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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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Example: rate of return oninvestment
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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Example: rate of return oninvestment
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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f
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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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