混合问题上的变量总和不匹配

Question

我正在尝试重现这种混合 example 但使用较少的变量，这部分工作得很好：

import pulp
from pulp import *

# Creates a list of the Ingredients
Ingredients = ['CHICKEN', 'BEEF', 'MUTTON', 'RICE']

# A dictionary of the costs of each of the Ingredients is created
costs = {'CHICKEN': 15, 
         'BEEF': 12, 
         'MUTTON': 17, 
         'RICE': 12
        }

# A dictionary of the protein percent in each of the Ingredients is created
proteinPercent = {'CHICKEN': 17, 
                  'BEEF': 2, 
                  'MUTTON': 16, 
                  'RICE': 8
                 }

# A dictionary of the fat percent in each of the Ingredients is created
fatPercent = {'CHICKEN': 10, 
              'BEEF': 14, 
              'MUTTON': 13, 
              'RICE': 16, 
              }

# Create the 'prob' variable to contain the problem data
prob = LpProblem("The Whiskas Problem", LpMinimize)

# A dictionary called 'ingredient_vars' is created to contain the referenced Variables
ingredient_vars = LpVariable.dicts("Ingr",Ingredients,0)

# The objective function is added to 'prob' first
prob += lpSum([costs[i]*ingredient_vars[i] for i in Ingredients]), "Total Cost of Ingredients per can"

# The  constraints are added to 'prob'
prob += lpSum([proteinPercent[i] * ingredient_vars[i] for i in Ingredients]) >= 15.5, "ProteinRequirement"
prob += lpSum([fatPercent[i] * ingredient_vars[i] for i in Ingredients]) >= 12.3, "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))

它计算每种成分所需的最佳量：

Status: Optimal
Ingr_BEEF = 0.0
Ingr_CHICKEN = 0.77916667
Ingr_MUTTON = 0.0
Ingr_RICE = 0.28177083
Total Cost of Ingredients per can =  15.068750009999999

但是当我将约束添加到代码中时，这并不等于 100%：

prob += lpSum([ingredient_vars[i] for i in Ingredients]) == 100, "PercentagesSum"

我得到这个结果：

Status: Optimal
Ingr_BEEF = 100.0
Ingr_CHICKEN = 0.0
Ingr_MUTTON = 0.0
Ingr_RICE = 0.0
Total Cost of Ingredients per can =  1200.0

这是错误的，因为它不满足其他约束。

编辑

看来我解释错了，我在想： 如果我想生产 3 个单位，输入的总和应该是 3。

我认为是这样的：

# The constraints are added to 'prob'
prob += lpSum([ingredient_vars[i] for i in Ingredients]) == 3, "PercentagesSum"
prob += lpSum(ingredient_vars["CHICKEN"]) <= 2, "CHICKEN"
prob += lpSum(ingredient_vars["BEEF"]) <= 1, "BEEF"
prob += lpSum(ingredient_vars["MUTTON"]) <= 1, "MUTTON"
prob += lpSum(ingredient_vars["RICE"]) <= 1, "RICE"

其中 2,1,1,1 是每种原材料的可用数量。

Answer 1

这是错误的，因为它不满足其他约束。[原文如此]

违反了哪个约束？如果你看看你是如何定义约束的，你会发现它们都被满足了。

prob += lpSum([proteinPercent[i] * ingredient_vars[i] for i in Ingredients]) >= 15.5, "ProteinRequirement"

给出 100 单位牛肉的解决方案意味着您有 2*100 = 200 单位的蛋白质 - 远远超过所需的 15.5。

prob += lpSum([fatPercent[i] * ingredient_vars[i] for i in Ingredients]) >= 12.3, "FatRequirement"

给出 100 单位 BEEF 的解决方案意味着您有 14*100 = 1400 单位的蛋白质 - 远远超过所需的 12.3。

真正的问题是我认为你有点混淆了单位。乘以百分比时，您需要除以 100。