Python script to solve Linear Programming problems using pulp library

simpleLP2.py

# simpleLP2.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/)
ChemCo produces two fertilizers, FastGro and ReallyFastGro. Each is made of
a mix of two growth agents. FastGro is a 50/50 mix of the two, and it sells
for $13 per pound. ReallyFastGro is a 25/75 mix (i.e., its 25% Agent 1 and 
75% Agent 2), and sells for $15/pound. Agent 1 costs $2/pound and Agent 2 costs 
$3/pound. ChemCo can purchase up to 250 pounds of Agent 1 and up to 350 pounds 
of Agent 2 each day. What is ChemCo's optimal production strategy? i.e., 
How much of each product should it produce in order to maximize its profits?

"""

import pulp as p

"""
What is the objective?
- To maximise profits, 
- how much of each products (pounds per day) should be produced

so variables are products here,
FG - volume of FastGro per day (pounds/ day)
RFG - volume of ReallyFastGro per day (pounds / day)

Objective function
Total Revenue = 13FG + 15RFG
Total cost = 
- Agent 1 = 2(0.5FG + 0.25RFG)
- Agent 2 = 3(0.5FG + 0.75RFG)

So total profit = 
13FG + 15RFG - 2(0.5FG + 0.25RFG) - 3(0.5FG + 0.75RFG)
= 10.5FG + 12.25RFG

"""

# declare all variables
X = p.LpVariable("X", 0)  # FG >= 0
Y = p.LpVariable("Y", 0)  # RFG >= 0

# Define the problem
# Objective is to maximize profit
prob = p.LpProblem("problem", p.LpMaximize)

# Define constraints
prob += 0.5 * X + 0.25 * Y <= 250
prob += 0.5 * X + 0.75 * Y <= 350
prob += X >= 0
prob += Y >= 0

# Define objective function
prob += 10.5 * X + 12.25 * Y

# 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 X and Y
print("To maximize profitability you have to sell {} FG and {} RFG".
      format(p.value(X), p.value(Y)))

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