Linear Programming

Using Python 

Install Anaconda:

Please install Anaconda and Jupytor Notebook

 

Install Python Packages using following syntax:

pip install numpy 

pip install PuLP 

Refer the following links for more details:

https://pypi.org/project/numpy/

https://pypi.org/project/PuLP/

Example 3: Maximization Problem

Max Z = 20X₁+60X₂

S.t.

30X₁+20X₂ ≤ 2700

5X₁+10X₂ ≤ 850

X₁+X₂ ≤ 95

X₁, X₂ ≥ 0

The python syntax is as follows:

# import the library pulp as p 

import pulp as p 

  

# Create a LP Maximization problem 

Lp_prob = p.LpProblem('Problem', p.LpMaximize)  

  

# Create problem Variables  

X1 = p.LpVariable("X1", lowBound = 0)   # Create a variable X1 >= 0 

X2 = p.LpVariable("X2", lowBound = 0)   # Create a variable X2 >= 0 

  

# Objective Function 

Lp_prob += 20 * X1 + 60 * X2    

  

# Constraints: 

Lp_prob += 30 * X1 + 20 * X2 <= 2700

Lp_prob += 5*X1 + 10*X2 <= 850

Lp_prob += X1 + X2 <= 95

  

# Display the problem 

print(Lp_prob) 

status = Lp_prob.solve()   # Solver 

print(p.LpStatus[status])   # The solution status 

  

# Printing the final solution 

print(p.value(X1), p.value(X2), p.value(Lp_prob.objective))

Solution is:

Problem:
MAXIMIZE
20*X1 + 60*X2 + 0
SUBJECT TO
_C1: 30 X1 + 20 X2 <= 2700

_C2: 5 X1 + 10 X2 <= 850

_C3: X1 + X2 <= 95

VARIABLES
X1 Continuous
X2 Continuous

Optimal
0.0 85.0 5100.0

Example 4: Minimization Problem

Min Z = 20X₁+60X₂

 S.t.

30X₁+20X₂ ≥ 2700

5X₁+10X₂ ≥ 850

X₁+X₂ ≥ 95

X₁, X₂ ≥ 0

The python syntax is as follows:

# Create a LP Minimization problem 

Lp_prob = p.LpProblem('Problem', p.LpMinimize)  

  

# Create problem Variables  

X1 = p.LpVariable("X1", lowBound = 0)   # Create a variable X1 >= 0 

X2 = p.LpVariable("X2", lowBound = 0)   # Create a variable X2 >= 0 

  

# Objective Function 

Lp_prob += 20 * X1 + 60 * X2    

  

# Constraints: 

Lp_prob += 30 * X1 + 20 * X2 >= 2700

Lp_prob += 5*X1 + 10*X2 >= 850

Lp_prob += X1 + X2 >= 95

  

# Display the problem 

print(Lp_prob) 

status = Lp_prob.solve()   # Solver 

print(p.LpStatus[status])   # The solution status 

  

# Printing the final solution 

print(p.value(X1), p.value(X2), p.value(Lp_prob.objective))

Solution is:

Problem:
MINIMIZE
20*X1 + 60*X2 + 0
SUBJECT TO
_C1: 30 X1 + 20 X2 >= 2700

_C2: 5 X1 + 10 X2 >= 850

_C3: X1 + X2 >= 95

VARIABLES
X1 Continuous
X2 Continuous

Optimal
170.0 0.0 3400.0

Sensitivity Analysis Using Python

import pulp

import pandas as pd

 

D = pulp.LpVariable('D', lowBound=0)

S = pulp.LpVariable('S', lowBound=0)

 

profit = pulp.LpProblem('maximum profit', pulp.LpMaximize)

 

profit += 10*S + 9*D, 'Objective Function'

profit += 7/10*S + D <= 630, 'Constraint for cutting and dying'

profit += 0.5*S + 5/6*D <=600, 'Constraint for sewing'

profit += S + 2/3*D<=708, 'Constraint for finishing'

profit += 1/10*S + 1/4*D<=135, 'Constraint for I&P'

 

profit.solve()

for variable in profit.variables():

     print(variable.name, '=', variable.varValue)

print('Maximum Profit :',pulp.value(profit.objective))

 

#Sensitivity Analysis

'''

cname = name of the constraint

cinfo = info about the constraint such as shadow price and slack variable

'''

SA = [{'Constraint Name ': cname, 'Slack Values':cinfo.slack, 'Shadow price': cinfo.pi} for cname, cinfo in profit.constraints.items()]

 

df = pd.DataFrame(SA)

 

print(df)

Solution is:

D = 252.0

S = 540.0

Maximum Profit : 7668.0

                   Constraint Name   Slack Values  Shadow price

0  Constraint_for_cutting_and_dying          -0.0        4.3750

1             Constraint_for_sewing         120.0       -0.0000

2          Constraint_for_finishing          -0.0        6.9375

3                Constraint_for_I&P          18.0       -0.0000

Using R 

install.packages("lpSolve")

Example 1: Maximization Problem

Max Z = 20X₁+60X₂

S.t.

30X₁+20X₂ ≤ 2700

5X₁+10X₂ ≤ 850

X₁+X₂ ≤ 95

X₁, X₂ ≥ 0

The syntax for R Software:

library(lpSolve)

obj.fun<-c(20,60)

constr<-matrix(c(30,20,5,10,1,1),ncol=2,byrow=TRUE)

constr.dir<-c("<=","<=","<=")

rhs<-c(2700,850,95)

prod.sol<-lp("max",obj.fun,constr,constr.dir,rhs,compute.sens=TRUE)

prod.sol$solution

lp("max", obj.fun, constr, constr.dir, rhs)

lp("max", obj.fun, constr, constr.dir, rhs, compute.sens=TRUE)$duals

Solution

Variables: X₁ = 0; X₂ = 85

The objective function is 5100 

Shadow Price: 0   6   0     Reduced Cost: -10   0

Example 2: Minimization Problem

Min Z = 20X₁+60X₂

 S.t.

30X₁+20X₂ ≥ 2700

5X₁+10X₂ ≥ 850

X₁+X₂ ≥ 95

X₁, X₂ ≥ 0

The syntax for R Software:

library(lpSolve)

obj.fun<-c(20,60)

constr<-matrix(c(30,20,5,10,1,1),ncol=2,byrow=TRUE)

constr.dir<-c(">=",">=",">=")

rhs<-c(2700,850,95)

prod.sol<-lp("min",obj.fun,constr,constr.dir,rhs,compute.sens=TRUE)

prod.sol$solution

lp("min", obj.fun, constr, constr.dir, rhs)

lp("min", obj.fun, constr, constr.dir, rhs, compute.sens=TRUE)$duals

Solution

Variables: X₁ = 170; X₂ = 0

The objective function is 3400 

Shadow Price: 0  4  0         Reduced Cost: 0  20

Acknowledgement: 

Sensitivity Analysis Using Python is contributed by Mr. Sushant Chougule, MBA (OM) Student, Symbiosis Institute of Operations Management, India. https://www.linkedin.com/in/sushantchougule/