting previously set objective.
  warnings.warn("Overwriting previously set objective.")
Status: Optimal
Total Optimum= None
__dummy = None
products_beer = 0.0
products_cheese = 0.0
products_cola = 0.0
products_peanuts = 0.0
The thread 'MainThread' (0xb30) has exited with code 0 (0x0).
The program '[10140] python.exe' has exited with code 0 (0x0).

谢谢。

      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)

Answer 1

我在这里重写了您的代码，提供了一个可行的解决方案。请查看 cmets，它提供了一些关于调试此类模型有用的提示。

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
# ~~> This is hard to maintain - I would select a different data structure
# it gets really complicated below as you will see!
itemset_dict = { "x1" : (("cola", "peanuts"),10),
           "x2" : (("peanuts","cheese"),20),
           "x3" : (("peanuts","beer"),30)
           }

# Good practice to first define your problem
my_lp_program = LpProblem('My LP Problem', LpMaximize)  

# ~~>You do not need bounds for binary variables, they are automatically 0/1
products_var=LpVariable.dicts("Products", products, cat='Binary')
itemsets_var=LpVariable.dicts("Itemsets", itemsets, cat='Binary')

# ~~> Not necessary - commended out
# defining itemsets variables
# for x in itemsets:
#     x = LpVariable(x, lowBound=0, upBound=1, cat='Binary')

'''
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')
'''

# ~~> Not necessary - commended out
# defining products variables
# for p in products:
#     p = LpVariable(p, lowBound=0, upBound=1, cat='Binary')

'''
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')
'''

# ~~> Not necessary - commended out, see below
#my_lp_program += 10*x1+20*x2+30*x3-5*p1-3*p2-1*p3-4*p4 , "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"

# ~~> Use an affine expression to define your objective.
# ~~> Even better, define two objects as LpAffineExpression and add them, 
# ~~> it keeps the code cleaner
my_lp_program += LpAffineExpression([(
    itemsets_var[x], itemset_dict[x][1])  for x in itemsets_var]) + \
    LpAffineExpression([(
    products_var[x], -costs[x])  for x in products_var])

# ~~> Not necessary - commended out
#my_lp_program +=cola+peanuts+cheese+beer<=3, "1Constained"
# my_lp_program +=lpSum([products_var[i] for i in products]) <= 3, "1Constaint"
# ~~> This is the right way to enter this constraint.
# ~~> I do not like the naming though..
my_lp_program += lpSum(products_var) <= 3, '1Constraint'

'''
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"
'''
# ~~> Here are your constraints
counter = 1
for a in itemset_dict.keys():
    item = itemsets_var[a]
    for b in itemset_dict[a][0]:
        product = products_var[b]
        counter +=1
        my_lp_program += product  >= item, "{}Constraint".format(counter)

# ~~> Great that you export the lp! If you look at the file you can
# ~~> spot a lot of issues with the model during debugging
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)

输出：

('Status:', 'Optimal')
('Total Optimum=', 42.0)
('Itemsets_x1', '=', 0.0)
('Itemsets_x2', '=', 1.0)
('Itemsets_x3', '=', 1.0)
('Products_beer', '=', 1.0)
('Products_cheese', '=', 1.0)
('Products_cola', '=', 0.0)
('Products_peanuts', '=', 1.0)