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CE 466 FE Exam Review Engineering Economics Spring 2013.

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1 CE 466 FE Exam Review Engineering Economics Spring 2013

2 Economic Equivalence Two cash flows are economically equivalent if, when the time value of money is considered, the two cash flows are identical. Analysis Methods Present and Future Worth Analysis Equivalent Uniform Cash Flow Analysis Rate of Return Analysis Benefit Cost Ratio Analysis Related Topics Capitalized Cost Bonds Break-Even Analysis vs Payback Period Depreciation and Taxes Inflation

3 Material Provided in Reference Manual BE FAMILIAR WITH THIS! Formulas for Cash Flow Manipulation Effective Interest Rate Capitalized Cost Depreciation Inflation Descriptions of Terms Break-Even analysis Inflation Taxation Capitalized Cost Bonds Rate of Return Benefit-Cost Analyisis Interest Tables

4 Cash Flow Formulas – from Reference Handbook, p. 114

5 NOT PROVIDED: Pictorial Representation CAN YOU DRAW THESE? Pictoral Representation – From Newnan, et. Al, front cover

6 Interest Rates r The nominal annual interest rate Possible wording 12%, compounded monthly iThe interest rate per interest period, i = r/m where m=# of compounding per year Possible wording 1% monthly interest i e The effective interest rate, i e = (1 + i) n -1 where n = # of compounding in the desired period Possible effective interest rates for r = 12%, compounded monthly (i = 1%) Effective quarterly interest rate: n = 3 Effective annual interest rate: n = 12

7 Example 1 – Effective Annual Interest A credit card company offers a credit card with an interest rate of 18%, compounded monthly. What is the effective annual interest rate for the card? (a)6.3% (b)18.0% (c)19.6% (d)21.6%

8 Solution to example 1 i e = (1 + r/m) m – 1 = ( /12) 12 – 1 = = 19.56% Answer: (c)

9 Example 2 – Effective quarterly Interest A bank is paying 0.25% monthly interest on a special savings account. What is the effective quarterly interest? (a)0.750% (b)0.752% (c)1.000% (d)1.010%

10 Solution to example 2 i quarterly = ( ) 3 – 1 = = 0.752% Answer: (b)

11 Present, Future Worth, and Equivalent Uniform Cash Flow Analyses Three Possible Goals Maximize BENEFITS or Minimize COSTS or Maximize the NET WORTH (BENEFITS – COSTS) Present Worth or Future Worth – Move each cash flow to a single point in time. Equivalent Uniform Cash Flow – Convert each to an equivalent uniform cash flow over the SAME TIME INTERVAL.

12 Example 3 – Present Worth Analysis An individual can afford monthly car payments of $450 for the next 5 years. Assuming interest on car loans is 6% compounded monthly, determine the greatest price of car he can afford. (a)$20,000 (b)$23,000 (c)$25,000 (d)$27,000

13 Solution to example 3 r = 6%  i = 6/12 = 0.5% monthly interest n = 5x12 = 60 months P = A(P/A, i, n) = $450(P/A, 0.5%, 60) =$450( ) = $23,277 Alternate solution: use formula P = [(1+i) n -1] / [(i)(1+i) n ] = $23,277 Answer: (b)

14 Example 4 – Equivalent Uniform Cash Flow Analysis A college savings account paying 6% annual interest is established for a 5 year old boy, with the objective of having $60,000 on his 18th birthday. If uniform deposits are made on each of the boy’s birthdays, starting with his 7th birthday and ending on his 18 th birthday, how much must each deposit be? (a)$3558 (b)$3988 (c)$4185 (d)$5000

15 Solution to example 4 There are 12 uniform deposits. A = F(A/F, 6%, 12) = 60,000 (0.0593) = $3,558 Answer: (a)

16 Example 5 – Net Present Worth A new milling machine will cost $150,000. The net benefits from the purchase are expected to be $40,000 the first year, increasing by $2,000 per year for the 8 year life of the machine. Interest is 10%. The net present worth of the investment is closest to: (a) $65,000 (b) $75,000 (c) $85,000 (d) $95,000

17 Solution to example 5 NPW = -150,000 + (40, (A/G,10%,8))(P/A,10%,8) = -150,000+(40,000*2000(3.0045))(5.3349) = $95,453 Answer: (d)

