which considers any inflationary effects in the cash flows.

Note 1: Unless otherwise stated, all cash flows given in the problems represent aftertax cash flows in actual dollars. The MARR also represents a market interest rate, which considers any inflationary effects in the cash flows. Note 2: Unless otherwise stated, all interest rates presented in this set of problems assume annual compounding. 6.1 Consider the following cash flows and compute the equivalent annual worth at i = 12%: An n Investment 0 1 2 3 4 5 6 -$10,000 $2,000 Revenue $2,000 $2,000 $3,000 $3,000 $1,000 $500 AE (12%) = [-$10,000 + $2,000(P/F, 12%, 1) + +$2,500(P/F, 12%, 6)] (A/P, 12%, 6) = -$180.96 1

6.2 (A) The following investment has a net present value of zero at i = 8%: X X $400 0 1 X X 5 6 $400 2 3 4 Years $2,145 Which of the following is the net equivalent annual worth at 8% interest? (a) (b) (c) (d) $400 $0 $500 $450 (b) NEAW = NPW * (A/P, 8, 6) 0 = -2145 + 400 * (P/F, 8, 1) + 400 * (P/F, 8, 4) + X * (P/A, 8, 2) (P/F, 8, 1) + X * (P/A, 8, 2) (P/F, 8, 4) = -2145 + 400 * [(P/F, 8, 1) + (P/F, 8, 4)] + X * (P/A, 8, 2) [(P/F, 8, 1) + (P/F, 8, 4)] = -2145 + 400 * [.9259 +.7350] + X * 1.7833 * [9259 +.7350] = -2145 + 400 * 1.6609 + X * 2.9619 1475.64 / 2.9619 = X = 498.21 2

6.3 Consider the following sets of investment projects: n 0 1 2 3 4 5 Project s Cash Flow A B C D -$2,000 -$4,000 -$3,000 -$9,000 $400 $3,000 -$2,000 $2,000 $500 $2,000 $4,000 $4,000 $600 $1,000 $2,000 $8,000 $700 $500 $4,000 $8,000 $800 $500 $2,000 $4,000 Compute the equivalent annual worth of each project at i = 10%, and determine the acceptability of each project. AE (10%) A = -$2,000(A/P, 10%, 5) + $400 +$100(A/G, 10%, 5) = $53.42 (Accept) AE (10%) B = -$4,000(A/P, 10%, 5) + $500 + [$2,500(P/F, 10%, 1) + $1,500(P/F, 10%, 2) + $500(P/F, 10%, 3)] (A/P, 10%, 5) = $470.47 (Accept) AE (10%) C = [-$3,000 - $2,000(P/F, 10%, 1) + + $2,000(P/F, 10%, 5)] (A/P, 10%, 5) = $1,045.73 (Accept) AE (10%) D = [-$9,000 + $2,000(P/F, 10%, 1) + + $4,000(P/F, 10%, 5)] (A/P, 10%, 5) = $2,659.68 (Accept) 3

6.4 (A) What is the annual-equivalence amount for the following infinite series at i = 12%? $1,200 $700 0 (a) (b) (c) (d) 1 2 3 4 5 6 7 8 Years 9 10 11 12 13 $950 $866 $926 None of the above (a) AE (12%) = $700 + [($500/0.12) (P/F, 12%, 6)] (0.12) = $700 + [($500/0.12) (.50663)] (0.12) = $953 NB AE (12%) = $700 +$500 * (P/F, 12%, 6) 4

6.5 Consider the following sets of investment projects: Period (n) 0 1 2 3 A -$3,500 $0 $0 $5,500 Project s Cash Flow B C -$3,000 $1,500 $1,800 $2,100 -$3,000 $3,000 $2,000 $1,000 D -$3,600 $1,800 $1,800 $1,800 Compute the equivalent annual worth of each project at i = 13%, and determine the acceptability of each project. AE (13%) A = -$3,500(A/P, 13%, 3) + $5,500(A/F, 13%, 3) = $132.04, Accept AE (13%) B = -$3,000(A/P, 13%, 3) + $1,500 + $300(A/G, 13%, 3) = $505.05, Accept AE (13%) C = -$3,000(A/P, 13%, 3) + $3,000 - $1,000(A/G, 13%, 3) = $810.71, Accept AE (13%) D = -$3,600(A/P, 13%, 3) + $1,800 = $275.32, Accept 5

