Chapter 3: Linear Programming Modeling Applications Jason C. H. Chen, Ph.D. Professor of MIS School...

Chapter 3:Linear Programming Modeling Applications

Jason C. H. Chen, Ph.D.Professor of MIS

School of Business AdministrationGonzaga UniversitySpokane, WA 99223

[email protected]

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


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


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


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


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


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


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


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


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


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


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


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


Budget is $8000800T + 925N + 290P + 380A < 8000

At Least 5 Radio Spots per WeekP + A > 5


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

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


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


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


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


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 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


Objective Function (in $ of interest earned)

Max 0.07T + 0.10B + 0.19G + 0.12P

+ 0.08M + 0.14C


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 2.0Use 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


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


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


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


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


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






Objective Function (in $ of personnel cost)

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


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 > 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)


Only 12 Full Time Tellers Available

F < 12

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


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


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


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


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


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


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 AvailableW1 < 5000 W4 < 3500W2 < 4500 W5 < 4000W3 < 3000 W6 < 3500

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


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


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






LHS Sign RHS

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


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


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


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


LP Model for a Blending Problem

Objective FunctionMin 0.33A + 0.47B + 0.38C


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


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


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

Month

Motor

A B

1 (January) 800 1000

2 (February) 700 1200

3 (March) 1000 1400

4 (April) 1100 1400


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


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


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)


(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 (month 4)

IB3 + PB4 – 1400 = IB4


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


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


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

Date post:	27-Dec-2015
Category:	Documents
Upload:	andrew-harris
View:	241 times
Download:	0 times

Chapter 3: Linear Programming Modeling Applications Jason C. H. Chen, Ph.D. Professor of MIS School...

Documents