Formulation of Linear Programming Problems: Notes and FAQs

Formulation is the step where a real-world story becomes mathematics. In a linear programming problem, a business wants the best outcome — maximum profit or minimum cost — within limits like time, money, material or space. Formulation means translating that story into three things: decision variables, an objective function and a set of constraints. Everything after is just solving; formulation is where the thinking happens.

A wrongly written constraint models the wrong problem, so formulation matters more than any solving technique. Every problem follows the same pattern: name what you control, write what to optimise, then list every limit as an inequality. Once internalised, diet plans, production schedules and transport problems all reduce to the same routine.

Key Concepts

• Decision variables: the quantities you choose, e.g. the numbers of tables and chairs to make.
• Objective function: the expression to maximise or minimise, such as profit Z = 60x + 30y.
• Constraints: the limits written as linear inequalities, such as 2x + y ≤ 100 for available machine hours.
• Non-negativity restrictions: x ≥ 0, y ≥ 0 — negative production is impossible.
• Feasible region: the set of (x, y) points satisfying all constraints at once; the answer always lies inside it.

How It Works

• Step 1 — Define the variables: let x and y stand for the quantities you control.
• Step 2 — Write the objective function: express profit or cost as a linear function of x and y, and state maximise or minimise.
• Step 3 — List every limit as an inequality: convert each resource limit (hours, material, budget) into an inequality in x and y.
• Step 4 — Add non-negativity: include x ≥ 0 and y ≥ 0 unless the problem genuinely allows negatives.
• Step 5 — Read it back: check every number in the story appears once and maximise/minimise matches the goal.

Worked Example

A workshop makes tables (x) and chairs (y). Each table needs 2 machine hours, each chair needs 1 machine hour, and 100 machine hours are available per week. Profit is ₹60 per table and ₹30 per chair. Formulate the linear programming problem to maximise profit.

1. Variables: let x = number of tables, y = number of chairs made per week.
2. Objective: maximise Z = 60x + 30y (total profit in rupees).
3. Machine-hour constraint: 2x + y ≤ 100.
4. Non-negativity: x ≥ 0, y ≥ 0.
5. Complete formulation: maximise Z = 60x + 30y subject to 2x + y ≤ 100, x ≥ 0, y ≥ 0. Sense-check: 40 tables and 10 chairs use 2(40) + 10 = 90 hours and earn ₹2700, so the model behaves sensibly.

Common Misconceptions

• Forgetting non-negativity: x ≥ 0 and y ≥ 0 are real constraints; omitting them admits absurd solutions like −5 chairs.
• Mixing up maximise and minimise: profit is maximised; cost, time and waste are minimised.
• Writing the constraint backwards: a resource limit is ≤; a minimum requirement is ≥. Swapping them models the opposite problem.

Real-World Applications

• Factory production planning: maximising profit under machine and labour limits.
• Diet problems: minimising food cost while meeting nutrition requirements.
• Transport: minimising shipping cost across routes with supply and demand limits.
• Investment: maximising return under budget and risk constraints.

Formulation turns a story into mathematics: define x and y, write the objective, convert every limit into an inequality, and add x ≥ 0, y ≥ 0. Check the goal word and each number, and the model is ready to solve.