将LP从Excel转换为Python

"" " Mining and metals We make steel with raw materials, we want to reduce the cost of producing this steel to make more money, but still respecting the minimum characteristics of quality steel "" " # Minimize the cost of metal alloys. # Characteristics of the steel to be made "" "Element %Minimum %Max %Real ( it is a var) Carbon 2 3 2.26 Copper 0.4 0.6 0.60 Manganese 1.2 1.65 1.20 "" " # Characteristics, stocks and purchase price of alloys "" " Alloy C% Cu% Mn% Stocks kg Price € / kg Iron alloy 2.50 0.00 1.30 4000 1.20 Iron alloy 3.00 0.00 0.80 3000 1.50 Iron alloy 0.00 0.30 0.00 6000 0.90 Copper alloy 0.00 90.00 0.00 5000 1.30 Copper alloy 0.00 96.00 4.00 2000 1.45 Aluminum alloy 0.00 0.40 1.20 3000 1.20 Aluminum alloy 0.00 0.60 0.00 2,500 1.00 "" " # Import the PuLP lib from pulp import * # Create the problem variable prob = LpProblem ("MinimiserLpAlliage", LpMinimize) # The 7 vars have a zero limit x1 = LpVariable ("Iron alloy 1", 0) x2 = LpVariable ("Iron alloy 2", 0) x3 = LpVariable ("Iron alloy 3", 0) x4 = LpVariable ("Copper alloy 1", 0) x5 = LpVariable ("Copper alloy 2", 0) x6 = LpVariable ("Aluminum alloy 1", 0) x7 = LpVariable ("Aluminum alloy 2", 0) # The objective function is to minimize the total cost of the alloys in EUROS for a given quantity in KGS prob + = 1.20 * x1 + 1.50 * x2 + 0.90 * x3 + 1.30 * x4 + 1.45 * x5 + 1.20 * x6 + 1.00 * x7, "AlliageCost" # Quantity constraint in KGS. prob + = x1 + x2 + x3 + x4 + x5 + x6 + x7 == 5000, "RequestedQuantity" # MIN constraints of% carbon, by alloy // ITS NOT WHAT I NEED prob + = x1> = 2.5, "MinCarboneRequirement1" prob + = x2> = 3, "MinCarboneRequirement2" prob + = x3> = 0, "MinCarboneRequirement3" prob + = x4> = 0, "MinCarboneRequirement4" prob + = x5> = 0, "MinCarboneRequirement5" prob + = x6> = 0, "MinCarboneRequirement6" prob + = x7> = 0, "MinCarboneRequirement7" # MIN constraints of% copper, by alloy // ITS WRONG ITS NOT WHAT I NEED prob + = x1> = 0, "MinCuivreRequirement1" prob + = x2> = 0, "MinCuivreRequirement2" prob + = x3> = 0.3, "MinCuivreRequirement3" prob + = x4> = 90, "MinCuivreRequirement4" prob + = x5> = 96, "MinCuivreRequirement5" prob + = x6> = 0.4, "MinCuivreRequirement6" prob + = x7> = 0.6, "MinCuivreRequirement7" # MIN constraints of% of Manganese, by alloy // ITS WRONG ITS NOT WHAT I NEED prob + = x1> = 1.3, "MinManganeseRequirement1" prob + = x2> = 0.8, "MinManganeseRequirement2" prob + = x3> = 0, "MinManganeseRequirement3" prob + = x4> = 0, "MinManganeseRequirement4" prob + = x5> = 4, "MinManganeseRequirement5" prob + = x6> = 1.2, "MinManganeseRequirement6" prob + = x7> = 0, "MinManganeseRequirement7" # MAX constraints of% of Manganese, by alloy // ITS WRONG ITS NOT WHAT I NEED prob + = x1 <= 1.3, "MaxManganeseRequirement1" prob + = x2 <= 0.8, "MaxManganeseRequirement2" prob + = x3 <= 0, "MaxManganeseRequirement3" prob + = x4 <= 0, "MaxManganeseRequirement4" prob + = x5 <= 4, "MaxManganeseRequirement5" prob + = x6 <= 1.2, "MaxManganeseRequirement6" prob + = x7 <= 0, "MaxManganeseRequirement7" # 5. MAX constraints from available stock, by alloy // I THINK IT IS OK prob + = x1 <= 4000, "MaxStock" prob + = x2 <= 3000, "MaxStock1" prob + = x3 <= 6000, "MaxStock2" prob + = x4 <= 5000, "MaxStock3" prob + = x5 <= 2000, "MaxStock4" prob + = x6 <= 3000, "MaxStock5" prob + = x7 <= 2500, "MaxStock6" # The problem data is written to an .lp file prob.writeLP ( "WhiskasModel.lp") # We use the solver prob.solve () # The status of the solution print ("Status:", LpStatus [prob.status]) # We magnify and display the optimums of each var for v in prob.variables (): print (v.name, "=", v.varValue) # The result of the objective function is here print ("Total", value (prob.objective))

