ch rate of return analysis

8/12/2019 Ch Rate of Return Analysis

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Engineering Economic Analysis

RATE OF RETURN ANALYSIS

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Three Major Methods ofEconomic Analysis

PW - Present Worth AW - Annual Worth IRR - Internal Rate of Return

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If PW = A(P/A,i,n)Then (P/A,i,n) = PW/ASolve for (P/A,i,n) and look up interest inCompound Interest Tables


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Calculating Rate ofReturn

The IRR is the interest rate at which the presentworth and equivalent uniform annual worthare equal to 0.

o PW Benefit - PW Cost = 0o PW Benefit/PW Cost = 1o NPW = 0o EUAB - EUAC = 0o PW Benefit = PW Cost

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Example: rate of return oninvestment

An $8200 investment returned $2000 per year over afive-year useful life. What was the rate of return?

PW of benefits/PW of costs = 1

2000(P/A), I, 5)/ 8200=1(P/A,I,5)=4.1i= 7%

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An investment resulted in thefollowing cash flow. Computethe rate of return.

EUAB-EUAC=0100+75(A/G,I,4)-700(A/P,I,4)=0i=5% EUAB-EUAC=208-197=11

i=8% EUAB-EUAC=205-211=-6i=7% EUAB-EUAC=206-206=0

YearCash

Flow

0 -$700

1 100

2 175

3 250

4 325

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An investmentresulted in thefollowing cashflow. Computethe rate ofreturn

B YearCash

Flow2 0 -$700

3 1 1004 2 1755 3 2506 4 325

IRR(B2:B6 7%

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Example A new corporate bond was initially sold by a

stockbroker to an investor for $1000. Theissuing corporation promised to pay the bondholder $40 interest on the $1000 face value ofthe bond every six month, and to repay the$1000 at the end of ten years. After one yearthe bond was sold by the original buyer for$950.

A) what rate of return did the original buyerreceived on his investment

B) What rate of return the new buyer expectto receive if e keeps the bond for its remainingnine-year life?

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Solution Try i=1.5% NPW= -1000+40(P/A,I,2)+950(P/F,I,2) NPW= -1000+ 40(1.956) + 950(0.9707= 0.41 So interest per six month is 1.5% means nominal

interest rate is 3% per year and effective annualinterest is (1+0.015)^2 -1 = 3.02%

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Rate of Return Analysis MARR: minimum attractive rate of return To evaluate an optional investment compare ROR

with MARR. If ROR>MARR then choose to do theinvestment. If ROR < MARR then do not accept theinvestment

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Rate of Return Analysis If you have more than one option then all feasible

options should have ROR>MARR. To choose thebest compute the incremental rate of return.

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Rate of Return Analysis You are given he choice o

selecting one of twomutually exclusivealternatives. Thealternatives are as follows:any money not investedhere may be investedelsewhere at the MARR of6%. If you can only chooseone alternative one time,which one would youselect?

Year Alt. 1 Alt. 20 -$10 -$20

1 15 28

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Rate of Return Analysis You have $20 in your wallet and two alternative ways of

lending Bill some money. A) lend Bill $10 with his promise of 50% return. That is,

he will pay you back $15 at the same agreed time. B) Lend bill $20 with his promise of a 40% return. He will

pay you back $28 at the same agreed time. You can select whether to lend bill $10 or $20. This is a

one-time situation and any money not lent Bill willremain in your wallet. Which alternative do you choose?

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Example 8.4You have available $70,000 to invest and have been

presented with 5 equal-lived, mutually exclusiveinvestment alternatives with cash flows as depictedbelow. Currently, you are earning 18% on yourinvestment of the $70,000. Hence, you will not choose to

invest in either of the alternatives if it does not provide areturn on investment greater than 18%.

Using the internal rate of return method, which (if either)would you choose? What is its rate of return?


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Data for Example 8.4

With an 18% MARR, which investment wouldyou choose?

Investment 1 2 3 4 5Initial Investment $15,000.00 $25,000.00 $40,000.00 $50,000.00 $70,000.00Annual Return $3,750.00 $5,000.00 $9,250.00 $11,250.00 $14,250.00Salvage Value $15,000.00 $25,000.00 $40,000.00 $50,000.00 $70,000.00Internal Rate of Return 25.00% 20.00% 23.13% 22.50% 20.36%


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Solution to Example 8.4Investment 1 2 - 1 3 - 1 4 - 3 5 - 4 Investment

, . , . , . , . , .

Annual Return, . , . , . , . , .

