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Mathematics · Ch 12 — Linear Programming

Linear Programming Problem and Its Mathematical Formulation

12.2

Linear Programming Problem and Its Mathematical Formulation

The Furniture Dealer Problem: A Concrete Starting Point

A dealer can invest in two products: tables and chairs. Each table costs ₹2500 and yields ₹250 profit; each chair costs ₹500 and yields ₹75 profit. He has at most ₹50,000 to invest and storage for at most 60 pieces. A few possibilities:

  • Only tables: with ₹50,000 he buys 500002500=20\frac{50000}{2500} = 20 tables. Profit =20×250=₹5000= 20 \times 250 = ₹5000.
  • Only chairs: he could buy 100 chairs but storage limits him to 60. Profit =60×75=₹4500= 60 \times 75 = ₹4500.
  • Mixed: 10 tables and 50 chairs (60 pieces). Cost =25000+25000=₹50000= 25000 + 25000 = ₹50000. Profit =2500+3750=₹6250= 2500 + 3750 = ₹6250.

The mixed option beats both extremes. Among all combinations satisfying both constraints, which one yields the maximum profit?

Mathematical Formulation: Translating the Problem into Equations

To answer this systematically, we convert the word problem into a mathematical model — its mathematical formulation.

Step 1: Identify the Decision Variables

Let xx = number of tables and yy = number of chairs the dealer buys.

Note

Decision variables are the quantities we can control. Here xx and yy are non-negative integers, but for mathematical convenience we often treat them as real numbers first.

Step 2: Express the Objective Function

The profit to be maximised is the objective function, denoted ZZ:

Z=250x+75yZ = 250x + 75y

Step 3: Identify the Constraints

Investment constraint: the total cost cannot exceed ₹50,000.

2500x+500y≤50000  ⇒  5x+y≤1002500x + 500y \leq 50000 \;\Rightarrow\; 5x + y \leq 100

Storage constraint: the total number of pieces cannot exceed 60.

x+y≤60x + y \leq 60

Non-negativity constraints:

x≥0,y≥0x \geq 0, \quad y \geq 0

Step 4: Assemble the Complete Mathematical Model

Maximize Z=250x+75yZ = 250x + 75y

Subject to constraints:

5x+y≤1005x + y \leq 100

x+y≤60x + y \leq 60

x≥0,y≥0x \geq 0, \quad y \geq 0

General Form of a Linear Programming Problem

A linear programming problem in two variables has:

  1. Decision variables: xx and yy.
  2. Objective function: a linear function Z=ax+byZ = ax + by to maximize or minimize.
  3. Constraints: linear inequalities (or equations) that restrict xx and yy.
  4. Non-negativity restrictions: x≥0x \geq 0, y≥0y \geq 0.
Important

"Linear" means every term in the objective function and constraints is of the first degree — no x2x^2, xyxy, sin⁡x\sin x, etc. "Programming" here means planning, not computer programming.

Key Terminology

TermMeaningExample
Decision variablesQuantities to be determinedxx (tables), yy (chairs)
Objective functionLinear function to optimizeZ=250x+75yZ = 250x + 75y
ConstraintsLinear inequalities limiting variables5x+y≤1005x + y \leq 100, x+y≤60x + y \leq 60
Non-negativity constraintsVariables cannot be negativex≥0x \geq 0, y≥0y \geq 0
Feasible solutionAny (x,y)(x, y) satisfying all constraints(10,50)(10, 50) is feasible
Optimal solutionFeasible solution that optimizes ZZTo be found

The Optimization Problem

In general, an LPP is either a maximization (e.g. profit, revenue) or a minimization (e.g. cost, time). The solution process finds all feasible solutions, then selects the one giving the best (maximum or minimum) value of ZZ. …