Chapter 5 Present-Worth Analysis

2 Loan versus Project Cash Flows Initial Project Screening Methods Present-Worth Analysis Methods to Compare Mutually Exclusive Alternatives

3 Bank Loan vs. Investment Project Bank Customer Loan Repayment Company Project Investment Return  Bank Loan  Investment Project

Payback Period

5 Federal Express: A Project Example 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

6 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?

7 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

8 Example: Conventional Payback Period NCash FlowCum. Flow $105,000+$10,000 $15,000 $25,000 $35,000 $45,000 $35,000 -$95,000 -$80,000 -$55,000 -$20,000 $25,000 $65,000 $100,000 Payback period should occur somewhere between N = 3 and N = 4.

9 Example: Conventional Payback Period NCash FlowCum. Flow $105,000+$10,000 $15,000 $25,000 $35,000 $45,000 $35,000 -$95,000 -$80,000 -$55,000 -$20,000 $25,000 $65,000 $100,000 20,000/45,000= 0.44 Payback period occurs at N =3.44

, , , , , Years (n) ~3.44 years Payback period $95,000 $15,000 $25,000 $35,000 $45,000 $35, Years Annual cash flow Cumulative cash flow ($)

11 Conventional Payback Period PeriodProject 1Project ,000 11,0009, ,000 3 Payback period 2 years Although the payback periods are the same, Project 2 is better because most investment is recovered at the end of year 1.

12 Discounted Payback Period  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 Example: Discounted Payback Period Calculation PeriodCash FlowCost of Funds (15%)* Cumulative Cash Flow 0-$95, ,000-$95,000(0.15) = -$14, , ,000-$94,250 (0.15) = -14, , ,000-$83,387 (0.15) = -12, , ,000-$60,895 (0.15) =-9, , ,000-$25,030(0.15) = -3,754 16, ,000$16,215 (0.15) = 2,432 53,648 * Cost of funds = (Unrecovered beginning balance) X (interest rate) -95,000-14,250+15,000= -94,250

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

Present Worth Analysis

16 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 Inflow Outflow 0 PW(i) inflow PW(i) outflow Net surplus PW(i) > 0

17 Evaluation of a Single Project Step 1: Determine the interest rate that the firm wishes to earn (Required rate of return or Minimum attractive rate of return-MARR) Step 2: Estimate the service life of the project Step 3: Estimate the cash inflow for each period over the service life Step 4: Estimate the cash outflow for each period over the service life Step 5: Determine the net cash flows for each period Net cash flow=Cash inflow-Cash outflow Step 6: Find the present worth of each net cash flow, add up these figures, find project’s PW Step 7: Accept the investment if PW>0, reject otherwise

18 Example - Tiger Machine Tool Company $75,000 $24,400 $27,340 $55, outflow inflow

19 Solution $75,000 $24,400 $27,340 $55, outflow inflow

20 Present Worth Amounts at Varying Interest Rates i (%)PW(i)i(%)PW(i) 0$32,50020-$3, , , , , , , , , , , , , , , , , *038-21, ,302 *Break even interest rate OR the internal rate of return

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

22 Future Worth Criterion  Given: Cash flows and MARR (i)  Find: The net equivalent worth at the end of project life $75,000 $24,400 $27,340 $55, Project life

23 Future Worth Criterion

24 What factors should the company consider in selecting a MARR in project evaluation? Rate of return that we expect to earn on an investment is a function of three components: to be rewarded for not being able to use our money inflation factor (decrease in purchasing power) risk premiums

25 Guideline for Selecting a MARR Real Return2% Inflation4% Risk premium0% Total expected return 6% Real Return2% Inflation4% Risk premium20% Total expected return 26% Risk-free real return Inflation Risk premium Treasury Bills Amazon.com Very safe Very risky

Can you explain what $3,553 really means? Meaning of Net Present Worth 1. Investment Pool Concept 2. Borrowed Funds Concept

27 Investment-Pool Concept There is an investment pool where the money earns MARR. The money can be taken from the pool for other investments. Suppose the company has $75,000. It has two options. (1)Take the money out and invest it in the project or (2) leave the money in the company. Let’s see what the consequences are for each option.

