# Chapter 3: Linear Programming Modeling Applications © 2007 Pearson Education.

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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

MaterialCost per yard Yards available per month Silk\$201,000 Polyester\$62,000 Cotton\$91,250 Resource Data Labor cost is \$0.75 per tie

Product Data Type of Tie SilkPolyesterBlend 1Blend 2 Selling Price (per tie) \$6.70\$3.55\$4.31\$4.81 Monthly Minimum 6,00010,00013,0006,000 Monthly Maximum 7,00014,00016,0008,500 Total material (yards per tie) 0.1250.080.10

Material Requirements (yards per tie) Material Type of Tie SilkPolyester Blend 1 (50/50) Blend 2 (30/70) Silk0.125000 Polyester00.080.050.03 Cotton000.050.07 Total yards0.1250.080.10

Decision Variables S = number of silk ties to make per month P = number of polyester ties to make per month B 1 = number of poly-cotton blend 1 ties to make per month B 2 = 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 + 2.81B 1 + 3.25B 2 Subject to the constraints: Material Limitations (in yards) 0.125S< 1,000 (silk) 0.08P + 0.05B 1 + 0.03B 2 < 2,000 (poly) 0.05B 1 + 0.07B 2 < 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

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 < 12P < 25 N < 5A < 20 Finally nonnegativity 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 Investment Interest RateRisk Score Trade credits7%1.7 Corp. bonds10%1.2 Gold stocks19%3.7 Platinum stocks12%2.4 Mortgage securities8%2.0 Construction loans14%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 < 0.25 (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 PeriodMin Num. Tellers 9 – 1010 10 – 1112 11 – 1214 12 – 116 1 – 218 2 - 317 3 – 415 4 – 510 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) P 1 = num. of part time tellers who work 9–1 P 2 = num. of part time tellers who work 10–2 P 3 = num. of part time tellers who work 11–3 P 4 = num. of part time tellers who work 12–4 P 5 = num. of part time tellers who work 1–5

Objective Function (in \$ of personnel cost) Min 90 F + 28 (P 1 + P 2 + P 3 + P 4 + P 5 ) Subject to the constraints: Part Time Hours Cannot Exceed 56 Hours 4 (P 1 + P 2 + P 3 + P 4 + P 5 ) < 56

Minimum Num. Tellers Needed By Hour Time of Day F + P 1 > 10 (9-10) F + P 1 + P 2 > 12 (10-11) 0.5 F + P 1 + P 2 + P 3 > 14 (11-12) 0.5 F + P 1 + P 2 + P 3 + P 4 > 16 (12-1) F + P 2 + P 3 + P 4 + P 5 > 18 (1-2) F + P 3 + P 4 + P 5 > 17 (2-3) F + P 4 + P 5 > 15 (3-4) F + P 5 > 10 (4-5)

Only 12 Full Time Tellers Available F < 12 finally nonnegativity: F, P 1, P 2, P 3, P 4, P 5 > 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 123456 Value \$15,500\$14,400\$10,350\$14,525\$13,000\$9,625 Pounds 500045003000350040003500 \$ / lb \$3.10\$3.20\$3.45\$4.15\$3.25\$2.75 Cu. ft. per lb 0.1250.0640.1440.4480.0480.018

Decision Variables W i = 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 3.10W 1 + 3.20W 2 + 3.45W 3 + 4.15W 4 + 3.25W 5 + 2.75W 6 Subject to the constraints: Weight Limit Of 15,000 Pounds W 1 + W 2 + W 3 + W 4 + W 5 + W 6 < 15,000

Volume Limit Of 1300 Cubic Feet 0.125W 1 + 0.064W 2 + 0.144W 3 + 0.448W 4 + 0.048W 5 + 0.018W 6 < 1300 Pounds of Each Item Available W 1 < 5000W 4 < 3500 W 2 < 4500 W 5 < 4000 W 3 < 3000W 6 < 3500 Finally nonnegativity: W i > 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

Grain Minimum Daily Requirement ABC \$ per pound\$0.33\$0.47\$0.38 Protein per pound 2228213 Riboflavin per pound 1614252 Phosphorus per pound 8791 Magnesium per pound 5060.425

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 P At = number of motor A to produce in month t (t=1,…,4) P Bt = number of motor B to produce in month t (t=1,…,4) I At = inventory of motor A at end of month t (t=1,…,4) I Bt = inventory of motor B at end of month t (t=1,…,4)

Sales Demand Data Month Motor AB 1 (January) 8001000 2 (February) 7001200 3 (March) 10001400 4 (April) 11001400

Production Data Motor (values are per motor) AB Production cost\$10\$6 Labor hours1.30.9 Production costs will be 10% higher in months 3 and 4 Monthly labor hours most be between 2240 and 2560

Inventory Data Motor AB Inventory cost (per motor per month) \$0.18\$0.13 Beginning inventory (beginning of month 1) 00 Ending Inventory (end of month 4) 450300 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 10P A1 + 10P A2 + 11P A3 + 11P A4 + 6P B1 + 6 P B2 + 6.6P B3 + 6.6P B4 + 0.18( I A1 + I A2 + I A3 + I A4 ) + 0.13( I B1 + I B2 + I B3 + I B4 ) Subject to the constraints: (see next slide)

Production & Inventory Balance 0 + P A1 – 800 = I A1 (month 1) 0 + P B1 – 1000 = I B1 I A1 + P A2 – 700 = I A2 (month 2) I B1 + P B2 – 1200 = I B2 I A2 + P A3 – 1000 = I A3 (month 3) I B2 + P B3 – 1400 = I B3 I A3 + P A4 – 1100 = I A4 (month 4) I B3 + P B4 – 1400 = I B4

Ending Inventory I A4 = 450 I B4 = 300 Maximum Inventory level I A1 + I B1 < 3300(month 1) I A2 + I B2 < 3300(month 2) I A3 + I B3 < 3300(month 3) I A4 + I B4 < 3300(month 4)

Range of Labor Hours 2240 < 1.3P A1 + 0.9P B1 < 2560 (month 1) 2240 < 1.3P A2 + 0.9P B2 < 2560 (month 2) 2240 < 1.3P A3 + 0.9P B3 < 2560 (month 3) 2240 < 1.3P A4 + 0.9P B4 < 2560 (month 4) finally nonnegativity: P Ai, P Bi, I Ai, I Bi > 0 Go to file 3-11.xls