纸浆求解器：如何使用循环设置最小变量生产

Question

我有这个玩具厂 LP：

# Import the PuLP lib
from pulp import *

# Products list
products = ["car", "bicycle"]

#Profit per product in $
profit = {"car": 8, "bicycle": 12}

# Used resources per product in kgs 
plasticAmount = {"car": 2, "bicycle": 4}
woodAmount    = {"car": 1, "bicycle": 1}
steelAmount   = {"car": 3, "bicycle": 2}


# Setting Problem variables dictionary
x = LpVariable.dicts("products ", products , 0)

# The Objective function : Maximising profit in $
prob += lpSum([profit[i] * x[i] for i in products ]), "Maximise"

# Total Stock amount Constraints in kgs
prob += lpSum([plasticAmount[i] * x[i] for i in  products]) <= 142 ,"MaxPlastic"
prob += lpSum([woodAmount [i]   * x[i] for i in  products]) <= 117 ,"MaxWood"
prob += lpSum([steelAmount[i]   * x[i] for i in  products]) <= 124 ,"MaxSteel"

# This constraints is not working : Minimal production amount should be at least 10 on each products ( need at least 10 Cars and 10 bicycles)
prob += lpSum([x[i] for i in produits]) >= 10 ,"MinProdObjs"

1. 我应该如何为每个产品设置 10 的最小产值？

2. 如果我有 200 种产品，我应该如何以更优雅的方式写这个？

3. Lp 是否正确？

最小生产约束：

prob += lpSum([x[i] for i in produits]) >= 10 ,"MinProdObjs"

简单的意思是（实际上，car 是“数量的汽车”，而自行车也是“数量的自行车”......也许变量名称不太好......）

prob += car + bicycle >= 10

或

prob += x1 + x2 >= 10

但它并没有按预期工作......

Answer 1

如果x[p] 是为产品p in P 生产的单位数，那么您只需添加以下形式的约束：

x[p] >= 10  forall p in P

代码翻译：

for p in products:
   prob += x[p] >= 10, f"min production units for product {p}"

有你的约束

prob += lpSum([x[i] for i in produits]) >= 10 ,"MinProdObjs"

您的意思是，您希望所有项目的总产值至少为 10。

另请注意，您的变量目前是小数，您可能希望使用整数。