{
 "cells": [
  {
   "cell_type": "code",
   "execution_count": 3,
   "id": "e922309a",
   "metadata": {},
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "status:1,Optimal\n",
      "objective:55000.0011\n",
      "A:0.0\n",
      "B:16.666667\n",
      "B1:13.333333\n",
      "B2:3.3333333\n",
      "C:16.666667\n",
      "Balance_B:-7.000000010748408e-07\n",
      "Maq_1:-1.9999999949504854e-06\n",
      "Maq_2:1.1999999998124622e-06\n",
      "Materia_Prima_M:4.000000011217253e-07\n",
      "Materia_Prima_N:-5.9999994\n",
      "Compromiso_de_Venta:6.666667\n"
     ]
    }
   ],
   "source": [
    "#EJERCICIO 2.4\n",
    "#importo librerias \n",
    "from pulp import LpMaximize, LpProblem, LpStatus, lpSum, LpVariable, LpMinimize\n",
    "import pandas as pd\n",
    "import matplotlib.pyplot as plt\n",
    "from scipy.optimize import linprog\n",
    "import numpy as np\n",
    "\n",
    "#Creo el modelo\n",
    "model= LpProblem (name=\"Pullovers\", sense=LpMaximize)\n",
    "\n",
    "#Inicializo las variables\n",
    "A=LpVariable(name=\"A\", \n",
    "             lowBound=0)#Siendo A: cantidad de pullovers A\n",
    "B=LpVariable(name=\"B\", \n",
    "             lowBound=0)#Siendo B: cantidad de pullovers B\n",
    "B1=LpVariable(name=\"B1\", \n",
    "             lowBound=0)#Siendo B1: cantidad de pullovers B aproducirse en la maquina 1\n",
    "B2=LpVariable(name=\"B2\", \n",
    "             lowBound=0)#Siendo B2: cantidad de pullovers B aproducirse en la maquina 2\n",
    "C=LpVariable(name=\"C\", \n",
    "             lowBound=0)#Siendo B1: cantidad de pullovers C\n",
    "\n",
    "\n",
    "#Determino las restricciones\n",
    "model+= (B1 + B2 == B, \"Balance B\")\n",
    "model+= (5*A+ 6*B1 <= 80, \"Maq 1\")\n",
    "model+= (4*B2+ 4*C <=80, \"Maq 2\")\n",
    "model+= (1.6*A+1.2*C <=20, \"Materia Prima M\")\n",
    "model+= (1.8*B <=36, \"Materia Prima N\")\n",
    "model+= (B >=10, \"Compromiso de Venta\")\n",
    "\n",
    "#Definimos la funcion objetivo la cual queremos maximizar\n",
    "model+=1000*A+1500*B+1800*C\n",
    "\n",
    "#Resolvemos el problema de programacion lineal\n",
    "status = model.solve()\n",
    "\n",
    "#Imprimo la solucion\n",
    "print(f\"status:{model.status},{LpStatus[model.status]}\")#tipo de solucion\n",
    "print(f\"objective:{model.objective.value()}\")#Beneficio maximo sujeto a las restricciones\n",
    "for var in model.variables():\n",
    "    print(f\"{var.name}:{var.value()}\")#Valores de solucion de las variables\n",
    "for name, constraint in model.constraints.items():\n",
    "    print(f\"{name}:{constraint.value()}\") \n",
    "    "
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 3",
   "language": "python",
   "name": "python3"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 3
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython3",
   "version": "3.8.5"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 5
}