Python script to solve Linear Programming problems using PuLP library

simpleLP1.py

# simpleLP1.py
# Python 2.7.6

"""
Python script to solve Linear Programming problems
Uses Pulp library

Problem statement: (src - http://fisher.osu.edu/~croxton.4/tutorial/)

1) Objective Function -
    mathematical expression of what we are trying to accomplish
    For eg. If you are producing 3 products P1, P2 and P3 each selling 
    for 2$, 3$ and 8$ respectively. 
    And if your objective is to maximise profit, then objective function is,

    Objective Function >> Maximise 2P1 + 3P2 + 8P3

2) Constraints - Linear expression which restricts our variables. 
    "Constraint sense" could be =<, >=, or =
    Strict inequalities such as < or > cannot be used
    "RHS" Right hand side - limiting value of expression

    For eg,
        a.  if we have a total production cost budget as $300 and the
        production cost of P1, P2, P3 are $1, $2 and $5 respectively
        b. Product P1 and product P2 should be made in a combination of
        atleast 50 units
        c. Product P1 takes 2 hours to produce and similarly product P2
        takes 4 hours and Product P3 takes 5 hours and the total production
        time is exactly 400 hours

    Constraints >> 
    1P1 + 2P2 + 5P3 <= 300
    1P1 + 1P2 >= 50
    2P1 + 4P2 + 5P3 = 400
    
    
3) Sign Restrictions - additional constraints, which are often forgotten
    e.g. The number of products cannot be negative, so
    P1 >= 0
    P2 >= 0
    P3 >= 0
"""


import pulp as p

# declare all variables
P1 = p.LpVariable("P1", 0)  # P1 >= 0
P2 = p.LpVariable("P2", 0)  # P2 >= 0
P3 = p.LpVariable("P3", 0)  # P3 >= 0


# define the problem
prob = p.LpProblem("problem", p.LpMaximize)

# define the constraints
prob += 1 * P1 + 2 * P2 + 5 * P3 <= 300
prob += 1 * P1 + 1 * P2 >= 50
prob += 2 * P1 + 4 * P2 + 5 * P3 == 400
prob += P1 >= 0
prob += P2 >= 0
prob += P3 >= 0


# Define the objective function to maximize
# Objective here is to maximize profits
prob += 2 * P1 + 3 * P2 + 8 * P3


# Solve the problem
status = prob.solve()
print("This is the {} solution".format(p.LpStatus[status]))

# make sure we got an optimal solution
assert status == p.LpStatusOptimal

# print the value of P1, P2 and P3
print("To maximize profitability you have to sell {} P1, {} P2 and {} P3".
      format(p.value(P1), p.value(P2), p.value(P3)))

print("The maximum profit to be derived is $ {}".format(
    prob.objective.value()))