6.6 (A) Consider an investment project with the following repeating cash flow pattern every four years for forever: $80 $80 $40 $40 0 1 2 $80 $80 3 4 $40 $40 5 6 Years 7 8 What is the annual-equivalence amount of this project at an interest rate of 12%? PW (12%) one-cycle = 80 * (P/F, 12, 1) + 80 * (P/ F, 12, 2) + 40 * (P/F, 12, 3) + 40 * (P/F, 12, 4) = $189.10 AE (12%) = 189.10(A/P, 12%, 4) = $62.25 (.3292) NB 189.10 *.12 = 22.69 6

6.7 The owner of a business is considering investing $55,000 in new equipment. He estimates that the net cash flows will be $5,000 during the first year and will increase by $2,500 per year each year thereafter. The equipment is estimated to have a 10-year service life and a net salvage value at the end of this time of $6,000. The firm s interest rate is 12%. (a) Determine the annual capital cost (ownership cost) for the equipment. (b) Determine the equivalent annual savings (revenues). (c) Determine whether this investment is wise. Given: I = $55,000, S = $6,000, N = 10 years, i = 12% (a) AE(12%)1 = ($55,000 - $6,000)(A/P,12%, 10) + $6,000(0.12) = $9,392 (b) AE (12%) 2 = $5,000 + $2,500 (A/G, 12%, 10) = $13,962 (c) AE(12%) = $13,962 - $9,392 = $4,570 This is a good investment. 7

Capital Recovery (Ownership) Cost 6.8 (A) Susan wants to buy a car that she will keep for the next four years. She can buy a Honda Civic at $15,000 and then sell it for $8,000 after four years. If she bought this car, what would be her annual ownership cost (capital recovery cost)? Assume that her interest rate is 14%. (Note in book it says interest rate is 6%) CR (14%) = ($15,000 - $8,000) (A/P, 14%, 4) + (.14) ($8,000) = $7,000 * (.3432) + (.14) ($8,000) = $2,402.40 + $1,120 = $3,522.40 6.9 Nelson Electronics Company just purchased a soldering machine to be used in its assembly cell for flexible disk drives. This soldering machine costs $250,000. Because of the specialized function it performs, its useful life is estimated to be five years. At the end of that time, its salvage value is estimated to be $40,000. What is the capital cost for this investment if the firm s interest rate is 18%? Given: I = $250,000, S = $40,000, N = 5 years, i = 18% CR (18%) = ($250,000 - $40,000) (A/P, 18%, 5) + (.18) ($40,000) = $74,353 8

6.10 A construction firm is considering establishing an engineering computing center. This center will be equipped with three engineering workstations that each would cost $25,000 and have a service life of five years. The expected salvage value of each workstation is $2,000. The annual operating and maintenance cost would be $15,000 for each workstation. At a MARR of 15%, determine the equivalent annual cost of operating the engineering center. Capital cost: CR (15%) = ($75,000 - $6,000) (A/P, 15%, 5) + (.15) ($6,000) = $21,484 Annual operating costs: $45,000 AE (15%) = $21,484 + $45,000 = $66,484 6.11 (A) Beginning next year, a foundation will support an annual seminar on campus by using the interest earnings on a $100,000 gift it received this year. It is determined that 8% interest will be realized for the first 10 years, but that plans should be made to anticipate an interest rate of only 6% after that time. What amount should be added to the foundation now in order to fund the seminar at a level of $10,000 per year into infinity? PW (i) = $10,000(P/A, 8%, 10) + ($10,000/0.06) (P/F, 8%, 10) = $77,199 + $67,101 = $144,300 The amount of additional funds should be $44,300. 9

