Ch8 RateofReturnMultipleAlternatives EL

8-*Lecture slides to accompanyEngineering Economy7th edition

Leland BlankAnthony TarquinChapter 8Rate of Return Multiple Alternatives 2012 by McGraw-Hill All Rights Reserved

2012 by McGraw-Hill All Rights Reserved

8-*LEARNING OUTCOMESExplain why incremental analysis is required in RORCompute incremental cash flow (CF) Interprete ROR on incremental CFSelect alternative by ROR based on PW relationSelect alternative by ROR based on AW relationSelect best from several alternatives using ROR method 2012 by McGraw-Hill All Rights Reserved


Independent projects and mutually Exclusive projects.

Which project to select in following situations;Projects are independentProject are mutually exclusiveME and capital is RM90,000

2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*

Alternative AAlternative BInvestment50,00085,000ROR35%29%

2012 by McGraw-Hill, New York, N.Y All Rights Reserved

8-*Why Incremental Analysis is Necessary Selecting the alternative with highest ROR may notyield highest return on available capitalMust consider weighted average of total capital available Capital not invested in a project is assumed to earn at MARRExample: Assume $90,000 is available for investment and MARR = 16% per year. If alternative A would earn 35% per year on investment of $50,000, and B would earn 29% per year on investment of $85,000, the weighted averages ROR are:Overall RORA = [50,000(0.35) + 40,000(0.16)]/90,000 = 26.6%Overall RORB = [85,000(0.29) + 5,000(0.16)]/90,000 = 28.3%

Which investment is better, economically? 2012 by McGraw-Hill All Rights Reserved


Why Incremental Analysis is Necessary 2012 by McGraw-Hill All Rights Reserved8-*If selection basis is higher ROR:Select alternative A (wrong answer)

If selection basis is higher overall ROR:Select alternative B

Conclusion: Must use an incremental ROR analysis to make a consistently correct selection

Unlike PW, AW, and FW values, if not analyzed correctly, ROR values can lead to an incorrect alternative selection. This is called the ranking inconsistency problem (discussed later)


8-*Incremental cash flow = cash flowB cash flowA where larger initial investment is Alternative BCalculation of Incremental CF Example: Either of the cost alternatives shown below can be used in a grinding process. Tabulate the incremental cash flows. BAFirst cost, $ -40,000 - 60,000 Annual cost, $/year -25,000 -19,000Salvage value, $ 8,000 10,000The ROR on the extra $20,000 investment in B determines which alternative to select (as discussed later) The incremental CF is shown in the (B-A) column 2012 by McGraw-Hill All Rights Reserved


Question 8-11Prepare a tabulation of incremental cash flows for the two machine alternatives below.

, 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*

Machine XMachine YFirst cost$ 35,000 90,000Annual operating cost, $ per year 31,60019,400

Salvage value, $ 0 8,000

Life, years 24


8-*Interpretation of ROR on Extra InvestmentFor independent projects, select all that have ROR MARR (no incremental analysis is necessary)Based on concept that any avoidable investment that does not yield at least the MARR should not be made.Once a lower-cost alternative has been economically justified, the ROR on the extra investment (i.e., additional amount of money associated with a higher first-cost alternative) must also yield a ROR MARR (because the extra investment is avoidable by selecting the economically-justified lower-cost alternative).

This incremental ROR is identified as i* 2012 by McGraw-Hill All Rights Reserved


8-*ROR Evaluation for Two ME AlternativesOrder alternatives by increasing initial investment cost(2) Develop incremental CF series using LCM of years(3) Draw incremental cash flow diagram, if needed(4) Count sign changes to see if multiple i* values exist(5) Set up PW, AW, or FW = 0 relation and find i*B-ANote: Incremental ROR analysis requires equal-service comparison. The LCM of lives must be used in the relation(6) If i*B-A < MARR, select A; otherwise, select B

If multiple i* values exist, find EROR using either MIRR or ROIC approach. 2012 by McGraw-Hill All Rights Reserved


8-*Example: Incremental ROR Evaluation B AFirst cost ,$ -40,000 -60,000 Annual cost, $/year -25,000 -19,000Salvage value, $ 8,000 10,000Life, years 5 5 2012 by McGraw-Hill All Rights ReservedInitial observations: ME, cost alternatives with equal life estimatesand no multiple ROR values indicated


