Download presentation

Presentation is loading. Please wait.

Published byJudah Ellard Modified about 1 year ago

1
Copyright Kaplan AEC Education, 2008 Engineering Economics Outline Overview CASH FLOW, p. 599 TIME VALUE OF MONEY, p. 600 EQUIVALENCE, p. 602

2
Copyright Kaplan AEC Education, 2008 COMPOUND INTEREST, p. 603 –Symbols and Functional Notation –Single Payment Formulas –Uniform Payment Series Formulas –Uniform Gradient –Continuous Compounding Engineering Economics Outline Overview Continued

3
Copyright Kaplan AEC Education, 2008 NOMINAL AND EFFECTIVE INTEREST, p. 610 –Non-Annual Compounding Engineering Economics Outline Overview Continued

4
Copyright Kaplan AEC Education, 2008 SOLVING ENGINEERING ECONOMICS PROBLEMS, p. 611 –Criteria –Present Worth Appropriate Problems Infinite Life and Capitalized Cost –Future Worth or Value Engineering Economics Outline Overview Continued

5
Copyright Kaplan AEC Education, 2008 SOLVING ENGINEERING ECONOMICS PROBLEMS –Annual Cost, p. 615 Criteria Application of Annual Cost Analysis –Rate of Return Analysis Two Alternatives –Benefit-Cost Analysis –Breakeven Analysis Engineering Economics Outline Overview Continued

6
Copyright Kaplan AEC Education, 2008 BONDS, p. 620 –Bond Value –Bond Yield PAYBACK PERIOD, p. 621 Engineering Economics Outline Overview Continued

7
Copyright Kaplan AEC Education, 2008 VALUATION AND DEPRECIATION, p. 622 –Notation –Straight Line Depreciation –Double Declining-Balance Depreciation –Modified Accelerated Cost Recovery System Depreciation Half-Year Convention Engineering Economics Outline Overview Continued

8
Copyright Kaplan AEC Education, 2008 INFLATION, p. 624 –Effect of Inflation on Rate of Return Engineering Economics Outline Overview Continued

9
Copyright Kaplan AEC Education, 2008 Time Value of Money An investment is estimated to return $5,000 per year for the next 12 years. If the investor wishes to obtain a 9% return per year, the most he should pay for this investment is closest to (a)$30,000 (b)$40,000 (c)$35,000 (d)$45,000

10
Copyright Kaplan AEC Education, 2008 Solution P = A (P/A, 9%, 12) = 5,000 (7.1607) = $35,803 Answer: (c)

11
Copyright Kaplan AEC Education, 2008 A college fund is established for a 5 year old boy, with the objective of having $60,000 upon his 18th birthday. If deposits into an account paying 5% per year are made on each birthday starting with his 6th birthday and ending with his 18th, how much must each deposit be? (a)$3400 (b)$3800 (c)$4200 (d)$4600 Time Value of Money

12
Copyright Kaplan AEC Education, 2008 Solution There are 13 equal deposits. A = F (A/F, 5%, 13) = 60,000 (0.05646) = $3,388 Answer: (a)

13
Copyright Kaplan AEC Education, 2008 A credit card company charges 12% compounded monthly on the unpaid balance. This is equivalent to an effective annual interest rate of most nearly (a)12% (b)12.3% (c)12.7% (d)12.9% Time Value of Money

14
Copyright Kaplan AEC Education, 2008 Solution i e = (1 + r/m) m – 1 = (1 + 0.12/12) 12 – 1 = 0.127 = 12.7% Answer: (c)

15
Copyright Kaplan AEC Education, 2008 An appliance store advertises that a refrigerator can be purchased for $900 cash, or no money down and 24 equal monthly payments of $40. If you paid for the refrigerator with the 24 monthly payment plan, the effective yearly interest rate on your purchase is closest to (a)5.5% (b)5.8% (c)6.2% (d)6.5% Time Value of Money

16
Copyright Kaplan AEC Education, 2008 Solution P = A (P/A, i, 24) 900 = 40 (P/A, i, 24) (P/A, i, n) = 22.50 Looking at the tables this is true for i approximately 0.5% per month i year = (1 + i month ) 12 – 1 = (1 + 0.005) 12 – 1 = 0.0617 = 6.17% Answer: (c)

17
Copyright Kaplan AEC Education, 2008 The annual maintenance costs for a drilling machine are estimated at $750 the first year, increasing by $50 every year to $800 the second year, $850 the third year, and so on. Assuming a useful life of 8 years and a 6% interest rate, the equivalent uniform annual cost (EUAC) for maintenance costs over the useful life is most nearly (a)$800 (b)$900 (c)$1000 (d)$1100 Time Value of Money Problems

18
Copyright Kaplan AEC Education, 2008 Solution EUAC = 750 + 50 (A/G, 6%, 8) = 750 + 50 (3.1952) = $910 Answer: (b)

19
Copyright Kaplan AEC Education, 2008 A store offers an extended warranty for $350 to cover all parts and labor repair costs on a plasma TV over a four year period. The standard warranty only covers repairs during the first year. At a 9% interest rate the minimum equal annual repair costs over years 2 through 4 that makes the extended warranty equally desirable is (a)$150 (b)$175 (c)$200 (d)$225 Time Value of Money

