Chapter 3:Linear Programming Modeling Applications

© 2007 Pearson Education

Linear Programming (LP) Can Be Used for Many Managerial Decisions:

• Product mix

• Make-buy

• Media selection

• Marketing research

• Portfolio selection

• Shipping & transportation

• Multiperiod scheduling

For a particular application we begin with

the problem scenario and data, then:

1) Define the decision variables

2) Formulate the LP model using the decision variables

• Write the objective function equation• Write each of the constraint equations

3) Implement the model in Excel

4) Solve with Excel’s Solver

Product Mix Problem: Fifth Avenue Industries

• Produce 4 types of men's ties

• Use 3 materials (limited resources)

Decision: How many of each type of tie to make per month?

Objective: Maximize profit

Material Cost per yardYards available

per month

Silk $20 1,000

Polyester $6 2,000

Cotton $9 1,250

Resource Data

Labor cost is $0.75 per tie

Product DataType of Tie

Silk Polyester Blend 1 Blend 2

Selling Price(per tie)

$6.70 $3.55 $4.31 $4.81

Monthly Minimum

6,000 10,000 13,000 6,000

Monthly Maximum 7,000 14,000 16,000 8,500

Total material(yards per tie) 0.125 0.08 0.10 0.10

Material Requirements(yards per tie)

Material

Type of Tie

Silk PolyesterBlend 1(50/50)

Blend 2(30/70)

Silk 0.125 0 0 0

Polyester 0 0.08 0.05 0.03

Cotton 0 0 0.05 0.07

Total yards 0.125 0.08 0.10 0.10

Decision Variables

S = number of silk ties to make per month

P = number of polyester ties to make per month

B1 = number of poly-cotton blend 1 ties to make per month

B2 = number of poly-cotton blend 2 ties to make per month

Profit Per Tie Calculation

Profit per tie =

(Selling price) – (material cost) –(labor cost)

Silk Tie

Profit = $6.70 – (0.125 yds)($20/yd) - $0.75

= $3.45 per tie

Objective Function (in $ of profit)

Max 3.45S + 2.32P + 2.81B1 + 3.25B2

Subject to the constraints:

Material Limitations (in yards)

0.125S < 1,000 (silk)

0.08P + 0.05B1 + 0.03B2 < 2,000 (poly)

0.05B1 + 0.07B2 < 1,250 (cotton)

Min and Max Number of Ties to Make

6,000 < S < 7,000

10,000 < P < 14,000

13,000 < B1 < 16,000

6,000 < B2 < 8,500

Finally nonnegativity S, P, B1, B2 > 0

Go to file 3-1.xls

Media Selection Problem:Win Big Gambling Club

• Promote gambling trips to the Bahamas

• Budget: $8,000 per week for advertising

• Use 4 types of advertising

Decision: How many ads of each type?

Objective: Maximize audience reached

Data

Advertising Options

TV Spot NewspaperRadio

(prime time)

Radio(afternoon)

AudienceReached(per ad)

5,000 8,500 2,400 2,800

Cost(per ad)

$800 $925 $290 $380

Max AdsPer week

12 5 25 20

Other Restrictions

• Have at least 5 radio spots per week

• Spend no more than $1800 on radio

Decision Variables

T = number of TV spots per week

N = number of newspaper ads per week

P = number of prime time radio spots per week

A = number of afternoon radio spots per week

Objective Function (in num. audience reached)

Max 5000T + 8500N + 2400P + 2800A

Subject to the constraints:

Budget is $8000800T + 925N + 290P + 380A < 8000

At Least 5 Radio Spots per WeekP + A > 5

No More Than $1800 per Week for Radio290P + 380A < 1800

Max Number of Ads per Week

T < 12 P < 25N < 5 A < 20

Finally nonnegativity T, N, P, A > 0

Go to file 3-3.xls

Portfolio Selection:International City Trust

Has $5 million to invest among 6 investments

Decision: How much to invest in each of 6 investment options?

