Chapter 5 Present-Worth Analysis

Transcription

1 Chapter 5 Present-Worth Analysis Describing Project Cash Flows: Loan versus Project Cash Flows Initial Project Screening Methods Present Worth Analysis Variations of Present Worth Analysis Comparing Mutually Exclusive Alternatives 1

2 Chapter Opening Story Federal Express Nature of Project: Equip 40,000 couriers with PowerPads Save 10 seconds per pickup stop Investment cost: $150 million Expected savings: $20 million per year Federal Express 2

3 Ultimate Questions Is it worth investing $150 million to save $20 million per year, say over 10 years? How long does it take to recover the initial investment? What kind of interest rate should be used in evaluating business investment opportunities? 3

4 Mr. Bracewell s Investment Problem Built a hydroelectric plant using his personal savings of $800,000 Power generating capacity of 6 million kwhs Estimated annual power sales after taxes - $120,000 Expected service life of 50 years Was Bracewell's $800,000 investment a wise one? How long does he have to wait to recover his initial investment, and will he ever make a profit? 4

5 Mr. Bracewell s Hydro Project 5

6 Bank Loan vs. Investment Project Bank Loan: Loan cash flow Loan Bank Repayment Customer Investment Project: Project cash flow Investment Company Project Return 6

7 Independent vs Mutually Exclusive Investment Projects Computer process control system, new warehouse, CAD system for engineering, etc: The decision on any one project has no effect on the decision made on another project. Such projects are independent. If there are alternative projects which resolve the same problem or meet the same need, then acceptance of one project means the rejection of all others. Such projects are mutually exclusive. Capital budgeting: Which projects should be selected with available funds? 7

8 Payback Method How fast can I recover my initial investment? Conventional payback method (time value of money ignored) Discounted payback method Payback screening - If the payback period is within acceptable range, formal project evaluation may start (a high-tech firm may have a short time limit since products rapidly become obsolete). 8

9 Conventional-Payback Method Principle: How fast can I recover my initial investment? Method: Based on cumulative cash flow 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 9

10 Example 5.1 Conventional Payback Period N Cash Flow Cum. Flow $105,000+$20,000 $15,000 $25,000 $35,000 $45,000 $45,000 $35,000 -$85,000 -$70,000 -$45,000 -$10,000 $35,000 $80,000 $115,000 Payback period should occur somewhere between N = 3 and N = 4. 10

11 Annual cash flow 0 $85,000 $15,000 $25,000 $35,000 $45,000 $45, Years $35,000 Cumulative cash flow ($) 150, ,000 50, , , years Payback period Years (n) 11

12 Payback Method Pitfall n Project 1 Payback:3 years Project 2 Payback:3.6 years Fails to measure profitability Ignores timing of cash flows 12

13 Discounted Payback Method 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 13

14 Discounted Payback Period Calculation Period Cash Flow Cost of Funds Cumulative (15%) Cash Flow 0 -$85, $85, ,000 -$85,000(0.15)= -$12,750-82, ,000 -$82,750(0.15)= -12,413-70, ,000 -$70,163(0.15)= -10,524-45, ,000 -$45,687(0.15)=-6,853-7, ,000 -$7,540(0.15)= -1,131 36, ,000 $36,329(0.15)= 5,449 76, (1.15)+15000=

15 Net Present Worth Measure Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i. Decision Rule: Accept the project if the net surplus is positive. Outflow PW(i) inflow PW(i) outflow Inflow Net surplus PW(i) > 0 15

16 MARR: Minimum Attractive Rate of Return The rate at which the firm can always invest the money in its investment pool. Possibly change over the life of project Risk-free return Inflation factor Risk premiums 16

17 Example Tiger Machine Tool Company inflow $24,400 $27,340 $55,760 outflow 0 $75, PW ( 15%) $24, 400( P / F, 15%,1) $27, 340( P / F, 15%,2) PW inflow ( 15%) $75, 000 outflow $55, 760( P / F, 15%,3) $78, 553 PW ( 15%) $78, 553 $75, 000 $3, 553 0, Accept 17

18 Present Worth Amounts at Varying Interest Rates: Choice of MARR is critical i (%) PW(i) i(%) PW(i) 0 $32, $3, , , , , , , , , , , , , , , , , * , ,302 *Break even interest rate 18

19 Present Worth Profile Accept Reject PW (i) ($ thousands) $ % Break even interest rate (or rate of return) i = MARR (%) 19

20 Meaning of Present Worth Two possible perspectives Investment pool: All funds in the firm s treasury can be placed in investments that yield a return equal to the MARR. Borrowed funds (project balance): If no funds are available for investment, the firm can borrow them at MARR from the capital market. 20