18 Example 6 – Equivalent Uniform Cash Flow Analysis Two alternatives are being considered: At 6% interest, the difference between the two Equivalent Uniform Cash Flows is most nearly: (a) $100 (b) $720 (c) $840 (d) $1000 Alternative AAlternative B 1 st Cost$25,000$32,000 Net Annual Benefits$9,500$10,300 Salvage Value$2,500$3,200 Useful Life, years10

19 Solution to example 6 Alternative A: EUA(B-C) = $9,500 + $2,500(A/F,6%,10) - $25,000(A/P,6%,10) = $6292 Alternative B: EUA(B-C) = $10,300 + $3,200(A/F,6%,10) - $32,000(A/P,6%,10) = $6194 Difference: $6292-$6194 = $98 Answer: (a)

20 Example 7 – Future Worth Analysis Each quarter for 40 years an individual deposits $600 into an IRA that pays 8% interest, compounded monthly. Determine the value of the retirement account at the end of the 40 years. (a)$554,000 (b)$621,000 (c)$693,000 (d)$852,000

21 Solution to example 7 r = 8%  i = % i quarterly = ( ) 3 -1 = o % n = 40*4 = 160 Using the formula, F = $600[(( ) )/(.02013)] = $693, Answer: (c)

22 Example 8 – Present Worth Analysis A food processing company is considering investing in new food packaging equipment that will cost $250,000. The equipment will save the company $60,000 the first year, decreasing by $5000 each year thereafter to $55,000 the second year, $50,000 the third year, and so on. At the end of its useful life of 8 years, the equipment will have a salvage value of $12,500. The company’s MARR is 8%. Determine the net present worth of the investment. (a)$10,550 (b)$12,520 (c)$16,580 (d)$21,470

23 Solution to example 8 NPW = -250, ,000(P/A, 8%,8) – 5,000(P/G, 8%,8) + 12,500(P/F, 8%,8) = -250,000+60,000(5.7466)-5000( )+12,500(.5403) = $12,520 Answer: (b)

24 Rate of Return Analysis Find the internal rate of return (IRR) and compare to the minimum acceptable rate of return (MARR) The MARR is the minimum interest rate or return on investment one is willing to accept. The IRR is the interest rate at which the present worth of costs equal the present worth of benefits To be acceptable the IRR must be ≥ the MARR Incremental analysis required to compare two alternatives

25 Example 9 – Rate of Return Analysis A company is considering investing in a piece of equipment that will cost $100,000 and have no salvage value. The company estimates the equipment will produce uniform benefits of $25,250 per year for the equipment’s useful life. The internal rate of return for this equipment purchase is most nearly: (a) 4% (b) 6% (c) 8% (d) 10%

26 Solution to example 9 PW of C = PW of B $100,000 = $25,250(P/A, irr, 5) 3.96 = (P/A, irr, 5) Look at 4%, 6%, 8%, and 10% to find the interest rate that gives the closest value. 6%: %: %: Answer: (c)

27 Example 10 – Rate of Return Analysis A firm will purchase a new milling machine for $25,000 that will save the company $4,000 each year for 10 years. At that time the machine will be salvaged for $2,500. The company’s rate of return on this investment is most nearly: (a) 6.5% (b) 8.4% (c) 10.5% (d) 12.6%

28 Solution to example 10 This is a trial and error solution. Suggestion: Work backward from the interest rates given to find the one that gives a NPW closest to ZERO. NPW = -25, [(1+i) 10 -1]/[i(1+i) 10 ] + 2,500(1+i) -10 Try i = 6.5%: NPW = $5087 i too small Try i = 8.4%: NPW = $2479 i too small Try i = 10.5%: NPW = -$19.79 CLOSE Answer: (c) -

29 Benefit/Cost Analysis Comparison of present worth of benefits to present worth of costs (or EUAB to EUAC) The benefit/cost ratio must be ≥ 1 for an option to be considered viable (benefits greater than costs) Salvage value is considered a REDUCTION IN COST INCREMENTAL ANALYSIS needed if comparing two or more options

30 Example 11 – Benefit/Cost Analysis A county is considering the following project Initial Cost$22,500,000 Maintenance$525,000 per year Savings$5,300,000 per year 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

31 Solution to example 11 PW cost = 22,500, ,000 (P/A, 8%, 12) = 22,500, ,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)

