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Chapter 13: Optimization Modeling 1. For each of the examples in this chapter, classify the constraints into the following categories. Are there any other types of constraints that do not fall into these categories? a. simple bounds b. limitations c. requirements d. proportional relationships e. balance constraints Product mix example: 6G + 6P 3,000 limitations 3G + 5P 1750 limitations Media Selection example: 500R + 2000T + 200N 40000 limitations 0 R 15 simple bounds T 12 simple bounds 6 N 30 simple bounds Process Selection example: D1 + P1 = 45000 requirements D2 + R2 + P2 = 76500 requirements D3 + R3 + P3 = 10000 requirements 0.213D1 + 0.192D2 + 0.227D3 6552 limitations 0.192R2 + 0.227R3 32760 limitations Blending example: X 1 + X 2 + X3 + X4 + X 5 + X6 + X7 + X8 = 1 requirements 0.169X1 + 0.12X2 + 0.085X3 + 0.154X4 + 0.085X5 +0 .12X6 + 0.18X7 + 0.119X8 0.13 relationships proportional 0.26X1 + 0.041X2 + 0.038X3 + 0.063X4 + 0.038X5 + 0.017X6 + 0.179X7 + 0.04X8 0.15relationships proportional 0.29X1 + 0.083X2 + 0.027X3 + 0.024X4 + 0.027X5 + 0.023X6 + 0.288X7 + 0.109X8 0.14 proportional relationships Production Planning example: PA + 0 = 150 + IA balance constraints PW + IA = 400 + IW balance constraints PS + IW = 50 + IS balance constraints Cash Management example: 200,000 – (A1 + B1 + C1 + 50,000) 100,000 requirements 1.005A1 + 12,000 – (A2 + B2) 100,000 requirements 100,000 1.005A2 – (A3 + B3 + 23,000) requirements 1.005A3 + 1.0175B1 + 20,000 - (A4 + B4 ) 100,000 requirements 1.005A4 + 1.0175B2 - (A5 + 41,000) 100,000 requirements 1.005A5 + 1.0175B3 + 13,000 - A6 100,000 requirements Transportation Problem example: X11 + X12 + X13 + X14 1200 limitations X21 + X22 + X23 + X24 800 limitations X11 + X21 = 150 requirements X12 + X22 = 350 requirements X13 + X23 = 500 requirements X14 + X24 = 1000 requirements 2. Explain the following terms: optimization, objective function, optimal solution, constraint, constraint function, feasible solution, infeasible solution. Optimization - the process of selecting values for decision variables that minimize or maximize some quantity of interest. Objective function - The quantity we seek to minimize or maximize through optimization. Optimal solution - the set of decision variable values that maximize or minimize the objective function. Constraint - limitations or requirements that decision variables must satisfy. Constraint function - Constraints are generally expressed mathematically as equations or inequalities. When represented mathematically, a constraint function is a function of the decision variables. Feasible solution - any solution that satisfies all constraints of a problem. Infeasible solution - an optimization problem that has no feasible solutions. 3. Describe the differences among linear, integer, and nonlinear optimization problems. Linear - has two basic properties. First, the objective function and all constraints are linear functions of the decision variables. This means that each function is simply a sum of terms, each of which is some constant multiplied by a decision variable and the second property of a linear optimization problem is that all variables are continuous, meaning that they may assume any real value (typically, nonnegative). Integer - some or all of the variables are restricted to being whole numbers. Nonlinear - the objective function and/or constraint functions are nonlinear functions of the decision variables; that is, terms cannot be written as a constant times a variable. 4. List and describe the major categories in which constraints generally fall. Simple Bounds. Simple bounds constrain the value of a single variable. Limitations. Limitations usually involve the allocation of scarce resources. Requirements. Requirements involve the specification of minimum levels of performance. Proportional Relationships. Proportional relationships are often found in problems involving mixtures or blends of materials or strategies. Balance Constraints. Balance constraints essentially state that “input = output” and ensure that the flow of material or money is accounted for at locations or between time periods. 5. List the important guidelines to follow for modeling optimization problems on spreadsheets. Put the objective function coefficients, constraint coefficients, and right-hand-side values in a logical format in the spreads Define a set of cells (either rows or columns) for the values of the decision variables. Define separate cells for the objective function and each constraint function (the left-hand side of a constraint). 6. What Excel functions should you avoid when implementing linear optimization models on spreadsheets? ABS MIN MAX INT ROUND IF COUNT 7. Explain the difference between general integer variables and binary variables. Decision variables that we force to be integers are called general integer variables Binary variables, are variables that are restricted to being either 0 or 1. 8. How can binary variables be used to model logical conditions? Provide several examples. Binary variables enable us to model logical decisions in optimization models. For example, binary variables can be used to model decisions such as whether (x = 1) or not (x = 0) to place a facility at a certain location, whether or not to run a production line, or whether or not to invest in a certain stock. One common example is project selection or capital budgeting, in which a subset of potential projects must be selected with limited resource constraints. es. Are there a logical format in the spreadsheet. side of a constraint). preadsheets? Chapter 13: Optimization Modeling 1. A manufacturer of office equipment produces two types of desks, standard and deluxe. Deluxe desks have oak tops and more expensive hardware, and require additional time for finishing and polishing. Standard desks require 80 square feet of pine wood, 12 hours of labor, while deluxe desks require 62 square feet of pine, 18 square feet of oak, and 18 hours of labor. For the next week, the company has 5000 square feet of pine, 700 square feet of oak, and 400 hours of labor available. Standard desks net a profit of $75 while deluxe desks net a profit of $160. All desks can be sold to national chains such as Staples or Office Depot. Develop a model to determine how many of each the company should make next week to maximize profit contribution. S = units of standard desk produced D = units of deluxe desk produced Max 75S + 160D subject to: 80S + 62D <= 5000 18D <= 700 12S + 18D <= 400 S, D >= 0 Chapter 13: Optimization Modeling 2. Metropolitan Airport Services, Inc. is considering purchasing cars for transportation service between the Municipal Airport and hotels in the downtown area. They are considering station wagons, minibuses, and large buses. They are considering station wagons, minibuses, and large buses. Purchase prices are $45,000 for each station wagon, $90,000 for each minibus, and $200,000 for each large bus. The Board of Directors has authorized a maximum budget of $5,000,000 for purchases. Because of the heavy air travel in the area, these vehicles would be utilized at maximum capacity regardless of the type of vehicle purchased. Expected net annual profit would be $1,500 per station wagon, $3,500 per minibus, and $5,000 per large bus. The company has hired 30 new drivers for these vehicles, each qualified to drive any one of the three vehicle types. The maintenance department has the capacity to handle an additional 