Business MathematicsBusiness Mathematicswww.uni-corvinus.hu/~u2w6ol

Rétallér Orsi

Graphical solutionGraphical solution

The problemThe problem

max z = 3x1 + 2x2

2x1 + x2 ≤ 100

x1 + x2 ≤ 80

x1 ≤ 40

x1 ≥ 0

x2 ≥ 0

Graphical solutionGraphical solution

Feasible region

Is there always one Is there always one solution?solution?

Possible LP solutionsPossible LP solutions

One optimumAlternative optimums (Infinite

solutions)InfeasibilityUnboundedness

Possible LP solutionsPossible LP solutions

One optimumAlternative optimums (Infinite

solutions)InfeasibilityUnboundedness

Possible LP solutionsPossible LP solutions

One optimumAlternative optimums (Infinite

solutions)InfeasibilityUnboundedness

Alternative optimumAlternative optimum

max z = 4x1 + x2

8x1 + 2x2 ≤ 16

5x1 + 2x2 ≤ 12

x1 ≥ 0

x2 ≥ 0

Alternative optimumAlternative optimum

Possible LP solutionsPossible LP solutions

One optimumAlternative optimums (Infinite

solutions)InfeasibilityUnboundedness

InfeasibilityInfeasibility

max z = x1 + x2

x1 + x2 ≤ 4

x1 - x2 ≥ 5

x1 ≥ 0

x2 ≥ 0

InfeasibilityInfeasibility

Possible LP solutionsPossible LP solutions

One optimumAlternative optimums (Infinite

solutions)InfeasibilityUnboundedness

UnboundednessUnboundedness

max z = -x1 + 3x2

x1 - x2 ≤ 4

x1 + 2x2 ≥ 4

x1 ≥ 0

x2 ≥ 0

UnboundednessUnboundedness

Sensitivity analysisSensitivity analysis

Sensitivity analysisSensitivity analysis

When is the yellow point the optimal solution?

Sensitivity analysisSensitivity analysis

The problemThe problem

max z = 3x1 + 2x2

2x1 + x2 ≤ 100

x1 + x2 ≤ 80

x1 ≤ 40

x1 ≥ 0

x2 ≥ 0

2x1 + x2 = 100

x1 + x2 = 80

Sensitivity analysisSensitivity analysis

2x1 + x2 = 100

x1 + x2 = 80Range of optimality:

[1;2]

Duality theoremDuality theorem

Problem – WinstonProblem – Winston

The Dakota Furniture Company manufactures desks, tables, and chairs. The manufacture of each type of furniture requires lumber and two types of skilled labor: finishing and carpentry. The amount of each resource needed to make each type of furniture is given in the following table.

Resource Desk Table Chair

Lumber (board ft)

8 6 1

Finishing(hours)

4 2 1,5

Carpentry(hours)

2 1,5 0,5

Problem – WinstonProblem – Winston

Problem – WinstonProblem – Winston

At present, 48 board feet of lumber, 20 finishing hours, and 8 carpentry hours are available. A desk sells for $60, a table for $30, and a chair for $20. Since the available resources have already been purchased, Dakota wants to maximize total revenue.

Formalizing the problemFormalizing the problem

8x1 + 6x2 + 1x3 ≤ 48

4x1 + 2x2 + 1,5x3 ≤ 20

2x1 + 1,5x2 + 0,5x3 ≤ 8

x1, x2, x3≥ 0

max z = 60x1 + 30x2 + 20x3

The new problemThe new problem

For how much could a company buy all the resources of the Dakota company?

(Dual task)

The prices for the resources are indicated as y1, y2, y3

Resource Desk Table Chair

Lumber (board ft)

8 6 1

Finishing(hours)

4 2 1,5

Carpentry(hours)

2 1,5 0,5

Problem – WinstonProblem – Winston

The primal problemThe primal problem

8x1 + 6x2 + 1x3 ≤ 48

4x1 + 2x2 + 1,5x3 ≤ 20

2x1 + 1,5x2 + 0,5x3 ≤ 8

x1, x2, x3≥ 0

max z = 60x1 + 30x2 + 20x3

The dual problemThe dual problem

min w = 48y1 + 20y2 + 8y3

Resource Desk Table Chair

Lumber (board ft)

8 6 1

Finishing(hours)

4 2 1,5

Carpentry(hours)

2 1,5 0,5

Problem – WinstonProblem – Winston

The primal problemThe primal problem

8x1 + 6x2 + 1x3 ≤ 48

4x1 + 2x2 + 1,5x3 ≤ 20

2x1 + 1,5x2 + 0,5x3 ≤ 8

x1, x2, x3≥ 0

max z = 60x1 + 30x2 + 20x3

The dual problemThe dual problem

min w = 48y1 + 20y2 + 8y3

8y1 + 4y2 + 2y3 ≥ 60

The dual problemThe dual problem

8y1 + 4y2 + 2y3 ≥ 60

6y1 + 2y2 + 1,5y3 ≥ 30

1y1 + 1,5y2 + 0,5y3 ≥ 20

y1, y2, y3 ≥ 0

min w = 48y1 + 20y2 + 8y3

Traditional minimum taskTraditional minimum task

2x1 + 3x2 ≥ 2

2x1 + x2 ≥ 4

x1 – x2 ≥ 6

x1, x2 ≥ 0

min z = 5x1 + 2x2

2y1 + 2y2 + y3 ≤ 5

3y1 + y2 – y3 ≤ 2

y1, y2, y3 ≥ 0

max w = 2y1 + 4y2 + 6y3

Traditional minimum taskTraditional minimum task

2x1 + 3x2 ≥ 2

2x1 + x2 ≥ 4

x1 – x2 ≥ 6

x1, x2 ≥ 0

min z = 5x1 + 2x2

2y1 + 2y2 + y3 ≤ 5

3y1 + y2 – y3 ≤ 2

y1, y2, y3 ≥ 0

max w = 2y1 + 4y2 + 6y3

A little help for dualityA little help for duality

Nontraditional minimum Nontraditional minimum tasktask

x1 + 2x2 + x3 ≥ 2

x1 – x3 ≥ 1

x2 + x3 = 1

2x1 + x2 ≤ 3

x1 ur, x2, x3 ≥ 0

min z = 2x1 + 4x2 + 6x3

Nontraditional minimum Nontraditional minimum tasktask

y1 + y2 + y4 = 2

2y1 + y3 + y4 ≤ 4

y1 – y2 + y3 ≤ 6

y1, y2 ≥ 0, y3 ur, y4 ≤ 0

max w = 2y1 + y2 + y3 + 3y4

Thank you for your Thank you for your attention!attention!