linear programming chapter 13 supplement. pottery example beaver creek pottery company is located on...

Linear Programming

Chapter 13 Supplement

Pottery Example

Beaver Creek Pottery Company is located on a Native American reservation. Each day, the company has available 40 hours of labor and 120 pounds of clay. The firm makes two products, bowls and mugs. A bowl requires 1 hour of labor and 4 pounds of clay. A mug requires 2 hours of labor and 3 pounds of clay. The firm's profit is $40 per bowl and $50 per mug. The company wants to maximize profit.

Pottery Example

Business Objective: Determine

• Number of bowls to make

• Number of mugs to make

to maximize profit.

Decision variables:

x = number of bowls to make

y = number of mugs to make

Objective Function

• Profit• $40 per bowl (x)• $50 per mug (y)

• Maximize Z = 40x + 50y

Constraint Table

Constrained Quantity

Bowls x

Mugsy Type

Max. orMinimum

Labor 1

4

2

3

<

<

40

120Clay

Constraints: x + 2y < 40 4x + 3y < 120

Nonnegativity Constraints

• We cannot make a negative amount of either product

• Add nonnegativity constraints

x > 0

y > 0

Problem Statement

Maximize Z = 40x + 50y

subject to

x + 2y < 40

4x + 3y < 120

x > 0

y > 0

Linear Programming Terminology

• Feasible solution: Any solution which satisfies all constraints.

• Feasible region: Set of points which satisfy all constraints.

• Optimal solution(s): A point which • Satisfies all constraints• Maximizes or minimizes the value of the

objective function.

Solving a Linear ProgrammingProblem by the Graphical Method

(1) Set up the problem:• Decision variables and their definitions.• Write objective function and state whether it

should be maximized or minimized.• Write the constraints as mathematical

inequalities or equalities.

(2) Draw the feasible region.

Solving a Linear ProgrammingProblem by the Graphical Method(2)

(3) Determine which point(s) in the feasible

region give an optimal solution• Point(s) which maximize or minimize the

objective function.

• Note: To use the graphical method, the problem must have only 2 variables.

To Draw the Feasible Region

• Convert each constraint into an equation.

• Draw the corresponding line.

• The set of points bounded by these lines is the feasible region.

Problem Statement

Maximize Z = 40x + 50y

subject to

x + 2y < 40

4x + 3y < 120

x > 0

y > 0

Feasible Region for This Problem

This region is bounded by the lines

x + 2y = 40 (Labor)

4x + 3y = 120 (Clay)

x = 0

y = 0

Plot x + 2y = 40

• x-intercept: Set y = 0.

x + 0 =40 ==> x = 40.

Point is (40,0)

• y-intercept: Set x=0.

(0)+2y=40 ==> y = 20.

Point is (0,20)

Bowls (x)10 20 30

10

20

30

40

Labor

Mu

gs

(y) Clay

FeasibleRegion

Points in feasibleregion satisfyall constraints.

40

Iso-Profit Lines

A set of points on which the objective function is constant

Example: The set of points satisfying

40x + 50y = 800

x-intercept: (20,0)

y-intercept: (0,16)

Bowls10 20 30

10

20

30

40

Labor

Mu

gs

Clay

40

Iso-profit Line

Profit = $800

Bowls10 20 30

10

20

30

40

Labor

Mu

gs

Clay

40

Iso-profit Linesfor profits of $800 and $1,000

Bowls10 20 30

10

20

30

40

Labor

Mu

gs

Clay

40

Iso-profit Lines

Optimal Point=(24,8)

Optimal Solution

• Satisfies x + 2y = 40 and 4x + 3y = 120

• Solution is (24,8): 24 bowls and 8 mugs

• Verification:

24 + 2(8) = 40

4(24) + 3(8) = 120

Maximum Profit

• Optimal solution is (24,8)

• Objective function is 40x + 50y

$40(24) + $50(8) = $1,360

• Maximum profit is $1,360

Fundamental Theorem of Linear Programming

In a linear programming problem, the optimal solution occurs at an extreme point (vertex or corner point) of the

feasible region.

Objective of Linear Programming

• Maximize the use of resources (or minimize costs) to achieve competitive priorities.

• Optimum solution is a base case which must be adjusted to reflect business realities.

Ch 11 Supp - 2© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e

Linear Programming Model

• Decision variables are mathematical symbols representing activity levels.

• Objective function is a mathematical, linear function which represents the organization’s objectives.• Used to compare alternative courses of

action.

Ch 11 Supp - 2© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e

Linear Programming Model (2)

• Constraints are mathematical, linear relationships representing restrictions on decision making• Resource constraints.• Policy or legal constraints.• Sales constraints.

• Constraints may be <, =, or >.

Linear programming maximizes or minimizes the objective function

subject to constraints.

Uses of Linear Programming

• Production Scheduling• Maximize profit• Minimize cost

Different objectives yield different schedules.

• Determine product or service mix• Maximize profit• Maximize revenue• Minimize costs

Uses of Linear Programming (2)

• Scheduling labor in services• Minimize cost

• Production-location problem• Allocate products and customers to plants

to minimize total cost of production and distribution

Uses of Linear Programming (3)

• Distribution• Minimize cost

• Facility location• Minimize transportation cost.

• Emergency response systems• Minimize average response time.

Conditions to Use Linear Programming

• Objective function must be linear

• Constraints must be linear inequalities or linear equations

Methods for Solving LP Problems

• Graphical method: limited to 2 variables

• Any linear programming problem• Simplex method• Karmarkar’s algorithm

• Transportation method:• Facility location problems• Production-location problem: minimize total

cost of production and shipment to customers

Slack Variables

• Slack = unused amount of any resource or constrained quantity

• S1 = Amount of unused labor

= 40 - (x + 2y) = 40 - x - 2y = 40 - 24 - 8(2)

= 0. Optimum solution uses all labor.

• S2 = Amount of unused clay

= 120 - (4x + 3y) = 120 - 4x - 3y

= 120 - 4(24) - 3(8) = 0.

Optimum solution uses all clay.

Sensitivity Analysis (Ranging)

• Dual value or shadow price• Incremental increase or decrease in the

objective function if one more unit of a resource is added.

• The amount we would pay to get one more unit of a resource.

• Valid only for resources which have no slack.

Shadow Price for Labor

• Dual value = shadow price = $16.

• If labor hours increase from 40 to 41, profit increases from $1360 to $1376.

• Amount to buy

= upper bound - original value

= 80 hours - 40 hours = 40 hours

Bowls10 20 30

10

20

30

40

Labor

Mu

gs

Clay

40 80

NewLabor

Optimal point = (0,40). Profit = $2,000