32 Example 12 – Benefit Cost Ratio Analysis A project will require an initial equipment cost of $9500 that will result in annual benefits of $2200 per year over the 15 year life of the equipment. At the end of the 15 years the equipment will be salvaged for $6000. The minimum acceptable rate of return is 12%. The Benefit/Cost Ratio for the project is nearest to: (a) 0.56 (b) 1.24 (c) 1.78 (d) 2.21

33 Solution to Example 12 EUAB = $2200/year for 15 years EUAC = $9500(A/P,12%,15)-$6000(A/F,12%,15) = 9500(0.1468)-6000(0.0268) = $ B/C Ratio = (2200/1233.8) = 1.78 Answer: (c)

34 Infinite Analysis Period: Capitalized Cost Capitalized Cost – The PRESENT cash amount that would need to be set aside now to cover a service indefinitely. Formula: P = A/i Infinite Uniform Cash Flow Formulas A= Pi A forever = A/cycle

35 Example 13 – Capitalized Cost Analysis The cleanup of an environmental disaster will cost the state $50,000 annually for perpetuity. Assuming interest = 4%, determine the state’s capitalized cost of this cleanup. (a) $1,250,000 (b) $3,575,000 (c) $5,000,000 (d) $8,250,000

36 Solution to example 13 For perpetuity  forever P=A/i = $50,000/.04 = $1,250,000 Answer: (a)

37 Example 14 – Infinite Life The initial cost of constructing a road is estimated to be $50 million. Annual maintenance is estimated to be $0.18 million per year. In addition, every 10 years the road will need resurfaced at a cost of $2 million. Interest is 6%. The equivalent uniform cost is most nearly: a) $6.33 million b) $5.33 million c) $4.33 million d) $3.33 million

38 Solution to Example 14 A = * (A/F, 6%,10) = (.0759) = Answer: (d)

39 Bonds Bond - A loan an investor makes to a corporation or government. Face Value (par) – The original purchase price of the bond. The bond holder will receive this amount when the bond reaches maturity. Coupon (stated interest) – The interest that the bond holder will receive while holding the bond. Maturity – The number of years the interest will be paid. Bonds can be resold for more or less than the face value. The rate of return you actually get is the yield.

40 Example 15 – Bonds A $5000 bond is being offered for sale. It has a stated interest rate of 7%, paid annually ($350 each year). The $5000 debt will be repaid at 8 years along with the last interest payment. If you want an 8% return on this investment (the bond yield), what is the most you would be willing to pay for the bond (the bond value)? (a) $3500 (b) $4700 (c) $5000 (d) $5200

41 Solution to example 15 Bond value = PW of all future benefits = 350 (P/A,8%,8) (P/F,8%,8) = 350 (5.747) (0.5403) = $4713 Answer: (b)

42 Example 16 – Bonds If the $5000 bond in the previous example could be purchased for $4200, what is the bond yield? (a) 5% (b) 8% (c) 10% (d) 12%

43 Solution to example 16 Bond value = PW of all future benefits $4200 = 350 (P/A, i %,8) (P/F, i %,8) By trial and error, try i = 10% 350 (5.335) (0.467) = $4202 – close enough Answer: (c)

44 Example 17 - Bonds An investor is considering purchasing a bond with a face value of $20,000 and 10 years left to mature. The bond pays 12% interest payable quarterly. If he wishes to get a 4% 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

45 Solution to example 17 Since the bond pays 12%, paid quarterly, its effective interest rate is 3% per quarter (every 3 months). Interest payment = i(Face value) = 0.03(20,000) = $600/quarter n = 10*4 = 40 quarters P = 600(P/A, 4%, 40) + 20,000(P/F, 4%, 40) = 600( ) + 20,000(0.2083) = $16,042 Answer: (b)

46 Break-Even Analysis Considers the time value of money. Single project Determines the value of a particular variable that makes the benefits = costs. Two projects Determines the value of a particular variable that makes the two projects equivalent.