Annual Equivalent Worth Criterion 6.12 The present price (year zero) of kerosene is $2.50 per gallon, and its cost is expected to increase by $.30 per year (e.g., kerosene at the end of year one will cost $2.80 per gallon). Mr. Garcia uses about 800 gallons of kerosene during a winter season for space heating. He has an opportunity to buy a storage tank for $700, and at the end of four years, he can sell the storage tank for $100. The tank has a capacity to supply four years of Mr. Garcia s heating needs, so he can buy four years of kerosene at its present price ($2.50). He can invest his money elsewhere at 8%. Should he purchase the storage tank? Assume that kerosene purchased on a pay-as-you-go basis is paid for at the end of the year. (However, kerosene purchased for the storage tank is purchased now.) Option 1 Cost of kerosene = $800 * 2.5 * 4 = $8,000 Option 1 Cost of kerosene = $800 * 2.8 = $2,240 Annual increase = $800 *.3= $240 n Option 1 Option 2 0 -$700,-$8,000 1 0 -$2,240 2 0 -$2,480 3 0 -$2,720 4 +$100 -$2,960 AE (8%) Option 1 = -$8,700(A/P, 8%, 4) + $100(A/F, 8%, 4) = -$2,604.52 AE (8%) Option 2 = -$2,240 - $240(A/G, 8%, 4) = -$2,576.96 Select Option 2. 10

6.13 (A) Consider the cash flows for the following investment projects: Project s Cash Flow A B -$4,000 $5,500 $1,000 -$1,400 $X -$1,400 $1,000 -$1,400 $1,000 -$1,400 n 0 1 2 3 4 (a) For project A, find the value of X that makes the equivalent annual receipts equal the equivalent annual disbursement at i = 13%. (b) Would you accept project B at i = 15%, based on the AE criterion? (a) AE(13%) = -$4,000(A/P, 13%, 4) + $1,000 + (X - $1,000) (P/F, 13%, 2) (A/P, 13%, 4) =0 AE (13%) = -$608.06 + 0.2638X = 0 X = $2,309.55 (b) AE(13%) = $5,500(A/P, 15%, 4) - $1,400 = $526.46 > 0 Accept project B. 11

6.14 An industrial firm can purchase a certain machine for $40,000. A down payment of $4,000 is required, and the balance can be paid in five equal year-end installments at 7% interest on the unpaid balance. As an alternative, the machine can be purchased for $36,000 in cash. If the firm s MARR is 10%, determine which alternative should be accepted, based on the annual-equivalence method. Option 1: Purchase-Borrow Option: Annual repayment of loan amount of $36,000: A = $36,000(A/P, 7%, 5) = $8,780 AE (10%) 1 = -$4,000(A/P, 10%, 5) - $8,780 = -$9,835 Option 2: Cash Purchase Option: AE (10%) 2 = -$36,000(A/P, 10%, 5) = -$9,497 Option 2 is a better choice. 12

6.15 (A) An industrial firm is considering purchasing several programmable controllers and automating their manufacturing operations. It is estimated that the equipment will initially cost $100,000 and the labor to install it will cost $35,000. A service contract to maintain the equipment will cost $5,000 per year. Trained service personnel will have to be hired at an annual salary of $30,000. Also estimated is an approximate $10,000 annual income-tax savings (cash inflow). How much will this investment in equipment and services have to increase the annual revenues after taxes in order to break even? The equipment is estimated to have an operating life of 10 years, with no salvage value, because of obsolescence. The firm s MARR is 10%. The total investment consists of the sum of the initial equipment cost and the installation cost, which is $135,000. Let R denote the break-even annual revenue. AE (10%) = -$135,000(A/P, 10%, 10) - $30,000 - $5,000 + $10,000 + R =0 Solving for R yields R = $46,971 13