80 station wagons. A minibus is equivalent to 1 2/3 station wagons for maintenance purposes, and a large bus is equivalent to 2 station wagons for maintenance purposes. Develop a linear optimization model to determine the optimal number of each type of vehicle to purchase in order to maximize profit. S = number of station wagons purchased M = number of mini buses purchased L = number of large buses purchased Max 1,500S + 3,500M + 5,000L subject to: 45,000S + 90,000M + 200,000L <= 5,000,000 S + M + L <= 30 S + 5/3M + 2L <= 80 S, M, L >= 0 Chapter 13: Optimization Modeling 3. The International Chef, Inc. markets three blends of oriental tea: premium, Duke Grey, and breakfast. The firm uses tea leaves from India, China, and new domestic California sources. Tea Leaves Quality Indian Chinese California Premium 40% 20% 40% Duke Grey 20% 30% 40% Breakfast 20% 40% 40% Net profit per pound for each blend is $0.50 for premium, $0.30 for Duke Grey, and $0.20 for breakfast. The firm's regular weekly supplies are 20,000 pounds of Indian tea leaves and 22,000 pounds of Chinese tea leaves. Because it is new, and until it proves itself, no more than 16,000 pounds of California tea is to be used in a week. The marketing research department reports that there is an almost unlimited market for premium and fine blends. However, the maximum expected sale for breakfast blend is 2,000 pounds. Develop a linear optimization model to determine the optimal mix to maximize profit. P = pounds of premium produced D = pounds of Duke Grey produced B = pounds of breakfast produced Max .50P + .30D + .20B subject to: .40P + .20D + .20B <= 20,000 .20P + .20D + .40B <= 22,000 .20P + .30D + .40B <= 16,000 B <= 2,000 P, D, B >= 0 Chapter 13: Optimization Modeling 4. You are in the business of producing and selling 100 pound bags of health food for farm animals. You plan to advertise that each bag will provide a farm animal its minimum weekly requirements of protein (200 grams), calcium (300 grams) and fiber (100 grams), and will contain no more than 500 calories. You have found supplies at reasonable cost for three possible ingredients. Cost Protein Calcium Fiber Calories Corn $.03/lb 100 g/lb 2 g/lb 1 g/lb 50/lb Fishbones $.005/lb 1 g/lb 50 g/lb none 2/lb Sawdust $.001/lb none none 200 g/lb 1/lb You plan to sell the bag for $1. Develop an optimization model for this problem. C = lb corn in bag F = lb fishbones in bag S = lb sawdust in bag Min .03C + .005F + .001S subject to: 100C + F >= 200 2C + 50F >= 300 C + 200S >= 100 50C + 2F + 1S <= 500 C + F + S =100 C, F, S >= 0 Chapter 13: Optimization Modeling 5. Wu’s Chinese Cuisine is a small restaurant located in a shopping mall. The restaurant is so popular because of its limited menu and fresh preparation that even if some items sell out, customers will usually substitute something else so that nearly all available food is sold each day. In planning a special “lunch for two” promotion, the owner wishes to maximize the profit associated with making three dishes: vegetable shrimp, broccoli beef, and chicken green bean. Relevant data on pricing, amount of ingredients used per serving, and cost of ingredients are given below. Number of pounds/serving Menu Item Price/serving Shrimp Broccoli Green beans Beef Mixed veggies Chicken Vegetable shrimp $8.89 0.3 0.2 0.1 0.4 0 0 Broccoli beef $7.59 0.4 0 0 0.3 0.3 0 Chicken green bean $7.29 0.05 0 0.45 0.1 0 0.4 Cost/lb $5.50 $1.20 $1.20 $1.20 $5.00 $4.00 The restaurant has available 55 hours of labor per day at a cost of $12/hour, and has a contract to purchase 30 lbs shrimp, 120 lbs broccoli, 120 lbs green beans, 120 lbs mixed veggies, 60 lbs beef, and 90 lbs chicken each day. Develop a linear optimization model to maximize profit. Max 8.89VS + 7.59BB + 7.29CG Subject to: VS = .3S+.2Broc+.1GB+.4MV BB = .4S+.3MV+.3Beef CG = .05S+.45GB+.1MV+.4C L<=55($12) S<=30($5.50) Broc<=120($1.20) GB<=120($1.20) MV<=120($1.20) Beef<=60($5.00) Ch<=90($4.00) Chapter 13: Optimization Modeling 6. A manufacturer of decorative candles has contracted with a national retailer to supply a set of special holiday candles to its stores. These include large jars, small jars, large pillars, small pillars, and a package of 4 votive candles. In negotiating the contract for the display, the manufacturer and retailer agreed that 8 feet would be designated for the display in each store, but that at least 2 feet be dedicated to large jars and large pillars, and at least one foot to the votive candle packages. At least as many jars as pillars must be provided. The manufacturer has obtained 200,000 pounds of wax, 250,000 feet of wick, and 100,000 ounces of holiday fragrance. The amount of materials and display size required for each product is shown in the table below: Large jar Small jar Large pillar Small pillar Votive pack Wax 0.5 0.25 0.5 0.25 0.3125 Fragrance 0.24 0.12 0.24 0.12 0.15 Wick 0.43 0.22 0.58 0.33 0.8 Display feet 0.48 0.24 0.23 0.23 0.26 Profit/unit $0.25 $0.20 $0.24 $0.21 $0.16 How many of each product should be made to maximize the profit? Max .25LJ+.2SJ+.24LP+.21SP+.16VP subject to: .48LJ+.24SJ+.23LP+.23SP+.26VP<=8ft LJ>=2ft LP>=2ft VP>=1ft (LJ+SJ)=(LP+SP) .5LJ+.25SJ+.5LP+.25SP+.3125VP<=200,000 .24LJ+.12SJ+.24LP+.12SP+.15VP<=100,000 .43LJ+.22SJ+.58LP+.33SP+.8VP<=250,000 Chapter 13: Optimization Modeling 7. A city council is reviewing housing proposals for a new development area. There is some dispute among various interest groups as to what the goals are. The zoning committee has recommended three types of housing: single-family houses, deluxe condominiums, and apartments, and have also provided the following data: Family Condos Apartment Land usage, acres/unit 0.25 0.4 0.125 Families housed/unit 1 4 6 Tax base generated/unit $50,000 $100,000 $25,000 Utility installation expense/unit $4,000 $8,000 $6,000 Twenty acres are available for zoning. Utility installation expense is to be held to no more than $1 million. A public opinion survey has been conducted, and the city council has reviewed this survey, important issues are to provide housing to families, generate tax base, and minimize installation costs. Develop linear optimization models to maximize families housed, and also for maximizing the tax base added F = number of single family houses C = number of deluxe condominiums A = number of apartments Max F + 4C + 6A subject to: .25F + .4C + .125 A <= 20 4,000F + 8,000C + 6,000A <= 1,000,000 F, C, A >= 0 F = number of single family houses C = number of deluxe condominiums A = number of apartments Max 50,000F + 100,000C + 25,000A subject to: .25F + .4C + .125 A <= 20 4,000F + 8,000C + 6,000A <= 1,000,000 F, C, A >= 0 Chapter 13: Optimization Modeling 8. A department store chain is planning opening a new store. They need to decide how to allocate the 100,000 square feet of available floor space among seven departments. Data on expected performance of each department per month, in terms of square feet (sf) are: expected Investment Risk as a % minimum maximum profit Department /sf of $ invested sf sf per sf Electronics $100 24% 6000 30000 $12.00 Furniture 50 12 10000 30000 6 Men's Clothing 30 5 2000 5000 2 Clothing 600 10 3000 40000 30 Jewelry 900 14 1000 1000 20 Books 50 2 1000 5000 1 Appliances 400 3 12000 40000 13 The company has gathered $20,000,000 to invest in floor stock. The risk element is a measure of risk associated with investment in floor stock. The idea is that electronics loses $24/$100 invested per month, based upon past records at other