Chapter.3 Linear programming Modeling applications
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
Resource Data
Material
Silk
Yards available
Cost per yard
per month
$20
1,000
Polyester
$6
2,000
Cotton
$9
1,250
Labor cost is $0.75 per tie
Product Data
Type of Tie
Silk
Selling Price
Polyester Blend 1 Blend 2
$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
0.125
0.08
0.10
0.10
(per tie)
Total material
(yards per tie)
Material Requirements
(yards per tie)
Type of Tie
Material
Silk
Blend 1
Polyester
(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
0.125
0.08
0.10
0.10
Total yards
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
Radio
Radio
TV Spot
Newspaper
(prime time)
(afternoon)
Audience
Reached
(per ad)
5,000
8,500
2,400
2,800
Cost
(per ad)
$800
$925
$290
$380
Max Ads
Per 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 $8000
800T + 925N + 290P + 380A < 8000
At Least 5 Radio Spots per Week
P+A>5
No More Than $1800 per Week for Radio
290P + 380A < 1800
Max Number of Ads per Week
T < 12
N< 5
Finally nonnegativity
P < 25
A < 20
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
Interest
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
Investment
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 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
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 2.0
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
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
9 – 10
10 – 11
11 – 12
Min Num. Tellers
10
12
14
12 – 1
1–2
2-3
3–4
4–5
16
18
17
15
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
F + P 1 + P2
0.5 F + P1 + P2 + P3
0.5 F + P1 + P2 + P3+ P4
F + P 2 + P 3 + P4 + P5
F + P3+ P4 + P5
F + P 4 + P5
F + P5
> 10
(9-10)
> 12
(10-11)
> 14
(11-12)
> 16(12-1)
> 18
(1-2)
> 17
(2-3)
> 15
(3-4)
> 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
6
Value $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 Available
W1 < 5000
W4 < 3500
W2 < 4500
W5 < 4000
W3 < 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
A
$ per pound
Grain
B
C
Minimum
$0.33 $0.47 $0.38
Daily
Requirement
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
Motor
A
B
Month
1 (January)
800
1000
2 (February)
700
1200
3 (March)
1000 1400
4 (April)
1100 1400
Production Data
Motor
(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
$0.18 $0.13
(per motor per month)
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
IB3 + PB4 – 1400 = IB4
(month 4)
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
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
Resource Data
Material
Silk
Yards available
Cost per yard
per month
$20
1,000
Polyester
$6
2,000
Cotton
$9
1,250
Labor cost is $0.75 per tie
Product Data
Type of Tie
Silk
Selling Price
Polyester Blend 1 Blend 2
$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
0.125
0.08
0.10
0.10
(per tie)
Total material
(yards per tie)
Material Requirements
(yards per tie)
Type of Tie
Material
Silk
Blend 1
Polyester
(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
0.125
0.08
0.10
0.10
Total yards
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
Radio
Radio
TV Spot
Newspaper
(prime time)
(afternoon)
Audience
Reached
(per ad)
5,000
8,500
2,400
2,800
Cost
(per ad)
$800
$925
$290
$380
Max Ads
Per 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 $8000
800T + 925N + 290P + 380A < 8000
At Least 5 Radio Spots per Week
P+A>5
No More Than $1800 per Week for Radio
290P + 380A < 1800
Max Number of Ads per Week
T < 12
N< 5
Finally nonnegativity
P < 25
A < 20
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
Interest
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
Investment
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 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
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 2.0
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
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
9 – 10
10 – 11
11 – 12
Min Num. Tellers
10
12
14
12 – 1
1–2
2-3
3–4
4–5
16
18
17
15
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
F + P 1 + P2
0.5 F + P1 + P2 + P3
0.5 F + P1 + P2 + P3+ P4
F + P 2 + P 3 + P4 + P5
F + P3+ P4 + P5
F + P 4 + P5
F + P5
> 10
(9-10)
> 12
(10-11)
> 14
(11-12)
> 16(12-1)
> 18
(1-2)
> 17
(2-3)
> 15
(3-4)
> 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
6
Value $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 Available
W1 < 5000
W4 < 3500
W2 < 4500
W5 < 4000
W3 < 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
A
$ per pound
Grain
B
C
Minimum
$0.33 $0.47 $0.38
Daily
Requirement
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
Motor
A
B
Month
1 (January)
800
1000
2 (February)
700
1200
3 (March)
1000 1400
4 (April)
1100 1400
Production Data
Motor
(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
$0.18 $0.13
(per motor per month)
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
IB3 + PB4 – 1400 = IB4
(month 4)
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