Solving Examples of Linear Programming Models

Prof. Yong Won SEO

(seoyw@cau.ac.kr)

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

(seoyw@cau.ac.kr)

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