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Linear Programming — chapter notes

6 sections 5 min readBeta

1. Terms of linear programming

A linear programming problem asks for the maximum or minimum of a linear function of the decision variables, subject to linear constraints.

  • Decision variables x,yx, y: the quantities to be chosen.
  • Objective function Z=ax+byZ = ax + by: profit, cost or time, to be optimised.
  • Constraints: linear inequalities from the limits of the problem.
  • Non-negativity: x≥0x \ge 0, y≥0y \ge 0.
  • Optimisation: finding the feasible plan with the best value of ZZ.
Every linear programming problem has these parts.
TermMeaningDairy example
Decision variablesQuantities we choosexx paneer batches, yy curd batches
Objective functionLinear quantity to maximise or minimiseZ=3000x+4000yZ = 3000x + 4000y (profit)
ConstraintsLinear limits on the variables2x+y≤142x + y \le 14, x+2y≤13x + 2y \le 13
Non-negativityVariables cannot be negativex≥0x \ge 0, y≥0y \ge 0
OptimisationFinding the best feasible planLargest profit

Formulas

  • General form
    Optimise Z=ax+by subject to linear constraints, x,y≥0\text{Optimise } Z = ax + by \ \text{subject to linear constraints},\ x, y \ge 0

Key terms

Feasible solution

A point satisfying every constraint, including non-negativity.

Optimal solution

A feasible solution at which Z takes its best value.

2. Advantages, limitations and application areas

Advantages: best use of scarce resources; clear, systematic decisions; shows bottlenecks and spare capacity; easy to re-solve when data change.

Limitations: assumes linearity (proportional profit and resource use); all coefficients must be known exactly; one objective at a time; answers may be fractional for whole items.

Application areas: production planning in factories, diet and feed mixing, transport of goods from depots to markets, allocation of advertising budgets, investment choices.

Know two of each for a 2-mark question.
AdvantagesLimitations
Best use of scarce resourcesAssumes everything is linear (proportional)
Clear, systematic decisionsCoefficients must be known exactly
Shows bottlenecks and spare capacityOnly one objective at a time
Easy to re-solve when data changeAnswers may be fractional for whole items

Key terms

Linearity

Doubling a variable doubles its contribution to Z and to each constraint.

Divisibility

Variables may take fractional values.

3. Types of problem and formulation

  • Manufacturing: maximise profit; constraints usually "at most" (≤\le) on machine hours, labour and material.
  • Diet: minimise cost; constraints usually "at least" (≥\ge) on nutrients.
  • Transportation-type: minimise transport cost; express every route through two variables using supplies and demands, then require each amount to be non-negative.

Steps: name the variables with units; write ZZ and its direction; turn each limit into an inequality; add x,y≥0x, y \ge 0.

The type tells you the direction of most inequalities.
TypeDecideTypical objectiveTypical constraints
ManufacturingHow much of each product to makeMaximise profitMachine hours, labour, raw material (≤\le)
DietHow much of each food to buyMinimise costMinimum nutrients (≥\ge)
TransportationHow much to send on each routeMinimise transport costSupplies and demands (route amounts ≥0\ge 0)
Read the table column by column into inequalities.
Paneer (xx)Curd (yy)Available
Pasteuriser hours22111414
Packing hours11221313
Profit (₹)3000300040004000
Four steps of formulation.
  • y=70y = 70
  • x+y=100x + y = 100
A transportation-type region in two variables.

Formulas

  • Transport route expressed in x, y
    P to C=100−x−y≥0  ⟺  x+y≤100\text{P to C} = 100 - x - y \ge 0 \iff x + y \le 100

Key terms

Diet problem

Choosing amounts of foods to meet minimum nutrient needs at least cost.

Transportation problem

Choosing amounts to send along routes to meet demands at least cost.

4. Graphing the feasible region

Each inequality ax+by≤cax + by \le c (or ≥c\ge c) is a half-plane bounded by the line ax+by=cax + by = c. Draw the line from its intercepts, then test a point not on it (usually the origin) to choose the side.

The feasible region is the common part of all the half-planes in the first quadrant. Its corner points are intersections of pairs of boundary lines (including the axes) that satisfy every constraint.

The region may be bounded, unbounded or empty. The ISC course uses at most three constraints besides non-negativity, so a region has at most five corners.

