Transcript 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
[email protected]
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
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3
Manufacturing Applications
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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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
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Product Data
Type of Tie
Silk
Selling Price
(per tie)
Monthly
Minimum
Monthly
Maximum
Total material
(yards per tie)
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Polyester Blend 1 Blend 2
$6.70
$3.55
$4.31
$4.81
6,000
10,000
13,000
6,000
7,000
14,000
16,000
8,500
0.125
0.08
0.10
0.10
6
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
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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
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< 1,250 (cotton)
10
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
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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
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Fifth Avenue Industries
S
P
B1
B2
All silk
All poly
Blend-1
Blend2
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:
Yards of silk
Cost/Yd
0.125
Yards of polyester
0.08
Yards of cotton
Maximum all silk
Minimum all poly
Minimum blend-1
<=
2000
$6
0.05
0.07
1250.00
<=
1250
$9
7000.00
<=
7000
13625.00
<=
14000
13100.00
<=
16000
8500.00
<=
8500
7000.00
>=
6000
13625.00
>=
10000
13100.00
>=
13000
8500.00
>=
6000
1
1
1
1
1
Minimum blend-2
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$20
2000.00
Maximum blend-2
Minimum all silk
1000
0.03
1
Maximum blend-1
<=
0.05
1
Maximum all poly
875.00
1
Go to file 3-1.xls
LHS
Sign
RHS
13
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
TV Spot
Advertising Options
Radio
Newspaper (prime time)
Radio
(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
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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 $8000
800T + 925N + 290P + 380A < 8000
At Least 5 Radio Spots per Week
P+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
N< 5
Finally nonnegativity
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P < 25
A < 20
T, N, P, A > 0
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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
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Win Big Gambling Club
T
N
P
A
TV
spots
Newspap
er ads
Primetime 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
Maximum newspaper
1
Max prime-time radio
1
Max afternoon radio
Total budget
Maximum radio $
Minimum radio spots
1
$800
$925
1.97
<=
12
5.00
<=
5
6.21
<=
25
0.00
<=
20
$290
$380
$8,000.00
<=
$8,000
$290
$380
$1,800.00
<=
$1,800
1
1
6.21
>=
5
LHS
Sign
RHS
Go to file 3-3.xls
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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
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
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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 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
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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 2.0
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
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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
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Go to file 3-5.xls
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International City Trust
T
B
G
P
M
C
Trade credits
Corp bonds
Gold
Platinum
Mortgages
Const loans
$1,250,000.00
$1,250,000.00
$250,000.00
$1,250,000.00
$500,000.00
$500,000.00
0.07
0.10
0.19
0.12
0.08
0.14
Total funds
1
1
1
1
1
1
Max trade credits
1
Dollars Invested
Interest
$520,000.00
Constraints:
Max corp bonds
1
Max gold
1
Max mortgages
$5,000,000
$1,250,000.00
<=
$1,250,000
$1,250,000.00
<=
$1,250,000
$250,000.00
<=
$1,250,000
$1,250,000.00
<=
$1,250,000
$500,000.00
<=
$1,250,000
$500,000.00
<=
$1,250,000
10,000,000.00
<=
10,000,000
$1,500,000.00
>=
$1,500,000
$2,500,000.00
>=
$2,250,000
LHS
Sign
RHS
1
Max const loans
1
1.7
1.2
Precious metals
Trade credits & bonds
<=
1
Max platinum
Risk score
$5,000,000.00
1
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1
3.7
2.4
1
1
2.0
2.9
Go to file 3-5.xls
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Employee Staffing Application
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
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
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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
F + P1 + P2
0.5 F + P1 + P2 + P3
0.5 F + P1 + P2 + P3+ P4
F + P2 + P3+ P4 + P5
F + P3+ P4 + P5
F + P4 + P5
F + P5
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> 10
> 12
> 14
> 16
> 18
> 17
> 15
> 10
(9-10)
(10-11)
(11-12)
(12-1)
(1-2)
(2-3)
(3-4)
(4-5)
37
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
F + P1 + P2
0.5 F + P1 + P2 + P3
0.5 F + P1 + P2 + P3+ P4
F + P2 + P3+ P4 + P5
F + P3+ P4 + P5
F + P4 + P5
F + P5
