chapter 3: linear programming modeling applications © 2007 pearson education
TRANSCRIPT
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Chapter 3:Linear Programming Modeling Applications
© 2007 Pearson Education
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Linear Programming (LP) Can Be Used for Many Managerial Decisions:
• Product mix
• Make-buy
• Media selection
• Marketing research
• Portfolio selection
• Shipping & transportation
• Multiperiod scheduling
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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)
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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
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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
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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
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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
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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)
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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
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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
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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)
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Only 12 Full Time Tellers Available
F < 12
finally nonnegativity: F, P1, P2, P3, P4, P5 > 0
Go to file 3-6.xls
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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)
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Sales Demand Data
Month
Motor
A B
1 (January) 800 1000
2 (February) 700 1200
3 (March) 1000 1400
4 (April) 1100 1400
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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
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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
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Production and Inventory Balance
(inventory at end of previous period)
+ (production the period)
- (sales this period)
= (inventory at end of this period)
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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)
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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
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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)
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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