Linear Programming Models: Graphical and Computer Methods

Using Excel’s Solver for LP Flair Furniture Example Other Examples Examples provided in the Excel LPExamplesFile

Using EXCEL Solver Setup the problem Determine the variables Determine the solution cells Objective function coefficients Define the objective function Constraint coefficiants Coefficient equations (Left-Hand Side, LHS) Coefficient signs Coefficient constraint values (RHS)

the Flair Furniture Example: Max 7T + 5C (profit) Subject to the constraints: 3T + 4C < 2400 (carpentry hrs) 2T + 1C < 1000 (painting hrs) C < 450 (max # chairs) T > 100 (min # tables) T, C > 0 (nonnegativity) Go to LPExamplesFile.xls

Flair Furniture EXCEL Ex. See the example in the Excel LPExamplesFile.

Linear Programming Modeling Applications

Linear Programming (LP) Can Be Used for Many Managerial Decisions: Product mix Make-buy Media selection Marketing research Portfolio selection Labor planning Shipping & transportation Blending Multiperiod scheduling

For a particular application we begin with the problem scenario and data, then: Define the decision variables Formulate the LP model using the decision variables Write the objective function equation Write each of the constraint equations Implement the model in Excel 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 Labor cost is $0.75 per tie Material Cost per yard Yards available per month Silk $20 1,000 Polyester $6 2,000 Cotton $9 1,250 Labor cost is $0.75 per tie

Product Data Type 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 Monthly Maximum 7,000 14,000 16,000 8,500 Total material (yards per tie) 0.125 0.08 0.10

Material Requirements (yards per tie) Type of Tie Silk Polyester Blend 1 (50/50) Blend 2 (30/70) 0.125 0.08 0.05 0.03 Cotton 0.07 Total yards 0.125 0.08 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 (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 Fifth Ave Industries in LPExamplesFile

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 Investment 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

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 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 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 International City in LPExamplesFile

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

Total minimum daily requirement is 112 hours 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 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

Work 4 consecutive hours (no lunch break) 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 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

Part Time Hours Cannot Exceed 56 Hours 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 > 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

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 $ / 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 0.448W4 + 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 Goodman Shipping in LPExamplesFile

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

Minimum Daily Requirement 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 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

Total Blend is 2 Ounces, or 0.125 Pounds 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 Whole Foods in LPExamplesFile

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 IAt = inventory of motor A at end of IBt = inventory of motor B at end of

Sales Demand Data Month Motor A B 1 (January) 800 1000 2 (February) 700 1200 3 (March) 1400 4 (April) 1100

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 must be between 2240 and 2560

Max inventory is 3300 motors Inventory Data Motor A B Inventory cost (per motor per month) $0.18 $0.13 Beginning inventory (beginning of month 1) 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 (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 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 Try to do this one on your own.