20
Copyright Kaplan AEC Education, 2008 Solution A = 350 (F/P, 9%, 1) (A/P, 9%, 3) = 350 (1.09) (0.39505) = $151 Answer: (a)

21
Copyright Kaplan AEC Education, 2008 Two alternatives are being considered: Time Value of Money AB Initial cost55,00060,000 Operating and Maintenance cost12,000 9,000 Salvage value 2,00015,000 Useful life46 At an interest rate of 8% the EUAC of Alternative A is most nearly (a)$28,200 (b)$29,200 (c)$30,200 (d)$31,200

22
Copyright Kaplan AEC Education, 2008 Solution EUAC = 55,000(A/P, 8%, 4) + 12,000 - 2,000(A/F, 8%, 4) = 55,000(.30192) + 12,000 - 2,000(0.22192) = $28,162 Answer: (a)

23
Copyright Kaplan AEC Education, 2008 A paving contractor can either purchase or lease a road grader. The purchase cost is $89,000. The lease plan is for five equal lease payments payable in advance (i.e., the first lease payment is at the start of the lease). Excluding all operating and maintenance costs and given a MARR of 14%, the maximum lease payment is closest to (a)$22,250 (b)$22,500 (c)$22,750 (d)$23,000 Time Value of Money

24
Copyright Kaplan AEC Education, 2008 Solution A = [89,000 (P/F, 14%, 1)} (A/P, 14%, 5) = {89,000 (0.8772)} (0.29128) = $22,740 Answer: (c)

25
Copyright Kaplan AEC Education, 2008 A company purchases a plastic injection system that will save the company $43,000 during the first year of operation, decreasing by $2,000 every year to $41,000 the second year, $39,000 the third year, and so forth. Given a MARR of 10% per year, the present worth of the savings over the 4-year life of the machine is closest to (a)$125,500 (b)$126,500 (c)$127,500 (d)$128,500 Time Value of Money

26
Copyright Kaplan AEC Education, 2008 Solution P = 43,000 (P/A, 10%, 4) – 2,000 (P/G, 10%, 4) = 43,000 (3.1699) – 2,000 (4.3781) = $127,550 Answer: (c)

27
Copyright Kaplan AEC Education, 2008 Interest Rate A credit card company charges 0.05% interest per day on the outstanding balance. The effective annual interest on charges made on this card is nearest to (a)16% (b)18% (c)20% (d)22%

28
Copyright Kaplan AEC Education, 2008 Solution i year = (1 + 0.0005) 365 – 1 = 0.2002 = 20.0% Answer: (c)

29
Copyright Kaplan AEC Education, 2008 Interest Rate A company is considering purchasing a new machine for $650,000 that will increase the firm’s net income by $150,000 per year over the next 5 years. If the company wishes to obtain a 15% return on its investment, the minimum salvage value of the machine at the end of the 5-year useful life should be closest to (a)$275,000 (b)$295,000 (c)$315,000 (d)$335,000

30
Copyright Kaplan AEC Education, 2008 Solution S = 650,000 (F/P, 15%, 5) – 150,000 (F/A, 10%, 5) = 650,000(2.0114) – 150,000(6.7424) = $296,050 Answer: (b)

31
Copyright Kaplan AEC Education, 2008 Comparison of Alternatives A company is considering two mutually exclusive alternative projects to enhance its production facility. The respective financial estimates for each project are as follows Project AProject B Initial Cost75,000105,000 Annual Savings16,000 24,000 Salvage Value 9,000 0 If the useful life of Project A is 4 years, with a MARR of 15%, the useful life in years of Project B that makes both projects equally desirable is most nearly (a)4 (b)5 (c)6 (d)7

32
Copyright Kaplan AEC Education, 2008 Solution EUAC A = 75,000(A/P, 15%, 4) - 16,000 - 9,000(A/F, 15%, 4) = 75,000(.35027) - 16,000 - 9,000(0.20027) = $8,468 EUAC B = 8,468 = 105,000(A/P, 15%, n) - 24,000 (A/P, 15%, n) = (8,468 + 24,000)/105,000 = 0.3092 From table look-up, the value of n that most nearly makes the above relation true is 5. Answer: (b)

33
Copyright Kaplan AEC Education, 2008 Bonds An investor is considering purchasing a bond with a face value of $20,000 and 10 years left to mature. The bond pays 8% interest payable quarterly. If he wishes to get a 3% per quarter return, the most he should pay for the bond is closest to (a)$15,400 (b)$16,000 (c)$16,400 (d)$16,800

34
Copyright Kaplan AEC Education, 2008 Solution Since the bond pays 8% compounded quarterly, its effective interest rate is 2% per 3 months. Interest payment = i(Face value) = 0.02(20,000) = $400/three months P = 400(P/A, 3%, 40) + 20,000(P/F, 3%, 40) = 400(23.1148) + 20,000(0.3066) = $15,377 Answer: (a)

