Taylor introms10 ppt_04
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Transcript of Taylor introms10 ppt_04
4-1Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Linear Programming:
Modeling Examples
Chapter 4
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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
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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.
■ $25,000 available for a production run.
■ 500 dozen blank T-shirts and sweatshirts in stock.
■ How many dozens (boxes) of each type of shirt to produce?
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A Product Mix Example (2 of 8)
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Processing Time (hr) Per dozen
Cost ($)
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 $25,000 x1 + x2 500 dozen sweatshirts
x3 + x4 500 dozen T-shirts
A Product Mix ExampleModel Construction (4 of 8)
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A Product Mix ExampleComputer Solution with Excel (5 of 8)
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Exhibit 4.1
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Exhibit 4.2
A Product Mix ExampleSolution with Excel Solver Window (6 of 8)
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Exhibit 4.3
A Product Mix ExampleSolution with QM for Windows (7 of 8)
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Exhibit 4.4
A Product Mix ExampleSolution with QM for Windows (8 of 8)
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Breakfast to include at least 420 calories, 5 milligrams of iron, 400 milligrams of calcium, 20 grams of protein, 12 grams of fiber, and must have no more than 20 grams of fat and 30 milligrams of cholesterol.
Breakfast Food Cal
Fat (g)
Cholesterol (mg)
Iron (mg)
Calcium (mg)
Protein (g)
Fiber (g)
Cost ($)
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 ExampleData 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 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 12xi 0, for all j
A Diet ExampleModel Summary (3 of 5)
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Exhibit 4.5
A Diet ExampleComputer Solution with Excel (4 of 5)
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4-15Exhibit 4.6
A Diet ExampleSolution with Excel Solver Window (5 of 5)
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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 ExampleModel Summary (1 of 4)
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An Investment ExampleComputer Solution with Excel (2 of 4)
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Exhibit 4.7
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Exhibit 4.8
An Investment ExampleSolution with Excel Solver Window (3 of 4)
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An Investment ExampleSensitivity Report (4 of 4)
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Exhibit 4.9
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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 ExampleData 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 = number of television commercials x2 = number of radio commercials x3 = number of newspaper ads
A Marketing ExampleModel Summary (2 of 6)
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Exhibit 4.10
A Marketing ExampleSolution with Excel (3 of 6)
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Exhibit 4.11
A Marketing ExampleSolution with Excel Solver Window (4 of 6)
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4-24Exhibit 4.13
A Marketing ExampleInteger Solution with Excel (5 of 6)
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Exhibit 4.12
4-25Exhibit
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 600Unit 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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Exhibit 4.15
A Transportation ExampleSolution with Excel (3 of 4)
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Exhibit 4.16
A Transportation ExampleSolution with Solver Window (4 of 4)
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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
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A Blend ExampleProblem Definition and Data (1 of 6)
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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 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 ExampleModel Summary (3 of 6)
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all xij 0
4-33Exhibit
4.17
A Blend ExampleSolution with Excel (4 of 6)
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4-34Exhibit 4.18
A Blend ExampleSolution with Solver Window (5 of 6)
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4-35Exhibit 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:
A Multi-Period Scheduling ExampleProblem Definition and Data (1 of 5)
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Week Computer Orders1 1052 1703 2304 1805 1506 250
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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 ExampleDecision 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 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 1r2 + o2 + i1 - i2 = 170 week 2r3 + o3 + i2 - i3 = 230 week 3r4 + o4 + i3 - i4 = 180 week 4r5 + o5 + i4 - i5 = 150 week 5r6 + o6 + i5 = 250week 6rj, oj, ij 0
A Multi-Period Scheduling ExampleModel Summary (3 of 5)
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A Multi-Period Scheduling ExampleSolution with Excel (4 of 5)
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Exhibit 4.20
4-40Exhibit
4.21
A 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 ratioInput 2 = supplementary funds/studentInput 3 = average educational level of parents
Output 1 = average reading SOL scoreOutput 2 = average math SOL scoreOutput 3 = average history SOL score
A Data Envelopment Analysis (DEA) ExampleProblem Definition (1 of 5)
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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, 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) ExampleDecision Variables and Model Summary (3 of 5)
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4-44Exhibit
4.22
A Data Envelopment Analysis (DEA) ExampleSolution with Excel (4 of 5)
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Exhibit 4.23
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 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 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)
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Example Problem SolutionSolution with QM for Windows (5 of 5)
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4-51Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall