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Solving Examples of Linear Programming Models
Prof. Yong Won SEO
College of Business Administration, CAU
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
Product Mix Example
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.
■ $25,000 available for a production run.
■ 500 dozen blank T-shirts and sweatshirts in stock.
■ How many dozens (boxes) of each type of shirt to produce?
Product Mix Example
Processing Time (hr) Per dozen
Cost ($)
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
Resource requirements for the product mix example.
Product Mix Example
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
Product Mix Example
Product Mix Example
• Model solution is:
• x1=175.56 boxes of front-only sweatshirts
• x2=57.78 boxes of front and back sweatshirts
• x3 = 500 boxes of front-only t-shirts
• Z=$45,522.22 profit
Product Mix Example: Sensitivity
• Additional 1 hour of processing time means $233.33 increased profit
Breakfast to include at least 420 calories, 5 milligrams of iron,
400 milligrams of calcium, 20 grams of protein, 12 grams of
fiber, and must have no more than 20 grams of fat and 30
milligrams of cholesterol.
Breakfast Food Cal
Fat (g)
Cholesterol (mg)
Iron (mg)
Calcium (mg)
Protein (g)
Fiber (g)
Cost ($)
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
Diet Example
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
Diet Example
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
Diet Example
Diet Example : Solution
• X3=1.025
• X8=1.241
• X10=2.975
• Z=$0.509 per meal
Diet Problem History
• The first problem tested on by Dantzig
• Originally formulated in 1945 by Nobel economist George Stigler– An important military issue during WW II.
• Consisted of 77 unknowns and 9 equations– 9 clerks using hand-operated desk calculators
– 120 MD to obtain the optimal Simplex solution
• Solution– $39.69 per year using wheat flour, cabbage, dried navy beans
(1939 prices)
– Simplex solution was better than Stiegler’s (using his own method), by 24 cents
An investor has $70,000 to divide among several instruments. Municipal
bonds have an 8.5% return, CD’s a 5% return, t-bills a 6.5% return, and
growth stock 13%.
The following guidelines have been established:
1.No more than 20% in municipal bonds
2.Investment in CDs should not exceed the other three alternatives
3.At least 30% invested in t-bills and CDs
4.More should be invested in CDs and t-bills than in municipal bonds and
growth stocks by a ratio of 1.2 to 1
5.All $70,000 should be invested.
Investment Example
Maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4subject 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
Investment Example
Total investment requirement,
=D10*B13+E10*B14+F10*B15+G10*B16
First guideline,
=D6*B13
Objective function, Z,
for total return
Investment Example
Shadow price for the
amount available to invest
Investment Example: Sensitivity
• Additional $1 investment yiels $0.095 (9.5%) increase in profit, with no upper bound
Investment Example: variation
• If the following guideline is removed, how to model?
5. All $70,000 should be invested.
[Case] Investment Application of GE Asset Mgt
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 available for 4 commercials
Radio time available for 10 commercials
Newspaper space available for 7 ads
Resources for no more than 15 commercials and/or ads
Marketing Example
Maximize Z = 20,000x1 + 12,000x2 + 9,000x3subject 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
Marketing Example
Marketing Example
• TV: 1.818, Radio: 10, Newspaper ads: 3.182 (????)– Total Exposures 185000
– Round down?
Exhibit 4.12
Decision variables
Click on “int” for
integer.
Marketing Example: Forcing Integer
Exhibit 4.14
Integer solution Better solution—17,000
more total exposures—than
rounded-down solution
Marketing Example: Integer Solution
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
Transportation Example
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
Transportation Example
=C5+C6+C7
=C5+D5+E5
Transportation Example: Solution
■ 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.
Blend Example
Component Maximum Barrels
Available/day Cost/barrel
1 4,500 $12
2 2,700 10
3 3,500 14
Blend Example
Grade Component Specifications Selling Price ($/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
Blend Example
• 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).
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 00.70x2s - 0.30x1s - 0.30x3s 00.60x1p - 0.40x2p - 0.40x3p 00.75x3p - 0.25x1p - 0.25x2p 00.40x1e- 0.60x2e- - 0.60x3e 00.90x2e - 0.10x1e - 0.10x3e 0
x1s + x2s + x3s 3,000 bbl.x1p+ x2p + x3p 3,000 bbl.x1e+ x2e + x3e 3,000 bbl.
all xij 0
Blend Example
=B7+B10+B13
Decision variables—B7:B15 =B7+B8+B9=0.5*B7-0.5*B8-0.5*B9
Blend Example: Solution
Blend Example: Sensitivity
• Component 1 seems the most critical resource– 1 barrel increase means $20 additional profit, up to 6200 barrels
[Case] Blending Problem of Petroleum Industry
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 1052 1703 2304 1805 1506 250
Multi-period Scheduling Example
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)
Multi-period Scheduling Example
Model summary:
Minimize Z = $190(r1 + r2 + r3 + r4 + r5 + r6) + $260(o1+o2+o3 +o4+o5+o6) + 10(i1 + i2 + i3 + i4 + i5)
subject to:
rj 160 computers in week j ( j = 1, 2, 3, 4, 5, 6)oj 150 computers in week j ( j = 1, 2, 3, 4, 5, 6)r1 + o1 - i1 = 105 week 1r2 + o2 + i1 - i2 = 170 week 2r3 + o3 + i2 - i3 = 230 week 3r4 + o4 + i3 - i4 = 180 week 4r5 + o5 + i4 - i5 = 150 week 5r6 + o6 + i5 = 250 week 6rj, oj, ij 0
Multi-period Scheduling Example
Exhibit 4.20
Decision variables for
regular production –
B6:B11
Decision variables for
overtime production –
D6:D11
B7+D7+I6; regular
production + overtime
production + inventory
from previous week
G7-H7
Multi-period Scheduling: Solution
[Case] Employee Scheduling Problem
DEA compares a number of decision making units(DMU) of the
same type based on their inputs (resources) and outputs.
The result indicates if a particular unit is less productive, or efficient,
than other units.
DEA (Data Envelopment Analysis)
[Case] DEA to compare ports’ efficiency
Comparing the efficiency
• Which department is more efficient?
• Comparing Sales/Labor cost
42
Department Input:Labor Cost($)
Output:Sales(# of contracts)
A 2000 1500
B 1500 1100
Department
Input:Labor Cost($)
Output:Sales(# of contracts)
Sales/Labor Cost
A 2000 1500 0.75
B 1500 1100 0.73
Comparing the efficiency is not easy
• Which department is more efficient?
• Measuring the efficiency(or performance) of different organizations is difficult because organizations have multiple input measures (number of workers, cost of labor, cost of machine operations, pay scale for employees, and cost of advertising) and multiple output measures (profit, sales, and market share).
43
Department
Input 1:Labor Cost($)
Input 2:Office Space(ft2)
Output:Sales(# of contracts)
A 2000 10000 1500
B 1500 6900 1100
Calculating Efficiency
• DEA offers a variety of models that use multiple inputs and outputs to compare the efficiency of two or more processes.
• The ratio model is based on the following definition of efficiency:
44
Weighted Sum of OutputsEfficiency =
Weighted Sum of Inputs
Calculating Efficiency (cont’d)
• Suppose we have the following input and output data:
• Defining efficiency based on the ratio model:
• The rationale is : “Choose whatever weights that would make your efficiency value maximum”
45
Department Labor Cost(input)
Sales(output 1)
Customer Satisfaction(output 2)
A 10 10 10
B 15 30 12
C 12 36 6
D 22 25 16
Laborv
CSwSaleswEfficiency
21
Building an LP for Department D
46
0,,
122
1625
112
636
115
1230
110
1010
22
1625
21
21
21
21
21
21
vww
v
ww
v
ww
v
ww
v
ww
toSubject
v
wwMaximize
0,,
221625
12636
151230
101010
122
1625
21
21
21
21
21
21
vww
vww
vww
vww
vww
v
toSubject
wwMaximize
LP
DEA Solution Meaning
• 4 DEA formulations can be made– To Evaluate A, B, C, D each
• If Z=1, the school is efficient one
• If Z
Graphical Analysis
48
Efficient frontier
0
0.2
0.4
0.6
0.8
1
1.2
0 0.5 1 1.5 2 2.5 3 3.5
Sales/Labor Cost
CS/Labor Cost
A
B
C
D
DEA Example
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
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
DEA Example: 4 schools’ data
DEA Example
• Objective: To compare 4 schools’ efficiency
– 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝑂𝑢𝑡𝑝𝑢𝑡
𝐼𝑛𝑝𝑢𝑡
• Multiple elements in I/O : weights needed
– 𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 =𝑂𝑢𝑡𝑝𝑢𝑡
𝐼𝑛𝑝𝑢𝑡=
𝑦1𝑂1+𝑦2𝑂2+𝑦3𝑂3
𝑥1𝐼1+𝑥2𝐼2+𝑥3𝐼3
• Evaluation can vary with weight settings– What weight values seems fair to you?
