紙漿求解器：如何使用循環設置最小變量產量

Question

我有這個玩具廠 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. 我應該如何為每個產品設置 10 的最小生產值？

2. 如果我有 200 個產品，我應該如何以更優雅的方式寫這個？

3. Lp 正確嗎？

最小生產約束：

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

簡單的意思（實際上，汽車是“汽車數量”，自行車也是“自行車數量”......也許變量名稱不太好......）

prob += car + bicycle >= 10

或者

prob += x1 + x2 >= 10

但它沒有按預期工作......

Answer 1

如果x[p]是為p in P的產品p in P生產的單位數，那么您可以添加以下形式的約束：

x[p] >= 10  forall p in P

翻譯成代碼：

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

有你的約束

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

您的意思是您希望所有項目的總產值至少為 10。

另請注意，您的變量目前是小數，您可能希望使用整數。