  • x+y=9x + y = 9
  • x+3y=15x + 3y = 15
A bounded region with four corners.
  • 3x+4y=243x + 4y = 24
Origin test: false, so shade the far side.
  • x+y=12x + y = 12
  • 2x+y=202x + y = 20
  • x+3y=30x + 3y = 30
Three non-trivial constraints: up to five corners.
  • x+y=8x + y = 8
  • x+3y=12x + 3y = 12
An unbounded region: it has corners but no outer edge.
  • x+y=4x + y = 4
  • 3x+2y=183x + 2y = 18
No common region: the problem is infeasible.
Classify the region before you optimise.
RegionExampleWhat it means
Boundedx+y≤9x + y \le 9, x+3y≤15x + 3y \le 15Fits inside a circle: both maximum and minimum exist
Unboundedx+y≥8x + y \ge 8, x+3y≥12x + 3y \ge 12Goes on without end: an optimum may or may not exist
Empty (infeasible)x+y≤4x + y \le 4, 3x+2y≥183x + 2y \ge 18No feasible point: no solution at all

Key terms

Corner point

A vertex of the feasible region: the meeting point of two boundary lines.

Unbounded region

A feasible region that cannot be enclosed in a circle.

Infeasible problem

A problem whose feasible region is empty.

5. The corner point method

Corner point theorem. If the feasible region is bounded and non-empty, Z=ax+byZ = ax + by has both a maximum and a minimum, each at a corner point.

Method: find every corner; evaluate ZZ at each; the largest and smallest values are the maximum and minimum.

Unbounded regions: let MM be the best corner value. For a maximum, if the open half-plane ax+by>Max + by > M has no point in common with the region, MM is the maximum; otherwise there is no maximum. (For a minimum use ax+by<Max + by < M.)

The corner point method.
  • x+y=9x + y = 9
  • x+3y=15x + 3y = 15
  • Z=20Z = 20
  • Z=39Z = 39
Parallel lines $4x + 5y = k$: the last point touched is a corner.
Eight sarees and four stoles a week.
CornerZ=500x+300yZ = 500x + 300y (₹)
(0,0)(0, 0)00
(10,0)(10, 0)50005000
(8,4)(8, 4)52005200 (maximum)
(3,9)(3, 9)42004200
(0,10)(0, 10)30003000
  • 2x+y=102x + y = 10
  • x+y=7x + y = 7
  • x+2y=10x + 2y = 10
  • 5x+4y=315x + 4y = 31
Half-plane test: nothing of the region lies below $5x + 4y = 31$.
Unbounded region: confirm with the half-plane.
CornerZ=5x+4yZ = 5x + 4y (₹)
(0,10)(0, 10)4040
(3,4)(3, 4)3131 (least)
(4,3)(4, 3)3232
(10,0)(10, 0)5050

Formulas

  • Half-plane test
    max⁡Z on an unbounded region exists  ⟺  {ax+by>M}∩R=∅\max Z \text{ on an unbounded region exists} \iff \{ax + by > M\} \cap R = \varnothing

Key terms

Half-plane test

Checking whether the open half-plane beyond the best corner value meets an unbounded region.

6. Special cases

  • No optimum: on an unbounded region, a feasible point that beats the best corner value shows that the optimum does not exist in that direction (for example Z=x+2yZ = x + 2y on x+y≥8x + y \ge 8, x+3y≥12x + 3y \ge 12 has no maximum).
  • Infeasible: an empty region (for example x+y≤4x + y \le 4 with 3x+2y≥183x + 2y \ge 18) has no solution.
  • Infinitely many optima: if ZZ has the same best value at two adjacent corners, every point of the edge between them is optimal; the objective line is parallel to that edge.
Five outcomes to recognise.
RegionObjectiveOutcome
Boundedany linear ZZMaximum and minimum both exist, at corners
Unboundedbest corner passes the half-plane testThat corner value is the optimum
Unboundeda feasible point beats the best cornerNo optimum in that direction
EmptyanyNo feasible solution
Anyequal best value at two adjacent cornersInfinitely many optimal solutions
One feasible point beyond the best corner is enough to prove there is no maximum.
CornerZ=x+2yZ = x + 2y
(0,8)(0, 8)1616
(6,2)(6, 2)1010
(12,0)(12, 0)1212
(0,20)(0, 20) (feasible, not a corner)40>1640 > 16
  • x+y=12x + y = 12
  • 2x+y=202x + y = 20
  • x+3y=30x + 3y = 30
  • Z=12Z = 12
The objective line is parallel to an edge: infinitely many optima.

Key terms

Multiple optimal solutions

The best value of Z is reached along a whole edge of the region.

Exam tips

How ISC marks are awarded

  • Definitions, advantages, limitations (2 marks): one mark per correct point.
  • Formulation (2–3 marks): variables and objective, constraints, non-negativity.
  • Solve graphically (3 marks): region, corner values, optimum.
  • Section D (5 marks): formulation, corners, values and the answer in the words of the problem.

Common mistakes

  • Reversing "at least" and "at most".
  • Missing the corner where two slanted lines meet.
  • Claiming a maximum on an unbounded region without the half-plane test.
  • Forgetting non-negativity, so that corners with negative coordinates are used.