• Rationale: “Choose whatever weights that would make your efficiency value maximum”
DEA to evaluate Delancey
0 variablesAll
11.1346006.0
697381
10.1234008.0
807981
15.1032005.0
677282
13.1126006.0
717586
1.1346006.0
697381
321
321
321
321
321
321
321
321
321
321
yyy
xxx
yyy
xxx
yyy
xxx
yyy
xxx
toSubject
yyy
xxxMaximize
Decision Variables:
xi = weight per unit of each output where i = 1, 2, 3
yi = weight per unit of each input where i = 1, 2, 3
Model Summary:
Maximize Z = 81x1 + 73x2 + 69x3subject to:
.06 y1 + 460y2 + 13.1y3 = 1
86x1 + 75x2 + 71x3 .06y1 + 260y2 + 11.3y382x1 + 72x2 + 67x3 .05y1 + 320y2 + 10.5y381x1 + 79x2 + 80x3 .08y1 + 340y2 + 12.0y381x1 + 73x2 + 69x3 .06y1 + 460y2 + 13.1y3
xi, yi 0
DEA to evaluate Delancey – LP form
Exhibit 4.22
Value of outputs, also in cell H8
=B5*B12+C5*B13+D5*B14
=E8*D12+F8*D13+G8*D14
DEA Solution: Z=Efficiency
Integer Programming (IP)
Prof. Yongwon Seo
College of Business Administration, CAU
Types of Integer Programming(IP) Problems
56
Total Integer Model: All decision variables required to have
integer solution values.
0-1 Integer Model: All decision variables required to have
integer values of zero or one.
Mixed Integer Model: Some of the decision variables (but not all) required to have integer values.
Total Integer Model
• Machine shop obtaining new presses and lathes.
• Marginal profitability: each press $100/day; each lathe $150/day.
• Resource constraints: $40,000 budget, 200 sq. ft. floor space.
• Machine purchase prices and space requirements:
57
Machine
Required Floor Space (ft.2)
Purchase Price
Press Lathe
15
30
$8,000
4,000
Total Integer Model
58
Maximize Z = $100x1 + $150x2
subject to:
$8,000x1 + 4,000x2 $40,000
15x1 + 30x2 200 ft2
x1, x2 0 and integer
0-1 Integer Model
• Recreation facilities selection to maximize daily usage by residents.
• Resource constraints: $120,000 budget; 12 acres of land.
• Selection constraint: either swimming pool or tennis center (not both).
59
Recreation
Facility
Expected Usage (people/day)
Cost ($)
Land Requirement (acres)
Swimming pool Tennis Center Athletic field Gymnasium
300 90 400 150
35,000 10,000 25,000 90,000
4 2 7 3
0-1 Integer Model
60
x1 = construction of a swimming pool
x2 = construction of a tennis center
x3 = construction of an athletic field
x4 = construction of a gymnasium
Maximize Z = 300x1 + 90x2 + 400x3 + 150x4
subject to:
$35,000x1 + 10,000x2 + 25,000x3 + 90,000x4 $120,000
4x1 + 2x2 + 7x3 + 3x4 12 acres
x1 + x2 1 facility
x1, x2, x3, x4 = 0 or 1
Mixed Integer Model
• $250,000 available for investments providing greatest return after one year.
• Data: – Condominium cost $50,000/unit; $9,000 profit if sold after one
year.
– Land cost $12,000/ acre; $1,500 profit if sold after one year.
– Municipal bond cost $8,000/bond; $1,000 profit if sold after one year.
– Only 4 condominiums, 15 acres of land, and 20 municipal bonds available.
61
Mixed Integer Model
62
x1 = condominiums purchased
x2 = acres of land purchased
x3 = bonds purchased
Maximize Z = $9,000x1 + 1,500x2 + 1,000x3
subject to:
50,000x1 + 12,000x2 + 8,000x3 $250,000
x1 4 condominiums
x2 15 acres
x3 20 bonds
x2 0
x1, x3 0 and integer
Why Integer Programming Complex
• Rounding non-integer solution values up to the nearest integer value can result in an infeasible solution.
• A feasible solution by rounding down non-integer solution values may result in a less than optimal (sub-optimal) solution.
63
Example of IP solution
64
Maximize Z = $100x1 + $150x2subject to:
8,000x1 + 4,000x2 $40,000
15x1 + 30x2 200 ft2
x1, x2 0 and integer
LP Optimal Solution:
Z = $1,055.56
x1 = 2.22 presses
x2 = 5.55 lathes
How to solve?
• Branch-and-Bound Method– Traditional approach to solving integer programming problems.
– Feasible solutions can be partitioned into smaller subsets
– Smaller subsets evaluated until best solution is found.
– Method is a tedious and complex mathematical process
65