8-*Example: ROR Evaluation of Two AlternativesBAFirst cost , $ -40,000 -60,000 Annual cost, $/year -25,000 -19,000Salvage value, $ 8,000 10,000Life, years 5 5Solution, using procedure:Order by first cost and find incremental cash flow B - A-20,000 +6000 +2000B - A0 = -20,000 + 6000(P/A,i*,5) + 2000(P/F,i*,5) i*B-A = 17.2% > MARR of 15%ROR on $20,000 extra investment is acceptable: Select BWrite ROR equation (in terms of PW, AW, or FW) on incremental CF Solve for i* and compare to MARR 2012 by McGraw-Hill All Rights Reserved


Testing your IQThe PW equation of B-A is;0 = -20,000 + 6000(P/A,i*,5) + 2000(P/F,i*,5)The cash flow for A is given as follow;

What is the cash flow for B? 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*

AFirst cost25,000AOC20,000Salvage10,000Life , years5


Alternatives with different life periodIn this situation we need to use the LCM life of both alternative. 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*


Example 8-2Two transformers; Type A and Type B


Type AType BInitial cost70,00090,000AOC9,0007,000Salvage5,00010,000Life812


Example 8-21-*

Sheet1

Incremental

Net cash flowscash flow

YearType AType B(B - A)

0-70,000-95,000-25,000

1-9,000-7,0002,000

2-9,000-7,0002,000

3-9,000-7,0002,000

4-9,000-7,0002,000

5-9,000-7,0002,000

6-9,000-7,0002,000

7-9,000-7,0002,000

8-74,000-7,00067,000

9-9,000-7,0002,000

10-9,000-7,0002,000

11-9,000-7,0002,000

12-9,000-92,000-83,000

13-9,000-7,0002,000

14-9,000-7,0002,000

15-9,000-7,0002,000

16-74,000-7,00067,000

17-9,000-7,0002,000

18-9,000-7,0002,000

19-9,000-7,0002,000

20-9,000-7,0002,000

21-9,000-7,0002,000

22-9,000-7,0002,000

23-9,000-7,0002,000

24-4,0003,0007,000

Totals-411,000-338,00073,000

Starting new life cycle for A = initial cost + AOC + salvage= -70,000 -9,000 +5,000

Check on summationsIncremental column should equal difference of columns

Sheet2

Sheet3

Practice QuestionTry Problem 8-11


Machine XMachine YFirst cost $-35,000-90,000Annual operating cost, $ per year-31,000-19,400Salvage values $08,000Life, years24MARR is 20%


Practice questionYear X Y Y - X0 -35,000 -90,000 -55,0001 -31,600 -19,400 +12,2002 -31,600-35,000 -19,400 +47,2003 -31,600 -19,400 +12,2004 -31,600 -19,400+8,000 +20,200

Write the PW equation for the incremental cash flow.

PW = -55,000 + 12,200(P/F,i,1) +47,200(P/F,i,2) +12,200(P/F,i,3) + 20,200(P/F,i,4) What is the ROR? 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*


Solving by hand (Trial and error) Try 22% 0 = -55,000 +12,200(0.8197) + 47,200(0.67190) + (12,200(0.5507) + 20,200(0.4514) = -55,000 +10,000 + 31,713 + 6718 + 9,118 = 2,549

Try 25%0 = -55,000 +12,200(0.8) + 47,200(0.64) + (12,200(0.512) + 20,200(0.4096) = -55,000 + 9,760 + 30,208 + 6,246 + 8,273 = -513

i = between 22% and 25%. Can you decide now which alternative to select ?

Do we need to do interpolation?



InterpolationDifferent in % = 25 22 = 3% Different in PW value = -513 - 2549 = 3062

The rate is less than 25% by (513 x 3/3062 ) = 0.5%The answer = 24.5%Using spreadsheet; i B-A = 24%

2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*25022-5132,549? %


Example 8-3Ford Motor selecting supplier based on their bid.

MARR is 12 %These are cost alternatives.


ABInitial cost-8,000-13,000Annual cost-3,500-1,600Salvage value02,000Life, years105


Incremental cash flow 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*= NPV(12%,D5:D14) + D4


PW of Incremental cash flow at various i 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*


Breakeven ROR Value An ROR at which the PW, AW or FW values:

Of cash flows for two alternatives are exactly equal. This is the i* value

Of incremental cash flows between two alternatives are exactly equal. This is the i* value 2012 by McGraw-Hill All Rights Reserved8-*If MARR > breakeven ROR, select lower-investment alternative


Example 8-4New filtration systems for commercial airliners are available that use an electric field to remove up to 99.9% of infectious diseases and pollutants from aircraft air. This is vitally important, as many of the flu germs, viruses, and other contagious diseases are transmitted through the systems that recirculate aircraft air many times per hour. Investments in the new filtration equipment can cost from $100,000 to $150,000 per aircraft, but savings in fuel, customer complaints, legal actions, etc., can also be sizable. Use the estimates below (in $100 units) from two suppliers provided to an international carrier to do the following, using a spreadsheet and an MARR of 15% per year. 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*

