1. Part II: Evaluating business & engineering assets Ch 5: Present worth analysis Ch 6: Annual equivalence analysis Ch 7: Rate-of-return analysis

2. (Conventional) Payback period Principle: How fast can I recover my initial investment? Method: Based on cumulative cash flow (or accounting profit) Screening guideline: If the payback period is less than or equal to some specified payback period, the project would be considered for further analysis. Weakness: Does not consider the time value of money Conventional-payback method is based on actual dollars Useful for initial screening

3. Example 5.1: Payback period NCash flowCum. flow 01234560123456 -$105,000+$20,000 $35,000 $45,000 $50,000 $45,000 $35,000 -$85,000 -$50,000 -$5,000 $45,000 $95,000 $140,000 $175,000 Payback period should occur somewhere between N = 2 and N = 3.

4. -100,000 -50,000 0 50,000 100,000 150,000 012 3 456 Years (n) 3.2 years payback period $85,000 $15,000 $25,000 $35,000 $45,000 $35,000 0 123456 Years Annual cash flow Cumulative cash flow ($)

5. Discounted payback period Discounted-payback method considers time value of money Principle: How fast can I recover my initial investment plus interest? Method: Based on cumulative discounted cash flow Screening guideline: If the discounted payback period (DPP) is less than or equal to some specified payback period, the project would be considered for further analysis. Weakness: Cash flows occurring after DPP are ignored Also used for initial screening

6. Example 5.2 Discounted payback period calculation PeriodCash flow Cost of funds (15%)* Cumulative cash flow 0-$85,0000 115,000-$85,000(0.15) = -$12,750-82,750 225,000-$82,750(0.15) = -12,413-70,163 335,000-$70,163(0.15) = -10,524-45,687 445,000-$45,687(0.15) =-6,853-7,540 545,000-$7,540(0.15) = -1,13136,329 635,000$36,329(0.15) = 5,44976,778 * Cost of funds = (unrecovered beginning balance) X (interest rate)

7. Summary: payback period Payback periods can be used as a screening tool for liquidity, but we need a measure of investment worth for profitability.

8. Net present worth measure Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i, or minimum attractive rate of return (MARR) Decision rule: Accept the project if the net surplus is positive. 2 3 4 5 0 1 inflow outflow 0 PW(i) inflow PW(i) outflow net surplus PW(i) > 0

9. Guideline for selecting a MARR real return2% inflation4% risk premium0% total expected return6% real return2% inflation4% risk premium20% total expected return26% Risk-free real return Inflation Risk premium U.S. Treasury bills Amazon.com very safe very risky

10. Example 5.3 - Tiger Machine Tool Company $75,000 $24,400 $27,340 $55,760 0 12 3 outflow inflow PW(15%) inflow PFPF$24,(/,)$27,(/,)40015%,134015%,2 =+ 760$55PF,(/,)15%,3 + $78, 553 = PW($75,15%)000 outflow = PW($78,$75,15%)553000 =- $3,,5530 Accept =>

11. -30 -20 -10 0 10 20 30 40 0 510152025303540 PW (i) ($ thousands) i = MARR (%) $3553 17.46% Break even interest rate (or rate of return) Accept Reject Present worth profile

12. Capitalized equivalent worth Principle: PW for a project with an annual receipt of A over infinite service life Equation: CE(i) = A(P/A, i,  ) = A/i A 0 P = CE(i) Ex 5.4: A = $2 million, i = 8%, N =  A/i = CE(i) = $2 million/0.08 = $25 million

13. Part II: Evaluating business & engineering assets Ch 5: Present worth analysis Ch 6: Annual equivalence analysis Ch 7: Rate-of-return analysis

14. Annual worth analysis Principle: Measure an investment worth on annual basis Benefit: By knowing the annual equivalent worth, we can: –seek consistency of report formats –determine unit costs (or unit profits) –facilitate unequal project life comparisons

15. Ex. 6.1: Computing equivalent annual worth $15 $3.5 $5 $12 $8 0 2 3 4 5 6 1 A = $1.835 2 3 4 5 6 1 AE(12%) = $6.946(A/P, 15%, 6) = $1.835 $6.946 0 0 PW(15%) = $6.946 $9.0 $10

16. Annual equivalent worth – repeating cash flow cycles $500 $700 $800 $400 $500 $700 $800 $400 $1,000 Repeating cycle First cycle: PW(10%) = -$1,000 + $500 (P/F, 10%, 1) +... + $400 (P/F, 10%, 5) = $1,155.68 AE(10%) = $1,155.68 (A/P, 10%, 5) = $304.87 Both cycles: PW(10%) = $1,155.68 + $1,155.68 (P/F, 10%, 5) AE(10%) = $1,873.27 (A/P, 10%,10) = $304.87

17. Annual equivalent cost When only costs are involved, the AE method is called the annual equivalent cost. Revenues must cover two kinds of costs: operating costs and capital costs. capital costs operating costs + annual equivalent costs