6.16 A certain factory building has an old lighting system, and lighting this building costs, on average, $20,000 a year. A lighting consultant tells the factory supervisor that the lighting bill can be reduced to $8,000 a year if $50,000 were invested in a new lighting system for the factory building. If the new lighting system is installed, an incremental maintenance cost of $3,000 per year must be considered. If the old lighting system has zero salvage value and the new lighting system is estimated to have a life of 20 years, what is the net annual benefit for this investment in new lighting? Consider the MARR to be 12%. Also consider that the new lighting system has zero salvage value at the end of its life. New lighting system cost: AE (12%) = $50,000(A/P, 12%, 20) + $8,000 + $3,000 = $17,694 Old lighting system cost: AE (12%) = $20,000 Annual savings from installing the new lighting system = $2,306 14

Unit-Profit or Unit-Cost Calculation 6.17 (A) Two 150-horsepower (HP) motors are being considered for installation at a municipal sewage-treatment plant. The first costs $4,500 and has an operating efficiency of 83%. The second costs $3,600 and has an operating efficiency of 80%. Both motors are projected to have zero salvage value after a life of 10 years. If all the annual charges, such as insurance and maintenance, amount to a total of 15% of the original cost of each motor, and if power costs are a flat 5 cents per kilowatt-hour, what is the minimum number of hours of full-load operation per year required in order to justify purchase of the more expensive motor at i = 6%? (A conversion factor you might find useful is 1 HP = 746 watts = 0.746 kilowatts.) Let T denote the total operating hours in full load. Motor I (Expensive): Annual power cost: (150/0.83)*(0.746)*(0.05)*T = $6.741T Equivalent annual cost of operating the motor: AE (6%) I = -$4,500(A/P, 6%, 10) - $675 [= 4,500 *.15] 6.741T = -$1,286.41 - $6.741T Motor II (Less Expensive): Annual power cost: (150/0.80)*(0.746)*(0.05)*T = $6.9938T Equivalent annual cost of operating the motor: AE (6%) II = -$3,600(A/P, 6%, 10) - $540 [= 3,600 *.15] 6.9938T = -$1,029.11-6.9938T Let AE (6%) I = AE (6%) II and solve for T. -$1,286.41 - $6.741T = -$1,029.11-6.9938T (6.9938 6.741) * T = 1286.41 1029.11 T = 257.3 /.2528 T = 1,017.8 hours per year 15

6.18 Two 180-horsepower water pumps are being considered for installation in a municipal work. Financial data for these pumps are given as follows: Item Initial cost Efficiency Useful life Annual operating cost Salvage value Pump I $6,000 86% 12 years $500 $0 Pump II $4,000 80% 12 years $440 $0 If power cost is a flat 6 cents per kwh over the study period, which of the following ranges includes the minimum number of hours of full-load operation per year required in order to justify the purchase of the more expensive pump at an interest rate of 8%? (1 HP = 746 watts = 0.746 kilowatts) (a) (b) (c) (d) 340 hours/year < minimum number of operation hours/year 390 hours/year 390 hours/year < minimum number of operation hours/year 440 hours/year 440 hours/year < minimum number of operation hours/year 490 hours/year 490 hours/year < minimum number of operation hours/year 540 hours/year Pump I: (180/.86)(.746)(.06)T = 9.368T AE (8%) I = $6,000(A/P, 8%, 12) + $500 + 9.368T = $1,296.2 + 9.368T Pump II: (180/.8)(.746)(.06)T = 10.071T AE (8%) II = $4,000(A/P, 8%, 12) + $440 + 10.071T = $970.8 + 10.071T $1,296.2 + 9.368T = $970.8 + 10.071T $1,296.2 - $970.8 = 10.071T - 9.368T 325.4 =.703T T = 463 hours Select (c). 16

