Post on 27-Dec-2015
transcript
Chapter 3:Linear Programming Modeling Applications
Jason C. H. Chen, Ph.D.Professor of MIS
School of Business AdministrationGonzaga UniversitySpokane, WA 99223
chen@jepson.gonzaga.edu
Dr. Chen, Decision Support Systems 2
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
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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
Dr. Chen, Decision Support Systems 4
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
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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
Dr. Chen, Decision Support Systems 12
LP Model for Product Mix Problem
Max 3.45S + 2.32P + 2.81B1 + 3.25B2Subject 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
Dr. Chen, Decision Support Systems 13
Fifth Avenue Industries
S P B1 B2
All silk All poly Blend-1Blend-
2
Number of units 7000.0 13625.0 13100.0 8500.0
Selling price $6.70 $3.55 $4.31 $4.81 $192,614.75
Labor cost $0.75 $0.75 $0.75 $0.75 $31,668.75
Material cost $2.50 $0.48 $0.75 $0.81 $40,750.00
Profit $3.45 $2.32 $2.81 $3.25 $120,196.00
Constraints: Cost/Yd
Yards of silk 0.125 875.00 <= 1000 $20
Yards of polyester 0.08 0.05 0.03 2000.00 <= 2000 $6
Yards of cotton 0.05 0.07 1250.00 <= 1250 $9
Maximum all silk 1 7000.00 <= 7000
Maximum all poly 1 13625.00 <= 14000
Maximum blend-1 1 13100.00 <= 16000
Maximum blend-2 1 8500.00 <= 8500
Minimum all silk 1 7000.00 >= 6000
Minimum all poly 1 13625.00 >= 10000
Minimum blend-1 1 13100.00 >= 13000
Minimum blend-2 1 8500.00 >= 6000
LHS Sign RHSGo to file 3-1.xls
Dr. Chen, Decision Support Systems 14
Marketing applications 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 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
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LP Model for Media Selection Problem
Objective FunctionMax 5000T + 8500N + 2400P + 2800A
Subject to the constraints:800T + 925N + 290P + 380A < 8000 P + A > 5290P + 380A < 1800T < 12P < 25N < 5A < 20 T, N, P, A > 0
Go to file 3-3.xls
Dr. Chen, Decision Support Systems 20
Win Big Gambling Club
T N P A
TV
spotsNewspap
er ads
Prime-time radio
spots
Afternoon radio spots
Number of units 1.97 5.00 6.21 0.00
Audience 5000 8500 2400 2800 67240.30
Constraints:
Maximum TV 1 1.97 <= 12
Maximum newspaper 1 5.00 <= 5
Max prime-time radio 1 6.21 <= 25
Max afternoon radio 1 0.00 <= 20
Total budget $800 $925 $290 $380 $8,000.00 <= $8,000
Maximum radio $ $290 $380 $1,800.00 <= $1,800
Minimum radio spots 1 1 6.21 >= 5
LHS Sign RHS
Go to file 3-3.xls
Dr. Chen, Decision Support Systems 21
Finance application 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
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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
Dr. Chen, Decision Support Systems 30
International City Trust
T B G P M C
Trade credits Corp bonds Gold Platinum Mortgages Const loans
Dollars Invested $1,250,000.00 $1,250,000.00 $250,000.00 $1,250,000.00 $500,000.00 $500,000.00
Interest 0.07 0.10 0.19 0.12 0.08 0.14 $520,000.00
Constraints:
Total funds 1 1 1 1 1 1 $5,000,000.00 <= $5,000,000
Max trade credits 1 $1,250,000.00 <= $1,250,000
Max corp bonds 1 $1,250,000.00 <= $1,250,000
Max gold 1 $250,000.00 <= $1,250,000
Max platinum 1 $1,250,000.00 <= $1,250,000
Max mortgages 1 $500,000.00 <= $1,250,000
Max const loans 1 $500,000.00 <= $1,250,000
Risk score 1.7 1.2 3.7 2.4 2.0 2.9 10,000,000.00 <= 10,000,000
Precious metals 1 1 $1,500,000.00 >= $1,500,000
Trade credits & bonds 1 1 $2,500,000.00 >= $2,250,000
LHS Sign RHS
Go to file 3-5.xls
Dr. Chen, Decision Support Systems 31
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
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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
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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 < 124 (P1 + P2 + P3 + P4 + P5) < 56F, P1, P2, P3, P4, P5 > 0
Go to file 3-6.xls
Dr. Chen, Decision Support Systems 40
Hong Kong Bank
F P1 P2 P3 P4 P5
FT
tellersPT
@9am
PT @10a
m
PT @11a
mPT
@NoonPT
@1pm
Number of tellers 10.0 0.0 7.0 2.0 5.0 0.0
Cost $90.00 $28.00 $28.00 $28.00 $28.00 $28.00 $1,292.00
Constraints:
9am-10am needs 1 1 10.0 >= 10
10am-11am needs 1 1 1 17.0 >= 12
11am-Noon needs 0.5 1 1 1 14.0 >= 14
Noon-1pm needs 0.5 1 1 1 1 19.0 >= 16