Salvage Value

, . , . , . , . , .

IR R 25.00% 12.50% 22.00% 20.00% 15.00%> M A R R ? Yes No Yes Yes NoDefender 1 1 3 4 4


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Present Worths with 10-Year Planning Horizon

Investment 1 2 3 4 5Initial Investment $15,000.00 $25,000.00 $40,000.00 $50,000.00 $70,000.00Annual Return $3,750.00 $5,000.00 $9,250.00 $11,250.00 $14,250.00

Salvage Value $15,000.00 $25,000.00 $40,000.00 $50,000.00 $70,000.00Present Worth $4,718.79 $2,247.04 $9,212.88 $10,111.69 $7,415.24


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Rate of Return Analysis A firm is considering which of two devices to

install to reduce costs in a particular situation. Both devices cost $1000, have useful lives of

five years and no salvage value. Device Acan be expected to result in $300 savingannually. Device B will provide cost savings of$400 the first year but will decline $50annually, making the second-year savings$350, the third year saving $300, and so forth.For a 7% MARR, which device should the firmpurchase?

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Solution

Year Device A Device B A-B

0 -1000 -1000 0

1 300 400 -100

2 300 350 -50

3 300 300 0

4 300 250 50

5 300 200 100

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Example 8.7Three mutually exclusive investment alternatives

are being considered; the cash flow profiles areshown below. Based on a 15% MARR, whichshould be chosen?

EOY CF(1) CF(2) CF(3)0 -$100,000 -$125,000 -$150,0001 $20,000 -$25,000 -$35,0002 $20,000 $75,000 $75,0003 $20,000 $70,000 $75,0004 $20,000 $60,000 $75,0005 $120,000 $55,000 $95,000


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Example 8.7 (Continued)

EOY CF(1) CF(2) CF(3) CF(2-1) CF(3-2)

0 -$100,000 -$125,000 -$150,000 -$25,000 -$25,0001 $20,000 -$25,000 -$35,000 -$45,000 -$10,0002 $20,000 $75,000 $75,000 $55,000 $03 $20,000 $70,000 $75,000 $50,000 $5,0004 $20,000 $60,000 $75,000 $40,000 $15,000

5 $120,000 $55,000 $95,000 -$65,000 $40,000IR R = 20.00% 19.39% 18.01% 16.41% 13.41%

PW 1(15%) =PV(0.15,5,-20000,-100000)-100000 = $16,760.78

PW 2(15%) =NPV(0.15,-25,75,70,60,55)*1000-125000 = $17,647.70

PW 3(15%) =NPV(0.15,-35,75,75,75,95)*1000-150000 = $15,702.99

Recommend Alternative 2


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-$2,000

-$1,500

-$1,000

-$500

$0

$500

$1,000

$1,500

$2,000

- 7 0 %

- 6 0 %

- 5 0 %

- 4 0 %

- 3 0 %

- 2 0 %

- 1 0 % 0 % 1 0

% 2 0 %

P W

( x $ 1 0 , 0 0 0 )

MARR

Incremental IRR Comparison of Alternatives

CF(2-1) CF(3-2)


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Incremental IRR Comparison of Alternatives

-$2

-$1

$0

$1

$2

$3

$4

$5

$6

$7

$8

- 1 0 % - 5

% 0 % 5 % 1 0 %

1 5 %

2 0 %

2 5 %

MARR

P W ( x $ 1 0 , 0 0 0 )

CF(2-1) CF(3-2)


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QuizTrue of False: If PW(A) > PW(B) > $0, then IRR(A) >

IRR(B) > MARR


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QuizTrue of False: If PW(A) > PW(B) > $0, then IRR(A) >

IRR(B) > MARR

Answer: FalseIRR is not a rank order method; it is an incrementalmethod. In Example 8.7, PW(2) > PW(1) > PW(3)and IRR(1) > IRR(2) > IRR(3). Knowing the rankorder of PW tells us nothing about the rank orderof IRR.


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Rate of Return Analysis Two machines are being considered

for purchase. If the MARR is 10%,which machine should be bought?

Machine X Machine YInitial Cost $200 $700

Uniform annual benefit 95 120End-of-useful-life salvage value 50 150Useful life in years 6 12

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Rate of Return (ROR)Analysis

Most frequently used measure of merit in industry More accurately called Internal Rate of Return (IRR)

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Analysis Period Just as in PW and AW analysis, the analysis period

must be considered:o Useful life of the alternative equals the analysis periodo Alternatives have useful lives different from the analysis periodo The analysis period is infinite, n =

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