Это тот фрейм данных, над которым я работаю над проблемой «выбора проекта»:
Return Sector Investment Project_name
0.290 Solar 228376120 Solar1
0.07 Solar 70021891 Solar2
0.25 Wind 6467237 Eolico1
0.3 Wind 417713440 Eolico2
0.16 Wind 377494250 Eolico3
0.28 Wind 230345712 Eolico4
0.29 CGHPCHBIO 35476862 CGH1
0.26 CGHPCHBIO 60671402 CGH2
0.07 CGHPCHBIO 349544333 PCH1
0.12 CGHPCHBIO 425442985 PCH2
0.29 CGHPCHBIO 66292734 PCH3
0.15 CGHPCHBIO 300677487 PCH4
0.25 CGHPCHBIO 409144798 Biomassa1
0.19 CGHPCHBIO 184123496 Biomassa2
0.08 CGHPCHBIO 61835863 Biomassa3
Моя цель:
Maximize the "Return"
Мои ограничения:
- Лимит 916000000 на общую сумму "Инвестиции";
- «Солнечный сектор» не может превышать 60% от общей суммы инвестиций;
- «Сектор ветра» не может превышать 60% изобщая сумма инвестиций;и
- «CGHPCHBIO Sector» не может превышать 25% от общей суммы инвестиций
Вот что я пытался до сих пор:
from pulp import *
import pandas as pd
import xlrd
#First, we create a LP problem with the method LpProblem in PuLP
prob = LpProblem("Selecao de Projetos",LpMaximize)
#Read the first rows dataset in a Pandas DataFrame
df = pd.read_excel("df.xlsx", encoding = 'unicode_escape')
#Create a list of the projects names
projects = list(df['Project_name'])
#Create a dictionary of investments for all the projects
investments = dict(zip(projects,df['Investment']))
#Create a dictionay of sectors for all the projects
sectors = dict(zip(projects,df['Sector']))
#Create a dictionary of Returns for all the projects
returns = dict(zip(projects,df['Return']))
#Create a dictionary of projects with lower bound = 0 and category continuous
project_vars = LpVariable.dicts("Project",projects,lowBound =0,cat='Continuous')
#Built the LP problem by assing the main objective function
prob += lpSum([returns[i]*project_vars[i] for i in projects])
#Add the constraints
prob += lpSum([investments[f] * project_vars[f] for f in projects]) <= 916000000
prob += lpSum([investments[f] * project_vars[f] for f in projects if sectors[f]=="Solar"]) <= lpSum([investments[f] * project_vars[f] for f in projects])*0.6
prob += lpSum([investments[f] * project_vars[f] for f in projects if sectors[f]=="Wind"]) <= lpSum([investments[f] * project_vars[f] for f in projects])*0.6
prob += lpSum([investments[f] * project_vars[f] for f in projects if sectors[f]=="CGHPCHBIO"]) <= lpSum([investments[f] * project_vars[f] for f in projects])*0.25
prob.solve()
#The status of the solution is printed to the screen
print("Status:", LpStatus[prob.status])
for v in prob.variables():
if v.varValue>0:
print(v.name, "=", v.varValue)
Результат, который я получил:
Project_CGH1 = 6.4549114
Project_Eolico1 = 84.982196
Project_Solar1 = 0.60163909
Результат, который мне нужен: какие проекты (Project_name) я бы выбрал, учитывая ограничения? Что-то вроде:
Projects chosen Investment
Solar1 228376120
Wind1 6467237
Wind3 377494250
CGH2 60671402
PCH4 300677487
Biomassa3 61835863