6.19 (A) You invest in a piece of equipment costing $20,000. The equipment will be used for two years, at the end of which time the salvage value of the machine is expected to be $10,000. The machine will be used for 6,000 hours during the first year and 8,000 hours during the second year. The expected annual net savings in operating costs will be $30,000 during the first year and $40,000 during the second year. If your interest rate is 10%, which of the following would be the equivalent net savings per machine hour? (a) (b) (c) (d) $4.00/hour $5.00/hour $6.00/hour $7.00/hour Capital cost CR (10%) = ($20,000 - $10,000) (A/P, 10%, 2) + $10,000(0.10) = $6,792 AEsavings (10%) = [$30,000(P/F, 10, 1) + $40,000(P/F, 10, 2)] (A/P, 10%, 2) = $34,762 Net annual savings = $34,762 - $6,792 = $27,970 Average Hours per Year = [(P/F, 10, 1)* (6,000) + (P/F, 10, 2) (8,000)] (A/P, 10%, 2) = 6,952 $27,970 = 6,952*C (net savings per machine hour) C = $4.02 per hour Select (a) 34,762/6,792 = 5 (see spreadsheet for accurate calculation) 17

6.20 A company is currently paying its employees $0.33 per mile to drive their own cars when on company business. However, the company is considering supplying employees with cars, which would involve the following cost components: car purchase at $22,000, with an estimated three-year life; a net salvage value of $5,000; taxes and insurance at a cost of $700 per year; and operating and maintenance expenses of $0.15 per mile. If the interest rate is 10% and the company anticipates an employee s annual travel to be 30,000 miles, what is the equivalent cost per mile (without considering income tax)? Option 1: Pay employee $0.33 per mile: Option 2: Provide a car to employee: AE (10%) capital cost = ($22,000 - $5,000) (A/P, 10%, 3) + (0.10) ($5,000) = $7,336 AE (10%) operating cost = $700 + ($0.15) (30,000) = $5,200 AE (10%) total cost = $7,336 + $5,200 = $12,536 Operating cost per mile = $12,536/30,000 = $0.42 Option 1 is a better choice. 18

6.22 (A) An electric automobile can be purchased for $25,000. The automobile is estimated to have a life of 12 years, with annual travel of 20,000 miles. Every three years, a new set of batteries will have to be purchased at a cost of $3,000. Annual maintenance of the vehicle is estimated to cost $700 per year. The cost of recharging the batteries is estimated at $0.015 per mile. The salvage value of the batteries and the vehicle at the end of 12 years is estimated at $2,000. Consider the MARR to be 7%. What is the cost per mile to own and operate this vehicle, based on the foregoing estimates? The $3,000 cost of the batteries is a net value, with the old batteries traded in for the new ones. Capital costs: AE (7%) 1 = ($25,000 - $2,000) (A/P, 7%, 12) + (0.07) ($2,000) = $3,036 Annual battery replacement costs: AE (7%) 2 = $3,000[(P/F, 7%, 3) + (P/F, 7%, 6) + (P/F, 7%, 9)] (A/P, 7%, 12) = $765.41 Annual recharging costs: AE (7%) 3 = ($0.015) (20,000) = $300 Total annual costs: AE (7%) = $3,036 + $765.41 +$300 + $700 = $4,801.41 Costs per mile: Cost/mile = $4,801.41/20,000 = $0.2401 19

6.23 A California utility firm is considering building a 50-megawatt geothermal plant that generates electricity from naturally occurring underground heat. The binary geothermal system will cost $85 million to build and $6 million to operate per year (including any income-tax effect). (Unlike a conventional fossil-fuel plant, this system will require virtually no fuel costs.) The geothermal plant is to last for 25 years. At the end of that time, the expected salvage value will be about the same as the cost to remove the plant. The plant will be in operation for 70% of the year (the plant-utilization factor or 70% of 8,760 hours per year). If the firm s MARR is 14%, determine the cost of generation electricity per kilowatt-hour. Annual total operating hours: (0.70)(8,760) = 6,132 hours per year Annual electricity generated: 50,000*6,132 = 306,600,000 kilowatt-hours Equivalent annual cost: AE (14%) = $85,000,000(A/P, 14%, 25) + $6,000,000 = $18,367,364 Cost per kilowatt-hour: $18,367,364/306,600,000 = $0.06 per kilowatt-hour 20