18. Capital (ownership) costs Two transactions are associated with owning equipment: initial cost (I) & salvage value (S). Capital costs: taking these items into consideration, we calculate the capital recovery costs as: CR(i) = I(A/P,i,N) – S(A/F,i,N) = (I – S)(A/P,i,N) + iS 0 1 2 3 N 0 N I S CR(i) Recall that (A/F,i,N) = (A/P,i,N) – i

19. 0 1 2 3 $24,400 $55,760 $27,340 $75,000 Operating hours per year 2,000 hr PW (15%) = $3553 AE (15%) = $3,553 (A/P, 15%, 3) = $1,556 Savings per machine hour = $1,556/2,000 = $0.78/hr Applying annual worth analysis Ex. 6.3 – Equivalent worth per unit of time Recall ex 5.3

20. Part II: Evaluating business & engineering assets Ch 5: Present worth analysis Ch 6: Annual equivalence analysis Ch 7: Rate-of-return analysis

21. Why RoR measure is so popular? Which statement is easier to understand? This project will bring in a 15% rate of return on investment. This project will result in a net surplus of $10,000 in NPW. Rate of return (RoR): a relative percentage method that measures the yield as a percent of investment over the life of a project

22. Return on investment Definition 1: Rate of return (RoR) is defined as the interest rate earned on the unpaid balance of an installment loan. Example: A bank lends $10,000 & receives annual payments of $4,021 over 3 years. The bank is said to earn a return of 10% on its loan of $10,000.

23. Rate of return Definition 2: Rate of return (RoR) is the break-even interest rate, i*, which equates the present worth of a project’s cash outflows to the present worth of its cash inflows. Mathematical relation: PW(i * ) = PW(i * ) cash inflows – PW(i * ) cash outflows = 0

24. Methods for finding rate of return –Using Excel’s financial command –Direct solution –Trial-and-error –“Cash Flow Analyzer” – online financial calculator Excel command to find the rate of return: =IRR(cell range, guess) e.g., =IRR(C0:C7, 10%)

25. Project AProject B Direct solution methods See example 7.2

26. Trial and error method – Project C Step 1: Guess an interest rate, say, i = 15% Step 2: Compute PW(i) at the guessed i value PW (15%) = $3,553 Step 3: If PW(i) > 0, then increase i. If PW(i) < 0, then decrease i. PW(18%) = -$749 Step 4: If you bracket the solution, use a linear interpolation to approximate the solution 3,553 0 -749 15% i 18% Based on example 6.3

27. Basic decision rule If ROR > MARR, accept This rule does not work for a situation where an investment has multiple rates of return

28. Return on invested capital Definition 3: Return on invested capital is defined as the interest rate earned on the unrecovered project balance of an investment project. It is commonly known as internal rate of return (IRR). Example: A company invests $10,000 in a computer and results in equivalent annual labor savings of $4,021 over 3 years. The company is said to earn a return of 10% on its investment of $10,000.

29. Project balance calculation 01230123 Beginning project balance Return on invested capital Payment received Ending project balance -$10,000 -$6,979 -$3,656 0 The firm earns a 10% rate of return on funds that remain internally invested in the project. Since the return is internal to the project, we call it internal rate of return. -$10,000-$6,979-$3,656 -$1,000 -$697 -$365 -$10,000 +$4,021+$4,021+$4,021

30. Multiple rates of return If a project has more than one rate of return, how would you make an accept/reject decision? Simple Investment Def: Initial cash flows are negative, and only one sign change occurs in the net cash flows series. Example: -$100, $250, $300 (-, +, +) RoR: A unique RoR Non-simple Investment Def: Initial cash flows are negative, but more than one sign changes in the remaining cash flow series. Example: -$100, $300, -$120 (-, +, -) RoR: A possibility of multiple RoR’s

31. Example 7.1 – Investment classification Period (N) Project AProject BProject C 0-$1,000 +$1,000 1-5003,900-450 2800-5,030-450 31,5002,145-450 42,000 Project A is a simple investment. Project B is a nonsimple investment. Project C is a simple borrowing.

32. How to proceed: if you encounter multiple rates of return Abandon the IRR analysis and use the PW criterion For MARR = 15% PW(15%) = –$1,000 + $2,300 (P/F,15%,1) – $1,320(P/F,15%,2 ) = $1.89 > 0 Accept the investment To find the true rate of return (or return on invested capital) for the project, see appendix 7A,

33. Incremental analysis (procedure) Step 1: Compute the cash flow for the difference between the projects (A,B) by subtracting the cash flow of the lower investment cost project (A) from that of the higher investment cost project (B). Step 2: Compute the IRR on this incremental investment (IRR B-A ). Step 3:Accept the investment B if and only if IRR B-A > MARR NOTE: Make sure that both IRR A and IRR B are greater than MARR.

34. Ex 7.7 - Incremental rate of return nB1B2B2 - B1 01230123 -$3,000 1,350 1,800 1,500 -$12,000 4,200 6,225 6,330 -$9,000 2,850 4,425 4,830 IRR25%17.43%15% Given MARR = 10%, which project is a better choice? Since IRR B2-B1 =15% > 10%, and also IRR B2 > 10%, select B2.