F
4 (P1 + P2 + P3 + P4 + P5)
F, P1, P2, P3, P4, P5 > 0
Dr. Chen, Decision Support Systems
> 10
> 12
> 14
> 16
> 18
> 17
> 15
> 10
< 12
< 56
(9-10)
(10-11)
(11-12)
(12-1)
(1-2)
(2-3)
(3-4)
(4-5)
Go to file 3-6.xls
39
Hong Kong Bank
F
P1
P2
P3
P4
P5
FT
tellers
PT
@9am
PT
@10am
PT
@11am
PT
@Noon
PT
@1pm
10.0
0.0
7.0
2.0
5.0
0.0
$90.00
$28.00
$28.00
$28.00
$28.00
$28.00
9am-10am needs
1
1
10am-11am needs
1
1
1
11am-Noon needs
0.5
1
1
1
Noon-1pm needs
0.5
1
1
1
1
1pm-2pm needs
1
1
1
1
2pm-3pm needs
1
1
3pm-4pm needs
1
4pm-5pm needs
1
Max full time
1
Number of tellers
Cost
$1,292.00
Constraints:
Part-time limit
4
4
4
10.0
>=
10
17.0
>=
12
14.0
>=
14
19.0
>=
16
1
24.0
>=
18
1
1
17.0
>=
17
1
1
15.0
>=
15
1
10.0
>=
10
10.0
<=
12
56.0
<=
56
4
4
LHS
Sign
RHS
Go to file 3-6.xls
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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
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
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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 Available
W1 < 5000
W4 < 3500
W2 < 4500
W5 < 4000
W3 < 3000
W6 < 3500
Finally nonnegativity: Wi > 0, i=1,…,6
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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 + W 2 + W3 + W4 + W 5 + W6
0.125W1 + 0.064W2 + 0.144W3
+0.448W4 + 0.048W5 + 0.018W6
Pounds of Each Item Available
W1
W2
W3
W4
W5
W6
Wi > 0, i=1,…,6
Dr. Chen, Decision Support Systems
< 15,000 (Weight Limit)
< 1300 (volume limit of truck)
< 5000 (item 1 availability)
< 4500 (item 2 availability)
< 3000 (item 3 availability)
< 3500 (item 4 availability)
< 4000 (item 5 availability)
< 3500 (item 6 availability)
Go to file 3-7.xls
46
Goodman Shipping
W1
W2
W3
W4
W5
W6
Item 1
Item 2
Item 3
Item 4
Item 5
Item 6
3,037.38
4,500.00
3,000.00
0.00
4,000.00
462.62
$3.10
$3.20
$3.45
$4.15
$3.25
$2.75
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
3037.38
<=
5000
4500.00
<=
4500
3000.00
<=
3000
0.00
<=
3500
4000.00
<=
4000
462.62
<=
3500
Weight in pounds
Load value
$48,438.08
Constraints:
Item 1 limit (pounds)
Item 2 limit (pounds)
Item 3 limit (pounds)
Item 4 limit (pounds)
Item 5 limit (pounds)
1
1
1
1
1
Item 6 limit (pounds)
1
LHS
Sign
RHS
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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48
$ per pound
Protein per
pound
Riboflavin per
pound
Phosphorus
per pound
Magnesium
per pound
Dr. Chen, Decision Support Systems
A
Grain
B
C
22
28
21
3
16
14
25
2
8
7
9
1
5
0
6
0.425
Minimum
Daily
$0.33 $0.47 $0.38
Requirement
49
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
Dr. Chen, Decision Support Systems
(lbs)
51
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
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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
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Whole Food Nutrition Center
A
B
C
Grain
A
Grain
B
Grain
C
Number of
pounds
0.025
0.050
0.050
Cost
$0.33
$0.47
$0.38
$0.05
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
0.125
=
0.125
LHS
Sign
RHS
Constraints:
1
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
Motor
A
B
Month
1 (January)
800
1000
2 (February)
700
1200
3 (March)
1000 1400
4 (April)
1100
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57
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
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Inventory Data
Motor
A
B
Inventory cost
$0.18 $0.13
(per motor per month)
Beginning inventory
0
0
(beginning of month 1)
Ending Inventory
450
300
(end of month 4)
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
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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)
Go to file 3-11.xls
PAi, PBi, IAi, IBi > 0
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Greenberg Motors
Number of Units
Cost
PA1
IA1
PA2
IA2
PA3
IA3
PA4
IA4
PB1
IB1
PB2
IB2
PB3
IB3
PB4
IB4
GM3A
Jan prod
GM3A
Jan inv
GM3A
Feb prod
GM3A
Feb inv
GM3A
Mar
prod
GM3A
Mar inv
GM3A
Apr
prod
GM3A
Apr inv
GM3B
Jan prod
GM3
B Jan
inv
GM3B
Feb prod
GM3
B
Feb
inv
GM3B
Mar prod
GM3
B
Mar
inv
GM3B
Apr prod
GM3B
Apr inv
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
$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
1
-1
1
-1
$76,301.62
Constraints:
GM3A Jan balance
GM3B Jan balance
GM3A Feb balance
1
1
-1
GM3B Feb balance
1
GM3A Mar balance
1
1
1
-1
-1
GM3B Mar balance
1
GM3A Apr balance
1
1
1
-1
-1
GM3B Apr balance
1
GM3A Apr Inventory
1
GM3B Apr Inventory
1
Jan storage cap
1
1
Feb storage cap
1
1
Mar storage cap
1
1
Apr storage cap
Jan labor max
1
1.3
Feb labor max
1
0.9
1.3
Mar labor max
0.9
1.3
Apr labor max
Jan labor min
-1
1
0.9
1.3
1.3
Feb labor min
Mar labor min
Apr labor min
0.9
0.9
1.3
0.9
1.3
0.9
1.3
0.9
800.00
=
800
1000.00
=
1000
700.00
=
700
1200.00
=
1200
1000.00
=
1000
1400.00
=
1400
1100.00
=
1100
1400.00
=
1400
450.00
=
450
300.00
=
300
476.92
<=
3300
915.38
<=
3300
757.69
<=
3300
750.00
<=
3300
2560.00
<=
2560
2560.00
<=
2560
2355.00
<=
2560
2560.00
<=
2560
2560.00
>=
2240
2560.00
>=
2240
2355.00
>=
2240
2560.00
>=
2240
Sig
n
RHS
LHS
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