Air Cleanser(Filter 1)Purely Heaven(Filter 2)Initial cost per aircraft, $ 10001500Estimated savings, $ per year375700 in year 1, decreasing by100 per year thereafterEstimated life, years 55


Two Mutually Exclusive AlternativesFiltration system to remove pollutant from aircraft air. 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*

Filter 1Filter 2YearCash flow,$000Cash flow, $0000-1,000-1,50013757002375600337550043754005375300i* value25.41%23.57%


Example 8-4Filtration system to remove pollutant from aircraft air. 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*

Filter 1Filter 1Filter 2YearCash flow, $100Rate, i%PW, $ 100Cash flow, $100PW, $100Incremental cash flow, $100Incremental PW, $1000-1,0000%875-1,5001,000-500125137510%42270046732546237520%121600103225-19337530%-87500-158125-71437540%-237400-35225-115537550%-349300-500-75-15160%-435-616-182i* value25.41%23.57%16.89%


8-*ROR Analysis Multiple AlternativesSix-Step Procedure for Mutually Exclusive Alternatives(1) Order alternatives from smallest to largest initial investment(2) For revenue alts, calculate i* (vs. DN) and eliminate all with i* < MARR; remaining alternative with lowest cost is defender. For cost alternatives, go to step (3)(3) Determine incremental CF between defender and next lowest-cost alternative (known as the challenger). Set up ROR relation(4) Calculate i* on incremental CF between two alternatives from step (3)(5) If i* MARR, eliminate defender and challenger becomes new defender against next alternative on list (6) Repeat steps (3) through (5) until only one alternative remains. Select it. For Independent ProjectsCompare each alternative vs. DN and select all with ROR MARR 2012 by McGraw-Hill All Rights Reserved


8-* Example: ROR for Multiple AlternativesThe five mutually exclusive alternatives shown below are under consideration for improving visitor safety and access to additional areas of a national park. If all alternatives are considered to last indefinitely, determine which should be selected on the basis of a rate of return analysis using an interest rate of 10%.

A B C D E_ First cost, $ millions -20 -40 -35 -90 -70 Annual M&O cost, $ millions -2 -1.5-1.9 -1.1 -1.3 Solution:Rank on the basis of initial cost: A,C,B,E,D; calculate CC valuesC vs. A: 0 = -15 + 0.1/0.1 i* = 6.7% (eliminate C)B vs. A: 0 = -20 + 0.5/0.1 i* = 25% (eliminate A)E vs. B: 0 = -30 + 0.2/0.1 i* = 6.7% (eliminate E)D vs. B: 0 = -50 + 0.4/0.1 i* = 8% (eliminate D)Select alternative B 2012 by McGraw-Hill All Rights Reserved


Solution to Example 8.6 2012 by McGraw-Hill, New York, N.Y All Rights Reserved1-*0 = -190,000 + 19,500( PA , i *,30)

CABDInitial cost-190,000-200,000-275,000-350,000Cash flow+19,500+20,000+35,000+42,000Alternative compared C to DNA to DNB to A D to BIncremental cost-190,000-200,000-75,000-75,000Incremental cash flow+19,000+22,000+13,000+7,000Calculated9.74369.09095.769210,7143i* ,%9.6310.4917.288.55Increment justified?NoYesYes NoAlternative selectedDN ABB


Question 8-34Five mutually exclusive revenue alternatives that have infinite lives are under consideration for increasing productivity in a manufacturing operation. The initial costs and cash flows of each project are shown. If the MARR is 14.9% per year, which alternative should be selected?


AlternativeABCDEInitial cost, -7,000-23,000-9,000-3,000 -16,000

Cash fl ow, $ per year1,0003,5001,4005002,200

Rate of return (vs DN), %14.315.215.616.713.8


8-*Must consider incremental cash flows for mutually exclusive alternativesSummary of Important PointsIncremental cash flow = cash flowB cash flowA where alternative with larger initial investment is Alternative BEliminate B if incremental ROR i* < MARR; otherwise, eliminate AFor multiple mutually exclusive alternatives, compare two at a timeand eliminate alternatives until only one remainsFor independent alternatives, compare each against DN and select all that have ROR MARR 2012 by McGraw-Hill All Rights ReservedBreakeven ROR is i* between project cash flows of two alternatives, or i* between incremental cash flows of two alternatives


*

********

Documents

Ch8 RateofReturnMultipleAlternatives EL