taylor introms10 ppt 041
Linear Programming:
Modeling Examples
Chapter 4
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-1
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
4-2
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.
■
$25,000 available for a production run.
■
500 dozen blank T-shirts and sweatshirts in stock.
■
How many dozens (boxes) of each type of shirt to produce?
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4-3
A Product Mix Example (2 of 8)
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4-4
A Product Mix Example
Data (3 of 8)
Processing
Cost
Profit
Time (hr)
($)
($)
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
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4-5
A Product Mix Example
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 $25,000
x1 + x2 500 dozen sweatshirts
x3 + x4 500 dozen T-shirts
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4-6
A Product Mix Example
Computer Solution with Excel (5 of 8)
Exhibit 4.1
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4-7
A Product Mix Example
Solution with Excel Solver Window (6 of 8)
Exhibit 4.2
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4-8
A Product Mix Example
Solution with QM for Windows (7 of 8)
Exhibit 4.3
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4-9
A Product Mix Example
Solution with QM for Windows (8 of 8)
Exhibit 4.4
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4-10
A Diet Example
Data and Problem Definition (1 of 5)
Breakfast Food
Cal
1. Bran cereal (cup)
90
2. Dry cereal (cup)
110
3. Oatmeal (cup)
100
4. Oat bran (cup)
90
5. Egg
75
6. Bacon (slice)
35
7. Orange
65
8. Milk-2% (cup)
100
9. Orange juice (cup) 120
10. Wheat toast (slice)
65
Fat Cholesterol Iron Calcium Protein Fiber Cost
(g)
(mg)
(mg)
(mg)
(g)
(g)
($)
0
0
6
20
3
5
0.18
2
0
4
48
4
2
0.22
2
0
2
12
5
3
0.10
2
0
3
8
6
4
0.12
5
270
1
30
7
0
0.10
3
8
0
0
2
0
0.09
0
0
1
52
1
1
0.40
4
12
0
250
9
0
0.16
0
0
0
3
1
0
0.50
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.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-11
A Diet Example
Model Construction – Decision Variables (2 of 5)
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
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4-12
A Diet Example
Model Summary (3 of 5)
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
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4-13
A Diet Example
Computer Solution with Excel (4 of 5)
Exhibit 4.5
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4-14
A Diet Example
Solution with Excel Solver Window (5 of 5)
Exhibit 4.6
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4-15
An Investment Example
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
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4-16
An Investment Example
Computer Solution with Excel (2 of 4)
Exhibit 4.7
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4-17
An Investment Example
Solution with Excel Solver Window (3 of 4)
Exhibit 4.8
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4-18
An Investment Example
Sensitivity Report (4 of 4)
Exhibit 4.9
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4-19
A Marketing Example
Data and Problem Definition (1 of 6)
Television Commercial
Exposure
(people/ad or
commercial)
20,000
Cost
$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
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4-20
A Marketing Example
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
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4-21
A Marketing Example
Solution with Excel (3 of 6)
Exhibit 4.10
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4-22
A Marketing Example
Solution with Excel Solver Window (4 of 6)
Exhibit 4.11
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4-23
A Marketing Example
Integer Solution with Excel (5 of 6)
Exhibit 4.12
Exhibit 4.13
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4-24
A Marketing Example
Integer Solution with Excel (6 of 6)
Exhibit 4.14
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4-25
A Transportation Example
Problem Definition and Data (1 of 3)
Warehouse supply of
Television Sets:
Retail store demand
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
1
2
3
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A
$16
14
13
To Store
B
$18
12
15
C
$11
13
17
4-26
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
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4-27
A Transportation Example
Solution with Excel (3 of 4)
Exhibit 4.15
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4-28
A Transportation Example
Solution with Solver Window (4 of 4)
Exhibit 4.16
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4-29
A Blend Example
Problem Definition and Data (1 of 6)
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
Extra
At least 60% of 1
At least 10% of 2
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20
18
4-30
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).
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4-31
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.
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.
all xij 0
x1e+ x2e + x3e 3,000 bbl.
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4-32
A Blend Example
Solution with Excel (4 of 6)
Exhibit 4.17
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4-33
A Blend Example
Solution with Solver Window (5 of 6)
Exhibit 4.18
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4-34
A Blend Example
Sensitivity Report (6 of 6)
Exhibit 4.19
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4-35
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
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4-36
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)
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4-37
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
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4-38
A Multi-Period Scheduling Example
Solution with Excel (4 of 5)
Exhibit 4.20
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4-39
A Multi-Period Scheduling Example
Solution with Solver Window (5 of 5)
Exhibit 4.21
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4-40
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
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4-41
A Data Envelopment Analysis (DEA) Example
Problem Data Summary (2 of 5)
Inputs
School
Outputs
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
.06
460
13.1
81
73
69
Delancey
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4-42
A Data Envelopment Analysis (DEA) Example
Decision Variables and Model Summary (3 of 5)
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
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4-43
A Data Envelopment Analysis (DEA) Example
Solution with Excel (4 of 5)
Exhibit 4.22
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4-44
A Data Envelopment Analysis (DEA) Example
Solution with Solver Window (5 of 5)
Exhibit 4.23
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4-45
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?
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4-46
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)
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4-47
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
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4-48
Example Problem Solution
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
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4-49
Example Problem Solution
Solution with QM for Windows (5 of 5)
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4-50
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4-51
Modeling Examples
Chapter 4
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-1
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
4-2
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.