1pm-2pm needs 1 1 1 1 1 24.0 >= 18
2pm-3pm needs 1 1 1 1 17.0 >= 17
3pm-4pm needs 1 1 1 15.0 >= 15
4pm-5pm needs 1 1 10.0 >= 10
Max full time 1 10.0 <= 12
Part-time limit 4 4 4 4 4 56.0 <= 56
LHS Sign RHS
Go to file 3-6.xls
Dr. Chen, Decision Support Systems 41
Transportation application and others
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
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LP Model for Vehicle Loading
Objective FunctionMax 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 AvailableW1 < 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
Dr. Chen, Decision Support Systems 47
Goodman Shipping
W1 W2 W3 W4 W5 W6
Item 1 Item 2 Item 3 Item 4 Item 5 Item 6
Weight in pounds 3,037.38 4,500.00 3,000.00 0.00 4,000.00 462.62
Load value $3.10 $3.20 $3.45 $4.15 $3.25 $2.75 $48,438.08
Constraints:
Weight limit 1 1 1 1 1 1 15000.00 <= 15000
Volume limit 0.125 0.064 0.144 0.448 0.048 0.018 1300.00 <= 1300
Item 1 limit (pounds) 1 3037.38 <= 5000
Item 2 limit (pounds) 1 4500.00 <= 4500
Item 3 limit (pounds) 1 3000.00 <= 3000
Item 4 limit (pounds) 1 0.00 <= 3500
Item 5 limit (pounds) 1 4000.00 <= 4000
Item 6 limit (pounds) 1 462.62 <= 3500
LHS Sign RHS
Go to file 3-7.xls
Dr. Chen, Decision Support Systems 48
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
Dr. Chen, Decision Support Systems 53
LP Model for a Blending Problem
Objective FunctionMin 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
Dr. Chen, Decision Support Systems 54
Whole Food Nutrition Center
A B C
Grain
AGrain
BGrain
C
Number of pounds 0.025 0.050 0.050
Cost $0.33 $0.47 $0.38 $0.05
Constraints:
Protein 22 28 21 3.00 >= 3
Riboflavin 16 14 25 2.35 >= 2
Phosphorus 8 7 9 1.00 >= 1
Magnesium 5 6 0.425 >= 0.425
Total Mix 1 1 1 0.125 = 0.125
LHS Sign RHSfile 3-9.xls
Dr. Chen, Decision Support Systems 55
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
Dr. Chen, Decision Support Systems 56
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)
Dr. Chen, Decision Support Systems 57
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
Dr. Chen, Decision Support Systems 59
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
Dr. Chen, Decision Support Systems 60
Production and Inventory Balance
(inventory at end of previous period)
+ (production the period)
- (sales this period)
= (inventory at end of this period)
Dr. Chen, Decision Support Systems 61
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)
Dr. Chen, Decision Support Systems 62
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
Dr. Chen, Decision Support Systems 63
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)
Dr. Chen, Decision Support Systems 64
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
Dr. Chen, Decision Support Systems 65
LP model for a Multiperiod SchedulingMin 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 = 1400IA4 = 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 > 0Go to file 3-11.xls
Dr. Chen, Decision Support Systems 66
Greenberg Motors
PA1 IA1 PA2 IA2 PA3 IA3 PA4 IA4 PB1 IB1 PB2 IB2 PB3 IB3 PB4 IB4
GM3A
Jan prodGM3A Jan inv
GM3A Feb prod
GM3A Feb inv
GM3A Mar prod
GM3A Mar inv
GM3A Apr prod
GM3A Apr inv
GM3B Jan prod
GM3B Jan
invGM3B
Feb prod
GM3B
Feb inv
GM3B Mar prod
GM3B
Mar inv
GM3B Apr prod
GM3B Apr inv
Number of Units 1,276.92 476.92 1,138.46 915.38 842.31 757.69 792.31 450.00 1,000.00 0.00 1,200.00 0.00 1,400.00 0.00 1,700.00 300.00
Cost $10.00 $0.18 $10.00 $0.18 $11.00 $0.18 $11.00 $0.18 $6.00 $0.13 $6.00 $0.13 $6.60 $0.13 $6.60 $0.13 $76,301.62
Constraints:
GM3A Jan balance 1 -1 800.00 = 800
GM3B Jan balance 1 -1 1000.00 = 1000
GM3A Feb balance 1 1 -1 700.00 = 700
GM3B Feb balance 1 1 -1 1200.00 = 1200
GM3A Mar balance 1 1 -1 1000.00 = 1000
GM3B Mar balance 1 1 -1 1400.00 = 1400
GM3A Apr balance 1 1 -1 1100.00 = 1100
GM3B Apr balance 1 1 -1 1400.00 = 1400
GM3A Apr Inventory 1 450.00 = 450
GM3B Apr Inventory 1 300.00 = 300
Jan storage cap 1 1 476.92 <= 3300
Feb storage cap 1 1 915.38 <= 3300
Mar storage cap 1 1 757.69 <= 3300
Apr storage cap 1 1 750.00 <= 3300
Jan labor max 1.3 0.9 2560.00 <= 2560
Feb labor max 1.3 0.9 2560.00 <= 2560
Mar labor max 1.3 0.9 2355.00 <= 2560
Apr labor max 1.3 0.9 2560.00 <= 2560
Jan labor min 1.3 0.9 2560.00 >= 2240
Feb labor min 1.3 0.9 2560.00 >= 2240
Mar labor min 1.3 0.9 2355.00 >= 2240
Apr labor min 1.3 0.9 2560.00 >= 2240
LHSSign RHS