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Chapter 4

Linear Programming Modelingand computer solution

Decision Science

Chapter Topics2

A Product Mix Example

A Diet Example

An Investment Example

A Marketing Example

A Transportation Example

A Blend Example

A Multi-Period Scheduling Example

A Data Envelopment Analysis Example

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A Product Mix ExampleProblem Definition (1 of 8) Four-product T-shirt/sweatshirt manufacturing company.

Must complete production within 72 hours

Truck capacity = 1,200 standard sized boxes.

Standard size box holds12 T-shirts.

One-dozen sweatshirts box is three times size of standard box.

$25,000 available for a production run.

500 dozen blank T-shirts and sweatshirts in stock.

How many dozens (boxes) of each type of shirt to produce?

Chapter 4 - Linear Programming:

Modeling Examples

4A Product Mix Example (2 of 8)

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Processing Time (hr) Per dozen

Cost ($) per

dozen

Profit ($) per

dozen

Sweatshirt - F 0.10 36 90

Sweatshirt B/F

0.25 48 125

T-shirt - F 0.08 25 45

T-shirt - B/F 0.21 35 65

A Product Mix ExampleData (3 of 8)

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Decision Variables:x1 = sweatshirts, front printingx2 = sweatshirts, back and front printingx3 = T-shirts, front printingx4 = T-shirts, back and front printing

Objective Function: Maximize Z = $90x1 + $125x2 + $45x3 + $65x4

Model Constraints:

0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72 hr 3x1 + 3x2 + x3 + x4 1,200 boxes

$36x1 + $48x2 + $25x3 + $35x4 $25,000 x1 + x2 500 dozen sweatshirts

x3 + x4 500 dozen T-shirts

A Product Mix ExampleModel Construction (4 of 8)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.1

A Product Mix ExampleComputer Solution with Excel (5 of 8)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.2

A Product Mix ExampleSolution with Excel Solver Window (6 of 8)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.3

A Product Mix ExampleSolution with QM for Windows (7 of 8)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.4

A Product Mix ExampleSolution with QM for Windows (8 of 8)

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Breakfast to include at least 420 calories, 5 milligrams of iron, 400 milligrams of calcium, 20 grams of protein, 12 grams of fiber, and must have no more than 20 grams of fat and 30 milligrams of cholesterol.

Breakfast Food Cal

Fat (g)

Cholesterol (mg)

Iron (mg)

Calcium (mg)

Protein (g)

Fiber (g)

Cost ($)

1. Bran cereal (cup) 2. Dry cereal (cup) 3. Oatmeal (cup) 4. Oat bran (cup) 5. Egg 6. Bacon (slice) 7. Orange 8. Milk-2% (cup) 9. Orange juice (cup)

10. Wheat toast (slice)

90 110 100

90 75 35 65

100 120

65

0 2 2 2 5 3 0 4 0 1

0 0 0 0

270 8 0

12 0 0

6 4 2 3 1 0 1 0 0 1

20 48 12

8 30

0 52

250 3

26

3 4 5 6 7 2 1 9 1 3

5 2 3 4 0 0 1 0 0 3

0.18 0.22 0.10 0.12 0.10 0.09 0.40 0.16 0.50 0.07

A Diet Example Data and Problem Definition (1 of 5)

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x1 = cups of bran cereal

x2 = cups of dry cereal

x3 = cups of oatmeal

x4 = cups of oat bran

x5 = eggs

x6 = slices of bacon

x7 = oranges

x8 = cups of milk

x9 = cups of orange juice

x10 = slices of wheat toast

A Diet ExampleModel Construction – Decision Variables (2 of 5)

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Minimize Z = 0.18x1 + 0.22x2 + 0.10x3 + 0.12x4 + 0.10x5 + 0.09x6

+ 0.40x7 + 0.16x8 + 0.50x9 + 0.07x10

subject to:90x1 + 110x2 + 100x3 + 90x4 + 75x5 + 35x6 + 65x7

+ 100x8 + 120x9 + 65x10 420

2x2 + 2x3 + 2x4 + 5x5 + 3x6 + 4x8 + x10 20

270x5 + 8x6 + 12x8 30

6x1 + 4x2 + 2x3 + 3x4+ x5 + x7 + x10 5

20x1 + 48x2 + 12x3 + 8x4+ 30x5 + 52x7 + 250x8

+ 3x9 + 26x10 400

3x1 + 4x2 + 5x3 + 6x4 + 7x5 + 2x6 + x7+ 9x8+ x9 + 3x10 20

5x1 + 2x2 + 3x3 + 4x4+ x7 + 3x10 12 xi 0, for all j

A Diet Example Model Summary (3 of 5)

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Exhibit 4.5

A Diet ExampleComputer Solution with Excel (4 of 5)

