Linear Programming Example
Main.LinearProgramming History
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- Generate a contour plot
- Import some other libraries that we'll need
- matplotlib and numpy packages must also be installed
import matplotlib import numpy as np import matplotlib.pyplot as plt
- Design variables at mesh points
x = np.arange(-1.0, 8.0, 0.02) y = np.arange(-1.0, 6.0, 0.02) x1, x2 = np.meshgrid(x,y)
- Equations and Constraints
profit = 100.0 * x1 + 125.0 * x2 A_usage = 3.0 * x1 + 6.0 * x2 B_usage = 8.0 * x1 + 4.0 * x2
- Create a contour plot
plt.figure()
- Weight contours
lines = np.linspace(100.0,800.0,8) CS = plt.contour(x1,x2,profit,lines,colors='g') plt.clabel(CS, inline=1, fontsize=10)
- A usage < 30
CS = plt.contour(x1,x2,A_usage,[26.0, 28.0, 30.0],colors='r',linewidths=[0.5,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- B usage < 44
CS = plt.contour(x1, x2,B_usage,[40.0,42.0,44.0],colors='b',linewidths=[0.5,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Container for 0 <= Product 1 <= 500 L
CS = plt.contour(x1, x2,x1 ,[0.0, 0.1, 4.9, 5.0],colors='k',linewidths=[4.0,1.0,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Container for 0 <= Product 2 <= 400 L
CS = plt.contour(x1, x2,x2 ,[0.0, 0.1, 3.9, 4.0],colors='k',linewidths=[4.0,1.0,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Add some labels
plt.title('Soft Drink Production Problem') plt.xlabel('Product 1 (100 L)') plt.ylabel('Product 2 (100 L)')
- Save the figure as a PNG
plt.savefig('contour.png')
- Show the plots
plt.show() (:sourceend:) (:divend:)
The linear program is solved with the APM model through a web-service while the contour plot is generated with the Python package Matplotlib.
(:toggle hide mycode2 button show="Show APM Python Source with Contours":) (:div id=mycode2:) (:source lang=python:)
- Import APM Python library
try:
from APMonitor import *
except:
# Automatically install APMonitor import pip pip.main(['install','APMonitor']) from APMonitor import *
- Select the server
server = 'https://byu.apmonitor.com'
- Give the application a name
app = 'production'
- Clear any previous applications by that name
apm(server,app,'clear all')
- Write the model file
fid = open('softdrink.apm','w') fid.write('Variables \n') fid.write(' x1 > 0 , < 5 ! Product 1 \n') fid.write(' x2 > 0 , < 4 ! Product 2 \n') fid.write(' profit \n') fid.write(' \n') fid.write('Equations \n') fid.write(' ! profit function \n') fid.write(' maximize profit \n') fid.write(' profit = 100 * x1 + 125 * x2 \n') fid.write(' 3 * x1 + 6 * x2 <= 30 \n') fid.write(' 8 * x1 + 4 * x2 <= 44 \n') fid.close() apm_load(server,app,'softdrink.apm')
- Solve on APM server
solver_output = apm(server,app,'solve')
- Display solver output
print(solver_output)
- Retrieve results
sol = apm_sol(server,app)
print ('') print (-- Results of the Optimization Problem --) print ('Product 1 (x1): ' + str(sol['x1'])) print ('Product 2 (x2): ' + str(sol['x2'])) print ('Profit: ' + str(sol['profit']))
- Display Results in Web Viewer
url = apm_web_var(server,app)
Linear programming (LP) is a mathematical optimization technique used to solve problems with a linear objective function and linear constraints. Linear Programming maximizes or minimizes a linear objective function of several variables subject to constraints that are also linear in the same variables.
Linear programming (LP) is a mathematical optimization technique used to solve problems with a linear objective function and linear constraints. Linear Programming maximizes or minimizes a linear objective function of several variables subject to constraints that are also linear in the same variables.
Linear programming (LP) is a mathematical optimization technique used to solve problems with a linear objective function and linear constraints. LP maximizes or minimizes a linear objective function of several variables subject to constraints that are also linear in the same variables.
Linear programming (LP) is a mathematical optimization technique used to solve problems with a linear objective function and linear constraints. Linear Programming maximizes or minimizes a linear objective function of several variables subject to constraints that are also linear in the same variables.
See Optimization with Python for examples with:
See Optimization with Python for examples of:
Linear programming (LP) is a mathematical optimization technique used to solve problems with a linear objective function and linear constraints. LP maximizes or minimizes a linear objective function of several variables subject to constraints that are also linear in the same variables.