35. Part III: Development of project cash flows Ch 8: Accounting for depreciation & income taxes –Accounting depreciation –Book depreciation methods –Tax depreciation methods –How to determine “accounting profit” –Corporate taxes

36. Asset depreciation: two concepts Depreciation Economic depreciation the gradual decrease in utility in an asset with use & time Accounting depreciation the systematic allocation of an asset’s value in portions over its depreciable life—often used in engineering economic analysis Physical depreciation Functional depreciation Book depreciation Tax depreciation

37. Assets used in business or held for production of income Assets having a definite useful life and a life longer than one year Assets that must wear out, become obsolete or lose value A qualifying asset for depreciation must satisfy all three conditions. Examples? What can be depreciated?

38. Ex 8.1 – Cost basis Cost of a new hole-punching machine (Invoice price) $62,500 + Freight725 + Installation labor2,150 + Site preparation3,500 Cost basis to use in depreciation calculation $68,875

39. Types of depreciation Book depreciation –In reporting net income to investors/stockholders –In pricing decisions Tax depreciation –In calculating income taxes for the IRS –In engineering economics, we use depreciation in the context of tax depreciation Methods of calculating depreciation –straight line –declining balance –unit production

40. Straight-line (SL) method Principle: A fixed asset provides its service in a uniform fashion over its life Annual depreciation: D n = (I – S) / N, and constant for all n. Book value B n = I – N (D) where I = cost basis S = salvage value N = depreciable life

41. Declining balance method Principle: A fixed asset provides its service in a decreasing fashion Annual depreciation: Book value: where 0 < α < 2(1/N) Note: if  is chosen to be the upper bound,  = 2(1/N), we call it a 200% DB or double declining balance method. B n = I(1 – α) n D n = αB n-1 = αI(1 – α) n-1

42. Ex. 8.3 – Declining balance method n012345n012345 D n $4,000 2,400 1,440 864 518 B n $10,000 6,000 3,600 2,160 1,296 778 I = $10,000 N = 5 yr S = $2,000 D n = αB n-1 = αI(1 – α) n-1 B n = i(1 – α) n B n must be > S n012345n012345 D n $4,000 2,400 1,440 160 0 B n $10,000 6,000 3,600 2,160 2,000

43. Principle: service units will be consumed in a non time- phased fashion Annual depreciation D n = Service units consumed for year total service units Units-of-production method (I - S)

44. Ex 8.5 – Units-of-production depreciation Given:I = $55,000 S = $5,000 total service units = 250,000 mi usage for this year = 30,000 mi Solution:

45. Modified accelerated cost recovery systems (MACRS) Personal property –definition: movable property; property of any kind except real property –depreciation based on DB method switching to SL –half-year convention (all assets placed in service mid year) –zero salvage value Real property –permanent fixtures –SL method –mid-month convention –zero salvage value

46. Calculation in % (0.5)(0.40)(100%)20% (0.4)(100%-20%)32% SL = (1/4.5)(80%)17.78% (0.4)(100%-52%)19.20% SL = (1/3.5)(48%)13.71% (0.4)(100%-71.20%) switch 11.52% SL = (1/2.5)(29.80%) to SL 11.52% SL = (1/1.5)(17.28%)11.52% SL = (0.5)(11.52%)5.76% Year (n) 1 2 3 4 5 6 MACRS (%) DDB SL

47. MACRS for real property 27.5-year (residential) 39-year (commercial) SL method zero salvage value mid-month convention Example: Placed a residential property in service in March. Find the depreciation allowance in year 1. D 1 = (9.5/12)(100%/27.5) = 2.879% Types of real property

48. Ex. 8.9 – Cash flow vs. net income ItemIncomeCash flow Gross income (revenue)$50,000 Expenses Cost of goods sold Depreciation Operating expenses 20,000 4,000 6,000 -20,000 -6,000 Taxable income20,000 Taxes (40%)8,000-8,000 Net income$12,000 Net cash flow$16,000

49. Ex 8.9 (cont.) – Net income versus net cash flow $0 $50,000 $40,000 $30,000 $20,000 $10,000 $8,000 $6,000 $20,000 net income depreciation income taxes operating expenses cost of goods sold net cash flow gross revenue $4,000 $12,000 net cash flows = net income + non-cash expense (depreciation)

50. Marginal & effective (average) tax rate for a taxable income of $16,000,000 Taxable income Marginal tax rateAmount of taxesCumulative taxes First $50,00015%$7,500 Next $25,00025%6,25013,750 Next $25,00034%8,50022,250 Next $235,00039%91,650113,900 Next $9,665,00034%3,286,1003,400,000 Next $5,000,00035%1,750,0005,150,000 Remaining $1,000,000 38%380,000$5,530,000

51. Capital gains & ordinary gains Cost basis Book valueSalvage value Capital gains Ordinary gains or depreciation recapture Total gains