Optimization with PuLP

Forecasting is never perfect. Therefore, there is need of re-balancing inventory in every retail or manufacturing organization. I will provide very basic example so someone can grasp some understanding.

Understanding the Transportation Problem

The transportation problem revolves around determining the most cost-effective distribution plan that meets all supply and demand constraints. Imagine we have a set of warehouses (suppliers) and a set of stores (consumers). Each warehouse has a certain supply capacity, each store has a demand that must be met, and there are costs associated with transporting goods from each warehouse to each store. The goal is to minimize the total transportation cost.

Installation: To solve this with PuLP, we first need to install the library if you haven’t

pip install pulp

Define constants: Now, let’s dive into a simple example. Suppose we have three warehouses or factories, A and B, C with inventories of 100, 300, and 150 units, respectively. We need to supply three stores, V, W, X, Y, and Z, with demands of 190, 40, 180, 160, and 60 units, respectively. The transportation costs per unit are given in a cost matrix.

# Creates a list of all the supply nodes
Warehouses = ["A", "B", "C"]

# Creates a dictionary for the number of units of supply for each supply node
supply = {"A": 100, "B": 500, "C": 150}

# Creates a list of all demand nodes
stores = ["V", "W", "X", "Y", "Z"]

# Creates a dictionary for the number of units of demand for each demand node
demand = {
    "V": 190,
    "W": 40,
    "X": 180,
    "Y": 160,
    "Z": 60,
}


# Costs matrix as provided
costs = [
    # V  W  X  Y  Z (Stores)
    [2, 4, 2, 2, 5],  # A (Warehouse)
    [3, 2, 1, 3, 4],  # B
    [1, 5, 3, 2, 2],  # C
]

 Converts the cost matrix into a dictionary for easier manipulation in PuLP. The makeDict utility function is used to map the costs matrix to the corresponding warehouses and stores, setting a default value of 0 for any missing entries.

# Convert costs to a dictionary
costs = makeDict([Warehouses, stores], costs, 0)

Initializes the optimization problem in PuLP. “LpMinimize” indicates that the goal is to minimize the total transportation cost.

# Creates the 'prob' variable to contain the problem data
prob = LpProblem("Transportation Problem", LpMinimize)

 Generates a list of tuples representing all possible routes for transportation, combining each warehouse with each store.

Routes = [(w, s) for w in Warehouses for s in stores]

 Creates a dictionary of decision variables representing the quantity of products to be transported along each route. These variables are integer (LpInteger) and must be greater than or equal to 0.

vars = LpVariable.dicts("Route", (Warehouses, stores), 0, None, LpInteger)

 Defines the objective function of the problem, which is to minimize the total transportation cost. This is achieved by summing up the product of the quantity transported along each route and its corresponding cost.

# The objective function is added to 'prob' first
prob += lpSum([vars[w][s] * costs[w][s] for (w, s) in Routes]), "Sum of Transporting Costs"

 Adds supply constraints for each warehouse, ensuring that the total quantity transported from a warehouse does not exceed its supply capacity.

# Supply maximum constraints are added to 'prob' for each supply node
for w in Warehouses:
    prob += lpSum([vars[w][s] for s in stores]) <= supply[w], f"Sum of Products out of Warehouse {w}"

 Adds demand constraints for each store, ensuring that the total quantity transported to a store meets or exceeds its demand requirement.

# Demand minimum constraints are added to 'prob' for each demand node
for s in stores:
    prob += lpSum([vars[w][s] for w in Warehouses]) >= demand[s], f"Sum of Products into Store {s}"

 Solve the optimization problem using PuLP’s default solver.

prob.solve()

 Prints the status of the solution (Optimal, Infeasible, Unbounded, etc.).

# The problem is solved using PuLP's choice of Solver
prob.solve()

Iterates through each decision variable (route) and prints its optimal value (quantity to be transported) if the problem was solved successfully.

for v in prob.variables():
    print(v.name, "=", v.varValue)

 

Prints the total minimum cost of transportation calculated by the solver.

# The optimized objective function value is printed to the screen
print("Total Cost of Transportation = ", value(prob.objective))

 

 

Done!

In conclusion, leveraging PuLP in Python to solve the transportation optimization problem offers a powerful and flexible approach to navigating the complexities of supply chain logistics. Through the thoughtful structuring of cost matrices and the careful application of linear programming principles, businesses can significantly reduce their transportation costs while ensuring that supply and demand constraints are met.