Other Analysis Techniques. Future Worth Analysis (FWA) Benefit-Cost Ratio Analysis (BCRA) Payback Period

Transcription

1 Other Analysis Techniques Future Worth Analysis (FWA) Benefit-Cost Ratio Analysis (BCRA) Payback Period 1

2 Techniques for Cash Flow Analysis Present Worth Analysis Annual Cash Flow Analysis Rate of Return Analysis Incremental Analysis Other Techniques: Future Worth Analysis Benefit-Cost Ration Analysis Payback Period Analysis Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 2

3 Future worth analysis Future worth analysis is equivalent to present worth analysis; the best alternative one way is also best the other way. There are many situations where we want to know what a future situation will be, if we take some particular course of action now. This is called future worth analysis. 3

4 Example 9-1 Ron Jamison, a 20-year-old college student, consumes about a carton of cigarettes a week. He wonders how much money he could accumulate by age 65 if he quit smoking now and put his cigarette money into a savings account. Cigarettes cost $35 per carton. Ron expects: that a savings account would earn 5% interest, compounded semiannually. Compute Ron's future worth at age 65. Semiannual saving $35/carton x 26 weeks = $910 Future worth (FW)= A(F/A,,21/2%, 90) = 910(329.2) = $299,572 4

5 Example 9-2 An east coast firm has decided to establish a second plant in Kansas City. There is a factory for sale for $850,000 that, with extensive remodeling, could be used. As an alternative, the company could buy vacant land for $85,000 and have a new plant constructed there. Either way, it will be 3 years before the company will be able to get a plant into production. The timing and cost of the various components for the factory are given in the following cash flow table. Year Construct New Plant 0 Buy land $ Design and $ initial cost 2 Balance of construction costs 1,200,000 3 Setup of production equipment 200,000 5

6 Remodel Available Factory 0 Purchase factory $850,000 1 Design and remodeling 250,000 2 Additional remodeling 250,000 3 Setup of production equipment 250,000 If interest is 8%, which alternative results in the lower equivalent cost when the firm begins production at the end of the third year? New Plant Future worth of cost (FW)= 85,000(F/ P,8%, 3) + 200,000(F/A, 8%,3) + 1,000,000 (F/P, 8%,1)= $1,836,000 Remodel Available Factory Future worth of cost (FW) = 850,ooo(F/ P, 8%, 3) + 250,000(F/A, 8%, 3) = $

7 Benefit-cost ratio analysis Since we can write PW of cost PW of benefit or EUAC EUAB we can equivalently write (PW of benefit)/pw of cost 1, or EUAB/EUAC 1. Economic analysis based on these ratios is called benefitcost ratio analysis. 7

8 Fixed input Benefit-cost ratio comparison criteria Situation Amount of money or other input resources are fixed Criterion Maximize B/C Fixed output Fixed benefit Maximize B/C Neither input nor output fixed Neither amount of money or other inputs not amount of benefits or other outputs are fixed Compute incremental benefitcost ratio ( B/ C) on the increment of investment between the alternatives. If B/ C 1, choose the highercost alternative; otherwise, choose the lower-cost alternative. 8

9 Example 9-3 Using an interest rate of 7%, choose the best alternative using the benefit-cost ratio analysis Year Device A Device B 0 $1000 $ We have solved this problem using the present worth analysis, the annual cash flow analysis, and the rate of return analysis. Now, we will use the Benefit-Cost Ratio analysis. Since this is a fixed cost case, the criterion is to maximize the B/C ratio. 9

10 Device A PW of cost = $1000 PW of benefits =300(P/A, 7%,5) =300(4.100) =$1230 B/C = PW of benefit/ PW of costs= 1230/ 1000 = 1.23 Device B PW of cost = $1000 PW of benefit = 400(P/ A, 7%, 5) - 50(P/G-, 7%, 5) =400(4.100)- 50(7.647) = = 1258 BPW of benefit 1258 B/C = BPW of benefit /PW of cost = 1258/1000= 1.26 Chose B 10

11 Example 9-4 Using an interest rate of 10%, choose the best machine using the benefit-cost ratio analysis Machine X Machine Y Initial cos $200 $700 Uniform annual benefit End of Useful live salvage value Useful life (years) 6 12 Machine X EUAC = 200(A/ P, 10%,6) - 50(A/ F, 10%,6) = 200(0.2296) - 50(0.1296) = 46-6 = $40 EUAB = $95 11

12 Machine Y EUAC = 700(A/ P, 10%, 12) - 150(A/ F, 10%, 12) = 700(0.1468) - 150(0.0468)= = $96 EUAB =$120 Machine Y - Machine X B / C= (120-95)/(96-40 )= 0.45 Since the incremental benefit-cost ratio is less than 1, it represents an undesirable increment of investment We therefore choose the lower-cost alternative-machine X 12

