Managerial decision modeling with spreadsheets by stair render chapter 03

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) 2) Define the decision variables Formulate the LP model using the decision variables • • 3) 4) 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 types of men's ties Use 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 Blend $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 Polyester (50/50) Blend (30/70) Silk 0.125 0 Polyester 0.08 0.05 0.03 Cotton 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 ties to make per month B2 = number of poly-cotton blend 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) Blending Problem: Whole Food Nutrition Center Making a natural cereal that satisfies minimum daily nutritional requirements Decision: How much of each of grains to include in the cereal? Objective: Minimize cost of a 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 Riboflavin per pound 16 14 25 Phosphorus per pound Magnesium per pound 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 ounce serving of cereal Objective Function (in $ of cost) Min 0.33A + 0.47B + 0.38C Subject to the constraints: Total Blend is Ounces, or 0.125 Pounds A + B + C = 0.125 (lbs) Minimum Nutritional Requirements 22A + 28B + 21C > (protein) 16A + 14B + 25C > (riboflavin) 8A + 7B + 9C > (phosphorus) 5A + 6C > 0.425 (magnesium) Finally nonnegativity: A, B, C > Go to file 3-9.xls Multiperiod Scheduling: Greenberg Motors Need to schedule production of electrical motors for each of the next 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 (January) 800 1000 (February) 700 1200 (March) 1000 1400 (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 and • 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 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 + 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 + PA1 – 800 = IA1 (month 1) + 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 > Go to file 3-11.xls ... scheduling For a particular application we begin with the problem scenario and data, then: 1) 2) Define the decision variables Formulate the LP model using the decision variables • • 3) 4) Write the objective...Linear Programming (LP) Can Be Used for Many Managerial Decisions: • • • • • • • Product mix Make-buy Media selection Marketing research Portfolio... Implement the model in Excel Solve with Excel’s Solver Product Mix Problem: Fifth Avenue Industries • • Produce types of men's ties Use materials (limited resources) Decision: How many of each type