Status: Optimal Aluminum_alloy_1 = 1.2 Aluminum_alloy_2 = 0.6 Copper_alloy_1 = 90.0 Alloy_of_copper_2 = 96.0 Alloy_of_fer_1 = 2.5 Alloy_of_fer_2 = 3.0 Iron_alloy_3 = 4806.7 Total 4,591.76,999,999,999,995

Mining and metals In the manufacture of steel with permeable materials, sur wants to reduce the cost of producing this steel to earn more money but still respecting the important characteristics of quality steel # Characteristics of the steel to be made """ Elément % minimal % Max Carbone 2 3 Cuivre 0.4 0.6 Manganèse 1.2 1.65 """ # Characteristics, stocks and purchase price of alloys at KILO """ Alliage C % Cu % Mn % Stocks kg Prix €/kg Alliage de fer 1 2,50 0,00 1,30 4000 1,20 Alliage de fer 2 3,00 0,00 0,80 3000 1,50 Alliage de fer 3 0,00 0,30 0,00 6000 0,90 Alliage de cuivre 1 0,00 90,00 0,00 5000 1,30 Alliage de cuivre 2 0,00 96,00 4,00 2000 1,45 Alliage d'alu 1 0,00 0,40 1,20 3000 1,20 Alliage d'alu 2 0,00 0,60 0,00 2500 1,00 """ # Importer la lib PuLP from pulp import * #Créer la variable du problème prob = LpProblem("MinimiserLpAlliage",LpMinimize) # The 7 vars have a zero limit, these decision variables are expressed in KILOS x1 = LpVariable("Alliage de fer 1",0) x2 = LpVariable("Alliage de fer 2",0) x3 = LpVariable("Alliage de fer 3",0) x4 = LpVariable("Alliage de cuivre 1",0) x5 = LpVariable("Alliage de cuivre 2",0) x6 = LpVariable("Alliage d'alu 1",0) x7 = LpVariable("Alliage d'alu 2",0) # The objective function is to minimize the total cost of the alloys in EUROS prob += 1.20 * x1 + 1.50 * x2 + 0.90 * x3 + 1.30 * x4 + 1.45 * x5 + 1.20 * x6 + 1.00 * x7, "CoutAlliages" # Quantity constraint in KGS. prob += x1 + x2 + x3 + x4 + x5 + x6 + x7 == 5000, "QuantitéDemandée" # Carbon stress. prob += (2.50 * x1 + 3.00 * x2 + x3 + x4 + x5 + x6 + x7 ) / 5000 <= 3,"carBmax" prob += (2.50 * x1 + 3.00 * x2 + x3 + x4 + x5 + x6 + x7 ) / 5000 >= 2,"carBmin" # Constraint cu . prob += (x1 + x2 + 0.30 * x3 + 90 * x4 + 96 * x5 + 0.40 * x6 + 0.60 * x7) / 5000 <= 0.6,"cuBmax" prob += (x1 + x2 + 0.30 * x3 + 90 * x4 + 96 * x5 + 0.40 * x6 + 0.60 * x7) / 5000 >= 0.4,"cuBmin" # Constraint Manganèse. prob += (1.30 * x1 + 0.80 * x2 + x3 + x4 + 4 * x5 + 1.20 * x6 + x7 ) / 5000 <= 1.65,"mgBmax" prob += (1.30 * x1 + 0.80 * x2 + x3 + x4 + 4 * x5 + 1.20 * x6 + x7 ) / 5000 >= 1.2,"mgBmin" # 5. MAX constraints from available stock, by alloy prob += x1 <= 4000 , "MaxStock" prob += x2 <= 3000 , "MaxStock1" prob += x3 <= 6000 , "MaxStock2" prob += x4 <= 5000 , "MaxStock3" prob += x5 <= 2000 , "MaxStock4" prob += x6 <= 3000 , "MaxStock5" prob += x7 <= 2500 , "MaxStock6" # The problem data is written to an .lp file prob.writeLP("acier.lp") # On utilise le solveur prob.solve() # The status of the solution print ("Status:", LpStatus[prob.status]) # We magnify and display the optimums of each var for v in prob.variables(): print (v.name, "=", v.varValue) # The result of the objective function is here print ("Total payable in euros", value(prob.objective)) """ Status: Infeasible Alliage_d'alu_1 = 0.0 Alliage_d'alu_2 = 0.0 Alliage_de_cuivre_1 = 0.0 Alliage_de_cuivre_2 = 0.0 Alliage_de_fer_1 = 0.0 Alliage_de_fer_2 = 0.0 Alliage_de_fer_3 = 10000.0 Total à payer en euros 9000.0 """