13 Example 9-5 Each of the five mutually exclusive alternatives presented below will last for 20 years and has no salvage value. MARR = 6%. A B C D E F Cost $4000 $2000 $6000 $1000 $9000 $10000 PWB $7330 $4700 $8730 $1340 $9000 $9500 B/C NPV (to check) The steps are the same as in incremental ROR, except that the criterion is now B/ C, and the cutoff is 1 instead of the MARR: 1) Be sure you identify all alternatives. 2) (Optional) Compute the B/C ratio for each alternative. Discard any with a B/C < 1. (We can discard F). 3) Arrange the remaining alternatives in ascending order of investment. 13

14 Example 9-5 D B A C E F Cost $1000 $2000 $4000 $6000 $9000 $10000 PWB $1340 $4700 $7330 $8730 $9000 $9500 B/C NPV ) Comparing B/ C with 1 for consecutive alternatives select the best alternative. B-D A-B C - A E-A Incremental Cost $1000 $2000 $2000 $5000 Incremental Benefit $3360 $2630 $1400 $1670 B/ C Comparison result B preferred A preferred A preferred A preferred Increment B-D is attractive; therefore B is preferred to D. Increment A-B is attractive; therefore B is preferred to A. Increment C-A is not attractive, as B/ C = 0.76 < 1 ; therefore A is preferred to C. Increment E-A is not attractive, as B/ C = 0.33 < 1 ; therefore A is preferred to E. Finally A is the best project. 14

15 Benefit/Cost Ratio Analysis Graphical representation A, B, C, and D are above the 45-degree line; their B/C ratio is > 1. F is below the line: B/C ratio is < 1. We can discard F if we wish. Examine each separable increment of investment. B/ C < 1 increment is not attractive B/ C 1 increment is desirable. PWB B A C E F PWB/PWC = 1 Compare D to B: Slope or B/ C > 1. B preferred Compare B to A: Slope or B/ C > 1. A preferred Compare A to C: Slope or B/ C < 1; discard C. Compare A to E: Slope or B/ C < 1; discard E. F was discarded earlier since its B/C is less than 1 Therefore A is the best alternative. Note: Despite the fact that Alt. B has the highest B/C ratio, we should not make the mistake of choosing B since the economic criterion here should be based on B/ C (since neither input nor output is fixed) D PWC B-D A-B C - A E-A Incremental Cost $1000 $2000 $2000 $5000 Incremental Benefit $3360 $2630 $1400 $1670 Incr.B/Incr. C

16 Payback Period Payback period is an approximate analysis method. For example, if a $1000 investment today generates $500 of benefits per year, we say its payback period is 1000/500 = 2 years. Four important points about the payback period 1. Payback period is an approximate, rather than an exact, analysis calculation. 2. All costs and all profits, or savings of the investment prior to payback, are included without considering differences in their timing. 3. All the economic consequences beyond the payback period are completely ignored. 4. Payback period may or may not select the same alternative as an exact economic analysis method. Payback period is used because 1. the concept can be readily understood, 2. the calculations can be readily made and understood by people unfamiliar with the use of the time value of money. The rate of return only measures the speed of return of the investment. The other analysis techniques we studied before measure the economic efficiency or the overall profitability of the investment. 16

17 Example 9-6 The cash flows for two alternatives are as follows: Year A B 0 -$1000 -$ You may assume the benefits occur throughout the year rather than just at the end of the year. Based on payback period which alternative should be selected? 17

18 Alternative A Payback period is the period of time required for the profit or other benefits of an investment to equal the cost of the investment. In the first 2 years, only $400 of the $1000 cost is recovered. The remaining $600 cost is recovered in the first half of Year3. Thus the payback period for Alt. A is 2.5 years. Alternative B Since the annual benefits are uniform, the payback period is simply $2783/$1200 per year = 2.3 years To minimize the payback period, choose Alt. B. 18

19 Example 9-7 A firm is trying to decide which of two weighing scales it should install to check a package-filling operation in the plant. If both scales have a 6-year life, which one should be selected? Assume an 8% interest rate. Alternative cost Uniform Annual Benefit End-of-Useful-Life Salvage Atlas $2000 $450 $100 Tom Thumb $ Atlas Scale Payback period = Cost /Uniform annual benefit 2000/450 = 4.4 years Tom Thumb Scale Payback period = Cost /Uniform annual benefit = 3000/600= 5 years 19

20 Example 9-8 A :firm is purchasing production equipment for a new plant. Two alternative machines are being considered for a particular operation. Tempo Machine Dura Machine Installed cost $30,000 $35000 Net annual benefit after $12,000 the first year, $1000 the first year, all annual expenses declining $3000 increasing $3000 per year thereafter per year thereafter have been deducted Useful life, in years 4 8 Neither machine has any salvage value. Compute the payback period for each of the alternatives 20

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22 Now, as a check on the payback period analysis, compute the rate of return for each alternative. Assume the minimum attractive rate of return is 10%. The cash flows for the two alternatives are as follows: 22

23 Year Tempo Machine Dura Machine sum

24 Tempo Machine Since the sum of the cash flows for the Tempo machine is zero, we see immediately that the $30,000 investment just equals the subsequent benefits. The resulting rate of return is 0%. 24

