Select 同一个物品在背包问题中多次出现[纸浆]

Question

我正在上课'discrete optimization course 其中，在课程中使用了一个名为 Minizinc 的工具来解决问题。

我想将 class 个示例翻译成 python，从这个开始：

我正在使用此示例代码重现结果：

v = {'hammer':6, 'wrench':10, 'screwdriver':8, 'towel':40}
w = {'hammer':13, 'wrench':21, 'screwdriver':17, 'towel':100}
q = {'hammer':1000, 'wrench':400, 'screwdriver':500, 'towel':150}
limit = 1000
items = list(sorted(v.keys()))

# Create model
m = LpProblem("Knapsack", LpMaximize)

# Variables
x = LpVariable.dicts('x', items, lowBound=0, upBound=1, cat=LpInteger)

# Objective
m += sum(v[i]*x[i] for i in items)

# Constraint
m += sum(w[i]*x[i] for i in items) <= limit


# Optimize
m.solve()

# Print the status of the solved LP
print("Status = %s" % LpStatus[m.status])

# Print the value of the variables at the optimum
for i in items:
    print("%s = %f" % (x[i].name, x[i].varValue))

# Print the value of the objective
print("Objective = %f" % value(m.objective))

但这是一个错误的答案，因为只采用了一种。如何将每个项目 (dict q) 的可用数量添加到约束中？

Answer 1

您需要对代码进行两处非常小的更改。首先，您需要删除在 x 变量上设置的上限。目前你有二进制变量 x[i]，它只能是一或零。

其次，您需要添加有效地为每个项目设置自定义上限的约束。下面的工作代码和结果解决方案 - 如您所见，选择了多个扳手（最高 v/w 比率），用一个锤子填满剩余的少量 space。

from pulp import *
v = {'hammer':6, 'wrench':10, 'screwdriver':8, 'towel':40}
w = {'hammer':13, 'wrench':21, 'screwdriver':17, 'towel':100}
q = {'hammer':1000, 'wrench':400, 'screwdriver':500, 'towel':150}
limit = 1000
items = list(sorted(v.keys()))

# Create model
m = LpProblem("Knapsack", LpMaximize)

# Variables
x = LpVariable.dicts('x', items, lowBound=0, cat=LpInteger)

# Objective
m += sum(v[i]*x[i] for i in items)

# Constraint
m += sum(w[i]*x[i] for i in items) <= limit

# Quantity of each constraint:
for i in items:
    m += x[i] <= q[i]


# Optimize
m.solve()

# Print the status of the solved LP
print("Status = %s" % LpStatus[m.status])

# Print the value of the variables at the optimum
for i in items:
    print("%s = %f" % (x[i].name, x[i].varValue))

# Print the value of the objective
print("Objective = %f" % value(m.objective))
print("Total weight = %f" % sum([x[i].varValue*w[i] for i in items]))

哪个returns:

状态 = 最佳

x_hammer = 1.000000
x_screwdriver = 0.000000
x_towel = 0.000000
x_wrench = 47.000000
Objective = 476.000000
Total weight = 1000.000000

Select 同一个物品在背包问题中多次出现[纸浆]

Select the same item several times in the knapsack problem [pulp]

python

optimization

linear-programming

pulp