Linear programming is widely used in engineering decision-making to optimize resource allocation, such as minimize costs, maximize profits, or maximize output, subject to a set of constraints. For example, a refinery might use linear programming to determine the optimal crude oils to purchase to maximize profit from their specific equipment that is subject to processing constraints and can make a certain amount of each refined product. Linear programming can also be used to solve complex logistical problems, such as a schedule production runs, determine optimal transportation routes, and minimize inventory costs.
An advantage of linear programming over nonlinear programming is the ability to handle a large number of variables and constraints, model complex situations, and provide an explainable, quantitative basis for decision-making. By using linear programming, engineers can make more informed decisions, reduce costs, and increase efficiency.
- Linear Programming (LP)
- Quadratic Programming (QP)
- Nonlinear Programming (NLP)
- Mixed Integer LP (MILP)
- Mixed Integer NLP (MINLP)
- 1️⃣ Linear Programming (LP)
- 2️⃣ Quadratic Programming (QP)
- 3️⃣ Nonlinear Programming (NLP)
- 4️⃣ Mixed Integer Linear Programming (MILP)
- 5️⃣ Mixed Integer Nonlinear Programming (MINLP)
Refinery Example LP Problem
Refinery Example LP Problem
See Optimization with Python for examples with:
- Linear Programming (LP)
- Quadratic Programming (QP)
- Nonlinear Programming (NLP)
- Mixed Integer LP (MILP)
- Mixed Integer NLP (MINLP)
Refinery Example LP Problem
See additional solution options for solving Linear Programming with Python.
m.Obj(-profit) # maximize
m.Maximize(profit)
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A simple production planning problem is given by the use of two ingredients A and B that produce products 1 and 2. In this case, it requires:
- 3 units of A and 6 units of B to produce Product 1
- 8 units of A and 4 units of B to produce Product 2
A simple production planning problem is given by the use of two ingredients A and B that produce products 1 and 2. The available supply of A is 30 units and B is 44 units. For production it requires:
- 3 units of A and 8 units of B to produce Product 1
- 6 units of A and 4 units of B to produce Product 2
m.Obj(-profit) # maximize
m.Maximize(profit) # maximize
Soft Drink Production Problem (Example 2)
Soft Drink Production Problem (Example 2)
Below are the source files for generating the contour plots in Python. The linear program is solved with the APM model through a web-service while the contour plot is generated with the Python package Matplotlib.
(:toggle hide mycode1 button show="Show Gekko Python Source":)
Below are the source files for generating the contour plots in GEKKO Python and APM Python.
(:toggle hide mycode1 button show="Show Gekko Python Source with Contours":)
(:toggle hide mycode2 button show="Show APM Python Source":)
The linear program is solved with the APM model through a web-service while the contour plot is generated with the Python package Matplotlib.
(:toggle hide mycode2 button show="Show APM Python Source with Contours":)
Python Solution
(:source lang=python:) from gekko import GEKKO
m = GEKKO() x1 = m.Var(lb=0,ub=5) x2 = m.Var(lb=0,ub=4) profit = m.Var()
m.Obj(-profit) # maximize m.Equation(profit==100*x1 + 125*x2) m.Equation(3*x1+6*x2<=30) m.Equation(8*x1+4*x2<=44)
m.solve()
print ('') print (-- Results of the Optimization Problem --) print ('Product 1 (x1): ' + str(x1.value)) print ('Product 2 (x2): ' + str(x2.value)) print ('Profit: ' + str(profit.value)) (:sourceend:)
Contour Plot
(:toggle hide mycode1 button show="Show Gekko Python Source":) (:div id=mycode1:)
- Import APM Python library
try:
from APMonitor import *
except:
# Automatically install APMonitor import pip pip.main(['install','APMonitor']) from APMonitor import *
- Select the server
server = 'https://byu.apmonitor.com'
- Give the application a name
app = 'production'
- Clear any previous applications by that name
apm(server,app,'clear all')