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Exhibit 4.6

A Diet Example Solution with Excel Solver Window (5 of 5)

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Budget limit $100,000

Television time for four commercials

Radio time for 10 commercials

Newspaper space for 7 ads

Resources for no more than 15 commercials and/or ads

Exposure (people/ad or commercial)

Cost

Television Commercial

20,000 $15,000

Radio Commercial

12,000 6,000

Newspaper Ad 9,000 4,000

A Marketing Example Data and Problem Definition (1 of 6)

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Maximize Z = 20,000x1 + 12,000x2 + 9,000x3

subject to:15,000x1 + 6,000x 2+ 4,000x3 100,000

x1 4 x2 10 x3 7 x1 + x2 + x3 15 x1, x2, x3 0 where x1 = Exposure from Television Commercial (people) x2 = Exposure from Radio Commercial (people) x3 = Exposure from Newspaper Ad (people)

A Marketing Example Model Summary (2 of 6)

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A Marketing Example Solution with Excel (3 of 6)

Chapter 4 - Linear Programming:

Modeling Examples

19A Marketing ExampleSolution with Excel Solver Window (4 of 6)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.12

Exhibit 4.13

A Marketing ExampleInteger Solution with Excel (5 of 6)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.14

A Marketing ExampleInteger Solution with Excel (6 of 6)

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Warehouse supply of Retail store demand Television Sets: for television sets:

1 - Cincinnati 300 A - New York 150

2 - Atlanta 200 B - Dallas 250

3 - Pittsburgh 200 C - Detroit 200

Total 700 Total 600

Unit Shipping Costs:

From Warehouse To Store

A B C

1 $16 $18 $11

2 14 12 13

3 13 15 17

A Transportation ExampleProblem Definition and Data (1 of 3)

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Minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C

subject to:x1A + x1B+ x1C 300

x2A+ x2B + x2C 200

x3A+ x3B + x3C 200

x1A + x2A + x3A = 150

x1B + x2B + x3B = 250

x1C + x2C + x3C = 200

All xij 0

A Transportation ExampleModel Summary (2 of 4)

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Modeling Examples

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Exhibit 4.15

A Transportation ExampleSolution with Excel (3 of 4)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.16

A Transportation ExampleSolution with Solver Window (4 of 4)

Chapter 4 - Linear Programming:

Modeling Examples

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Determine the optimal mix of the three components in each grade of motor oil that will maximize profit. Company wants to produce at least 3,000 barrels of each grade of motor oil.

Decision variables: The quantity of each of the three components used in each grade of gasoline (9 decision variables); xij = barrels of component i used in motor oil grade j per day, where i = 1, 2, 3 and j = s (super), p (premium), and e (extra).

A Blend ExampleProblem Statement and Variables (2 of 6)

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Maximize Z = 11x1s + 13x2s + 9x3s + 8x1p + 10x2p + 6x3p + 6x1e + 8x2e + 4x3e

subject to: x1s + x1p + x1e 4,500 x2s + x2p + x2e 2,700 x3s + x3p + x3e 3,500 0.50x1s - 0.50x2s - 0.50x3s 0 0.70x2s - 0.30x1s - 0.30x3s 0 0.60x1p - 0.40x2p - 0.40x3p 0 0.75x3p - 0.25x1p - 0.25x2p 0 0.40x1e- 0.60x2e- - 0.60x3e 0 0.90x2e - 0.10x1e - 0.10x3e 0 x1s + x2s + x3s 3,000 x1p+ x2p + x3p 3,000 x1e+ x2e + x3e 3,000 xij 0

A Blend ExampleModel Summary (3 of 6)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.17

A Blend ExampleSolution with Excel (4 of 6)

Chapter 4 - Linear Programming:

Modeling Examples

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Exhibit 4.18

A Blend ExampleSolution with Solver Window (5 of 6)

Chapter 4 - Linear Programming:

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Exhibit 4.19

A Blend ExampleSensitivity Report (6 of 6)

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Production Capacity: 160 computers per week 50 more computers with overtime

Assembly Costs: $190 per computer regular time; $260 per computer overtime

Inventory Holding Cost: $10/computer per week

Order schedule: Week Computer Orders 1 105 2 170 3 230 4 180 5 150 6 250

A Multi-Period Scheduling ExampleProblem Definition and Data (1 of 5)

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Decision Variables:

rj = regular production of computers in week j(j = 1, 2, …, 6)

oj = overtime production of computers in week j(j = 1, 2, …, 6)

ij = extra computers carried over as inventory in week j(j = 1, 2, …, 5)

A Multi-Period Scheduling Example Decision Variables (2 of 5)

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Model summary:

