在混合优化问题中添加数量作为约束

Question

我正在复制这个混合问题的例子：https : //www.coin-or.org/PuLP/CaseStudies/a_blending_problem.html

使用以下数据：

import pulp

from pulp import *

import pandas as pd

food = ["f1","f2","f3","f4"]

KG = [10,20,50,80]

Protein =       [18,12,16,18]

Grass = [13,14,13,16]

price_per_kg =  [15,11,10,12]

##            protein,carbohydrates,kg

df = pd.DataFrame({"tkid":food,"KG":KG,"Protein":Protein,"Grass":Grass,"value":price_per_kg})

这是代码：

deposit =  df["tkid"].values.tolist()

factor_volumen = 1

costs =  dict((k,v) for k,v in zip(df["tkid"],df["value"]))

Protein =  dict((k,v) for k,v in zip(df["tkid"],df["Protein"]))

Grass =  dict((k,v) for k,v in zip(df["tkid"],df["Grass"]))

KG =  dict((k,v) for k,v in zip(df["tkid"],df["KG"]))

prob = LpProblem("The Whiskas Problem", LpMinimize)

deposit_vars = LpVariable.dicts("Ingr",deposit,0)

prob += lpSum([costs[i]*deposit_vars[i] for i in deposit]), "Total Cost of Ingredients per can"

prob += lpSum([deposit_vars[i] for i in deposit]) == 1.0, "PercentagesSum"

prob += lpSum([Protein[i] * deposit_vars[i] for i in deposit]) >= 17.2, "ProteinRequirement"

prob += lpSum([Grass[i] * deposit_vars[i] for i in deposit]) >= 11.8, "FatRequirement"

prob.writeLP("WhiskasModel.lp")

prob.solve()

# The status of the solution is printed to the screen

print ("Status:", LpStatus[prob.status])

# Each of the variables is printed with it's resolved optimum value

for v in prob.variables():

    print (v.name, "=", v.varValue)

# The optimised objective function value is printed to the screen

print ("Total Cost of Ingredients per can = ", value(prob.objective))

这部分正在工作，但我需要再添加一个限制，即我想生产多少公斤。

我试过做这两个限制：

## total KG produced == 14

prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) == 14, "KGRequirement"

### Can´t not use more that 8KG from deposit 1

prob += lpSum([KG[i] * deposit_vars[i] for i in deposit[0:1]]) <= 8, "KGRequirement1" 

我收到此错误：

Status: Infeasible

Ingr_f1 = 0.83636364

Ingr_f2 = 0.11818182

Ingr_f3 = 0.045454545

Ingr_f4 = 0.0

Total Cost of Ingredients per can =  14.30000007

但使用存款4来满足这一点应该是可能的，所以我认为约束是不正确的。

Answer 1

我意识到百分比限制是错误的，我只需要添加我想要生产多少的限制：prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) == 14, "KGRequirement"并且成分的加权平均值也满足条件。prob += lpSum([Protein[i] *KG[i] * deposit_vars[i] for i in deposit]) >= 17.2*14, "ProteinRequirement"这是严格的限制：prob += lpSum([Protein[i] *KG[i] * deposit_vars[i] for i in deposit]) >= 17.2*14, "ProteinRequirement"prob += lpSum([Grass[i] *KG[i] * deposit_vars[i] for i in deposit]) >= 11.8*14, "FatRequirement"prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) == 14, "KGRequirement"prob += lpSum([KG[i] * deposit_vars[i] for i in deposit[0:1]]) <= 8, "KGRequirement1"