Break-Even Analysis 6.24 The estimated cost of a completely installed and ready-to-operate 40-kilowatt generator is $30,000. Its annual maintenance costs are estimated at $500. The energy that can be generated annually, at full load, is estimated to be 100,000 kilowatt-hours. If the value of the energy generated is considered to be $0.08 per kilowatt-hour, how long will it be before this machine becomes profitable? Consider the MARR to be 9% and the salvage value of the machine to be $2,000 at the end of its estimated life of 15 years. Minimum operating hours: AE (9%) = ($30,000 - $2,000) (A/P, 9%, 15) + (0.09) ($2,000) + $500 = $4,153.65 Let T denote the annual operating hours. Then the total kilowatt-hours generated would be 40T. Since the value of the energy generated is $0.08 per kilowatt-hour, we establish the following relationship: $0.08 *40T = $4,153.65 Solving for T yields T = 1,298 hours [per year for 15 years] Annual worth of the generator at full load operation: AE (9%) = (0.08) (100,000) - $4,153.65 = $3,846.35 Discounted payback period at full load of operation: n 0 1 1 5 Investment Revenue Maintenance cost -$30,000 +$2,000 $8,000 $8,000 -$500 -$500 Net cash flow -$30,000 $7,500 $9,500 $30,000 = $7,500(P/A, 9%, n) Solving for n yields 21

n = 5.185 years 22

6.25 (A) A large state university, currently facing a severe parking shortage on its campus, is considering constructing parking decks off campus. A shuttle service composed of minibuses could pick up students at the off-campus parking deck and quickly transport them to various locations on campus. The university would charge a small fee for each shuttle ride, and the students could be quickly and economically transported to their classes. The funds raised by the shuttle would be used to pay for minibuses, which cost about $150,000 each. Each minibus has a 12-year service life, with an estimated salvage value of $3,000. To operate each minibus, the following additional expenses must be considered: Item Driver Maintenance Insurance Annual Expenses $40,000 $7,000 $2,000 If students pay 10 cents for each ride, determine the annual ridership (i.e., the number of shuttle rides per year) required to justify the shuttle project, assuming an interest rate of 6%. Capital cost: AE (6%) 1 = ($150,000 - $3,000) (A/P, 6%, 12) + (0.06) ($3,000) = $17,714 Annual operating costs: AE (6%) 2 = $40,000 + $7,000 + $2,000 = $49,000 Total annual system costs: AE (6%) = $17,714 + $49,000 = $66,714 Number of rides required per year: Number of rides = $66,714/ ($0.10) = 667,140 rides 23

6.26 Eradicator Food Prep, Inc. has invested $7 million to construct a food irradiation plant. This technology destroys organisms that cause spoilage and disease, thus extending the shelf life of fresh foods and the distances over which they can be shipped. The plant can handle about 200,000 pounds of produce in an hour, and it will be operated for 3,600 hours a year. The net expected operating and maintenance costs (considering any income-tax effects) would be $4 million per year. The plant is expected to have a useful life of 15 years, with a net salvage value of $700,000. The firm s interest rate is 15%. (a) If investors in the company want to recover the plant investment within six years of operation (rather than 15 years), what would be the equivalent after-tax annual revenues that must be generated? (b) To generate the annual revenues determined in part (a), what minimum processing fee per pound should the company charge to its customers? Given: Investment cost = $7 million Plant capacity = 200,000 1bs/hour Plant operating hours = 3,600 hours per year O&M cost = $4 million per year Useful life = 15 years Salvage value = $700,000, and MARR = 15%. (a) PW (15%) = -$7,000,000 + (R - $4,000,000) (P/A, 15%, 6) = 3.7844R - $22,137,600 =0 Solving for R yields R = $5,849,700 per year (b) Minimum processing fee per 1b (after-tax): $5,849,700/ [(200,000) (3,600)] = $0.0081 per 1b Comments: The minimum processing fee per lb should be higher on a before-tax basis. 24

