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Topic 6 Additional Problems.doc

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Department
Economics
Course
1000
Professor
Tom Haffie
Semester
Winter

Description
Note 1: Unless otherwise stated, all cash flows given in the problems represent after- tax 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 Revenue 0 -$10,000 1 $2,000 2 $2,000 3 $3,000 4 $3,000 5 $1,000 6 $2,000 $500 Answer 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 X X $400 $400 0 1 2 3 4 5 6 Years $2,145 Which of the following is the net equivalent annual worth at 8% interest? (a) $400 (b) $0 (c) $500 (d) $450 Answer (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: Project’s Cash Flow n A B C D 0 -$2,000 -$4,000 -$3,000 -$9,000 1 $400 $3,000 -$2,000 $2,000 2 $500 $2,000 $4,000 $4,000 3 $600 $1,000 $2,000 $8,000 4 $700 $500 $4,000 $8,000 5 $800 $500 $2,000 $4,000 Compute the equivalent annual worth of each project at i = 10%, and determine the acceptability of each project. Answer 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 1 2 3 4 5 6 7 8 9 10 11 12 13 … Years (a) $950 (b) $866 (c) $926 (d) None of the above Answer (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 Project’s Cash Flow (n) A B C D 0 -$3,500 -$3,000 -$3,000 -$3,600 1 $0 $1,500 $3,000 $1,800 2 $0 $1,800 $2,000 $1,800 3 $5,500 $2,100 $1,000 $1,800 Compute the equivalent annual worth of each project at i = 13%, and determine the acceptability of each project. Answer 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 $80 $80 ∞ $40 $40 $40 $40 0 1 2 3 4 5 6 7 8 Years What is the annual-equivalence amount of this project at an interest rate of 12%? Answer PW (12%) one-cycle * (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. Answer 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%) = 25,000 + $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%) Answer 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%? Answer 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. Answer • 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? Answer 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.) Answer 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 n A B 0 -$4,000 $5,500 1 $1,000 -$1,400 2 $X -$1,400 3 $1,000 -$1,400 4 $1,000 -$1,400 (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? Answer (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. Answer • 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%) = -$36,000(A/P, 10%, 5) 2 = -$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%. Answer 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. Answer • 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.) Answer 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%) = II3,600(A/P, 6%, 10) - $540 [= 3,600 * .15] – 6.9938T = -$1,029.11 - 6.9938T Let AE (6%) = IE (6%) and sIIve 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 Pump I Pump II Initial cost $6,000 $4,000 Efficiency 86% 80% Useful life 12 years 12 years Annual operating cost $500 $440 Salvage value $0 $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) 340 hours/year < minimum number of operation hours/year ≤ 390 hours/year (b) 390 hours/year < minimum number of operation hours/year ≤ 440 hours/year (c) 440 hours/year < minimum number of operation hours/year ≤ 490 hours/year (d) 490 hours/year < minimum number of operation hours/year ≤ 540 hours/year Answer 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%) =II4,000(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) $4.00/hour (b) $5.00/hour (c) $6.00/hour (d) $7.00/hour Answer Capital cost CR (10%) = ($20,000 - $10,000) (A/P, 10%, 2) + $10,000(0.10) = $6,792 AE (10%) = [$30,000(P/F, 10, 1) + $40,000(P/F, 10, 2)] (A/P, 10%, 2) savings = $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
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