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## CH_04_LP_MODELING_EXAMPLE

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# CH_04_LP_MODELING_EXAMPLE

### CH_04_LP_MODELING_EXAMPLE

Linear Programming:
Modeling Examples

Chapter 4 4-1

Chapter Topics 4-2

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

A Product Mix Example
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?

A Product Mix Example (2 of 8)

A Product Mix Example
Data (3 of 8)

Processing Cost Profit

Time (hr) (\$) (\$)

Per dozen per dozen 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 Example 4-6
Model Construction (4 of 8)

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

A Product Mix Example
Computer Solution with Excel (5 of 8)

Exhibit 4.1 4-7

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

Exhibit 4.2 4-8

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

Exhibit 4.3

A Product Mix Example
Solution with QM for Windows (8 of 8)

Exhibit 4.4

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

Breakfast Food Fat Cholesterol Iron Calcium Protein Fiber Cost

Cal (g) (mg) (mg) (mg) (g) (g) (\$)

1. Bran cereal (cup) 90 0 0 6 20 3 5 0.18

2. Dry cereal (cup) 110 2 0 4 48 4 2 0.22

3. Oatmeal (cup) 100 2 0 2 12 5 3 0.10

4. Oat bran (cup) 90 2 0 3 86 4 0.12

5. Egg 75 5 270 1 30 7 0 0.10

6. Bacon (slice) 35 3 8 0 02 0 0.09

7. Orange 65 0 0 1 52 1 1 0.40

8. Milk-2% (cup) 100 4 12 0 250 9 0 0.16

9. Orange juice (cup) 120 0 0 0 31 0 0.50

10. Wheat toast (slice) 65 1 0 1 26 3 3 0.07

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.

A Diet Example
Model Construction – Decision Variables (2 of 5)

x1 = cups of bran cereal 4-12
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 Summary (3 of 5)

Minimize Z = 0.18x1 + 0.22x2 + 0.10x3 + 0.12x4 + 0.10x5 + 0.09x6
subject to: + 0.40x7 + 0.16x8 + 0.50x9 + 0.07x10

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
Computer Solution with Excel (4 of 5)

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Exhibit 4.5

4-14

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

Exhibit 4.6

An Investment Example 4-16
Model Summary (1 of 4)

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

An Investment Example
Computer Solution with Excel (2 of 4)

Exhibit 4.7 4-17

An Investment Example
Solution with Excel Solver Window (3 of 4)

Exhibit 4.8

An Investment Example
Sensitivity Report (4 of 4)

Exhibit 4.9

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

Television Commercial Exposure Cost
commercial)

20,000

Budget limit \$100,000
Television time for four commercials
Resources for no more than 15 commercials and/or ads

A Marketing Example 4-21
Model Summary (2 of 6)

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
Solution with Excel (3 of 6)

Exhibit 4.10

A Marketing Example
Solution with Excel Solver Window (4 of 6)

Exhibit 4.11 4-23

A Marketing Example
Integer Solution with Excel (5 of 6)

Exhibit 4.12

Exhibit 4.13

A Marketing Example
Integer Solution with Excel (6 of 6)

Exhibit 4.14

A Transportation Example
Problem Definition and Data (1 of 3)

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 A To Store C
\$16 B \$11
1 \$18
2 14 12 13
3 13 15 17

A Transportation Example
Model Summary (2 of 4)

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 Example
Solution with Excel (3 of 4)

Exhibit 4.15 4-28

A Transportation Example
Solution with Solver Window (4 of 4)

Exhibit 4.16

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

Component Maximum Barrels Cost/barrel
Available/day
1 4,500 \$12
2 10
3 2,700 14
3,500

Grade Component Specifications Selling Price (\$/bbl)

Super At least 50% of 1 \$23
Premium Not more than 30% of 2
Extra 20
At least 40% of 1 18
Not more than 25% of 3

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

A Blend Example
Problem Statement and Variables (2 of 6)

■ 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
Model Summary (3 of 6)

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

subject to:

x1s + x1p + x1e ≤ 4,500 bbl. all xij ≥ 0
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
Solution with Excel (4 of 6)

Exhibit 4.17

A Blend Example
Solution with Solver Window (5 of 6)

Exhibit 4.18

A Blend Example
Sensitivity Report (6 of 6)

Exhibit 4.19

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

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 Example
Decision Variables (2 of 5)

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
Model Summary (3 of 5)

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

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

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Exhibit 4.20

4-39

A Multi-Period Scheduling Example
Solution with Solver Window (5 of 5)

Exhibit 4.21

A Data Envelopment Analysis (DEA) Example
Problem Definition (1 of 5)

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 Data Summary (2 of 5)

School Inputs Outputs
Alton 12 3 12 3
Beeks .06 \$260 11.3 86 75 71
Carey .05 320 10.5 82 72 67
Delancey
.08 340 12.0 81 79 80
.06 460 13.1 81 73 69

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

Decision Variables: 4-43

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

A Data Envelopment Analysis (DEA) Example
Solution with Excel (4 of 5)

Exhibit 4.22

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

Exhibit 4.23

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?

Example Problem Solution
Model Formulation (2 of 5)

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 Solution
Model Formulation (3 of 5)

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

Example Problem Solution 4-49
Model Summary (4 of 5)

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