[英]Pulp solver :How to set a minimal variable production using a loop
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"
我应该如何为每个产品设置 10 的最小生产值?
如果我有 200 个产品,我应该如何以更优雅的方式写这个?
Lp 正确吗?
最小生产约束:
prob += lpSum([x[i] for i in produits]) >= 10 ,"MinProdObjs"
简单的意思(实际上,汽车是“汽车数量”,自行车也是“自行车数量”......也许变量名称不太好......)
prob += car + bicycle >= 10
或者
prob += x1 + x2 >= 10
但它没有按预期工作......
如果x[p]
是为p in P
的产品p in P
生产的单位数,那么您可以添加以下形式的约束:
x[p] >= 10 forall p in P
翻译成代码:
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。
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