Chapter 8 Linear Programming: Modeling Examples
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Transcript of Chapter 8 Linear Programming: Modeling Examples
1Chapter 8 - Linear Programming: Modeling Examples
Chapter 8
Linear Programming: ModelingExamples
Introduction to Management Science
8th Edition
by
Bernard W. Taylor III
2Chapter 8 - Linear Programming: Modeling Examples
Chapter Topics
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
3Chapter 8 - Linear Programming: Modeling Examples
A Product Mix ExampleProblem Definition (1 of 7)
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.
$25,000 available for a production run.
500 dozen blank T-shirts and sweatshirts in stock.
How many dozens (boxes) of each type of shirt to produce?
4Chapter 8 - Linear Programming: Modeling Examples
Processing Time (hr) Per dozen
Cost ($)
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 (2 of 7)
5Chapter 8 - Linear Programming: Modeling Examples
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 $25,000 x1 + x2 500 doz. sweatshirts
x3 + x4 500 doz. T-shirts
A Product Mix ExampleModel Construction (3 of 7)
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Exhibit 8.1
A Product Mix ExampleComputer Solution with Excel (4 of 7)
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Exhibit 8.2
A Product Mix ExampleSolution with Excel Solver Window (5 of 7)
8Chapter 8 - Linear Programming: Modeling Examples
Exhibit 8.3
A Product Mix ExampleSolution with QM for Windows (6 of 7)
9Chapter 8 - Linear Programming: Modeling Examples
Exhibit 8.4
A Product Mix ExampleSolution with QM for Windows (7 of 7)
10Chapter 8 - Linear Programming: Modeling Examples
Breakfast to include at least 420 calories, 5 milligrams of iron, 400 milligrams of calcium, 20 grams of protein, 12 grams of fiber, and must have no more than 20 grams of fat and 30 milligrams of cholesterol.
Breakfast Food Cal
Fat (g)
Cholesterol (mg)
Iron (mg)
Calcium (mg)
Protein (g)
Fiber (g)
Cost ($)
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
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, i=1,…,10.
A Diet ExampleModel Summary (3 of 5)
13Chapter 8 - Linear Programming: Modeling ExamplesExhibit 8.5
A Diet ExampleComputer Solution with Excel (4 of 5)
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Exhibit 8.6
A Diet ExampleSolution with Excel Solver Window (5 of 5)
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Maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4
subject to:x1 14,000
– x1 + x2 – x3 – x4 0 x2 + x3 21,000 –1.2x1 + x2 + x3 – 1.2 x4 0 x1 + x2 + x3 + x4 = 70,000 x1, x2, x3, x4 0 where x1 = amount invested in municipal bonds ($) 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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Exhibit 8.7
An Investment ExampleComputer Solution with Excel (2 of 4)
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Exhibit 8.8
An Investment ExampleSolution with Excel Solver Window (3 of 4)
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Exhibit 8.9
An Investment ExampleSensitivity Report (4 of 4)
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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 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 = Exposure from Television Commercial (people) x2 = Exposure from Radio Commercial (people) x3 = Exposure from Newspaper Ad (people)
A Marketing ExampleModel Summary (2 of 6)
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Exhibit 8.10
A Marketing ExampleSolution with Excel (3 of 6)
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Exhibit 8.11
A Marketing ExampleSolution with Excel Solver Window (4 of 6)
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Exhibit 8.12
Exhibit 8.13
A Marketing ExampleInteger Solution with Excel (5 of 6)
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Exhibit 8.14
A Marketing ExampleInteger Solution with Excel (6 of 6)
25Chapter 8 - Linear Programming: Modeling Examples
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
To Store From Warehouse
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
xij 0, i=1,2,3 and j= A,B,C.
A Transportation ExampleModel Summary (2 of 4)
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Exhibit 8.15
A Transportation ExampleSolution with Excel (3 of 4)
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Exhibit 8.16
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
18
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 motor oil (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 0.50·(x1s – x2s – x3s) 0 0.70x2s – 0.30·(x1s + x3s) 0 0.60x1p – 0.40·(x2p + x3p) 0 0.25·(3·x3p – x1p – x2p) 0 0.20·(2·x1e – 3·x2e- – 3·x3e) 0 0.10·(9·x2e – x1e – x3e) 0 x1s + x2s + x3s 3,000 x1p+ x2p + x3p 3,000 x1e+ x2e + x3e 3,000 xij 0, i=1,2,3, and j=s,p,e.
A Blend ExampleModel Summary (3 of 6)
component availability
blend specification; e.g.:
x1s/(x1s+x2s+x3s) ≥ 0.50
x1s ≥ 0.50·(x1s+x2s+x3s)
0.50·(x1s–x2s–x3s) ≥ 0
production requirement
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Exhibit 8.17
A Blend ExampleSolution with Excel (4 of 6)
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Exhibit 8.18
A Blend ExampleSolution with Solver Window (5 of 6)
34Chapter 8 - Linear Programming: Modeling ExamplesExhibit 8.19
A Blend ExampleSensitivity Report (6 of 6)
35Chapter 8 - Linear Programming: Modeling Examples
Production Capacity: 160 computers per week 50 more computers with overtime
Assembly Costs: $190 per computer regular time; $260 per computer overtime
Inventory Cost: $10/comp. per week
Order schedule: Week (j) Computer Orders (dj) 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 per week j
(j = 1,…,6) -- decision variable
oj = overtime production of computers per week j
(j = 1,…,6) -- decision variable
ij = extra computers carried over as inventory in week j
(j = 1,…,5) — computed as ij = rj + oj + ij–1 – dj, where
dj is the demand (orders) as tabulated on p.35.
A Multi-Period Scheduling ExampleDecision Variables (2 of 5)
Note: the ij variables are being calculated, so that they are not independent from the rj and oj decision variables: the ij’s are functions of rj and oj.
37Chapter 8 - Linear Programming: Modeling Examples
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, …, 6)oj 150 (j = 1, …, 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 ExampleModel Summary (3 of 5)
Overall constraints
Individual constraints
Nontrivial for the ij!
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Exhibit 8.20
A Multi-Period Scheduling ExampleSolution with Excel (4 of 5)
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Exhibit 8.21
A Multi-Period Scheduling ExampleSolution with Solver Window (5 of 5)
40Chapter 8 - Linear Programming: Modeling Examples
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)
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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 = 81·x1 + 73·x2 + 69·x3
subject to: .06y1 + 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, i=1,2,3.
A Data Envelopment Analysis (DEA) ExampleDecision Variables and Model Summary (3 of 5)
43Chapter 8 - Linear Programming: Modeling ExamplesExhibit 8.22
A Data Envelopment Analysis (DEA) ExampleSolution with Excel (4 of 5)
44Chapter 8 - Linear Programming: Modeling Examples
Exhibit 8.23
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 (1/2) fish;Bow Chow at least half (1/2) 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?
46Chapter 8 - Linear Programming: Modeling Examples
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)
47Chapter 8 - Linear Programming: Modeling Examples
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)
48Chapter 8 - Linear Programming: Modeling Examples
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, i=h,f,m, and j=m,b.
Example Problem SolutionModel Summary (4 of 5)
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Example Problem SolutionSolution with QM for Windows (5 of 5)
Exhibit 8.24
50Chapter 8 - Linear Programming: Modeling Examples