■
$25,000 available for a production run.
■
500 dozen blank T-shirts and sweatshirts in stock.
■
How many dozens (boxes) of each type of shirt to produce?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-3
A Product Mix Example (2 of 8)
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-4
A Product Mix Example
Data (3 of 8)
Processing
Cost
Profit
Time (hr)
($)
($)
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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-5
A Product Mix Example
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 $25,000
x1 + x2 500 dozen sweatshirts
x3 + x4 500 dozen T-shirts
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4-6
A Product Mix Example
Computer Solution with Excel (5 of 8)
Exhibit 4.1
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4-7
A Product Mix Example
Solution with Excel Solver Window (6 of 8)
Exhibit 4.2
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4-8
A Product Mix Example
Solution with QM for Windows (7 of 8)
Exhibit 4.3
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4-9
A Product Mix Example
Solution with QM for Windows (8 of 8)
Exhibit 4.4
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4-10
A Diet Example
Data and Problem Definition (1 of 5)
Breakfast Food
Cal
1. Bran cereal (cup)
90
2. Dry cereal (cup)
110
3. Oatmeal (cup)
100
4. Oat bran (cup)
90
5. Egg
75
6. Bacon (slice)
35
7. Orange
65
8. Milk-2% (cup)
100
9. Orange juice (cup) 120
10. Wheat toast (slice)
65
Fat Cholesterol Iron Calcium Protein Fiber Cost
(g)
(mg)
(mg)
(mg)
(g)
(g)
($)
0
0
6
20
3
5
0.18
2
0
4
48
4
2
0.22
2
0
2
12
5
3
0.10
2
0
3
8
6
4
0.12
5
270
1
30
7
0
0.10
3
8
0
0
2
0
0.09
0
0
1
52
1
1
0.40
4
12
0
250
9
0
0.16
0
0
0
3
1
0
0.50
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.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-11
A Diet Example
Model Construction – Decision Variables (2 of 5)
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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
4-12
A Diet Example
Model Summary (3 of 5)
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
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4-13
A Diet Example
Computer Solution with Excel (4 of 5)
Exhibit 4.5
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4-14
A Diet Example
Solution with Excel Solver Window (5 of 5)
Exhibit 4.6
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4-15
An Investment Example
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
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4-16
An Investment Example
Computer Solution with Excel (2 of 4)
Exhibit 4.7
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4-17
An Investment Example
Solution with Excel Solver Window (3 of 4)
Exhibit 4.8
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4-18
An Investment Example
Sensitivity Report (4 of 4)
Exhibit 4.9
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4-19
A Marketing Example
Data and Problem Definition (1 of 6)
Television Commercial
Exposure
(people/ad or
commercial)
20,000
Cost
$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
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4-20
A Marketing Example
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
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4-21
A Marketing Example
Solution with Excel (3 of 6)
Exhibit 4.10
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4-22
A Marketing Example
Solution with Excel Solver Window (4 of 6)
Exhibit 4.11
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4-23
A Marketing Example
Integer Solution with Excel (5 of 6)
Exhibit 4.12
Exhibit 4.13
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4-24
A Marketing Example
Integer Solution with Excel (6 of 6)
Exhibit 4.14
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4-25
A Transportation Example
Problem Definition and Data (1 of 3)
Warehouse supply of
Television Sets:
Retail store demand
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
1
2
3
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A
$16
14
13
To Store
B
$18
12
15
C
$11
13
17
4-26
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
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4-27
A Transportation Example
Solution with Excel (3 of 4)
Exhibit 4.15
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4-28
A Transportation Example
Solution with Solver Window (4 of 4)
Exhibit 4.16
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4-29
A Blend Example
Problem Definition and Data (1 of 6)
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
Extra
At least 60% of 1
At least 10% of 2
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20
18
4-30
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).
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4-31
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.
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.
all xij 0
x1e+ x2e + x3e 3,000 bbl.
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4-32
A Blend Example
Solution with Excel (4 of 6)
Exhibit 4.17
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4-33
A Blend Example
Solution with Solver Window (5 of 6)
Exhibit 4.18
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4-34
A Blend Example
Sensitivity Report (6 of 6)
Exhibit 4.19
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4-35
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
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4-36
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)
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4-37
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
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4-38
A Multi-Period Scheduling Example
Solution with Excel (4 of 5)
Exhibit 4.20
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4-39
A Multi-Period Scheduling Example
Solution with Solver Window (5 of 5)
Exhibit 4.21
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4-40
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
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4-41
A Data Envelopment Analysis (DEA) Example
Problem Data Summary (2 of 5)
Inputs
School
Outputs
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
.06
460
13.1
81
73
69
Delancey
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4-42
A Data Envelopment Analysis (DEA) Example
Decision Variables and Model Summary (3 of 5)
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
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4-43
A Data Envelopment Analysis (DEA) Example
Solution with Excel (4 of 5)
Exhibit 4.22
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4-44
A Data Envelopment Analysis (DEA) Example
Solution with Solver Window (5 of 5)
Exhibit 4.23
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4-45
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?
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4-46
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)
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4-47
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
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4-48
Example Problem Solution
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
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4-49
Example Problem Solution
Solution with QM for Windows (5 of 5)
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4-50
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4-51