21 Investment Pool Concept The company has $75000 available for investment. Two choices: Do not invest, keep the money in the investment pool and earn at MARR. Invest in the project 21

22 Meaning of Net Present Worth Investment pool $75,000 N = 3 How much would you have if the Investment is made? $24,400(F/P,15%,2) = $32,269 $27,340(F/P,15%,1) = $31,441 $55,760(F/P,15%,0) = $55,760 $119,470 Project $55,760 $27,340 $24, How much would you have if the investment was not made? $75,000(F/P,15%,3) = $114,066 What is the net gain from the investment? $119,470 - $114,066 = $5,404 PW(15%) = $5,404(P/F,15%,3) = $3,553 Net future worth of the project 22

23 Project Balance Concept (Borrowed funds) N Beginning Balance -$75,000 -$61,850 -$43,788 Interest -$11,250 -$9,278 -$6,568 Payment -$75,000 +$24,400 +$27,340 +$55,760 Project Balance -$75,000 -$61,850 -$43,788 +$5,404 Net future worth, FW(15%) PW(15%) = $5,404 (P/F, 15%, 3) = $3,553 23

24 Project Balance Diagram Project balance ($) 60,000 40,000 20, ,000-40,000-60,000 -$75,000 -$61,850 Terminal project balance (net future worth, or project surplus) Discounted payback period -$43,788 $5,404-80, , , Year(n) 24

25 Example: You need to know if the building of a new warehouse is justified under the following conditions: The proposal is for a warehouse costing $ The warehouse has an expected useful life of 35 years and a net salvage value of $ Annual receipts of $17000 are expected, annual maintenance and administrative costs will be $4000, and annual income taxes are $2000. Given this data, which of the following are correct? (a) The proposal is justified for a MARR of 9%. (b) The proposal has a net present worth of $ , when 6% is used as the interest rate. (c) The proposal is acceptable as long as MARR 10.77%. 25

26 26 (True) 10.77% gives Solving for 0,35), 25000(,35), 11000( (c) (b)true (True) ,9%,35) 25000(,9%,35) 11000( (9%) (a) i i i F P A i P F P A P PW

27 Future Worth Criterion Given: Cash flows and MARR (i) Find: The net equivalent worth at a time period other than 0 (typically at the end of project life) 0 $55,760 $24,400 $27, $75,000 Project life 27

28 Future Worth Criterion FW( 15%) $24, 400( F/ P, 15%,2 ) $27, 340( F/ P, 15%,1 ) inflow $55, 760( F/ P, 15%,0 ) $119, 470 FW( 15%) $75, 000( F / P, 15%,3 ) outflow $114, 066 FW( 15%) $119, 470 $114, 066 $5, 404 0, Accept 28

29 Capitalized Equivalent Worth Principle: PW for a project with an annual receipt of A over infinite (or extremely long) service life Equation: CE(i) = A(P/A, i, ) = A/i 0 P = CE(i) A Computing PW: capitalization of project cost Cost is called capitalized cost N 29

30 Given: i = 10%, N = Find: P or CE (10%) $2,000 $1,000 0 P = CE (10%) =? 10 $1, 000 $1, 000 CE( 10%) ( P/ F, 10%,10) $10, 000( ) $13,

31 Example: Bridge Construction Construction cost = $2,000,000 Annual Maintenance cost = $50,000 Renovation cost = $500,000 every 15 years Planning horizon = infinite period Interest rate = 5% 31

32 $50,000 $2,000,000 $500,000 $500,000 $500,000 $500,000 32

33 Solution: Construction Cost P 1 = $2,000,000 Maintenance Costs P 2 = $50,000/0.05 = $1,000,000 Renovation Costs P 3 = $500,000(P/F, 5%, 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60)... = {$500,000(A/F, 5%, 15)}/0.05 = $463,423 Total Present Worth P = P 1 + P 2 + P 3 = $3,463,423 33

34 Alternate way to calculate P 3 Concept: Find the effective interest rate per payment period $500,000 $500,000 $500,000 $500,000 Effective interest rate for a 15-year cycle i = ( ) 15-1 = % Capitalized equivalent worth P 3 = $500,000/ = $463,423 34

35 Comparing Mutually Exclusive Projects Mutually Exclusive Project Fundamental principle in comparing mutually exclusive projects (alternatives) is that they must be compared over an equal time span (or planning horizon) Do-Nothing Alternative 35

36 Analysis Period The time span over which the economic effects of an investment will be evaluated (study period or planning horizon). Required Service Period The time span over which the service of an equipment (or investment) will be needed. Useful life of a project Compare projects with different useful lives over an equal time span 36

37 Comparing Mutually Exclusive Projects Principle: Projects must be compared over an equal time span. Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period. 37

38 How to Choose An Analysis Period Case 1 Analysis period equals project lives Finite Analysis = Required period service period Case 2 Case 3 Analysis period is shorter than project lives Analysis period is longer than project lives Required service period Case 4 Analysis period is longest of project life in the group Infinite Project repeatability likely Analysis period is lowest common multiple of project lives Project repeatability unlikely Analysis period equals one of the project lives 38

39 Case 1: Analysis Period Equals Project Lives Compute the PW for each project over its life $2,075 $600 $450 $500 $1,400 0 $2,110 A $1,000 $4,000 PW (10%) = $283 A PW (10%) = $579 B B 39

40 Comparing projects requiring different levels of investment Assume that the unused funds will be invested at MARR. $450 $600 $500 $2,110 $1,000 Project A 3,993 $1,400 $2,075 $450 $600 $500 $1,000 Modified Project A $4,000 Project B This portion of investment will earn 10% return on investment. $3,000 PW(10%) A = $283 PW(10%) B = $579 40

41 Example: Consider the following two mutually exclusive investment projects. Assume that MARR=12%. Which alternative would be selected by using the PW criterion? Which alternative would be selected by using the net-future-worth criterion? Project cash flow n A B

42 Present worth : 2610 PWA(12%) PWB (12%) Select A Net - future - worth : FW (or simply (1.12) FW A B Select A. (12%) 4500(1.12) (12%) 2900(1.12) (1.12) ) 1210(1.12) (1.12) (1.12)

43 Often, project lives do not match the required analysis period and/or do not match each other. For example, two machines may perform the same functions, but one will last longer, and both of them last longer than the analysis period for which they are being considered. 43

44 Case 2: Analysis Period Shorter than Project Lives Example: 5-year production project when all the alternative equipment have useful lives of 7 years. Estimate the salvage value at the end of the required service period. Salvage value is the amount of money for which the equipment could be sold after its service to the project or the dollar measure of its remaining usefulness. Compute the PW for each project over the required service period. 44

45 Comparison of unequal-lived service projects when the required service period is shorter than the individual project life (Required service period =2 years) PW(15%)A = -$362,000 PW(15%)B = -$364,000 45

46 PW(15%) A = - $362 PW(15%) B = - $364 Model A would be preferred. 46

47 Case 3: Analysis Period Longer than Project Lives Come up with replacement projects that match or exceed the required service period. Same alternative or new technology? Purchase, lease or subcontract? Compute the PW for each project over the required service period. 47

48

49 Comparison for Service Projects with Unequal Lives when the required service period is longer than the individual project life PW(15%) A = - $34359 PW(15%) B = - $

50 Case 4: Analysis period coincides with longest project life Consider two mutually exclusive processes for natural resource extraction (e.g. coal): 10 years vs 8 years. No need to continue the project if 8-year process is chosen (all coal extracted). The two processes can be compared over an analysis period of 10 years (longest of 8 and 10). Revenues must be included even if total (undiscounted) revenue is the same for both processes, since the faster process has larger present worth. 50

51 Case 4: Analysis period coincides with longest project life Example: Natural resource extraction PW(15%) drill = $2,208,470 PW(15%) lease = $2,180,210 51

52 Analysis Period is Not Specified Project Repeatability Unlikely Use common service (revenue) period. Project Repeatability Likely Use the lowest common multiple of project lives. 52

53 Project Repeatability Likely PW(15%) A =-$53,657 Model A: 3 Years Model B: 4 years LCM (3,4) = 12 years PW(15%) B =-$48,534 53

54 Summary Present worth is an equivalence method of analysis in which a project s cash flows are discounted to a lump sum amount at present time. The MARR or minimum attractive rate of return is the interest rate at which a firm can always earn or borrow money. MARR is generally dictated by management and is the rate at which NPW analysis should be conducted. Two measures of investment, the net future worth and the capitalized equivalent worth, are variations to the NPW criterion. 54

55 The term mutually exclusive means that, when one of several alternatives that meet the same need is selected, the others will be rejected. Revenue projects are those for which the income generated depends on the choice of project. Service projects are those for which income remains the same, regardless of which project is selected. The analysis period (study period) is the time span over which the economic effects of an investment will be evaluated. The required service period is the time span over which the service of an equipment (or investment) will be needed. 55

56 The analysis period should be chosen to cover the required service period. When not specified by management or company policy, the analysis period to use in a comparison of mutually exclusive projects may be chosen by an individual analyst. 56