places for outdated inventory, pilferage, breakage, etc. Electronics is the highest risk item. Expected profit is after covering risk. Develop a linear optimization model to maximize profit. E = sf allocated to electronics F = sf allocated to furniture M = sf allocated to men's clothing C = sf allocated to clothing J = sf allocated to jewelry B = sf allocated to books A = sf allocated to appliance R = dollars at risk Max 12E + 6F + 2M + 30C + 20J + B + 13A subject to: 100E + 50F + 30M + 600C + 900J + 50B + 400A <= 20,000,000 E + F + M + C + J + B + A <= 100,000 (.24*100)E + (.12*50)F + (.05*30)M + (.10*60)C + (.14*900)J + (.02*50)B + (.03*400)A = R E >= 6,000 F >= 10,000 M >= 2,000 C >= 3,000 J >= 1,000 B >= 1,000 A >= 12,000 E <= 30,000 F <= 30,000 M <= 5,000 C <= 40,000 J <= 1000 B <= 5,000 A <= 40,000 E, F, M, C, J, B, A >= 0 Chapter 13: Optimization Modeling 9. An energy company blends different types of coal to customer specifications. Their current inventory consists of five types of coal; the table below shows the properties of each type: BTU rating; percent of sulfur, ash, and moisture; cost/ton; and availability of each type: BTU Sulfur Ash Moisture Cost/ton Tons available Eastern High Ash 9500 0.55 31 8 $24 100,000 Northern Appalachian 13000 2.75 8.5 7 $45 75000 Central Appalachian 12500 0.75 12 7 $54 25000 Big Sandy 11500 1 15 8 $47 25000 Southern River Basin 10400 0.6 7 14 $40 12000 a. A utility has placed an order for a barge load of 1550 tons with a BTU requirement of at least 11,900, and maximum of 0.715 percent sulfur, 13.5 percent ash, and 8 percent moisture. Develop a linear optimization model to find the least cost way of meeting this order. b. Another utility wants 5 barges with a BTU rating of at least 12000, and maximum of 1.6 percent sulfur, 15 percent ash, and 11 percent moisture. Develop a linear optimization model to find the least cost way of meeting this order (independent of part a). a. Minimum Cost 1550Tons = EHA(24)+NA(45)+CA(54)+BS(47)+SR(40) subject to: (9,500)EHA+(13,000)NA+(12,500)CA+(11,500)BS+(10,400)SR >=11,900 (.55)EHA+(2.75)NA+(.75)CA+(1.0)BS+(.6)SR <=.715 (31)EHA+(8.5)NA+(12)CA+(15)BS+(7)SR <=13.5 (8)EHA+(7)NA+(7)CA+(8)BS+(14)SR <=8 b. Minimum Cost 5,750 Tons = EHA(24)+NA(45)+CA(54)+BS(47)+SR(40) subject to: (9,500)EHA+(13,000)NA+(12,500)CA+(11,500)BS+(10,400)SR >=12,000 (.55)EHA+(2.75)NA+(.75)CA+(1.0)BS+(.6)SR <= 1.6 (31)EHA+(8.5)NA+(12)CA+(15)BS+(7)SR <=15 (8)EHA+(7)NA+(7)CA+(8)BS+(14)SR <= 11 Chapter 13: Optimization Modeling 10. A South American honey farm makes five types of honey: cream, filtered, pasteurized, mélange (a mixture of several types), and strained, which are sold in 1 or 0.5 kilogram glass containers, one and 0.75 kg plastic containers, or in bulk. Key data are shown below Selling prices (Chilean pesos) .75 kg plastic 1 kg plastic .5 kg glass 1 kg glass bulk/kg cream (a) 744 880 760 990 616 filtered (b) 635 744 678 840 521 pasteurized (c) 696 821 711 930 575 mélange (d) 669 787 683 890 551 strained (e) 683 804 697 910 563 Minimum demand .75 kg plastic 1 kg plastic .5 kg glass 1 kg glass cream (a) 300 250 350 200 filtered (b) 250 240 300 180 pasteurized (c) 230 230 350 300 mélange (d) 350 300 250 350 strained (e) 360 350 250 380 Maximum demand .75 kg plastic 1 kg plastic .5 kg glass 1 kg glass cream (a) 550 350 470 310 filtered (b) 400 380 440 300 pasteurized (c) 360 390 490 400 mélange (d) 530 410 390 430 strained (e) 480 420 380 500 Package costs (Chilean pesos) .75 kg plastic 1 kg plastic .5 kg glass 1 kg glass 91 112 276 351 Harvesting and production costs for each product per kilogram in pesos are cream (a) 322 filtered (b) 255 pasteurized (c) 305 mélange (d) 272 strained (e) 287 Develop a linear optimization model to maximize profit. Maximize Profit = Sales - (Packaging + Harvesting) Cream Sales = a(744)+a(880)+a(760)+a(990)+a(616) Filtered Sales = b(635)+b(744)+b(678)+b(840)+b(521) Pasteurized Sales = c(696)+c(821)+c(711)+c(930)+c(575) Melange Sales = d(669)+d(787)+d(683)+d(890)+d(551) Strained Sales = e(683)+e(804)+e(697)+e(910)+e(563) Package Costs Cream = a(91)+a(112)+a(276)+a(351)+a(0) Filtered = b(91)+b(112)+b(276)+b(351)+b(0) Pasteurized = c(91)+c(112)+c(276)+c(351)+c(0) Melange = d(91)+d(112)+d(276)+d(351)+d(0) Strained = e(91)+e(112)+e(276)+e(351)+e(0) Harvest and Production Costs = a(322) + b(255) + c(305) + d(272) + e(287) Subject to: Minimum Demand Maximum Demand .75 kg plastic .75 kg plastic 300 cream (a) 550 250 filtered (b) 400 230 pasteurized (c) 360 350 mélange (d) 530 360 strained (e) 480 Minimum Demand Maximum Demand 1 kg plastic 1 kg plastic 250 cream (a) 350 240 filtered (b) 380 230 pasteurized (c) 390 300 mélange (d) 410 350 strained (e) 420 Minimum Demand Maximum Demand .5 kg glass .5 kg glass 350 cream (a) 470 300 filtered (b) 440 350 pasteurized (c) 490 250 mélange (d) 390 250 strained (e) 380 Minimum Demand Maximum Demand 1 kg glass 1 kg glass 200 cream (a) 310 180 filtered (b) 300 300 pasteurized (c) 400 350 mélange (d) 430 380 strained (e) 500 Chapter 13: Optimization Modeling Jeremy Haller of Haller’s Pub & Brewery, has compiled data describing the amount of different ingredients and labor resources needed to brew six different types of beers that the brewery makes. He also gathered financial information and estimated demand over a 26-week forecast horizon. These data are shown in the table below. The profits for each batch of each type of beer are: Light Ale: $3,925.78 Golden Ale: $4,062.75 Freedom Wheat: $3,732.34 Berry Wheat: $3,704.49 Dark Ale: $3,905.79 Hearty Stout: $3,490.22 These values incorporate fixed overhead costs of $7500 per batch. Use the data to validate the profit figures and develop a linear optimization model to maximize profit. Amounts for one batch (14 Barrels - 30 kegs - 4350 pints) of beer Berry Cost per Light Ale Golden Ale Freedom Wheat Wheat Dark Ale Hearty Stout Availability Unit A G FW BW DA HS Percent Demand 27.00 22.00 19.00 10.00 11.00 11.00 American 2- Row Grain (lb.) 525.00 400.00 375.00 350.00 450.00 375.00 30,000 $0.35 American 6- Row Grain (lb.) 125.00 125.00 150.00 250.00 225.00 8,000 $0.40 American Crystal Grain (lb.) 175.00 175.00 5,000 $0.42 German Vienna Grain (lb.) 125.00 200.00 175.00 50.00 5,000 $0.45 Flaked Barley (lb.) 75.00 150.00 150.00 75.00 5,000 $0.47 Light Dry Malt Extract (lb.) 35.00 45.00 50.00 25.00 2,000 $0.37 Hallertauer Hops (lb.) 4.00 3.00 2.00 2.00 8.00 500 $0.32 Kent Golding Hops (lb.) 1.00 4.00 4.00 500 $0.29 Tettnanger Hops (lb.) 4.00 2.00 4.00 2.00 500 $0.31 Brewing Labor (hrs.) 70.00 72.00 81.00 83.00 75.00 96.00 4,032 $18.00 Average # of Pints per batch 4350 4350 4350 4350 4350 4350 Beer Price (per pint) $3.00 $3.00 $3.00 $3.00 $3.00 $3.00 Average Demand (pints per week) 2,153 1,755 1,515 798 877 877 Average Demand (batches per week) 0.495 0.403 0.348 0.183 0.202 0.202 Average Demand (batches per 26 weeks) 12.870 10.487 9.057 4.767 5.243 5.243 Yearly Gross A G FW BW DA HS $3,925.78 $4,062.75 $3,732.34 $3,705.11 $3,905.79 $3,490.22 This equals the data provided. Maximize .27A+.22G+.19FW+.1BW+.11DA+.11HS Constraints (525*A)+(400*G)+(375*FW)+(350*BW)+(450*DA)+(375*HS)<=30,000 (125*G)+(125*FW)+(150*BW)+(250*DA)+(225*HS)<=8000 (175*A)+(175*HS)<=5000 (125*A)+(200*FW)+(175*BW)+(50*DA)<=5000 (75*A)+(150*FW)+(150*BW)+(75*HS)<=5000 (35*A)+(45*FW)+(50*BW)+(25*DA)<=2000 (4*A)+(3*G)+(2*FW)+(2*BW)+(8*DA)<=500 (1*G)+(4*DA)+(4*HS)<=500 (4*A)+(2*FW)+(4*DA)+(2*HS)<=500 (70*A)+(72*G)+(81*FW)+(83*BW)+(75*DA)+(96*HS)<=4032 Chapter 13: Optimization Modeling 