6.27 A corporate executive jet with a seating capacity of 20 has the following cost factors: Item Cost Initial cost $12,000,000 Service life 15 years Salvage value $2,000,000 Crew costs per year $225,000 Fuel cost per mile $1.10 Landing fee $250 Maintenance costs per year $237,500 Insurance costs per year $166,000 Catering cost per passenger round trip $75 The company flies three round-trips from Boston to London per week, a distance of 3,280 miles one way. How many passengers must be carried on an average round trip in order to justify the use of the jet if the alternative commercial airline first-class round-trip fare is $3,400 per passenger? The firm s MARR is 15%. Let X denote the average number of round-trip passengers per year. Capital costs: CR (15%) = ($12,000,000 - $2,000,000) (A/P, 15%, 15) + (0.15) ($2,000,000) = $2,010,171 Annual crew costs: $225,000 Annual fuel costs for round trips: ($1.10) (3,280) (2) (3) (52) = $1,125,696 Annual landing fees: ($250) (2) (3) (52) = $78,000 Annual maintenance, insurance, and catering costs: $237,500 + $166,000 + $75X = $403,500 + $75X 25

Total equivalent annual costs: AE (15%) = $2,010,171 + $225,000 + $1,125,696 + $78,000 + $403,500 + $75*X = 3,842,367 +75*X Solving for X yields $3,400X = 3,842,367 +75*X X = 1,156 passengers per year Or 1,156 / (3) (52) = 7.41 8 passengers per round trip 26

Comparing Mutually Exclusive Alternatives by Using the AE Method 6.29 You are considering two types of electric motors for your paint shop. Financial information and operating characteristics are summarized as follows: Summary Info and Characteristics Brand X Brand Y Price O&M cost per year Salvage value Capacity Efficiency $4,500 $300 $250 150 HP 83% $3,600 $500 $100 150 HP 80% O&M = Operating & Maintenance If you plan to operate the motor for 2,000 hours annually, which of the given options represents the total cost savings per operating hour associated with the more efficient brand (Brand X) at an interest rate of 12%? The motor will be needed for 10 years. Assume that power costs are five cents per kilowatt-hour (1 HP = 0.746 kw). (a) (b) (c) (d) Less than 10 cents/hr Between 10 cents/hr and 20 cents/hr, inclusive Between 20.1 cents/hr and 30 cents/hr, inclusive Greater than or equal to 30.1 cents/hr AE(12%)X = (4,500 250)(A/P, 12%, 10) + (.12) (250) + 300 + [150(.746)/ (.83)] (2,000) (.05) = $14,564 Unit cost = $14,564/2,000 = $7.28 per hour AE(12%)Y = (3,600 100)(A/P, 12%, 10) + (.12) (100) + 500 + [150(.746)/ (.80)] (2,000) (.05) = $15,119 Unit cost = $15,119/2,000 = $7.56 per hour Select (c) Difference = $.28 = 28c 27

6.30 (A) The following cash flows represent the potential annual savings associated with two different types of production processes, each of which requires an investment of $12,000 and has zero salvage value: n 0 1 2 3 4 Process A -$12,000 $9,120 $6,840 $4,560 $2,280 Process B -$12,000 $6,350 $6,350 $6,350 $6,350 Assuming an interest rate of 15%, complete the following tasks: (a) Determine the equivalent annual savings for each process. (b) Determine the hourly savings for each process, assuming 2,000 hours of operation per year. (c) Determine which process should be selected. (a) AE (15%) A = $12,000(A/P, 15%, 4) + [$9,120 $2,280(A/G, 15%, 4)] = $1,892.95 AE (15%) B = $12,000(A/P, 15%, 4) + $6,350 = $2,146.82 (b) Process A: $1,892.95/2,000 = $0.9465/hour Process B: $2,146.82/2,000 = $1.0734/hour (c) Process B is a better choice. 28