- Write the model file
fid = open('softdrink.apm','w') fid.write('Variables \n') fid.write(' x1 > 0 , < 5 ! Product 1 \n') fid.write(' x2 > 0 , < 4 ! Product 2 \n') fid.write(' profit \n') fid.write(' \n') fid.write('Equations \n') fid.write(' ! profit function \n') fid.write(' maximize profit \n') fid.write(' profit = 100 * x1 + 125 * x2 \n') fid.write(' 3 * x1 + 6 * x2 <= 30 \n') fid.write(' 8 * x1 + 4 * x2 <= 44 \n') fid.close() apm_load(server,app,'softdrink.apm')
- Solve on APM server
solver_output = apm(server,app,'solve')
- Display solver output
print(solver_output)
- Retrieve results
sol = apm_sol(server,app)
from gekko import GEKKO
m = GEKKO() x1 = m.Var(lb=0,ub=5) x2 = m.Var(lb=0,ub=4) profit = m.Var()
m.Obj(-profit) # maximize m.Equation(profit==100*x1 + 125*x2) m.Equation(3*x1+6*x2<=30) m.Equation(8*x1+4*x2<=44)
m.solve()
print ('Product 1 (x1): ' + str(sol['x1'])) print ('Product 2 (x2): ' + str(sol['x2'])) print ('Profit: ' + str(sol['profit']))
- Display Results in Web Viewer
url = apm_web_var(server,app)
print ('Product 1 (x1): ' + str(x1.value)) print ('Product 2 (x2): ' + str(x2.value)) print ('Profit: ' + str(profit.value))
(:divend:)
(:toggle hide mycode2 button show="Show APM Python Source":) (:div id=mycode2:) (:source lang=python:)
- Import APM Python library
try:
from APMonitor import *
except:
# Automatically install APMonitor import pip pip.main(['install','APMonitor']) from APMonitor import *
- Select the server
server = 'https://byu.apmonitor.com'
- Give the application a name
app = 'production'
- Clear any previous applications by that name
apm(server,app,'clear all')
- Write the model file
fid = open('softdrink.apm','w') fid.write('Variables \n') fid.write(' x1 > 0 , < 5 ! Product 1 \n') fid.write(' x2 > 0 , < 4 ! Product 2 \n') fid.write(' profit \n') fid.write(' \n') fid.write('Equations \n') fid.write(' ! profit function \n') fid.write(' maximize profit \n') fid.write(' profit = 100 * x1 + 125 * x2 \n') fid.write(' 3 * x1 + 6 * x2 <= 30 \n') fid.write(' 8 * x1 + 4 * x2 <= 44 \n') fid.close() apm_load(server,app,'softdrink.apm')
- Solve on APM server
solver_output = apm(server,app,'solve')
- Display solver output
print(solver_output)
- Retrieve results
sol = apm_sol(server,app)
print ('') print (-- Results of the Optimization Problem --) print ('Product 1 (x1): ' + str(sol['x1'])) print ('Product 2 (x2): ' + str(sol['x2'])) print ('Profit: ' + str(sol['profit']))
- Display Results in Web Viewer
url = apm_web_var(server,app)
- Generate a contour plot
- Import some other libraries that we'll need
- matplotlib and numpy packages must also be installed
import matplotlib import numpy as np import matplotlib.pyplot as plt
- Design variables at mesh points
x = np.arange(-1.0, 8.0, 0.02) y = np.arange(-1.0, 6.0, 0.02) x1, x2 = np.meshgrid(x,y)
- Equations and Constraints
profit = 100.0 * x1 + 125.0 * x2 A_usage = 3.0 * x1 + 6.0 * x2 B_usage = 8.0 * x1 + 4.0 * x2
- Create a contour plot
plt.figure()
- Weight contours
lines = np.linspace(100.0,800.0,8) CS = plt.contour(x1,x2,profit,lines,colors='g') plt.clabel(CS, inline=1, fontsize=10)
- A usage < 30
CS = plt.contour(x1,x2,A_usage,[26.0, 28.0, 30.0],colors='r',linewidths=[0.5,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- B usage < 44
CS = plt.contour(x1, x2,B_usage,[40.0,42.0,44.0],colors='b',linewidths=[0.5,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Container for 0 <= Product 1 <= 500 L
CS = plt.contour(x1, x2,x1 ,[0.0, 0.1, 4.9, 5.0],colors='k',linewidths=[4.0,1.0,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Container for 0 <= Product 2 <= 400 L
CS = plt.contour(x1, x2,x2 ,[0.0, 0.1, 3.9, 4.0],colors='k',linewidths=[4.0,1.0,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Add some labels
plt.title('Soft Drink Production Problem') plt.xlabel('Product 1 (100 L)') plt.ylabel('Product 2 (100 L)')
- Save the figure as a PNG
plt.savefig('contour.png')
- Show the plots
plt.show() (:sourceend:) (:divend:)
(:source lang=python:)
- Import APM Python library
try:
from APMonitor import *
except:
# Automatically install APMonitor import pip pip.main(['install','APMonitor']) from APMonitor import *
- Select the server
server = 'https://byu.apmonitor.com'
- Give the application a name
app = 'production'
- Clear any previous applications by that name
apm(server,app,'clear all')
- Write the model file
fid = open('softdrink.apm','w') fid.write('Variables \n') fid.write(' x1 > 0 , < 5 ! Product 1 \n') fid.write(' x2 > 0 , < 4 ! Product 2 \n') fid.write(' profit \n') fid.write(' \n') fid.write('Equations \n') fid.write(' ! profit function \n') fid.write(' maximize profit \n') fid.write(' profit = 100 * x1 + 125 * x2 \n') fid.write(' 3 * x1 + 6 * x2 <= 30 \n') fid.write(' 8 * x1 + 4 * x2 <= 44 \n') fid.close() apm_load(server,app,'softdrink.apm')
- Solve on APM server
solver_output = apm(server,app,'solve')
- Display solver output
print(solver_output)
- Retrieve results
sol = apm_sol(server,app)
print ('') print (-- Results of the Optimization Problem --) print ('Product 1 (x1): ' + str(sol['x1'])) print ('Product 2 (x2): ' + str(sol['x2'])) print ('Profit: ' + str(sol['profit']))
- Display Results in Web Viewer
url = apm_web_var(server,app)
- Generate a contour plot
- Import some other libraries that we'll need
- matplotlib and numpy packages must also be installed
import matplotlib import numpy as np import matplotlib.pyplot as plt
- Design variables at mesh points
x = np.arange(-1.0, 8.0, 0.02) y = np.arange(-1.0, 6.0, 0.02) x1, x2 = np.meshgrid(x,y)
- Equations and Constraints
profit = 100.0 * x1 + 125.0 * x2 A_usage = 3.0 * x1 + 6.0 * x2 B_usage = 8.0 * x1 + 4.0 * x2
- Create a contour plot
plt.figure()
- Weight contours
lines = np.linspace(100.0,800.0,8) CS = plt.contour(x1,x2,profit,lines,colors='g') plt.clabel(CS, inline=1, fontsize=10)
- A usage < 30
CS = plt.contour(x1,x2,A_usage,[26.0, 28.0, 30.0],colors='r',linewidths=[0.5,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- B usage < 44
CS = plt.contour(x1, x2,B_usage,[40.0,42.0,44.0],colors='b',linewidths=[0.5,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Container for 0 <= Product 1 <= 500 L
CS = plt.contour(x1, x2,x1 ,[0.0, 0.1, 4.9, 5.0],colors='k',linewidths=[4.0,1.0,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Container for 0 <= Product 2 <= 400 L
CS = plt.contour(x1, x2,x2 ,[0.0, 0.1, 3.9, 4.0],colors='k',linewidths=[4.0,1.0,1.0,4.0]) plt.clabel(CS, inline=1, fontsize=10)
- Add some labels
plt.title('Soft Drink Production Problem') plt.xlabel('Product 1 (100 L)') plt.ylabel('Product 2 (100 L)')
- Save the figure as a PNG
plt.savefig('contour.png')
- Show the plots
plt.show() (:sourceend:)
<iframe width="560" height="315" src="//www.youtube.com/embed/i8WS6HlE8qM?list=UU2GuY-AxnNxIJFAVfEW0QFA" frameborder="0" allowfullscreen></iframe>
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Refinery Optimization with Linear Programming
- Solve Refinery Optimization Problem with Integer Variables
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Refinery Optimization with Mixed Integer Linear Programming
- Solve Refinery Optimization Problem with Integer Variables
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Linear Programming Example 1
Linear Programming Example 2
Soft Drink Production Problem (Example 2)
Linear Programming Example
Linear Programming Example 1
A refinery must produce 100 gallons of gasoline and 160 gallons of diesel to meet customer demands. The refinery would like to minimize the cost of crude and two crude options exist. The less expensive crude costs $80 USD per barrel while a more expensive crude costs $95 USD per barrel. Each barrel of the less expensive crude produces 10 gallons of gasoline and 20 gallons of diesel. Each barrel of the more expensive crude produces 15 gallons of both gasoline and diesel. Find the number of barrels of each crude that will minimize the refinery cost while satisfying the customer demands.
- Solve Refinery Optimization Problem with Continuous Variables
- Solve Refinery Optimization Problem with Integer Variables
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