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 pulpfrom pulp import *import pandas as pdfood = ["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,kgdf = 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 = 1costs =  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 screenprint ("Status:", LpStatus[prob.status])# Each of the variables is printed with it's resolved optimum valuefor v in prob.variables():    print (v.name, "=", v.varValue)# The optimised objective function value is printed to the screenprint ("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 == 14prob += lpSum([KG[i] * deposit_vars[i] for i in deposit]) == 14, "KGRequirement"### Can´t not use more that 8KG from deposit 1prob += lpSum([KG[i] * deposit_vars[i] for i in deposit[0:1]]) <= 8, "KGRequirement1" 我收到此错误：Status: InfeasibleIngr_f1 = 0.83636364Ingr_f2 = 0.11818182Ingr_f3 = 0.045454545Ingr_f4 = 0.0Total 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"