linear programming: modeling...

4-1 Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Linear Programming:

Modeling Examples

Chapter 4

4-2

Chapter Topics

A Product Mix Example

A Diet Example

An Investment Example

A Marketing Example

A Transportation Example

A Blend Example

A Multiperiod Scheduling Example

A Data Envelopment Analysis Example

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

4-3


Problem 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?


4-4

A Product Mix Example (2 of 8)


4-5

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


Data (3 of 8)


4-6

Decision Variables:

x1 = sweatshirts, front printing

x2 = sweatshirts, back and front printing

x3 = T-shirts, front printing

x4 = 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


Model Construction (4 of 8)


4-7


Computer Solution with Excel (5 of 8)


Exhibit 4.1

4-8

Exhibit 4.2


Solution with Excel Solver Window (6 of 8)


4-9

Exhibit 4.3


Solution with QM for Windows (7 of 8)


4-10

Exhibit 4.4




4-11

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)


4-12

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 Example

Model Construction – Decision Variables (2 of 5)


4-13

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 calories

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

270x5 + 8x6 + 12x8 30 mg cholesterol

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

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

+ 3x9 + 26x10 400 mg of calcium

3x1 + 4x2 + 5x3 + 6x4 + 7x5 + 2x6 + x7

+ 9x8+ x9 + 3x10 20 g protein

5x1 + 2x2 + 3x3 + 4x4+ x7 + 3x10 12

xi 0, for all j

A Diet Example

Model Summary (3 of 5)


4-14

Exhibit 4.5

A Diet Example



4-15

Exhibit 4.6

A Diet Example



4-16

Maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4

subject to:

x1 $14,000

x2 - x1 - x3- x4 0

x2 + x3 $21,000

-1.2x1 + x2 + x3 - 1.2 x4 0

x1 + x2 + x3 + x4 = $70,000

x1, x2, x3, x4 0

where

x1 = amount ($) invested in municipal bonds

x2 = amount ($) invested in certificates of deposit

x3 = amount ($) invested in treasury bills

x4 = amount ($) invested in growth stock fund




4-17




Exhibit 4.7

4-18

Exhibit 4.8




4-19


Sensitivity Report (4 of 4)


Exhibit 4.9

4-20

Exposure (people/ad or commercial)

Cost

Television Commercial 20,000 $15,000

Radio Commercial 2,000 6,000

Newspaper Ad 9,000 4,000

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

A Marketing Example

Data and Problem Definition (1 of 6)


4-21

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 = number of television commercials

x2 = number of radio commercials

x3 = number of newspaper ads

A Marketing Example



4-22

Exhibit 4.10

A Marketing Example

Solution with Excel (3 of 6)


4-23

Exhibit 4.11

A Marketing Example



4-24

Exhibit 4.13

A Marketing Example

Integer Solution with Excel (5 of 6)


Exhibit 4.12

4-25

Exhibit 4.14

A Marketing Example

Integer Solution with Excel (6 of 6)


4-26

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


Problem Definition and Data (1 of 3)


4-27

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




4-28

Exhibit 4.15




4-29

Exhibit 4.16


Solution with Solver Window (4 of 4)


4-30

Component Maximum Barrels

Available/day Cost/barrel

1 4,500 $12

2 2,700 10

3 3,500 14

Grade Component Specifications Selling Price ($/bbl)

Super At least 50% of 1

Not more than 30% of 2 $23

Premium At least 40% of 1

Not more than 25% of 3

20

Extra At least 60% of 1 At least 10% of 2

18

A Blend Example



4-31

■ 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 Example

Problem Statement and Variables (2 of 6)


4-32

Maximize Z = 11x1s + 13x2s + 9x3s + 8x1p + 10x2p + 6x3p + 6x1e + 8x2e + 4x3e subject to: x1s + x1p + x1e 4,500 bbl. x2s + x2p + x2e 2,700 bbl. x3s + x3p + x3e 3,500 bbl. 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 bbl. x1p+ x2p + x3p 3,000 bbl. x1e+ x2e + x3e 3,000 bbl.

A Blend Example



all xij 0

4-33

Exhibit 4.17

A Blend Example



4-34

Exhibit 4.18

A Blend Example



4-35

Exhibit 4.19

A Blend Example

Sensitivity Report (6 of 6)


4-36

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:

A Multi-Period Scheduling Example



Week Computer Orders

1 1052 1703 2304 1805 1506 250

4-37

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)


Decision Variables (2 of 5)


4-38

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 computers in week j (j = 1, 2, 3, 4, 5, 6)

oj 150 computers in week j (j = 1, 2, 3, 4, 5, 6)

r1 + o1 - i1 = 105 week 1

r2 + o2 + i1 - i2 = 170 week 2

r3 + o3 + i2 - i3 = 230 week 3

r4 + o4 + i3 - i4 = 180 week 4

r5 + o5 + i4 - i5 = 150 week 5

r6 + o6 + i5 = 250 week 6

rj, oj, ij 0




4-39




Exhibit 4.20

4-40

Exhibit 4.21




4-41

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

Input 2 = supplementary funds/student

Input 3 = average educational level of parents

Output 1 = average reading SOL score

Output 2 = average math SOL score

Output 3 = average history SOL score

A Data Envelopment Analysis (DEA) Example

Problem Definition (1 of 5)


4-42

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


Problem Data Summary (2 of 5)


4-43

Decision Variables:

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

yi = 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


Decision Variables and Model Summary (3 of 5)


4-44

Exhibit 4.22




4-45

Exhibit 4.23




4-46

Example Problem Solution

Problem 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?


4-47

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)


Model Formulation (2 of 5)


4-48

Step 3: Formulate the Model Constraints

Amount of each ingredient available each week:

xhm + xhb 9,600 ounces of horse meat

xfm + xfb 12,800 ounces of fish

xcm + xcb 16,000 ounces of cereal additive

Recipe requirements:

Meow Chow: xfm/(xhm + xfm + xcm) 1/2 or - xhm + xfm- xcm 0

Bow Chow: xhb/(xhb + xfb + xcb) 1/2 or xhb- xfb - xcb 0

Can Content: xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces


Model Formulation (3 of 5)


4-49

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 meat

xfm + xfb 12,800 ounces of fish

xcm + xcb 16,000 ounces of cereal additive

- xhm + xfm- xcm 0

xhb- xfb - xcb 0

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

xij 0




4-50




linear programming: modeling...

Documents