25 Dura Machine 35,000 = 1000(P/A, i, 8) +3000(P/G, i, 8) Try i = 20%: 35,000 =~ 1000(3.837)+ 3000(9.883) =~ ,649 = 33,486 The 20% interest rate is too high. Try i = 15%: 35,000 =~ 1000(4.487)+ 3000(12.481) =~ ,443 = 41,930 This time, the interest rate is too low. Linear interpolation would show that the rate of return is approximately 19%. Using an exact calculation-rate of return. It is clear that the Tempo is not very attractive economicly 25

26 Yet it was this alternative,and not "the Dura Machine, that was preferred based On the payback period calculations. On the other hand, the shorter payback period for the Tempo does give a measure of the speed of the return of the investment not found in the Dura. The conclusion to be drawn is that liquidity and profitability may be two quite different criteria. 26

27 From the discussion and the examples, we see that payback period can be helpful in providing a measure of the speed of the return of the investment. This might be quite important, for example, for a company that is short of working capital or for a firm in an industry experiencing rapid changes in technology. Calculation of payback period alone, however, must not be confused with a careful economic analysis. We have shown that a short payback period does not always mean that the associated investment is desirable. Thus, payback period should not be considered a suitable replacement for accurate economic analysis calculations. 27

28 SENSITIVITY AND BREAKEVEN ANALYSIS The variation to a particular estimate that would be necessary to change a particular decision."this is called sensitivity analysis. An analysis of the sensitivity of a problem's decision to its various parameters highlights the important and significant aspects of that problem. Breakeven analysis is a form of sensitivity analysis. To illustrate the sensitivity of a decision between alternatives to particular estimates, breakeven analysis is often presented as a breakeven chart. 28

29 Example 9-9 Consider a project that may be constructed to full capacity now or may be constructed in two stages. Construction Costs Two-stage construction Construct first stage now $ Construct second stage $ n years from now Full-capacity construction Construct full capacity now $

30 Other Factors 1. All facilities will last for 40 years regardless of when they are installed; after 40 years, they will have zero salvage value. 2. The annual cost of operation and maintenance is the same for both two-stage construction and full-capacity construction. 3. Assume an 8% interest rate. Plot "age when second stage is constructed" versus "costs for both alternatives." Mark the breakeven point on your graph. What is the sensitivity of the decision to secondstage construction 16 or more years in the future? 30

31 Since we are dealing with a common analysis period, the calculations may be either annual cost or present worth. Present worth calculations appear simpler and are used here. Construct Full Capacity Now PW of cost = $140,000 31

32 Two-Stage Construction First stage constructed now and the second stage to be constructed n years hence...compute the PW of cost for several values of n (years). PW of cost = 100, ,000(P/F, 8%, n) n=5 PW= 100, ,000(0.6806)=$181,700 n = 10 PW = 100, ,000(0.4632)=155,600 n =20 PW= 100, ,000(0.2145)=125,700 n=30 PW= 100, ,000(0.0994)=111,900 These data are plotted in the form of a breakevell chart in Figure

33 The breakeven point on the graph is the point at which both alternatives have equivalent costs. 33

34 Example 9-10 Example 8-3 posed the following situation. Three mutually exclusive alternatives are given, each with a 20-year life and no salvage value. The minimum attractive rate of return is 6%. A B C Initial Cost $2000 $4000 $5000 Uniform Annual Benefit In Example 8-3 we found that Alt. B was the preferred alternative. Here we would like to know how sensitive the decision is to our estimate of the initial cost of B. If B is preferred at an initial cost of $4000, it will continue to be preferred at any smaller initial cost. But how much higher than $4000 can the initial cost be and still have B the preferred alternative? The computations may be done several different ways. With neither input nor output fixed, maximizing net present worth is a suitable criterion. 34

35 Alternative A NPW = PW of benefit - PW of cost Alternative B =41O(P / A, 6%, 20) = 410(11.470) = $2703 Let x = initial cost of B. NPW = 639(P/ A, 6%, 20) - x = 639(11.470)- x = x Alternative C NPW = 700(P/A, 6%,20) = 700(11.470) = $3029 For the three alternatives, we see that B will only maximize NPW as long as its NPW is greater than = x x = = $4300 Therefore, B is the preferred alternative if its initial cost does not exceed $

36 Initial cost above $4300, C is preferred. We have a breakeven point at $4300. When B has an initial cost of $4300, B and C are equally desirable. 36

37 HW Chapter Chapter

38 Lesson from Example 9-8: liquidity and profitability can be very different criteria. We discussed the definitions of liquidity and profitability in class. 38

39 Summary Future Worth. A future worth calculation occurs when the point in time at which the comparison between alternatives will be made is in the future. The best alternative according to future worth should also be best according to present worth. Benefit-Cost Ratio Analysis. We compute a ratio of benefits to costs, using either PW or ACF calculations. Graphically, the method is similar to PW analysis. When neither input nor output is fixed, we use incremental benefit-cost analysis ( B/ C). Payback Period. The payback period is the period of time needed for the profit or other benefits of an investment to equal its cost. This method is simple to use and understand, but is a poor analysis technique for ranking alternatives. It provides a measure of the speed of return of the investment, but is not an accurate measure of its profitability. 39