12. Larry Doyle was recently named by Governor McGraw as campaign director for his upcoming reelection campaign. Governor McGraw thinks that if he can get his message to 2 million people in the state, he has a good chance to win a large chunk of votes at the state Libertarian Party Convention. Larry has obtained the following information about advertising media availability and cost: Voter Exposure Cost per Maximum Medium per $1,000 spent Insertion Units Prime time TV 10,000 $5,000 60 Non-prime time TV 4,000 $4,000 60 Radio 3,500 $3,000 100 Newspaper 1,500 $2,000 120 Billboards 750 $1,000 150 Governor McGraw has a campaign fund of $6 million available, which according to state election law, cannot be exceeded. As a traditionalist, Governor McGraw has specified that he wants to place at least 1 ad in each of the 50 largest newspapers in the state (just before it is time for them to make their editorial recommendations to voters). Formulate a linear model to maximize voter exposure. PTV = number of prime time TV ads NTV = number of non-prime time TV ads R = number of radio ads N = number of newspaper ads B = number of billboard ads Max 10,000PTV + 4,000NTV + 3,500R + 1,500N + 750B subject to: 5,000PTV + 4,000NTV + 3,000R + 2,000N + 1,000B <= 6,000,000 PTV <= 60 NTV <= 60 R <= 100 N <= 120 B <= 150 N >= 50 PTV, NTV, R, N, B >= 0 Chapter 13: Optimization Modeling 13. An international graduate student will receive a $28,000 foundation scholarship and a partial tuition scholarship. He must pay $1500 in tuition for each of the autumn, winter, and spring quarters, and $500 in the summer, which are due on the first day of September, December, March, and May, respectively. Living expenses are estimated to be $1500 per month, payable on the first day of the month. The foundation will pay him $18000 on August 1, and the remainder on May 1. To earn as much interest as possible, the student wishes to invest the money. Three types of investments are available at his bank: a 3-month CD, earning 0.75 percent (net 3-month rate); a 6-month CD, earning 1.9 percent; and a 12-month CD, earning 4.2 percent. Develop a linear optimization model to determine how can he best invest the money and meet his financial obligations? Data: 3 Month CD 6 Month CD 12 Month CD Return 0.75% 1.90% 4.20% Net Month Expenditures Cash available Aug 1 1,500 16,500 Sep 2 3,000 13,500 Oct 3 1,500 12,000 Nov 4 1,500 10,500 Dec 5 3,000 7,500 Starting Cash 18,000 Jan 6 1,500 6,000 mid cash 10,000 Feb 7 1,500 4,500 Min allowed 0 Mar 8 3,000 1,500 Apr 9 1,500 0 May 10 2,000 8,000 Jun 11 1,500 6,500 Jul 12 1,500 5,000 Model: Cash Int Earned Month 3 Month CD 6 Month CD 12 Month CD Balance 3 Month CD 1 1,000.00 1,500.00 5,000.00 9,000.00 2 1,000.00 5,000.00 3 3,500.00 1007.5 4 3,007.50 1007.5 5 1,000.00 15.00 0 6 500.00 1,000.00 7,015.00 0 7 7,043.50 1007.5 8 1,000.00 4,051.00 503.75 9 3,054.75 0 10 1,054.75 1007.5 11 562.25 0 12 81.25 0 Ending 5,291.25 Model assumes complete withdrawal after investment (Total interest earned means investment interest) Total 6 Month CD 12 Month CD Int Earned 1007.5 1007.5 0 1528.5 1528.5 0 1007.5 0 503.75 0 0 0 1007.5 1019 1019 0 5210 5210 stment interest) Chapter 13: Optimization Modeling 14. You have been given $1 million by a deceased relative. You plan to invest this money (in varying amounts) in up to four long-term investment instruments (stocks, bonds, savings, and real estate). You evaluate investments in stocks and bonds at the beginning of each of the next six years. Each dollar invested in stocks at the beginning of the year is expected to return an average of $1.15 in time for immediate reinvestment one year later. Each dollar invested in bonds at the beginning of each year is expected to return $1.30 three years later (a profit of 30% in three years). Credit Union Savings returns $1.05 one year later, and each dollar invested in real estate is expected to return $1.30 four years later. A maximum of $200,000 can be invested in real estate in any one year. You want to diversify your investments to minimize risk. The total invested in stocks in a given year should not exceed 30 percent of the total investment in the other alternatives for that year. At least 25 percent of the total investment should be in credit union savings certificates. You also want to have $150,000 available in cash (which can be immediately reinvested) at the end of the third year as leverage in negotiating with your relatives. Develop a model to maximize the cash on hand at the end of the sixth year. Si = $ invested in stocks at beginning of year i, i = 1-6 Bi = $ invested in bonds at beginning of year i, i = 1-4 Ci = $ invested in credit union savings at beginning of year i, i = 1-6 Ri = $ invested in real estate at beginning of year i, i = 1-3 Max 1.15S6 + 1.30B4 + 1.05C6 + 1.30R3 subject to: S1 + B1 + C1 + R1 = 1,000,000 S2 + B2 + C2 + R2 = 1.15S1 + 1.05C1 S3 + B3 + C3 + R3 = 1.15S2 + 1.05C2 S4 + B4 + C4= 1.15S3 +1.30B1+ 1.05C3 S5 + C5 = 1.15S4 +1.30B2+ 1.05C4 + 1.30R1 S6+ C6 = 1.15S5 +1.30B3+ 1.05C5 + 1.30R2 R1 <= 200,000 R2 <= 200,000 R3 <= 200,000 S1 <= .3 (B1 + C1 + R1) S2 <= .3 (B2 + C2 + R2) S3 <= .3 (B3 + C3 + R3) S4 <= .3 (B4 + C4) S5 <= .3 (C5 ) S6 <= .3 (C6 ) C1 >= .25(S1 + B1 + C1 + R1) C2 >= .25(S2 + B2 + C2 + R2) C3 >= .25(S3 + B3 + C3 + R3) C4 >= .25(S4 + B4 + C4) C5 >= .25(S5 + C5) C6 >= .25(S6 + C6) 1.15S3 +1.30B1+ 1.05C3 >= 150,000 all Si, Bi, Ci, Ri >= 0 Chapter 13: Optimization Modeling 15. Liquid Gold, Inc. transports radioactive waste from nuclear power points to disposal sites around and about the country. Each plant has an amount of material that must be moved each period. Each site has a limited capacity per period. The cost of transporting between sites is given below (some combinations of plants and storage sites are not to be used, and no figure is given). cost to site: Plant Material S51 S62 S73 S87 Site Capacity P1 20,876 105 86 x 23 S51 285,922 P2 50,870 86 58 41 x S62 308,578 P3 38,652 93 46 65 38 S73 111,995 P4 28,951 116 27 94 x S87 208,555 P5 87,423 88 56 82 89 P6 76,190 111 36 72 x P7 58,237 169 65 48 x Develop a transportation model for this problem. xij = amount shipped from plant Pi to site j Site j S51 1 S62 2 S73 3 S87 4 Min 105x11 + 86x12 + 23x14+ 86x21 + 58x22 + 41x23 + 93x31 + 46x32 + 65x33 + 38x34 + 116x41 + 27x42 + 94x43 + 88x51 + 56x52 + 82x53 + 89x54 + 111x61 + 36x62 + 72x63 + 169x71 + 65x72 + 48x73 subject to: x11 + x12 + x14 = 20,876 x21 + x22 + x23 = 50,870 x31 + x32 + x33 + x34 = 38,652 x41 + x42 + x43 = 28,951 x51 + x52 + x53 + x54 = 87,453 x61 + x62 + x63 = 76,190 x71 + x72 + x73 = 58,237 x11 + x21 + x31 + x41 + x51 + x61 + x71 <= 285,922 x12 + x22 + x32 + x42 + x52 + x62 + x72 <= 308,578 1 x23 + x33 + x43 + x53 + x63 + x73 <= 111,95511,995 x14 + x34 + x54 <= 208,555 all xij >= 0 Chapter 13: Optimization Modeling 16 A company in Victoria, TX produces bottles of aspirin products. Sales prices are given below: PRODUCT SALES PRICE Super Seltzer $3.00 Capsules $3.50 Cheap Seltzer $2.00 Tablets $2.50 The company ships these products to two distributors, located at Hearne, TX and Cuero, TX. There is unlimited demand at each distributor. Shipping costs per bottle and contracted minimum quantities for each distributor are: Shipping cost/bottle Hearne Cuero $0.21 $0 .22 Minimum demand Super Seltzer 700 1000 Capsules 800 1500 Cheap Seltzer 1000 800 Tablets 1800 5000 Raw material costs and maximum available quantities are given below: Cost/ounce Maximum ounces Acetylsalicylic acid $0.60 50,000 Sodium $0.30 25,000 The production costs per bottle and raw materials required per bottle are: Production Ounces Ounces cost/bottle Acet. Acid Sodium Super Seltzer $0.25 2 3 Capsules 0.35 4 0 Cheap Seltzer 0.15 2 2 Tablets 0.1 3 0 Total number of bottles of capsules produced is to be no more than 20% of the total number of bottles produced. SSH = bottles of super seltzer shipped to Hearne CH = bottles of capsules shipped to Hearne CSH = bottles of cheap seltzer shipped to Hearne TH = bottles of tablets shipped to Hearne SSC = bottles of super seltzer shipped to Cuero CC = bottles of capsules shipped to Cuero CSC = bottles of cheap seltzer shipped to Cuero TC = bottles of tablets shipped to Cuero Max .44SSH + .54CH - .16 CS + .39TH + .43SSC + .53CC - .17CSC + .38TC subject to: 2(SSH + SSC) + 4(CH + CC) + 2(CSH + CSC) + 3(TH + TC) <= 50,000 3(SSH + SSC) + 2(CSH + CSC) <= 25,000 CH + CC <= .2 (SSH + SSC + CH + CC + CSH + CSC + TH + TC) SSH >= 700 CH >=800 CSH >= 1,000 TH >= 1,800 SSC >= 1,000 CC >= 1,500 CSC >= 800 TC >= 5,000 SSH, SSC, CH, CC, CSH, CSC, TH, TC >= 0 Note: the objective function coefficient are computed by subtracting raw material, production, and shipping costs from the sale price. Chapter 13: Optimization Modeling 17. Mirza Manufacturing makes four electronic products, each of which is comprised of 3 main materials: magnet, wiring, and casing. The products are shipped to three distribution centers in North America, Europe, and Asia. Marketing has specified that no location should receive more than 3 times the demand. The material cost/unit are: magnet – $0.59, wire – $0.29, and casing – $0.31. The table below shows the number of units of each material required in each unit of end product and the production cost per unit. Product Production Cost/unit Magnets Wire Casing A $0.25 4 2 2 B $0.35 3 1 3 C $0.15 2 2 1 D $0.10 8 3 2 Additional information is provided below. Min Demand Product NA EU Asia A 850 900 100 B 700 200 500 C 1100 800 600 D 1500 3500 2000 Max Demand Product NA EU Asia A 2550 2700 300 B 2100 600 1500 C 3300 2400 1800 D 4500 10500 6000 Shipping cost/unit Product NA EU Asia A 0.2 0.25 0.35 B 0.2 0.2 0.3 C 0.2 0.2 0.3 D 0.2 0.2 0.25 Unit Sales Revenue Product NA EU Asia A 4 4.5 4.55 B 3.7 3.9 3.95 C 2.7 2.9 2.4 D 6.8 6.5 6.9 Available Raw Material Magnet 120000 Wire 50000 Casing 40000 Develop an appropriate linear optimization model to maximize net profit. Product A Needs 4M+2W+2C Product B Needs 3M+1W+3C Product C needs 2M+2W+1C Product D needs 8M+3W+2C A(4M)+B(3M)+C(2M)+D(8M)<=120000 A(2W)+B(1W)+C(2W)+D(3W)<=5000 A(2C)+B(3C)+(1C)+(2C)<=40000 A NA <= 3*2550 A EU <=3*2700 A ASIA <= 3*300 B NA <= 3*2100 B EU <=3*600 B ASIA <= 3*1500 C NA <= 3*3300 C EU <=3*2400 C ASIA <= 3*1800 D NA <= 3*4500 D EU <=3*10500 D ASIA <= 3*6000 A NA > 850 A EU > 900 A ASIA>100 B NA >700 B EU >200 B ASIA > 500 C NA> 1100 C EU>800 C ASIA >600 D NA > 1500 D EU > 3500 D ASIA > 2000 Minimize A(PC)+B(PC)+C(PC)+D(PC) Minimize A(SC)+B(SC)+C(SC)+D(SC) Maximize A(SR)+B(SR)+C(SR)+D(SR) Chapter 13: Optimization Modeling 18. Thermal transfer printing transfers ink from a ribbon onto paper through a combination of heat and pressure. Different types of printers use different sizes of ribbons. A ribbon manufacturer has forecasted demands for seven different ribbon sizes: Size Forecast (rolls) 60 mm 1620 83 mm 512 102 mm 840 110 mm 2640 120 mm 502 130 mm 755 165 mm 680 The rolls from which ribbons are cut are 880 mm in length. Scrap is valued at $0.05 per mm. Generate 10 different cutting patterns so that each size can be cut from at least one pattern. Use your data to construct and solve an optimization model for finding the number of patterns to cut to meet demand and minimize trim loss. Number of Rolls Pattern 60 83 102 110 120 130 165 Scrap A 1 1 1 2 1 1 1 0 B 2 0 0 0 0 2 3 5 C 2 1 1 5 0 0 0 25 D 3 0 0 5 0 1 0 20 E 13 1 0 0 0 0 0 17 F 5 3 2 0 1 0 0 7 G 6 0 0 0 0 0 3 25 H 8 0 0 0 1 2 0 20 I 8 1 0 1 0 0 1 42 J 9 1 0 0 2 0 0 17 Number of Feet of Product Pattern 60 83 102 110 120 130 165 Scrap A 60 83 102 220 120 130 165 0 B 120 0 0 0 0 260 495 5 C 120 83 102 550 0 0 0 25 D 180 0 0 550 0 130 0 20 E 780 83 0 0 0 0 0 17 F 300 249 204 0 120 0 0 7 G 360 0 0 0 0 0 495 25 H 480 0 0 0 120 260 0 20 I 480 83 0 110 0 0 165 42 J 540 83 0 0 240 0 0 17 Minimize 0A+5B+25C+20D+17E+7F+25G+20H+42I+17J Constraints 60 mm 1A+2B+2C+3D+13E+5F+6G+8H+8I+9J >= 1620 83 mm 1A+0B+1C+0D+1E+3F+0G+0H+1I+1J >= 512 102 mm 1A+0B+1C+0D+0E+2F+0G+0H+0I+0J >= 840 110 mm 2A+0B+5C+5D+0E+0F+0G+0H+1I+0J >= 2640 120 mm 1A+0B+0C+0D+0E+1F+0G+1H+0I+2J >= 502 130 mm 1A+2B+0C+1D+0E+0F+0G+2H+0I+09 >= 755 165 mm 1A+3B+0C+0D+0E+0F+3G+0H+1I+0J >= 680 Chapter 13: Optimization Modeling 19. The personnel director of a company that recently absorbed another firm, and is now downsizing, must relocate four information systems analysts from recently closed locations. Unfortunately, there are only three positions available for five people. Salaries are fairly uniform among this group (those with higher pay were already given the opportunity to begin anew). Moving expenses will be used as the means of determining who will be sent where. Estimated moving expenses are: Analyst moving cost to: Gary Salt Lake City Fresno Arlene $8,000 $7,000 $5,000 Bobby $5,000 $8,000 $12,000 Charlene $9,000 $15,000 $16,000 Douglas $4,000 $8,000 $13,000 Emory $7,000 $3,000 $4,000 Model this as an integer optimization model to minimize cost and determine which analysts to relocate to the three locations. Xij = 1: relocate analyst i to city j 0: do not relocate analyst i to city j Analyst i Arlene 1 Bobby 2 Charlene 3 Douglas 4 Emory 5 City j Gary 1 Salt Lake City 2 Fresno 3 Min 8,000X11 + 7,000X12 + 5,000X13 + 5,000X21 + 8,000x22 + 12,000X23 + 9,000X31 + 15,000X32 + 16,000X33 + 4,000X41 + 8,000X42 + 13,000X43 + 7,000X51 + 3,000X52 + 4,000X53 subject to: X11 + X12 + X13 <= 1 X21 + X22 + X23 <= 1 X31 + X32 + X33 <= 1 X41 + X42 + X43 <= 1 X51 + X52 + X53 <= 1 X11 + X21 + X31 + X 41 + X51 = 1 X12 + X22 + X32 + X 42 + X52 = 1 X13 + X23 + X33 + X 43 + X53 = 1 all Xij = 1 or 0 Arlene to Fresno Douglas to Gary Emory to Salt Lake City Chapter 13: Optimization Modeling 20. A medical device company is allocating next year’s budget among its divisions. As a result, the R&D Division needs to determine which research and development projects to fund. Each project requires various software and hardware and consulting expenses, along with internal human resources. Internal Additional Project Estimate NPV engineers costs 1 $600,000 9 $196,000 2 $680,000 12 $400,000 3 $550,000 7 $70,000 4 $400,000 4 $180,000 5 $350,000 8 $225,000 6 $725,000 10 $275,000 7 $340,000 8 $130,000 A budget allocation of $1,250,000 has been approved, and 32 engineers are available to work on the projects. The R&D group has determined that at most one of projects 1 and 2 should be pursued, and that if project 4 is chosen, then project 2 must also be chosen. Develop a model to select the best projects within the budget. NPV Eng Costs Decision 2,3,4,5 $1,980,000 31 $875,000 1 Yes 2,3,4,7 $1,970,000 31 $780,000 1 Yes 1,5,6 $1,675,000 27 $696,000 1 Yes 1,6,7 $1,665,000 27 $601,000 1 Yes 2,3,4 $1,630,000 23 $650,000 1 Yes 1,6 $1,325,000 19 $471,000 1 Yes 2,4 $1,080,000 16 $580,000 1 Yes 1,5 $950,000 17 $421,000 1 Yes 1,7 $940,000 17 $326,000 1 Yes 1 $3,645,000 58 $1,476,000 0 No 2,3,4,5,6,7 $3,045,000 49 $1,280,000 0 No 2,3,4,5,6 $2,705,000 41 $1,150,000 0 No 2,3,4,6,7 $2,695,000 41 $1,055,000 0 No 1,3,5,6,7 $2,565,000 42 $896,000 0 No 2,4,5,6,7 $2,495,000 42 $1,210,000 0 No 2,4,5,6 $2,155,000 34 $1,080,000 0 No 1,5,6,7 $2,015,000 35 $826,000 0 No Project 1 Project 2 Project 3 Project 4 Project 5 NPV $ 600,000 $ 680,000 $ 550,000 $ 400,000 $ 350,000 Additional Costs 196000 400000 70000 180000 225000 Engineers 9 12 7 4 8 Model 1 Project selection decisions 0 1 1 1 1 Costs 0 400000 70000 180000 225000 Engineers 0 12 7 4 8 NPV $ - $ 680,000 $ 550,000 $ 400,000 $ 350,000 Project 6 Project 7 Resources $ 725,000 $ 340,000 275000 130000 1250000 10 8 32 0 0 Total 0 0 875000 0 0 31 $ - $ - $ 1,980,000 Chapter 13: Optimization Modeling 21. A call center has five categories of employees: Level 1 managers who spend 20 percent of their time in the call center and the remainder doing project work; Level 2 managers who spend 50 percent of their time in the call center and the rest on project work; Technical Specialist 1, who spend 50 percent of their time in the call center; 30 percent on project work, and the remainder on other tasks; Technical Specialist 2, who spend 80 percent in the call center, and 10 percent on project work; and outsourced contractors, who spend 100 percent of their time on call center work. Based on analysis of data at a call center, the call center manager has determined that she needs at least 5 full time equivalent people on project work and 30 on call center work. Level 1 managers earn $100,000 per year; Level 2 managers, $90,000; Technical Specialists earn $65,000 for both types; and contractors are paid $75,000. What is the best staffing mix for this operation? need 5 for project need 30 for Call Center Call Project Other Salary L1M 0.2 0.8 100000 L2M 0.5 0.5 90000 TS1 0.5 0.3 0.2 65000 TS2 0.8 0.1 0.1 65000 Cont 1 75000 L1M L2M TS1 TS2 Cont #needed 1 1 1 34 2 Call 0.2 0.5 0.5 27.2 2 30.4 Project 0.8 0.5 0.3 3.4 0 5 Salaries 100000 90000 65000 2210000 150000 2615000 Chapter 13: Optimization Modeling 22. A software support division of a major corporation has 8 projects that can be performed. Each project requires different amounts of development time and testing time. In the coming planning period, 1190 hours of development time and 1000 hours of testing time are available, based on the skill mix of the staff. The internal transfer price (revenue to the support division) and the times required for each project are shown below. Which projects should be selected to maximize revenue? Project Development Time Testing Time Transfer Price 1 80 67 $23,520 2 248 208 $72,912 3 41 34 $12,054 4 110 92 $32,340 5 240 202 $70,560 6 195 164 $57,232 7 269 226 $79,184 8 110 92 $32,340 Project 1 Project 2 Project 3 Project 4 Transfer Price $ 23,520 $ 72,912 $ 12,054 $ 32,340 Development Time 80 248 41 110 Testing Time 67 208 34 92 Model Project selection decisions 0 1 0 1 Development Time 0 248 0 110 Testing Time 0 208 0 92 Transfer Price $ - $ 72,912 $ - $ 32,340 Project 5 Project 6 Project 7 Project 8 Resources $ 70,560 $ 57,232 $ 79,184 $ 32,340 240 195 269 110 1190 202 164 226 92 1000 1 1 1 1 Total 240 195 269 110 1172 202 164 226 92 984 $ 70,560 $ 57,232 $ 79,184 $ 32,340 $ 344,568 Chapter 13: Optimization Modeling 23. Toward the end of the fiscal year, a company division has $75,000 remaining in its budget, with a “use it or lose it” policy. Four potential projects are available for consideration. Each project requires a fixed amount of cash outlay, project management labor time, client labor time, and analyst time. Relevant data are shown in the table below. Sixty five hours of project management time, 30 hours of client labor time, and 85 hours of analyst time are available. Determine which projects should be selected to maximize the total return. Project 1 Project 2 Project 3 Project 4 Return $80,000 70,000 50,000 40,000 Cash outlay $25,000 40,000 12,000 27,500 Project management labor 25 20 15 12 Client labor 15 12 10 8 Analyst labor 60 50 35 30 Data Available Project 1 Project 2 Project 3 Project 4 Resources Expected Return (NPV) $ 80,000 $ 70,000 $ 50,000 $ 40,000 Cash requirements $ 25,000 $ 40,000 $ 12,000 $ 27,500 $ 75,000 Project management labor 25 20 15 12 65 Client labor 15 12 10 8 30 Analyst labor 60 50 35 30 85 Model Project selection decisions 0 1 1 0 Total Cash Used $ - $ 40,000 $ 12,000 $ - $ 52,000 Project management labor 0 20 15 0 35 Client labor 0 12 10 0 22 Analyst labor 0 50 35 0 85 Return $ - $ 70,000 $ 50,000 $ - $ 120,000 Chapter 13: Optimization Modeling 24. Many high technology products such as crystals and alloys can be manufactured more efficiently in the weightless environment of earth orbit. You are planning production operations for a space flight. Five products are being considered. There are unit profits, volumes, weights, labor hours per unit, and maximum demands. Model this problem as an integer optimization problem. Products Alloy1 Alloy2 Crystal 1 Crystal2 Interferon Profit ($/unit) 10 1.7 3.5 1.6 2.6 Volume (CF/unit) 9 3 10 7 13 Weight (lb/unit) 59 18 26 26 10 Labor (/unit) 2.2 0.5 0.7 0.2 1.1 Demand () 22 69 90 40 85 Volume available: 600 CF Weight allowable: 2100 lb Labor available: 40 hr A1 = units of Alloy1 A2 = units of Alloy2 C1 = units of Crystal1 C2 = units of Crystal2 I = units of Interferon Max 10A1 + 1.7A2 + 3.5C1 + 1.6C2 + 2.6I subject to: 9A1 + 3A2 + 10C1 + 7C2 + 13I <= 600 59A1 + 18A2 + 26C1 + 26C2 + 10I <= 2100 2.2A1 + 0.5A2 + 0.7C1 + 0.2C2 + 1.1I <= 40 A1 <= 22 A2 <= 69 C1 <= 90 C2 <= 40 I <= 85 A1, A2, C1, C2, I >= 0 and integer Chapter 13: Optimization Modeling 25. Soapbox is a local band that plays classic and contemporary rock. The band members charge $600 for a three-hour gig. They would like to play at least 30 gigs per year, but need to determine the best way to promote themselves. The most they are willing to spend on promotion is $5000. The possible promotion options are: · Playing free gigs · Making a demo CD · Hiring an agent · Handing out fliers · Creating a web site Each free gig costs them about $150 for travel and equipment, but generates about 4 paying gigs. A high-quality studio demo CD should help the band book about 20 gigs, but will cost about $750. A demo CD made on home recording equipment will only cost $250, but may only result in 6 bookings. A good agent will get the band about 15 gigs, but agents typically charge 20 percent of all revenue, whether booked by the agent or not. The band can create a web site for $150 and would expect to generate 6 gigs from this exposure. They also estimate that they may book one gig for every 100 fliers they hand out, which would cost 8 cents each. Develop an optimization model to find the best promotion strategy to maximize their profit. Gigs Gen Promotion Budget 5000 30 "Free" Costs 150 4 Pro CD 750 20 Home CD 250 6 Agent 20% 15 Web 150 6 Fliers 8 1 Free Gigs Pro CD Home CD Agent Web Fliers Number 1 1 1 0 1 1 Costs 150 750 250 0 150 8 1308 Gig Gen 4 20 6 0 6 1 37 Revenue 22200 Chapter 13: Optimization Modeling 26. Dannenfelser Design works with clients in three major project categories: architecture, interior design, and combined. Each type of project requires an estimated number of hours for different categories of employees as shown in the table below. Architecture Interior Design Combined Hourly Rate Principal 15 5 18 $115 Sr. Designer 25 35 40 $85 Draftsman 40 30 60 $60 Administrator 5 5 8 $40 In the coming planning period, 184 hours of Principal time, 414 hours of Sr. Designer time, 588 hours of draftsman time, and 72 hours of administrator time are available. Profit per project average $1290 for architecture, $1110 for interior design, and $1878 for combined projects. The firm would like to work on at least one of each type of project for exposure among clients. Assuming that the firm has more demand than they can possibly handle, find the best mix of projects to maximize profit. Architecture Interior Design Combined Hourly Rate Time Available Principal 15 5 18 $115 184 Sr. Designer 25 35 40 $85 414 Draftsman 40 30 60 $60 588 Administrator 5 5 8 $40 72 Profit 1290 1110 1878 Architecture Interior Design Combined Totals Number of Projects 10 2 1 13 Principal 150 10 18 178 Sr. Designer 250 70 40 360 Draftsman 400 60 60 520 Administrator 50 10 8 68 Profit 12900 2220 1878 16998 Chapter 13: Optimization Modeling 27. A young entrepreneur has invented a new an air--adjustable basketball shoe with pump, similar to those advertised widely by more expensive brand names. He contacted a supplier of Victor basketball shoes, a little-known brand with low advertising. This supplier would provide shoes at the nominal price of $6 per pair of shoes. He needs to know the best price at which to sell these shoes. As a business student with strong economics training, he remembered that the volume sold is affected by the product’s price—the higher the price, the lower the volume. He asked his friends and acquaintances what they would pay for a premium pair of basketball shoes that were a “little off-brand.” Based on this data, he developed the formula Volume = 1000 - 20 Price There are some minor expenses involved, including a $50 fee for selling shoes in the neighborhood (a fixed cost), as well as his purchase price of $6 per shoe. Develop an appropriate objective function to find the optimal price level. Profit = (Price - 6) x (1000 - 20Price) - 50 2 = -20Price + 1120Price - 6050 Chapter 13: Optimization Modeling 28. The entrepreneur in the previous problem did very well selling Victor shoes. His shoe supplier told him of a new product,Top Notch, that was entering the market. This shoe would be a product substitute for Victors, so that the higher the price of either shoe, the greater the demand for the other. He interviewed more potentialclients to determine price response and cross-elasticities. This yielded the following relationships: Volume of Victors = 1000 - 20 Pv + 1 Pa Volume of Top Notch Shoes = 800 + 2 Pv - 18 Pa where Pv = price of Victors, and Pa = price of Abibas. Develop a new profit function to include this information. Profit Victors = (Pv - 6) x (1000 - 20 Pv + 1 Pa) - 50 2 = -20Pv + 1120Pv + PvPa - 6Pa - 6050 Profit Abibas = (Pa - 6) x (800 + 2Pv - 18Pa) - 50 2 = -18Pa + 908Pa + 2PvPa - 12Pv - 4850 2 2 Total Profit = -20Pv - 18Pa + 1108Pv + 902Pa + 3PvPa - 10,900 Chapter 13: Optimization Modeling 29. The Hal Chase Investment Planning Agency is in business to help investors optimize their return from investment, to include consideration of risk. Hal deals with three investment mediums: a stock fund, a bond fund, and his own Sports and Casino Investment Plan (SCIP). The stock fund is a mutual fund investing in openly traded stocks. The bond fund focuses on the bond market, which has a much stabler but lower expected return. SCIP is a high-risk scheme, often resulting in heavy losses, but occasionally coming through with spectacular gains. Average returns, their variances, and covariances, are given below. Stock Bond SCIP Average return 0.148 0.06 0.152 Variance 0.014697 0.000155 0.160791 Covariance with Stock 0.000468 -0.002222 Covariance with Bond -0.000227 Negative covariance indicates that SCIP tends to move in the opposite direction as stocks or bonds. Develop a portfolio optimization model for this situation. Stock = percentage of Stock in portfolio Bond = percentage of Bond in portfolio SCIP = percentage of SCIP in portfolio Min .014697Stock2 + .000155Bond2 + .160791SCIP2 + 2(.000468)StockBond + 2(-.002222)StockSCIP + 2(-.000227)BondSCIP subject to: Stock + Bond + SCIP = 1 .148Stock + .06Bond + .152SCIP >= .10 Stock, Bond, SCIP >= 0 Note: the second constraint assumes a minimum yield of .10. Chapter 13: Optimization Modeling 30. Develop a spreadsheet model for the media selection problem described in this chapter. Data: Radio TV Magazine Budget Cost/ad 500 2,000 200 50,000 Exposure value/ad 2,000 3,500 2,700 Minimum Units 0 12 6 Maximum Units 15 - 30 Model: Radio TV Magazine Number of ads 15 18 30 Total Budget 7,500 36,500 6,000 50,000 Total Exposure 30,000 63,875 81,000 174,875 Note: use decision variables (changing cells) for Max/Min Units. Chapter 13: Optimization Modeling 31. Develop a spreadsheet model for the process selection problem described in this chapter. Outsourcing Demand Mill cost cost Data: Fabric (yards) ($/yard) ($/yard) 1 45,000 0.65 0.85 2 76,500 0.61 0.75 3 10,000 0.50 0.65 Dobbie Dobbie Regular Regular loom loom loom loom capacity capacity capacity capacity Fabric (yards/hour) (hours/yard) (yards/hour) (hours/yard) 1 4.7 0.213 0.0 99,999 2 5.2 0.192 5.2 0.192 3 4.4 0.227 4.4 0.227 Hours available 6,552 32,760 Model: Fabric Dobbie Regular Outsourcing 1 30,794 0 14,205 2 0 76,500 0 3 0 10,000 0 Total Fabric 1: yards produced 30,794 0 14,205 45,000 Fabric 2: yards produced 0 76,500 0 76,500 Fabric 3: yards produced 0 10,000 0 10,000 Dobbie: hours used 6,552 Regular: hours used 32,760 Fabric 1: cost 20,016 0 12,075 32,091 Fabric 2: cost 0 46,665 0 46,665 Fabric 3: cost 0 5,000 0 5,000 Total cost 83,756 Chapter 13: Optimization Modeling 32. Develop a spreadsheet model for the blending problem described in this chapter. Data: Ingredient Protein Fat Fiber Cost/lb. Sunflower seeds 0.169 0.260 0.290 0.22 White millet 0.120 0.041 0.083 0.19 Kibble corn 0.085 0.038 0.027 0.10 Oats 0.154 0.063 0.024 0.10 Cracked corn 0.085 0.038 0.027 0.07 Wheat 0.120 0.017 0.023 0.05 Safflower 0.180 0.179 0.288 0.26 Canary grass seed 0.119 0.040 0.109 0.11 Requirement 0.13 0.15 0.14 Total weight 1 Model: Pounds in Mixture Protein Fat Fiber Cost Sunflower seeds 0.547 0.09 0.14 0.16 0.12 White millet 0.000 0.00 0.00 0.00 0.00 Kibble corn 0.000 0.00 0.00 0.00 0.00 Oats 0.000 0.00 0.00 0.00 0.00 Cracked corn 0.000 0.00 0.00 0.00 0.00 Wheat 0.453 0.05 0.01 0.01 0.02 Safflower 0.000 0.00 0.00 0.00 0.00 Canary grass seed 0.000 0.00 0.00 0.00 0.00 Total 1.000 0.15 0.15 0.17 0.14 Chapter 13: Optimization Modeling 33. Develop a spreadsheet model for the production planning problem described in this chapter. Data: Forecasted Labor Labor Sales per Hour per Unit Autumn 150 5.50 11.00 Winter 400 7.00 14.00 Spring 50 6.25 12.50 Holding cost 1.20 Model: Production Inventory Available Cost Autumn 550 400 150 6,530 Winter 0 0 400 0 Spring 50 0 50 625 7,155 Chapter 13: Optimization Modeling 34. Develop a spreadsheet model for the cash management problem described in this chapter. Data: 1 Month CD 3 Month CD 6 Month CD Return 0.50% 1.75% 2.30% Net Month Expenditures 1 50,000 2 -12,000 3 23,000 4 -20,000 5 41,000 Starting Cash 200,000 6 -13,000 Minimum Cash 100,000 Model: Cash Month 1 Month CD 3 Month CD 6 Month CD Balance 1 50,000.00 0.00 0.00 100,000.00 2 22,885.57 39,364.43 0.00 100,000.00 3 0.00 0.00 0.00 100,000.00 4 941.98 19,058.02 0.00 100,000.00 5 0.00 0.00 0.00 100,000.00 6 13,000.00 0.00 0.00 100,000.00 Ending 132,456.53 Note: in the model above, the end of month cash balance is re-invested in the next month. Chapter 13: Optimization Modeling 35. Develop a spreadsheet model for the distribution center location problem described in this chapter. Data: Plant/D.C. Cleveland Baltimore Chicago Phoenix Capacity Marietta 12.60 14.35 11.52 17.58 1,200 Minneapolis 9.75 12.63 8.11 15.88 800 Fayetteville 10.41 11.54 9.87 8.32 1,500 Chico 13.88 16.95 12.51 11.64 1,500 Demand 300 500 700 1,800 Maximum new plants 1 Model: Total Plant/D.C. Cleveland Baltimore Chicago Phoenix Shipped Capacity Marietta 200 500 0 300 1,000 1,200 Minneapolis 100 0 700 0 800 Build 800 Fayetteville 0 0 0 1,500 1,500 1 1,500 Chico 0 0 0 0 0 0 0 Demand met 300 500 700 1,800 1 Total Cost = 34,101 Chapter 13: Optimization Modeling 36. Develop a spreadsheet model for the cutting stock problem described in this chapter. Size of end item Pattern 12” 15” 30” Scrap 1 0 6 0 10 2 0 0 3 10 3 8 0 0 4 4 2 1 2 1 5 7 1 0 1 Demands this week are 870 12” rolls, 450 15” rolls, and 650 30’ rolls. The problem is to develop a model that will determine how many 100” rolls to cut into each of the five patterns in order to meet demand and scrap. 