The book says the result with the excel solver is : iron_1 : 4000 kgs iron_2 : 0 kgs iron_3 : 397.76kgs cu_1 : 0 kgs cu_2 : 27.61kgs al_1 : 574.62kgs al_2 : 0kgs Cost in euros 5887.57 Steel contains 2% carb, 0.6 % cu, 1.2 %

1条回答

网友

1楼 · 发布于 2024-09-30 20:28:08

你的部分问题是你如何理解/应用百分比。我的建议是尽早将百分比[0-100]转换为小数[0-1.0]。你知道吗

在excel中，当单元格显示50%时，单元格的数值实际上是0.5。以这种方式处理百分比意味着你不必一直除以100，可以用一个百分比乘以另一个百分比，一切都很正常。你知道吗

下面的代码符合您的要求：

"""
 Mining and metals

We make steel with raw materials, we want to reduce the cost of producing this steel
to make more money, but still respecting the minimum characteristics of quality steel

"""

# Minimize the cost of metal alloys.
# Characteristics of the steel to be made

"""Element      %Minimum  %Max   %Real (it is a var)
   Carbon       2         3      2.26
   Copper       0.4       0.6    0.60
   Manganese    1.2       1.65   1.20

"""

# Characteristics, stocks and purchase price of alloys
"""
Alloy          C%   Cu%   Mn%     Stocks kg Price € / kg
Iron alloy     2.50 0.00  1.30    4000      1.20
Iron alloy     3.00 0.00  0.80    3000      1.50
Iron alloy     0.00 0.30  0.00    6000      0.90
Copper alloy   0.00 90.00 0.00    5000      1.30
Copper alloy   0.00 96.00 4.00    2000      1.45
Aluminum alloy 0.00 0.40  1.20    3000      1.20
Aluminum alloy 0.00 0.60  0.00    2500      1.00
"""