47 Example 18 – Break-Even Analysis A company is planning to update its production equipment and is considering two different options. Each is anticipated to have a 15 year life. The initial costs and salvage values of each are shown. In addition, the annual saving for Option A is predicted to be $14,000 per year, but the company is unsure of the annual savings for Option B. Option AOption B Initial Cost$84,000$140,000 Annual Savings$14,000 ?????? Salvage Value $8,000 $12,000 If interest is 10%, determine the breakeven point between the two projects. (a) 18,720 (b) 21,240 (c) 23,210 (d) 26,930

48 Solution to example 18 EUA(B-C) A = -84,000(A/P, 10%, 15) + 14, ,000(A/F, 10%, 15) = 3208 EUA(B-C) B = 3208 = -140,000(A/P, 10%, 15) + A + 12,000(A/F,10%,15) A = 21,240 Answer: (b)

49 Example 19 – Break-Even and Benefit /Cost Analyses 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

50 Solution to example 19 In order that B/C = 1.0 PW cost = PW benefit PW benefit = 150,000, ,000,000 (P/A, 10%, 8) = 150,000, ,000,000 (5.3349) = $283,372,500 EUAC benefit = $283,372,500 (A/P, 10%, 8) = $283,372,500 ( ) = $53,115,341 $ equivalent /crash = 53,115,341/9,200 = $5,773 Answer: (a)

51 Example 20 –Break-Even Analysis A company is considering two alternative forklifts with equal useful lives and the following characteristics 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

52 Solution to example 20 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 = From table look-up, the value of n that most nearly makes the above relation true is 5. Answer: (a)

53 Example 21 – Break-Even Analysis 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 12% 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)$120,000 (b)$193,000 (c)$256,000 (d)$307,000

54 Solution to example 21 S = 650,000 (F/P, 12%, 5) – 150,000 (F/A, 12%, 5) = 650,000(1.7623) – 150,000(6.3528) = $192,575 Answer: (b)

55 Payback Period Does not consider the time value of money. Determines the time required to recover the initial cost of a project or an investment.

56 Example 22 – Payback Period 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

57 Solution to example 22 Costs = 55,000 The payback period is the time when total income to date is equal to the total costs. Costs - Income = 0 55,000-25,000-20,000=10,000 to pay back at end of year 2. Since the income is stated as $25,000 per year, one can assume that savings occur uniformly throughout the year. 15,000 comes in during year 3. Therefore: Payback period =2 + (10,000/15,000) = 2.67 Answer: (b)

58 Depreciation Costs of capital assets (e.g., major equipment or facilities) are allocated (depreciated) over time for tax purposes Taxable income is total income less depreciation and ordinary expenses Book value at a point in time is the original cost of an asset minus depreciation to that point in time

59 Example 23 - 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

60 Solution to example 23 D t = D 2 = (95,000 – 10,000)/5 = $17,000 Answer: (c)

61 Example 24 – Depreciation A depreciable asset costs $25,000 and has an estimated salvage value of $2500. It has a MACRS class life of 5 years. At the end of three years its book value is closest to: (a) $4520 (b) $4800 (c) $7200 (d) $8980

62 Solution to example 24 MACRS depreciation ignores salvage value. Depreciation the 1 st year = (20/100)(25,000) = $5,000 Depreciation the 2 nd year = (32/100)(25,000) = $8,000 Depreciation the 3 rd year = (19.2/100)(25,000) = $4,800 Book value = cost – sum of depreciations = $25,000 - $5,000 - $8,000 - $4,800 = $7,200 Answer: (c)

63 Inflation Inflation - increase in the price of goods/services. Actual dollars Dollars in the actual market inflate over time by D n = D o (1+f) n f = annual inflation rate Real Dollars Buying power dollars Move through time with real interest rate

64 Inflation - continued Equation in reference manual: d=i +f + (i)(f) d = market interest rate, for actual dollars i = real interest rate, for buying power dollars f = annual inflation rate

65 Example 25 - 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%

66 Solution to example 25 21,000 = 15,000 (1 + f) 10 f = (21,000/15,000) 0.1 – 1 = = 3.4% Note the similarity to the equation F = P(1+i) n Answer: (c)

67 Example 26 – Inflation An annuity pays $2500 per year for the next 10 years. Inflation is expected to be 4% per year. After inflation, a real interest rate of 5% is desired. The amount that should be paid for the annuity is nearest to: (a) $10,200 (b) $15,900 (c) $18,600 (d) $25,000

68 Solution to example 26 Step #1: Find the desired market interest rate, d. d = (0.04)(0.05) = or 9.2% Step #2: Find the present worth of the annuity, at market interest = 9.2%. Answer: (b).

69 Sources Fundamentals of Engineering Supplied-Reference Handbook, 8 th Edition, Revised – pg Engineering Economic Analysis, 11 th Edition, D.G. Newnan, J.P Lavelle, & T.G. Eschenbach, Oxford University Press, 2012


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