35
Copyright Kaplan AEC Education, 2008 Benefit to Cost Analysis A county is considering the following project Given a useful life of 12 years and an interest rate of 8%, the benefit to cost ratio is closest to (a)0.67 (b)1.01 (c)1.51 (d)1.67 Initial Cost$22,500,000 Maintenance$525,000 per year Savings$5,300,000 per year

36
Copyright Kaplan AEC Education, 2008 Solution PW cost = 22,500,000 + 525,000 (P/A, 8%, 12) = 22,500,000 + 525,000 (7.5361) = $26,456,453 PW benefit = 5,300,000 (P/A, 8%, 12) = 5,300,000 (7.5361) = $39,941,330 B/C = PW benefit / PW cost = 39,941,330/26,456,453 = 1.51 Answer: (c)

37
Copyright Kaplan AEC Education, 2008 Benefit to Cost Analysis A proposed change to highway design standards is expected to reduce the number of vehicle crashes by 9,200 per year, but have initial cost of $150,000,000 and annual costs of $25,000,000. Given an interest rate of 10% and a study period of 8 years, the average cost of each vehicle crash in order that the benefit-to-cost ratio be 1.0 is closest to (a)$5700 (b)$6700 (c)$8700 (d)$9700

38
Copyright Kaplan AEC Education, 2008 Solution In order that B/C = 1.0 PW cost = PW benefit PW benefit = 150,000,000 + 25,000,000 (P/A, 10%, 8) = 150,000,000 + 25,000,000 (5.3349) = $283,372,500 EUAC benefit = $283,372,500 (A/P, 10%, 8) = $283,372,500 (0.18744) = $53,115,341 $ equivalent /crash = 53,115,341/9,200 = $5,773 Answer: (a)

39
Copyright Kaplan AEC Education, 2008 Payback, Breakeven Analysis, and Capitalized Cost An engineering department is considering purchase of an advanced computational fluid dynamics software system to enhance productivity. The initial cost of the software is $55,000 but is expected to result in efficiency savings of $25,000 the first year, with this amount decreasing by $5,000 per year thereafter. The payback period for the software is closest to (a)2 (b)2.67 (c)3 (d)3.67

40
Copyright Kaplan AEC Education, 2008 Solution Costs = 55,000 The payback period is the time when total income to date is equal to the total costs. Costs = Income = 25,000 + 20,000 + 15,000 + … Since the income is stated as $25,000 per year, one can assume that savings occur uniformly throughout the year. Therefore, the payback period is 2.67 years. Answer: (b)

41
Copyright Kaplan AEC Education, 2008 Payback, Breakeven Analysis and Capitalized Cost A company is considering two alternative forklifts with equal useful lives and the following characteristics AB Initial Cost25,00032,000 Total Annual Costs 4350 2500 Given an interest rate of 10%, the service life in years at which both machines have the same equivalent uniform annual cost (EUAC) is most nearly (a)5 (b)7 (c)9 (d)11

42
Copyright Kaplan AEC Education, 2008 Solution EUAC A = 25,000 (A/P, 10%, n) + 4,350 EUAC B = 32,000 (A/P, 10%, n) + 2,500 EUAC A = EUAC B 25,000 (A/P, 10%, n) + 4,350 = 32,000 (A/P, 10%, n) + 2,500 (A/P, 10%, n) = (4,350 – 2,500) / 7,000 = 0.2643 From table look-up, the value of n that most nearly makes the above relation true is 15. Answer: (a)

43
Copyright Kaplan AEC Education, 2008 Payback, Breakeven Analysis and Capitalized Cost An alumnus wishes to endow a scholarship to her alma mater that will provide $12,000 per year in perpetuity. Although the donation will be given today, the first scholarship will be given in 3 years (i.e., at time t = 3 years). At an interest rate of 8%, the amount she will need to donate is closest to (a)$119,600 (b)$122,600 (c)$125,600 (d)$128,600

44
Copyright Kaplan AEC Education, 2008 Solution P = (A/i) (P/F, 8%, 2) = (12,000/0.08) (0.8573) = (150,000) (0.8573) = $128,595 Answer: (d)

45
Copyright Kaplan AEC Education, 2008 Depreciation A company purchases a plastic extrusion machine for $95,000. If this machine has an estimated salvage value of $10,000 at the end of its five-year useful life and recovery period, the second year straight line depreciation is closest to (a)$13,000 (b)$15,000 (c)$17,000 (d)$19,000

46
Copyright Kaplan AEC Education, 2008 Solution D t = D 2 = (95,000 – 10,000)/5 = $17,000 Answer: (c)

47
Copyright Kaplan AEC Education, 2008 Inflation A compact car costs approximately $21,000 today. If a comparable car cost $15,000 ten years ago, the average annual inflation in compact car prices over the past ten years is closest to (a)2.6% (b)3.0% (c)3.4% (d)3.8%

48
Copyright Kaplan AEC Education, 2008 Solution 21,000 = 15,000 (1 + f) 10 f = (21,000/15,000) 0.1 – 1 = 0.034 = 3.4% Answer: (c)

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google