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CE 466 FE Exam Review

CE 466 FE Exam Review . Engineering Economics Spring 2013. 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

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CE 466 FE Exam Review

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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 RepresentationCAN 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 i The interest rate per interest period, i = r/m where m=# of compounding per year Possible wording 1% monthly interest ie The effective interest rate,  ie= (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 ie = (1 + r/m)m – 1 = (1 + 0.18/12)12 – 1 = 0.1956 = 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 iquarterly = (1 + 0.0025)3 – 1 = 0.00752 = 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(51.7256) = $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 18th 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: • $65,000 • $75,000 • $85,000 • $95,000

  17. Solution to example 5 NPW = -150,000 + (40,000+2000(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

  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 = 0.6667% iquarterly = (1.006667)3 -1 = 0.02013 o4 2.013% n = 40*4 = 160 Using the formula, F = $600[((1.02013)160 -1)/(.02013)] = $693,280.81 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 + 60,000(P/A, 8%,8) – 5,000(P/G, 8%,8) + 12,500(P/F, 8%,8) = -250,000+60,000(5.7466)-5000(17.8061)+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: • 4% • 6% • 8% • 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%: 4.212 8%: 3.993 10%: 3.791 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 = -25,000+4000[(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 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 PWcost = 22,500,000 + 525,000 (P/A, 8%, 12) = 22,500,000 + 525,000 (7.5361) = $26,456,453 PWbenefit = 5,300,000 (P/A, 8%, 12) = 5,300,000 (7.5361) = $39,941,330 B/C = PWbenefit/ PWcost = 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: • 0.56 • 1.24 • 1.78 • 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) = $1233.8 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: • $6.33 million • $5.33 million • $4.33 million • $3.33 million

  38. Solution to Example 14 A = 0.18 + 50*.06 + 2(A/F, 6%,10) = .18+3+2(.0759) = 3.3318 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)? • $3500 • $4700 • $5000 • $5200

  41. Solution to example 15 Bond value = PW of all future benefits = 350 (P/A,8%,8) + 5000 (P/F,8%,8) = 350 (5.747) + 5000 (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? • 5% • 8% • 10% • 12%

  43. Solution to example 16 Bond value = PW of all future benefits $4200 = 350 (P/A,i%,8) + 5000 (P/F,i%,8) By trial and error, try i= 10% 350 (5.335) + 5000 (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(19.7928) + 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 A Option 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. • 18,720 • 21,240 • 23,210 • 26,930

  48. Solution to example 18 EUA(B-C)A = -84,000(A/P, 10%, 15) + 14,000 + 8,000(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 PWcost = PWbenefit PWbenefit = 150,000,000 + 25,000,000 (P/A, 10%, 8) = 150,000,000 + 25,000,000 (5.3349) = $283,372,500 EUACbenefit= $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)

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