Minimize Z = $190(r1 + r2 + r3 + r4 + r5 + r6) + $260(o1 + o2 + o3

+ o4 + o5 +o6) + 10(i1 + i2 + i3 + i4 + i5)

subject to:

rj 160 (j = 1, 2, 3, 4, 5, 6)oj 150 (j = 1, 2, 3, 4, 5, 6)r1 + o1 - i1 = 105r2 + o2 + i1 - i2 = 170r3 + o3 + i2 - i3 = 230 r4 + o4 + i3 - i4 = 180r5 + o5 + i4 - i5 = 150r6 + o6 + i5 = 250rj, oj, ij 0

A Multi-Period Scheduling Example Model Summary (3 of 5)

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Exhibit 4.20

A Multi-Period Scheduling ExampleSolution with Excel (4 of 5)

Chapter 4 - Linear Programming:

Modeling Examples

35A Multi-Period Scheduling ExampleSolution with Solver Window (5 of 5)

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DEA compares a number of service units of the same type based on their inputs (resources) and outputs. The result indicates if a particular unit is less productive, or efficient, than other units.

Elementary school comparison:

Input 1 = teacher to student ratio Output 1 = average reading SOL score

Input 2 = supplementary $/student Output 2 = average math SOL score

Input 3 = parent education level Output 3 = average history SOL score

A Data Envelopment Analysis (DEA) ExampleProblem Definition (1 of 5)

Chapter 4 - Linear Programming:

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Inputs Outputs

School 1 2 3 1 2 3

Alton .06 $260 11.3 86 75 71

Beeks .05 320 10.5 82 72 67

Carey

.08

340

12.0

81

79

80

Delancey

.06

460

13.1

81

73

69

A Data Envelopment Analysis (DEA) ExampleProblem Data Summary (2 of 5)

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Decision Variables:

xi = a price per unit of each output where i = 1, 2, 3yi = a price per unit of each input where i = 1, 2, 3

Model Summary:

Maximize Z = 81x1 + 73x2 + 69x3

subject to: .06 y1 + 460y2 + 13.1y3 = 1 86x1 + 75x2 + 71x3 .06y1 + 260y2 + 11.3y3

82x1 + 72x2 + 67x3 .05y1 + 320y2 + 10.5y3

81x1 + 79x2 + 80x3 .08y1 + 340y2 + 12.0y3

81x1 + 73x2 + 69x3 .06y1 + 460y2 + 13.1y3

xi, yi 0

A Data Envelopment Analysis (DEA) ExampleDecision Variables and Model Summary (3 of 5)

39A Data Envelopment Analysis (DEA) ExampleSolution with Excel (4 of 5)

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A Data Envelopment Analysis (DEA) ExampleSolution with Solver Window (5 of 5)

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Example Problem SolutionProblem Statement and Data (1 of 5) Canned cat food, Meow Chow; dog food, Bow Chow.

Ingredients/week: 600 lb horse meat; 800 lb fish; 1000 lb cereal.

Recipe requirement: Meow Chow at least half fish; Bow Chow at least half horse meat.

2,250 sixteen-ounce cans available each week.

Profit /can: Meow Chow $0.80; Bow Chow $0.96.

How many cans of Bow Chow and Meow Chow should be produced each week in order to maximize profit?

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Step 1: Define the Decision Variables

xij = ounces of ingredient i in pet food j per week, where i = h (horse meat), f (fish) and c (cereal), and j = m (Meow chow) and b (Bow Chow).

Step 2: Formulate the Objective Function

Maximize Z = $0.05(xhm + xfm + xcm) + 0.06(xhb + xfb + xcb)

Example Problem SolutionModel Formulation (2 of 5)

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Step 3: Formulate the Model Constraints

Amount of each ingredient available each week:xhm + xhb 9,600 ounces of horse meatxfm + xfb 12,800 ounces of fishxcm + xcb 16,000 ounces of cereal additive

Recipe requirements:Meow Chowxfm/(xhm + xfm + xcm) 1/2 or - xhm + xfm- xcm 0Bow Chow

xhb/(xhb + xfb + xcb) 1/2 or xhb- xfb - xcb 0Can Content Constraint

xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces

Example Problem SolutionModel Formulation (3 of 5)

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Step 4: Model Summary

Maximize Z = $0.05xhm + $0.05xfm + $0.05xcm + $0.06xhb + 0.06xfb + 0.06xcb

subject to:xhm + xhb 9,600 ounces of horse meatxfm + xfb 12,800 ounces of fishxcm + xcb 16,000 ounces of cereal additive- xhm + xfm- xcm 0 xhb- xfb - xcb 0xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces

xij 0

Example Problem SolutionModel Summary (4 of 5)

Chapter 4 - Linear Programming:

Modeling Examples

45Example Problem SolutionSolution with QM for Windows (5 of 5)