28 $75, $24,400 $27,340 $55,760 Investment pool 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 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 Project Return to investment pool N = 3, MARR = 15% Meaning of Net Present Worth PW(MARR) = $5,404(P/F,15%,3) = $3,553

29 Borrowed-Funds Concept The firm does not have the money required for investment Suppose that the firm borrows the required money from a bank at an interest rate of MARR

30 Borrowed-Funds and Project Balance Concepts N0123N0123N0123N0123 Beginning Balance Interest PaymentProjectBalance -$75,000 -$11,250 +$24,400 -$61,850 -$9,278 +$27,340 -$43,788 -$6,568 +$55,760 +$5,404 Net future worth, FW(15%) PW(15%) = $5,404 (P/F, 15%, 3) = $3,553

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

32 Capitalized-Equivalent Method Principle: PW for a project with an annual receipt of A over infinite service life Example: Public projects such as bridges, waterway constructions, hydroelectric dams. Equation: CE(i) = A(P/A, i,  ) = A/i A 0 P = CE(i) : Capitalized-equivalent worth N ∞N ∞

33 Practice Problem 10 $1,000 $2,000 P = CE (10%) = ? 0 Given: i = 10%, N = ∞ Find: P or CE (10%) ∞

34 Solution 10 $1,000 $2,000 P = CE (10%) = ? 0 ∞ OR

35 A Bridge Construction Project  Construction cost = $2,000,000  Annual Maintenance cost = $50,000  Renovation cost = $500,000 every 15 years  Planning horizon = infinite period  Interest rate = 5%

36 $500,000 $2,000,000 $50, Years

37 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 (Capitalized-Equivalent) P = P 1 + P 2 + P 3 = $3,463,423

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

Comparing Mutually Exclusive Alternatives

40 Comparing Mutually Exclusive Projects Any one of the alternatives will fulfill the need, others will be excluded. Mutually exclusive alternatives must be compared over an equal time span. If the existing system still works, “do- nothing” may also be an alternative. Cash flows of the new proposals are generated relative to those of do-nothing: Incremental costs and returns will be used.

41  Revenue Projects Projects whose revenues depend on the choice of alternatives Alternative with the largest net gains (output-input) is selected (largest net present value)  Service Projects Projects whose revenues do not depend on the choice of alternative, they must produce the same amount of output Alternative with the least cost (lower present value) is selected

42  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.

43 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.

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

45 Example: Buying versus Lease Decision Option 1 Debt Financing Option 2 Lease Financing Price$14,695 Down payment$2,000$ APR (%)6% compounded monthly Monthly payment$372.55$ Length36 months Cash due at lease end$300 Selling Price at lease end$

46 Which Interest Rate to Use to Compare These Options? APR = 6% compounded monthly

47 Your Earning Interest Rate = 6%  Effective Monthly Interest Rate =0.5% Debt Financing: P debt = $2,000 + $372.55(P/A, 0.5%, 36) - $8,673.10(P/F, 0.5%, 36) = $6, Lease Financing: P lease = $ $236.45(P/A, 0.5%, 35) + $300(P/F, 0.5%, 36) = $8,556.90

48 Case 2: Analysis Period Shorter than Project Lives Ex: In the construction industry, the building project has a shorter completion time than the useful time of equipment purchased.  Estimate the salvage value at the end of the required service period.  Compute the PW for each project over the required service period.

49 Example - Comparison of unequal-lived service projects when the required service period is shorter than the individual project life Required Service Period = 2 years

50 PW(15%)B = -$364,000 PW(15%)A = -$362,000

51 Case 3: Analysis Period Longer than Project Lives  Come up with replacement projects that match or exceed the required service period  We may assume that the replacement project will be same as the initial project  Compute the PW for each project over the required service period  If the analysis period is not given: Select it as the LCM (lowest common multiple) of project lives.

52 Example - Comparison for Service Projects with Unequal Lives when the required service period is longer than the individual project life Required Service Period = 5 years Model A Model B $15,000 $12,500 $4,000 $4,500 $5,000 $5,500 $1,500 $5,000 $5,500 $6,000 $2,

53 Lease Model A to serve the remaining service life PW(15%)A = -$ PW(15%)B = -$33,173

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.

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.

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 (as the lowest common multiple of project lives).