Objective: Maximize interest earned

Data

InvestmentInterest

Rate Risk Score

Trade credits 7% 1.7

Corp. bonds 10% 1.2

Gold stocks 19% 3.7

Platinum stocks 12% 2.4

Mortgage securities 8% 2.0

Construction loans 14% 2.9

Constraints

• Invest up to $ 5 million

• No more than 25% into any one investment

• At least 30% into precious metals

• At least 45% into trade credits and corporate bonds

• Limit overall risk to no more than 2.0

Decision VariablesT = $ invested in trade credit

B = $ invested in corporate bonds

G = $ invested gold stocks

P = $ invested in platinum stocks

M = $ invested in mortgage securities

C = $ invested in construction loans

Objective Function (in $ of interest earned)

Max 0.07T + 0.10B + 0.19G + 0.12P

+ 0.08M + 0.14C

Subject to the constraints:

Invest Up To $5 Million

T + B + G + P + M + C < 5,000,000

No More Than 25% Into Any One Investment

T < 0.25 (T + B + G + P + M + C)

B < 0.25 (T + B + G + P + M + C)

G < 0.25 (T + B + G + P + M + C)

P < 0.25 (T + B + G + P + M + C)

M < 0.25 (T + B + G + P + M + C)

C < 0.25 (T + B + G + P + M + C)

At Least 30% Into Precious Metals

G + P > 0.30 (T + B + G + P + M + C)

At Least 45% Into

Trade Credits And Corporate Bonds

T + B > 0.45 (T + B + G + P + M + C)

Limit Overall Risk To No More Than 2.0Use a weighted average to calculate portfolio risk

1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C < 2.0

T + B + G + P + M + C

OR

1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C <

2.0 (T + B + G + P + M + C)

finally nonnegativity: T, B, G, P, M, C > 0

Go to file 3-5.xls

Labor Planning:Hong Kong Bank

Number of tellers needed varies by time of day

Decision: How many tellers should begin work at various times of the day?

Objective: Minimize personnel cost

Time Period Min Num. Tellers

9 – 10 10

10 – 11 12

11 – 12 14

12 – 1 16

1 – 2 18

2 - 3 17

3 – 4 15

4 – 5 10

Total minimum daily requirement is 112 hours

Full Time Tellers

• Work from 9 AM – 5 PM

• Take a 1 hour lunch break, half at 11, the other half at noon

• Cost $90 per day (salary & benefits)

• Currently only 12 are available

Part Time Tellers

• Work 4 consecutive hours (no lunch break)

• Can begin work at 9, 10, 11, noon, or 1

• Are paid $7 per hour ($28 per day)

• Part time teller hours cannot exceed 50% of the day’s minimum requirement

(50% of 112 hours = 56 hours)

Decision Variables

F = num. of full time tellers (all work 9–5)

P1 = num. of part time tellers who work 9–1

P2 = num. of part time tellers who work 10–2

P3 = num. of part time tellers who work 11–3

P4 = num. of part time tellers who work 12–4

P5 = num. of part time tellers who work 1–5

Objective Function (in $ of personnel cost)

Min 90 F + 28 (P1 + P2 + P3 + P4 + P5)

Subject to the constraints:

Part Time Hours Cannot Exceed 56 Hours

4 (P1 + P2 + P3 + P4 + P5) < 56

Minimum Num. Tellers Needed By Hour Time of Day

F + P1 > 10 (9-10)

F + P1 + P2 > 12 (10-11)

0.5 F + P1 + P2 + P3 > 14 (11-12)

0.5 F + P1 + P2 + P3+ P4 > 16 (12-1)

F + P2 + P3+ P4 + P5 > 18 (1-2)

F + P3+ P4 + P5 > 17 (2-3)

F + P4 + P5 > 15 (3-4)

F + P5 > 10 (4-5)

Only 12 Full Time Tellers Available

F < 12

finally nonnegativity: F, P1, P2, P3, P4, P5 > 0

Go to file 3-6.xls

Vehicle Loading:Goodman Shipping

How to load a truck subject to weight and volume limitations

Decision: How much of each of 6 items to load onto a truck?

Objective: Maximize the value shipped

Data

Item

1 2 3 4 5 6Value $15,500 $14,400 $10,350 $14,525 $13,000 $9,625

Pounds 5000 4500 3000 3500 4000 3500

$ / lb $3.10 $3.20 $3.45 $4.15 $3.25 $2.75

Cu. ft. per lb

0.125 0.064 0.144 0.448 0.048 0.018

Decision Variables

Wi = number of pounds of item i to load onto truck, (where i = 1,…,6)

Truck Capacity

• 15,000 pounds

• 1,300 cubic feet

Objective Function (in $ of load value)

Max 3.10W1 + 3.20W2 + 3.45W3 + 4.15W4 + 3.25W5 + 2.75W6

Subject to the constraints:

Weight Limit Of 15,000 Pounds

W1 + W2 + W3 + W4 + W5 + W6 < 15,000

Volume Limit Of 1300 Cubic Feet

0.125W1 + 0.064W2 + 0.144W3 +0.448W4 + 0.048W5 + 0.018W6 < 1300

Pounds of Each Item AvailableW1 < 5000 W4 < 3500W2 < 4500 W5 < 4000W3 < 3000 W6 < 3500

Finally nonnegativity: Wi > 0, i=1,…,6

Go to file 3-7.xls

Blending Problem:Whole Food Nutrition Center

Making a natural cereal that satisfies minimum daily nutritional requirements

Decision: How much of each of 3 grains to include in the cereal?

