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Chapter 3: Linear Programming Modeling Applications. Jason C. H. Chen, Ph.D. Professor of MIS School of Business Administration Gonzaga University Spokane, WA 99223 chen@jepson.gonzaga.edu. Linear Programming (LP) Can Be Used for Many Managerial Decisions:. 1. Manufacturing applications

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chapter 3 linear programming modeling applications

Chapter 3:Linear Programming Modeling Applications

Jason C. H. Chen, Ph.D.

Professor of MIS

School of Business Administration

Gonzaga University

Spokane, WA 99223

chen@jepson.gonzaga.edu

linear programming lp can be used for many managerial decisions
Linear Programming (LP) Can Be Used for Many Managerial Decisions:
  • 1. Manufacturing applications
    • Product mix
    • Make-buy
  • 2. Marketing applications
    • Media selection
    • Marketing research
  • 3. Finance application
    • Portfolio selection
  • 4. Transportation application and others
    • Shipping & transportation
    • Multiperiod scheduling
slide3
For a particular application we begin with

the problem scenario and data, then:

  • Define the decision variables
  • Formulate the LP model using the decision variables
    • Write the objective function equation
    • Write each of the constraint equations
  • Implement the model in Excel
  • Solve with Excel’s Solver
manufacturing applications product mix problem fifth avenue industries
Manufacturing ApplicationsProduct 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

slide5

Resource Data

Labor cost is $0.75 per tie

decision variables
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

slide9
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

slide10

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)

slide11
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

lp model for product mix problem
LP Model for Product Mix Problem

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

Subject to the constraints:

0.125S < 1,000 (yards of silk)

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

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

6,000 < S < 7,000

10,000 < P < 14,000

13,000 < B1 < 16,000

6,000 < B2 < 8,500

S, P, B1, B2 > 0

Go to file 3-1.xls

marketing applications media selection problem win big gambling club
Marketing applicationsMedia 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

other restrictions
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

slide17
Objective Function (in num. audience reached)

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

Subject to the constraints:

Budget is $8000

800T + 925N + 290P + 380A < 8000

At Least 5 Radio Spots per Week

P + A > 5

slide18
No More Than $1800 per Week for Radio

290P + 380A < 1800

Max Number of Ads per Week

T < 12 P < 25

N < 5 A < 20

Finally nonnegativity T, N, P, A > 0

lp model for media selection problem
LP Model for Media Selection Problem

Objective Function

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

Subject to the constraints:

800T + 925N + 290P + 380A < 8000

P + A > 5

290P + 380A < 1800

T < 12

P < 25

N < 5

A < 20

T, N, P, A > 0

Go to file 3-3.xls

finance application portfolio selection international city trust
Finance applicationPortfolio 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

constraints
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 variables1
Decision Variables

T = $ 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

slide25
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

slide26
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)

slide27
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)

slide28
Limit Overall Risk To No More Than 2.0

Use 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

lp model for portfolio selection
LP Model for Portfolio Selection

Max 0.07T + 0.10B + 0.19G + 0.12P+ 0.08M + 0.14C

Subject to the constraints:

T + B + G + P + M + C < 5,000,000 (total funds)

T < 0.25 (T + B + G + P + M + C) (Max trade credits)

B < 0.25 (T + B + G + P + M + C) (Max corp bonds)

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

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

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

C < 0.25 (T + B + G + P + M + C) (Max const loans)

1.7T + 1.2B + 3.7G + 2.4P + 2.0M + 2.9C < 2.0 (T + B + G + P + M + C) (Risk score)

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

T + B > 0.45 (T + B + G + P + M + C) (Trade credits & bonds)

T, B, G, P, M, C > 0

Go to file 3-5.xls

employee staffing application labor planning hong kong bank
Employee Staffing ApplicationLabor 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

slide33
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
slide34

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)
slide35
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

slide36
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

slide37
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)

slide38
Only 12 Full Time Tellers Available

F < 12

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

lp model for labor planning
LP Model for Labor Planning

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

Subject to the constraints:

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)

F < 12

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

F, P1, P2, P3, P4, P5> 0

Go to file 3-6.xls

transportation application and others vehicle loading goodman shipping
Transportation application and othersVehicle 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

slide43
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
slide44
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

slide45
Volume Limit Of 1300 Cubic Feet

0.125W1 + 0.064W2 + 0.144W3 +

0.448W4 + 0.048W5 + 0.018W6< 1300

Pounds of Each Item Available

W1< 5000 W4< 3500

W2< 4500 W5< 4000

W3< 3000 W6< 3500

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

lp model for vehicle loading
LP Model for Vehicle Loading

Objective Function

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

Subject to the constraints:

W1 + W2 + W3 + W4 + W5 + W6< 15,000 (Weight Limit)

0.125W1 + 0.064W2 + 0.144W3

+0.448W4 + 0.048W5 + 0.018W6< 1300 (volume limit of truck)

Pounds of Each Item Available

W1< 5000 (item 1 availability)

W2< 4500 (item 2 availability)

W3< 3000 (item 3 availability)

W4< 3500 (item 4 availability)

W5< 4000 (item 5 availability)

W6< 3500 (item 6 availability)

Wi > 0, i=1,…,6

Go to file 3-7.xls

blending problem whole food nutrition center
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

decision variables2
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

slide51
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)

slide52
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

lp model for a blending problem
LP Model for a Blending Problem

Objective Function

Min 0.33A + 0.47B + 0.38C

Subject to the constraints:

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

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

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

5A + 6C > 0.425 (magnesium units)

A + B + C = 0.125 (lbs of total mix)

A, B, C > 0

Go to file 3-9.xls

multiperiod scheduling greenberg motors
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 variables3
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)

production data
Production Data
  • Production costs will be 10% higher in months 3 and 4
  • Monthly labor hours most be between
  • 2240 and 2560
inventory data
Inventory Data

Max inventory is 3300 motors

production and inventory balance
Production and Inventory Balance

(inventory at end of previous period)

+ (production the period)

- (sales this period)

= (inventory at end of this period)

slide61
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)

slide62
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

slide63
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)

slide64
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

lp model for a multiperiod scheduling
LP model for a Multiperiod Scheduling

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:

PA1 – IA1- = 800 (P&I balance month 1)

PB1 – IB1 = 1000

IA1 + PA2 – IA2 =700 (month 2)

IB1 + PB2 – IB2 =1200

IA2 + PA3 – IA3 = 1000 (month 3)

IB2 + PB3 –IB3 = 1400

IA3 + PA4 – IA4 = 1100 (month 4)

IB3 + PB4 – IB4 = 1400

IA4 = 450 (Ending Inventory)

IB4 = 300 (Ending Inventory)

IA1 + IB1< 3300 (maximal inventory level month 1)

IA2 + IB2< 3300 (month 2)

IA3 + IB3< 3300 (month 3)

IA4 + IB4< 3300 (month 4)

2240 < 1.3PA1 + 0.9PB1< 2560 (range of labor hours 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)

PAi, PBi, IAi, IBi> 0

Go to file 3-11.xls