6.31 Travis Wenzel has $2,000 to invest. Usually, he would deposit the money in his savings account, which earns 6% interest, compounded monthly. However, he is considering three alternative investment opportunities: Option 1: Purchasing a bond for $2,000. The bond has a face value of $2,000 and pays $100 every six months for three years. The bond matures in three years. Option 2: Buying and holding a growth stock that grows 11% per year for three years. Option 3: Making a personal loan of $2,000 to a friend and receiving $150 interest per year for three years. Determine the equivalent annual cash flows for each option, and select the best option. Given: i = 6% interest compounded monthly. Monthly nominal =.5%. The effective annual interest = (1.005)12 1 = 6.17% per year and The effective semiannual interest = (1.005)6 1 = 3.04% per semiannual Option 1: Buying a bond AE (3.04%) 1 = -$2,000(A/P, 3.04%, 6) + $100 + $2,000(A/F, 3.04%, 6) = $39.20 per semiannual AE (6.17%) = $39.20(F/A, 3.04%, 2) = $79.59 per year Option 2: Buying and holding a growth stock for 3 years AE (6.17%) 2 = -$2,000(A/P, 6.17%, 3) + $2,735.26(A/F, 6.17%, 3) = $107.17 Option 3: Receiving $150 interest per year for 3 years AE (6.17%) 3 = -$2,000(A/P, 6.17%, 3) + $150 + $2,000(A/F, 6.17%, 3) = $126.60 29

Making the personal loan is the best option. 30

6.32 A chemical company is considering two types of incinerators to burn solid waste generated by a chemical operation. Both incinerators have a burning capacity of 20 tons per day. The following data have been compiled for comparison of the two incinerators: Item Incinerator A Incinerator B $1,200,000 $750,000 Annual O&M costs $50,000 $80,000 Service life 20 years 10 years Salvage value $60,000 $30,000 Income taxes $40,000 $30,000 Installed cost If the firm s MARR is know to be 13%, determine the processing cost per ton of solid waste for each incinerator. Assume that incinerator B will be available in the future at the same cost. Equivalent annual cost: AE (13%) A = ($1,200,000 - $60,000) (A/P, 13%, 20) + (0.13) ($60,000) + $50,000 +$40,000 = $260,083 AE (13%) B = ($750,000 - $30,000) (A/P, 13%, 10) + (0.13) ($30,000) + $80,000 +$30,000 = $216,395 Processing cost per ton: C1 = $260,083/ (20) (365) = $35.63 per ton C2 = $216,395/ (20) (365) = $29.64 per ton Incinerator B is a better choice. 31

6.34 (A) Norton Auto Parts, Inc., is considering two different forklift trucks for use in its assembly plant: Truck A costs $15,000 and requires $3,000 annually in operating expenses. It will have a $5,000 salvage value at the end of its 3-year service life. Truck B costs $20,000, but requires only $2,000 annually in operating expenses; its service life is four years, after which its expected salvage value is $8,000. The firm s MARR is 12%. Assuming that the trucks are needed for 12 years and that no significant changes are expected in the future price and functional capacity of both trucks; select the most economical truck, based on AE analysis. Since the required service period is 12 years and the future replacement cost for each truck remains unchanged, we can easily find the equivalent annual cost over 12-year period by simply finding the annual equivalent cost of the first replacement cycle for each truck. Truck A: Four replacements are required AE (12%) A = ($15,000 - $5,000) (A/P, 12%, 3) + (0.12) ($5,000) + $3,000 = $7,763.50 Truck B: Three replacements are required AE (12%) B = ($20,000 - $8,000) (A/P, 12%, 4) + (0.12) ($8,000) + $2,000 = $6,910.80 Truck B is a more economical choice. 32