Objective: Minimize cost of a 2 ounce serving of cereal

Grain

Minimum Daily

Requirement

A B C

$ per pound $0.33 $0.47 $0.38

Protein per pound

22 28 21 3

Riboflavin per pound

16 14 25 2

Phosphorus per pound

8 7 9 1

Magnesium per pound

5 0 6 0.425

Decision Variables

A = pounds of grain A to use

B = pounds of grain B to use

C = pounds of grain C to use

Note: grains will be blended to form a 2 ounce serving of cereal

Objective Function (in $ of cost)

Min 0.33A + 0.47B + 0.38C

Subject to the constraints:

Total Blend is 2 Ounces, or 0.125 Pounds

A + B + C = 0.125 (lbs)

Minimum Nutritional Requirements

22A + 28B + 21C > 3 (protein)

16A + 14B + 25C > 2 (riboflavin)

8A + 7B + 9C > 1 (phosphorus)

5A + 6C > 0.425 (magnesium)

Finally nonnegativity: A, B, C > 0

Go to file 3-9.xls

Multiperiod Scheduling:Greenberg Motors

Need to schedule production of 2 electrical motors for each of the next 4 months

Decision: How many of each type of motor to make each month?

Objective: Minimize total production and inventory cost

Decision Variables

PAt = number of motor A to produce in month t (t=1,…,4)

PBt = number of motor B to produce in month t (t=1,…,4)

IAt = inventory of motor A at end of month t (t=1,…,4)

IBt = inventory of motor B at end of month t (t=1,…,4)

Sales Demand Data

Month

Motor

A B

1 (January) 800 1000

2 (February) 700 1200

3 (March) 1000 1400

4 (April) 1100 1400

Production DataMotor

(values are per motor)

A B

Production cost $10 $6

Labor hours 1.3 0.9

• Production costs will be 10% higher in months 3 and 4

• Monthly labor hours most be between 2240 and 2560

Inventory Data

Motor

A B

Inventory cost

(per motor per month)$0.18 $0.13

Beginning inventory

(beginning of month 1)0 0

Ending Inventory

(end of month 4)450 300

Max inventory is 3300 motors

Production and Inventory Balance

(inventory at end of previous period)

+ (production the period)

- (sales this period)

= (inventory at end of this period)

Objective Function (in $ of cost)

Min 10PA1 + 10PA2 + 11PA3 + 11PA4

+ 6PB1 + 6 PB2 + 6.6PB3 + 6.6PB4

+ 0.18(IA1 + IA2 + IA3 + IA4)

+ 0.13(IB1 + IB2 + IB3 + IB4)

Subject to the constraints:

(see next slide)

Production & Inventory Balance

0 + PA1 – 800 = IA1 (month 1)

0 + PB1 – 1000 = IB1

IA1 + PA2 – 700 = IA2 (month 2)

IB1 + PB2 – 1200 = IB2

IA2 + PA3 – 1000 = IA3 (month 3)

IB2 + PB3 – 1400 = IB3

IA3 + PA4 – 1100 = IA4 (month 4)

IB3 + PB4 – 1400 = IB4

Ending Inventory

IA4 = 450

IB4 = 300

Maximum Inventory level

IA1 + IB1 < 3300 (month 1)

IA2 + IB2 < 3300 (month 2)

IA3 + IB3 < 3300 (month 3)

IA4 + IB4 < 3300 (month 4)

Range of Labor Hours

2240 < 1.3PA1 + 0.9PB1 < 2560 (month 1)

2240 < 1.3PA2 + 0.9PB2 < 2560 (month 2)

2240 < 1.3PA3 + 0.9PB3 < 2560 (month 3)

2240 < 1.3PA4 + 0.9PB4 < 2560 (month 4)

finally nonnegativity: PAi, PBi, IAi, IBi > 0

Go to file 3-11.xls