B - 1© 2011 Pearson Education, Inc. publishing as Prentice Hall

BB Linear ProgrammingLinear Programming

PowerPoint presentation to accompany Heizer and Render Operations Management, 10e Principles of Operations Management, 8e

PowerPoint slides by Jeff Heyl

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Outline Requirements of a Linear Programming Problem Formulating Linear Programming Problems

Shader Electronics Example

Graphical Solution to a Linear Programming Problem Graphical Representation of Constraints Iso-Profit Line Solution Method Corner-Point Solution Method

Solving Minimization Problems Linear Programming Applications

Production-Mix Example Diet Problem Example

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Learning Objectives

1. Formulate linear programming models, including an objective function and constraints

2. Graphically solve an LP problem with the iso-profit line method

3. Graphically solve an LP problem with the corner-point method

4. Construct and solve a minimization problem

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Why Use Linear Programming?

A mathematical technique to help plan and make decisions relative to the trade-offs necessary to allocate resources

Will find the minimum or maximum value of the objective

Guarantees the optimal solution to the model formulated

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LP Applications

1. Scheduling school buses to minimize total distance traveled

2. Allocating police patrol units to high crime areas in order to minimize response time to 911 calls

3. Scheduling tellers at banks so that needs are met during each hour of the day while minimizing the total cost of labor

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LP Applications

4. Selecting the product mix in a factory to make best use of machine- and labor-hours available while maximizing the firm’s profit

5. Picking blends of raw materials in feed mills to produce finished feed combinations at minimum costs

6. Determining the distribution system that will minimize total shipping cost

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Requirements of an LP Problem

1. LP problems seek to maximize or minimize some quantity

2. The presence of restrictions, or constraints

3. There must be alternative courses of action to choose from

4. The objective and constraints in linear programming problems must be expressed in terms of linear equations or inequalities

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Formulating LP Problems

The product-mix problem at Shader Electronics

Two products

1. Shader x-pod, a portable music player

2. Shader BlueBerry, an internet-connected color telephone

Determine the mix of products that will produce the maximum profit

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Formulating LP Problems

x-pods BlueBerrys Available HoursDepartment (X1) (X2) This Week

Hours Required to Produce 1 Unit

Electronic 4 3 240

Assembly 2 1 100

Profit per unit $7 $5

Decision Variables:X1 = number of x-pods to be producedX2 = number of BlueBerrys to be produced

Table B.1

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Formulating LP Problems

Objective Function:

Maximize Profit = $7X1 + $5X2

There are three types of constraints Upper limits where the amount used is ≤

the amount of a resource Lower limits where the amount used is ≥

the amount of the resource Equalities where the amount used is =

the amount of the resource

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Formulating LP Problems

Second Constraint:

2X1 + 1X2 ≤ 100 (hours of assembly time)

Assemblytime available

Assemblytime used is ≤

First Constraint:

4X1 + 3X2 ≤ 240 (hours of electronic time)

Electronictime available

Electronictime used is ≤

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Graphical Solution Can be used when there are two

decision variables1. Plot the constraint equations at their

limits by converting each equation to an equality

2. Identify the feasible solution space

3. Create an iso-profit line based on the objective function

4. Move this line outwards until the optimal point is identified

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Graphical Solution

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

–| | | | | | | | | | |

0 20 40 60 80 100

Nu

mb

er o

f B

lueB

erry

s

Number of x-pods

X1

X2

Assembly (Constraint B)

Electronics (Constraint A)Feasible region

Figure B.3

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Graphical Solution

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

–| | | | | | | | | | |

0 20 40 60 80 100

Nu

mb

er o

f B

lueB

erry

s

Number of x-pods

X1

X2

Assembly (Constraint B)

Electronics (Constraint A)Feasible region

Figure B.3

Iso-Profit Line Solution Method

Choose a possible value for the objective function

$210 = 7X1 + 5X2

Solve for the axis intercepts of the function and plot the line

X2 = 42 X1 = 30

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Graphical Solution

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

–| | | | | | | | | | |

0 20 40 60 80 100

Nu

mb

er o

f B

lueB

erry

s

Number of x-pods

X1

X2

Figure B.4

(0, 42)

(30, 0)

$210 = $7X1 + $5X2

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Graphical Solution

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

–| | | | | | | | | | |

0 20 40 60 80 100

Nu

mb

er o

f B

lueB

erry

s

Number of x-pods

X1

X2

Figure B.5

$210 = $7X1 + $5X2

$420 = $7X1 + $5X2

$350 = $7X1 + $5X2

$280 = $7X1 + $5X2

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Graphical Solution

100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

–| | | | | | | | | | |

0 20 40 60 80 100

Nu

mb

er o

f B

lueB

erry

s

Number of x-pods

X1

X2

Figure B.6

$410 = $7X1 + $5X2

Maximum profit line

Optimal solution point(X1 = 30, X2 = 40)

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100 –

–

80 –

–

60 –

–

40 –

–

20 –

–

–| | | | | | | | | | |

0 20 40 60 80 100

Nu

mb

er o

f B

lueB

erry

s

Number of x-pods

X1

X2

Corner-Point Method

Figure B.7

1

2

3

4

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Corner-Point Method

The optimal value will always be at a corner point

Find the objective function value at each corner point and choose the one with the highest profit

Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0

Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400

Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

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Corner-Point Method

The optimal value will always be at a corner point

Find the objective function value at each corner point and choose the one with the highest profit

Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0

Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400

Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Solve for the intersection of two constraints

2X1 + 1X2 ≤ 100 (assembly time)

4X1 + 3X2 ≤ 240 (electronics time)

4X1 + 3X2 = 240

- 4X1 - 2X2 = -200

+ 1X2 = 40

4X1 + 3(40) = 240

4X1 + 120 = 240

X1 = 30

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Corner-Point Method

The optimal value will always be at a corner point

Find the objective function value at each corner point and choose the one with the highest profit

Point 1 : (X1 = 0, X2 = 0) Profit $7(0) + $5(0) = $0

Point 2 : (X1 = 0, X2 = 80) Profit $7(0) + $5(80) = $400

Point 4 : (X1 = 50, X2 = 0) Profit $7(50) + $5(0) = $350

Point 3 : (X1 = 30, X2 = 40) Profit $7(30) + $5(40) = $410

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Solving Minimization Problems

Formulated and solved in much the same way as maximization problems

In the graphical approach an iso-cost line is used

The objective is to move the iso-cost line inwards until it reaches the lowest cost corner point

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Minimization Example

X1 = number of tons of black-and-white picture chemical produced

X2 = number of tons of color picture chemical produced

Minimize total cost = 2,500X1 + 3,000X2

Subject to:X1 ≥ 30 tons of black-and-white chemical

X2 ≥ 20 tons of color chemical

X1 + X2 ≥ 60 tons total

X1, X2 ≥ $0 nonnegativity requirements

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Minimization ExampleTable B.9

60 –

50 –

40 –

30 –

20 –

10 –

–| | | | | | |

0 10 20 30 40 50 60X1

X2

Feasible region

X1 = 30X2 = 20

X1 + X2 = 60

b

a

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Minimization Example

Total cost at a = 2,500X1 + 3,000X2

= 2,500 (40) + 3,000(20)= $160,000

Total cost at b = 2,500X1 + 3,000X2

= 2,500 (30) + 3,000(30)= $165,000

Lowest total cost is at point a

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LP ApplicationsProduction-Mix Example

Department

Product Wiring Drilling Assembly Inspection Unit Profit

XJ201 .5 3 2 .5 $ 9XM897 1.5 1 4 1.0 $12TR29 1.5 2 1 .5 $15BR788 1.0 3 2 .5 $11

Capacity MinimumDepartment (in hours) Product Production Level

Wiring 1,500 XJ201 150Drilling 2,350 XM897 100Assembly 2,600 TR29 300Inspection 1,200 BR788 400

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LP ApplicationsX1 = number of units of XJ201 produced

X2 = number of units of XM897 produced

X3 = number of units of TR29 produced

X4 = number of units of BR788 produced

Maximize profit = 9X1 + 12X2 + 15X3 + 11X4

subject to .5X1 + 1.5X2 + 1.5X3 + 1X4 ≤ 1,500 hours of wiring

3X1 + 1X2 + 2X3 + 3X4 ≤ 2,350 hours of drilling

2X1 + 4X2 + 1X3 + 2X4 ≤ 2,600 hours of assembly

.5X1 + 1X2 + .5X3 + .5X4 ≤ 1,200 hours of inspection

X1 ≥ 150 units of XJ201

X2 ≥ 100 units of XM897

X3 ≥ 300 units of TR29

X4 ≥ 400 units of BR788

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LP ApplicationsDiet Problem Example

A 3 oz 2 oz 4 ozB 2 oz 3 oz 1 ozC 1 oz 0 oz 2 ozD 6 oz 8 oz 4 oz

Feed

Product Stock X Stock Y Stock Z

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LP ApplicationsX1 = number of pounds of stock X purchased per cow each month

X2 = number of pounds of stock Y purchased per cow each month

X3 = number of pounds of stock Z purchased per cow each month

Minimize cost = .02X1 + .04X2 + .025X3

Ingredient A requirement: 3X1 + 2X2 + 4X3 ≥ 64

Ingredient B requirement: 2X1 + 3X2 + 1X3 ≥ 80

Ingredient C requirement: 1X1 + 0X2 + 2X3 ≥ 16

Ingredient D requirement: 6X1 + 8X2 + 4X3 ≥ 128

Stock Z limitation: X3 ≤ 80

X1, X2, X3 ≥ 0Cheapest solution is to purchase 40 pounds of grain X

at a cost of $0.80 per cow

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In-Class Problems from the Lecture Guide Practice Problems

Problem 1:Chad’s Pottery Barn has enough clay to make 24 small vases or 6 large vases. He has only enough of a special glazing compound to glaze 16 of the small vases or 8 of the large vases. Let X1 = the number of small vases and X2 = the number of large vases. The smaller vases sell for $3 each, and the larger vases would bring $9 each.(a) Formulate the problem(b) Solve the problem graphically

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In-Class Problems from the Lecture Guide Practice Problems

Problem 2:A fabric firm has received an order for cloth specified to contain at least 45 pounds of cotton and 25 pounds of silk. The cloth can be woven out of any suitable mix of two yarns A and B. They contain the proportions of cotton and silk (by weight) as shown in the following table:

Material A costs $3 per pound, and B costs $2 per pound. What quantities (pounds) of A and B yarns should be used to minimize the cost of this order?

  Cotton SilkA 30% 50%B 60% 10%

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