Solving LPs in Python with pulp
Linear programming (LP) problems frequently appear in analytics when you want to optimize an objective—such as minimizing costs or maximizing profit—while satisfying real-world constraints. Python’s PuLP library provides a simple and intuitive way to formulate and solve LPs without requiring any low-level solver knowledge.
What Is PuLP?
PuLP is a Python library that allows you to create optimization problems in a readable, model-oriented way. It automatically connects to solvers such as CBC (default), GLPK, or commercial ones if available.
Working with PuLP usually involves:
- Defining a problem object
- Declaring decision variables
- Writing the objective function
- Adding constraints
- Solving the model and retrieving the results
Basic Structure of an LP in PuLP
Below is the general template of an LP model:
# 1. Create problem
problem = LpProblem("My_Problem", LpMinimize) # or LpMaximize
# 2. Create variables
x = LpVariable("x", lowBound=0)
y = LpVariable("y", lowBound=0)
# 3. Objective function
problem += 3*x + 5*y
# 4. Constraints
problem += x + 2*y <= 10
problem += 3*x + y <= 12
# 5. Solve
problem.solve()
Example: Resource Allocation Problem
Suppose a company produces two products using limited machine hours.
- Product A requires 2 hours
- Product B requires 3 hours
- We have 40 total machine hours
- Profit is 30 for A and $50 for B
- Objective: maximize profit
Modeling the Problem
Let:
- x - number of units of Product A
- y - number of units of Product B
Constraints:
2x+3y≤40Objective:
max(30x+50y)12345678910111213141516171819202122from pulp import LpProblem, LpVariable, LpMaximize, value # Create optimization model model = LpProblem("Production_Optimization", LpMaximize) # Decision variables x = LpVariable("Product_A", lowBound=0) y = LpVariable("Product_B", lowBound=0) # Objective function model += 30 * x + 50 * y # Constraint: machine hours model += 2 * x + 3 * y <= 40 # Solve model.solve() print("Optimal Production Plan:") print("Product A:", value(x)) print("Product B:", value(y)) print("Max Profit:", value(model.objective))
PuLP prints the optimal values:
- How many units of each product should be produced
- The maximum achievable profit
- Which constraints are binding (tight) vs non-binding
The results give actionable insights: "Produce more of the item with higher marginal profit until the machine constraint becomes binding".
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Can you explain what a binding constraint means in this context?
How would I add more constraints, like material limits, to the model?
What if I want to minimize costs instead of maximizing profit?
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Solving LPs in Python with pulp
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Linear programming (LP) problems frequently appear in analytics when you want to optimize an objective—such as minimizing costs or maximizing profit—while satisfying real-world constraints. Python’s PuLP library provides a simple and intuitive way to formulate and solve LPs without requiring any low-level solver knowledge.
What Is PuLP?
PuLP is a Python library that allows you to create optimization problems in a readable, model-oriented way. It automatically connects to solvers such as CBC (default), GLPK, or commercial ones if available.
Working with PuLP usually involves:
- Defining a problem object
- Declaring decision variables
- Writing the objective function
- Adding constraints
- Solving the model and retrieving the results
Basic Structure of an LP in PuLP
Below is the general template of an LP model:
# 1. Create problem
problem = LpProblem("My_Problem", LpMinimize) # or LpMaximize
# 2. Create variables
x = LpVariable("x", lowBound=0)
y = LpVariable("y", lowBound=0)
# 3. Objective function
problem += 3*x + 5*y
# 4. Constraints
problem += x + 2*y <= 10
problem += 3*x + y <= 12
# 5. Solve
problem.solve()
Example: Resource Allocation Problem
Suppose a company produces two products using limited machine hours.
- Product A requires 2 hours
- Product B requires 3 hours
- We have 40 total machine hours
- Profit is 30 for A and $50 for B
- Objective: maximize profit
Modeling the Problem
Let:
- x - number of units of Product A
- y - number of units of Product B
Constraints:
2x+3y≤40Objective:
max(30x+50y)12345678910111213141516171819202122from pulp import LpProblem, LpVariable, LpMaximize, value # Create optimization model model = LpProblem("Production_Optimization", LpMaximize) # Decision variables x = LpVariable("Product_A", lowBound=0) y = LpVariable("Product_B", lowBound=0) # Objective function model += 30 * x + 50 * y # Constraint: machine hours model += 2 * x + 3 * y <= 40 # Solve model.solve() print("Optimal Production Plan:") print("Product A:", value(x)) print("Product B:", value(y)) print("Max Profit:", value(model.objective))
PuLP prints the optimal values:
- How many units of each product should be produced
- The maximum achievable profit
- Which constraints are binding (tight) vs non-binding
The results give actionable insights: "Produce more of the item with higher marginal profit until the machine constraint becomes binding".
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