12" 870 15" 450 30" 650 Change the number cell value Pattern 12” 15” 30” Scrap Number 1 0 342 0 570 57 2 0 0 630 2100 210 3 800 0 0 400 100 4 80 57 20 10 10 5 0 57 0 0 Demand 880 456 650 3080 Chapter 13: Optimization Modeling 37. Develop a spreadsheet model for the computer configuration described in this chapter. Option Variable Utility slower processor Xp1 0.2 faster processor Xp2 0.7 fastest processor Xp3 0.9 2 year warranty Xw1 0.5 3 year warranty Xw2 0.55 512MB memory Xm1 0.7 1 GB memory Xm2 0.8 30 GB hard drive Xd1 0.1 60 GB hard drive Xd2 0.3 80 GB hard drive Xd3 1 CD only Xc1 0 CDRW Xc2 0.95 DVD/RW Xc3 0.45 8 cell battery Xb 0.15 music and photo software Xs 0.85 Base Unit 1000 The following options are available, along with price changes: Processor: slower (subtract $179), faster (add $100), fastest (add $300) Warranty: 2 year (add $129), 3 year (add $269), Memory: 512MB (add $50), 1 GB (add $500) Hard drive: 30GB (subtract $29), 60GB (add $39), 80GB (add $79) Modular bay: CD-ROM only (subtract $39): CD/RW (add $79), DVD/RW (add $179) Battery: 8-cell Lithium ion (add $59) Enhanced music and photo software (add $79) 1 2 3 Processor p -179 100 300 Warranty w 0 129 269 Memory m 0 50 500 HD d -29 39 79 Mod Bay c -39 79 179 Batt b 59 0 0 Music s 79 0 0 Data Base 1000 Change Data (1, 2, 3) Processor 2 Warranty 1 Memory 1 HD 3 Mod Bay 3 Batt 2 Mus 2 Model Base 1000 Processor 100 Warranty 0 Memory 0 HD 79 Mod Bay 179 Batt 0 Music 0 Total Price 1358 Chapter 13: Optimization Modeling 38. Develop a spreadsheet model for the distribution center location described in this chapter. SAME AS 35 Data: Plant/D.C. Cleveland Baltimore Chicago Phoenix Capacity Marietta 12.60 14.35 11.52 17.58 1,200 Minneapolis 9.75 12.63 8.11 15.88 800 Fayetteville 10.41 11.54 9.87 8.32 1,500 Chico 13.88 16.95 12.51 11.64 1,500 Demand 300 500 700 1,800 Maximum new plants 1 Model: Total Plant/D.C. Cleveland Baltimore Chicago Phoenix Shipped Capacity Marietta 200 500 0 300 1,000 1,200 Minneapolis 100 0 700 0 800 Build 800 Fayetteville 0 0 0 1,500 1,500 1 1,500 Chico 0 0 0 0 0 0 0 Demand met 300 500 700 1,800 1 Total Cost = 34,101 Chapter 13: Optimization Modeling 39. Develop a spreadsheet model for the direct marketing problem described in this chapter. Data: Magazine 1 2 3 4 5 Customer Cost 900 1000 1100 1500 1500 1 10 0 0 0 0 0 2 1,4 1 0 0 1 0 3 1 1 0 0 0 0 4 5,6 0 0 0 0 1 5 5 0 0 0 0 1 6 10 0 0 0 0 0 7 2,9 0 1 0 0 0 8 5,8 0 0 0 0 1 9 1,5,10 1 0 0 0 1 10 4,6,8,10 0 0 0 1 0 11 6 0 0 0 0 0 12 3 0 0 1 0 0 13 5 0 0 0 0 1 14 2,6 0 1 0 0 0 15 8 0 0 0 0 0 16 6 0 0 0 0 0 17 4,5 0 0 0 1 1 18 7 0 0 0 0 0 19 5,6 0 0 0 0 1 20 2,8 0 1 0 0 0 21 7,9 0 0 0 0 0 22 6 0 0 0 0 0 23 3,6,10 0 0 1 0 0 24 NONE 0 0 0 0 0 25 5,8 0 0 0 0 1 26 3,10 0 0 1 0 0 27 2,8 0 1 0 0 0 28 4,7 0 0 0 1 0 29 6 0 0 0 0 0 30 3,4,5,10 0 0 1 1 1 31 4 0 0 0 1 0 32 8 0 0 0 0 0 33 1,3,10 1 0 1 0 0 34 4,5 0 0 0 1 1 35 1,5,6 1 0 0 0 1 36 1,3 1 0 1 0 0 37 3,5,8 0 0 1 0 1 38 3 0 0 1 0 0 39 2,7 0 1 0 0 0 40 2,7 0 1 0 0 0 41 7 0 0 0 0 0 42 4,5,6 0 0 0 1 1 43 NONE 0 0 0 0 0 44 5,10 0 0 0 0 1 45 1,2 1 1 0 0 0 46 7 0 0 0 0 0 47 1,5,10 1 0 0 0 1 48 3 0 0 1 0 0 49 1,3,4 1 0 1 1 0 50 NONE 0 0 0 0 0 51 2,6 0 1 0 0 0 52 NONE 0 0 0 0 0 53 2,5,8,9,10 0 1 0 0 1 Budget Limit 2500 Model: Magazine x y Budget 1 1 9 900 2 0 0 0 3 0 0 0 4 0 0 0 5 1 16 1500 6 0 0 0 7 0 0 0 8 0 0 0 9 0 0 0 10 1 11 1100 Maximize 36 3500 6 7 8 9 10 1500 1000 1200 500 1100 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 Chapter 13: Optimization Modeling CASE: DISTRIBUTION CENTER LOCATION FOR TRACWAY Tracway produces its most popular model of lawn tractor in its St. Louis, Greenwood, and Camarillo plants and ships these units to major distribution centers in Atlanta, Chicago, and San Diego. To support its growing presence in the Pacific Rim, Tracway is considering adding two distribution centers. Locations being considered are Taiwan; Singapore; Sydney, Australia; and Auckland, New Zealand. Tracway anticipates locating distribution centers in two of these new locations. Data acquired from accounting, marketing, and production are shown in Figure 13.10. Total demand in the Pacific Rim is expected to be 5000 units. Develop an appropriate optimization model to identify the best location for the new DCs and transportation allocations to meet demand. Distribution Center Plant Atlanta Chicago San Diego Taiwan Singapore Sydney Auckland St Louis 35 40 60 130 120 148 145 Greenwood 30 30 45 136 130 160 170 Camarillo 60 65 50 115 100 120 170 Demand 9000 3000 9500 Need two plants to handle 5000 increase assuming equal facilities of 2500 Model: Plant/D.C. Atlanta Chicago San Diego Taiwan Singapore Sydney Auckland St Louis 9,000 Greenwood 2,000 7,000 Camarillo 1,000 2,500 2,500 2500 Demand met 9,000 3,000 9,500 2,500 2,500 0 0 Total Cost = 1,417,500 Plant/D.C. Atlanta Chicago San Diego Taiwan Singapore Sydney Auckland St Louis 9,000 2,000 Greenwood 1,000 7,000 Camarillo 2,500 2,500 2500 Demand met 9,000 3,000 9,500 2,500 0 2,500 0 Total Cost = 1,452,500 Plant/D.C. Atlanta Chicago San Diego Taiwan Singapore Sydney Auckland St Louis 9,000 2,500 Greenwood 3,000 5,000 Camarillo 4,500 2,500 Demand met 9,000 3,000 9,500 2,500 0 0 2,500 Total Cost = 1,505,000 Plant/D.C. Atlanta Chicago San Diego Taiwan Singapore Sydney Auckland St Louis 9,000 2,000 Greenwood 1,000 7,000 Camarillo 2,500 2500 2500 Demand met 9,000 3,000 9,500 0 2,500 2,500 0 Total Cost = 1,415,000 Plant/D.C. Atlanta Chicago San Diego Taiwan Singapore Sydney Auckland St Louis 9,000 2,500 Greenwood 3,000 5,000 Camarillo 4,500 2500 Demand met 9,000 3,000 9,500 0 2,500 0 2,500 Total Cost = 1,467,500 Plant/D.C. Atlanta Chicago San Diego Taiwan Singapore Sydney Auckland St Louis 9,000 2,500 Greenwood 3,000 5,000 Camarillo 4,500 2500 Demand met 9,000 3,000 9,500 0 0 2,500 2,500 Total Cost = 1,517,500 amarillo plants Capacity 12000 8000 7500 Total Shipped Capacity 9,000 12000 9,000 8000 8,500 7500 Total Shipped Capacity 11,000 12000 8,000 8000 7,500 7500 Total Shipped Capacity 11,500 12000 8,000 8000 7,000 7500 Total Shipped Capacity 11,000 12000 8,000 8000 7,500 7500 Total Shipped Capacity 11,500 12000 8,000 8000 7,000 7500 Total Shipped Capacity 11,500 12000 8,000 8000 7,000 7500

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Discrete Mathematical Structures, Bernard Kolman, science and engineering, calculus-based physics course, discrete mathematics, quantum physics, physics concepts, fundamental physics, computer science, Hans C. Ohanian

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posted: | 4/19/2010 |

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