# Import the PuLP lib
from pulp import *

# Create the problem variable
prob = LpProblem ("MinimiserLpAlliage", LpMinimize)

# Problem Data
input_mats = ["iron_1", "iron_2", "iron_3",
              "cu_1", "cu_2",
              "al_1", "al_2"]

input_costs = {"iron_1": 1.20, "iron_2": 1.50, "iron_3": 0.90,
               "cu_1":   1.30, "cu_2": 1.45,
               "al_1":   1.20, "al_2":   1.00}

#                               C%     Cu%   Mn%
input_composition = {"iron_1": [0.025, 0.000,  0.013],
                     "iron_2": [0.030, 0.000,  0.008],
                     "iron_3": [0.000, 0.003,  0.000],
                     "cu_1":   [0.000, 0.900,  0.000],
                     "cu_2":   [0.000, 0.960,  0.040],
                     "al_1":   [0.000, 0.004,  0.012],
                     "al_2":   [0.000, 0.006,  0.000]}

input_stock = {"iron_1": 4000, "iron_2": 3000, "iron_3": 6000,
               "cu_1": 5000, "cu_2":  2000,
               "al_1": 3000, "al_2": 2500}

request_quantity = 5000

Carbon_min = 0.02
Carbon_max = 0.03

Cu_min = 0.004
Cu_max = 0.006

Mn_min = 0.012
Mn_max = 0.0165

# Problem variables - amount in kg of each input
x = LpVariable.dicts("input_mat", input_mats, 0)

# The objective function is to minimize the total cost of the alloys in EUROS for a given quantity in KGS
prob += lpSum([input_costs[i]*x[i] for i in input_mats]), "AlliageCost"

# Quantity constraint in KGS.
prob += lpSum([x[i] for i in input_mats]) == request_quantity, "RequestedQuantity"

# MIN/MAX constraint of carbon in resultant steel
prob += lpSum([x[i]*input_composition[i][0] for i in input_mats]) >= Carbon_min*request_quantity, "MinCarbon"
prob += lpSum([x[i]*input_composition[i][0] for i in input_mats]) <= Carbon_max*request_quantity, "MaxCarbon"

# MIN/MAX constraints of copper in resultant steel
prob += lpSum([x[i]*input_composition[i][1] for i in input_mats]) >= Cu_min*request_quantity, "MinCu"
prob += lpSum([x[i]*input_composition[i][1] for i in input_mats]) <= Cu_max*request_quantity, "MaxCu"

# MIN/MAX constraints of manganese in resultant steel
prob += lpSum([x[i]*input_composition[i][2] for i in input_mats]) >= Mn_min*request_quantity, "MinMn"
prob += lpSum([x[i]*input_composition[i][2] for i in input_mats]) <= Mn_max*request_quantity, "MaxMn"


# MAX constraints of available stock
for i in input_mats:
    prob += x[i] <= input_stock[i], ("MaxStock_" + i)

# Solve the problem
prob.solve()

# The status of the solution
print ("Status:", LpStatus [prob.status])

# Dislay the optimums of each var
for v in prob.variables ():
    print (v.name, "=", v.varValue)

# Display mat'l compositions
Carbon_value = sum([x[i].varValue*input_composition[i][0] for i in input_mats])/request_quantity
Cu_value = sum([x[i].varValue*input_composition[i][1] for i in input_mats])/request_quantity
Mn_value = sum([x[i].varValue*input_composition[i][2] for i in input_mats])/request_quantity

print ("Carbon content: " + str(Carbon_value))
print ("Copper content: " + str(Cu_value))
print ("Manganese content: " + str(Mn_value))

# The result of the objective function is here
print ("Total", value (prob.objective))

从中我得到：

Status: Optimal
input_mat_al_1 = 574.62426
input_mat_al_2 = 0.0
input_mat_cu_1 = 0.0
input_mat_cu_2 = 27.612723
input_mat_iron_1 = 4000.0
input_mat_iron_2 = 0.0
input_mat_iron_3 = 397.76302
Carbon content: 0.02
Copper content: 0.006000000036
Manganese content: 0.012000000008
Total 5887.57427835

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