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8/14/2019 Chapter 7x
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CHAPTER 7
Rate of ReturnAnalysis: Single
AlternativeMcGrawHill
ENGINEERING ECONOMY,SixthEdition
by Blank and
Tarquin
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Chapter 7 Learning Objectives
1. Definition of Rate of Return (ROR)
2. ROR using PW and AW
3. Calculations about ROR
4. Multiple RORs
5. Composite ROR
6. ROR of Bonds
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Section 7.1INTERPRETATION OF A RATE OF
RETURN VALUE
The IRR method is one of the popular
time-discounted measures of investment
worth that is related to the NPV approach.
DEFINITION follows
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7.1 INTERPRETATION OF A RATEOF RETURN VALUE
DEFINITION
ROR is either the interest rate paid onthe unpaid balance of a loan, or theinterest rate earned on theunrecovered investment balance of aninvestment such that the final
payment or receipt brings theterminal value to equal 0.
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7.1 INTERPRETATION OF A RATEOF RETURN VALUE
In rate of return problems you seekan unknown interest rate (i*) thatsatisfies the following:
PWi*(+ cash flows) PWi*( - cash
flows) = 0
This means that the interest rate i*, is an unknown parameter and mustbe solved or approximated.
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7.1 Unrecovered InvestmentBalance
ROR is the interest rateearned/charged on the unrecoveredinvestment balance of a loan or
investment project
ROR is not the interest rate earnedon the original loan amount or
investment amount
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7.1 The Loan ScheduleYear BOY Bal Payment Interest
Amount
Prin. Red.
Amount
UnPaidBalance
0 $1,000 0 -- --- $1,000
1 1,000 315.47 100.00 215.47 784.53
2 784.53 315.47 78.45 237.02 547.51
3 547.51 315.47 54.75 260.72 286.79
4 286.79 315.47 28.68 286.79 0
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7.1 Unrecovered InvestmentBalance
For this loan the unpaid loanbalances at the end of each year are:
$1,000
784.53
547.51
286.79
0
0
1
2
3
4
Unpaid loan
balance is now0 at the end of
the life of theloan
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7.1 Unrecovered InvestmentBalance
The URBt =4 is exactly 0 at a 10%
rate$1,000
784.53
547.51
286.79
0
0
1
2
3
4
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7.1 Reconsider the following
Assume you invest $1000 over 4years
The investment generates$315.47/year
Draw the cash flow diagram0 1 2 3
4
P=$-1,000
A =+315.47
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7.1 Investment Problem
What interest rate equates thefuture positive cash flows to theinitial investment?
We can state:
-$1000= 315.47(P/A, i*,4)
Where i* is the unknown interestrate that makes the PW(+) = PW(-)
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7.1 Investment Problem
$1000= 315.47(P/A, i*,4)
Solve the above for the i* rate
(P/A,i*,4) = 1000/315.47 = 3.16987
Given n = 4 what value of i* yields aP/A factor value = 3.16987?
Interest Table search yields i*=10%
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7.1 Investment Problem
i*=10% per year
Just like the loan problem, we cancalculate the unrecovered investmentbalances that are similar to theunpaid loan balances
See the next slide.
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7.1 Unrecovered InvestmentBalances (UIB)
We set up the following table
t C.F(t) Future Value for 1 year UIBt
0 -1,000 ---- -1,000
1 +315.47 -1000(1.10)+315.47= -784.53
2 +315.47 -784.53(1.10)+315.47= -547.51
3 +315.47 -547.51(1.10)+315.47= -286.79
4 +315.47 -286.79(1.10)+315.47= 0
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7.1 UIBs for the Example
See Figure 7.1t C.F(t) UIBt
0 -1,000 -1,000
1 +315.47 -784.53
2 +315.47 -547.51
3 +315.47 -286.79
4 +315.47 0
The unrecoveredinvestmentbalances have beencalculated at a 10%interest rate.
Note, the
investment is fullyrecovered at theend of year 4
The UIB = 0 at a
10% interest rate
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7.1 UIBs for the Example
See Figure 7.1t C.F(t) UIBt
0 -1,000 -1,000
1 +315.47 -784.53
2 +315.47 -547.51
3 +315.47 -286.79
4 +315.47 0
The 10% rate is theonly interest rate
that will cause theUIB at the end ofthe projects life toequal exactly 0
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7.1 UIBs for the Example
See Figure 7.1t C.F(t) UIBt
0 -1,000 -1,000
1 +315.47 -784.53
2 +315.47 -547.51
3 +315.47 -286.79
4 +315.47 0
Note, all of UIBsare negative at the
10% rate.
This means thatthe investment isunrecovered
throughout the life.
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7.1 Pure Investment
The basic definition of ROR is theinterest rate that will cause theinvestment balance at the end of the
project to exactly equal 0
If there is only one such interestrate that will cause this, the
investment is said to be a PUREinvestment or, Conventionalinvestment
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7.1 UIBs for the Example
Unrecovered means that the
investment balance for a given year is
negative.
If a projects UIBs are all negative
(under-recovered) then that
investment will possess one unique
interest rate to cause the UIBt=n to
equal 0
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7.1 ROR - Explained
ROR is the interest rate earned onthe unrecovered investment balancesthroughout the life of the investment.
ROR is not the interest rate earnedon the original investment
ROR (i*) rate will also cause the
NPV(i*) of the cash flow to = 0.
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Section 7.2ROR using Present Worth
PW definition of ROR
PW(-CFs) = PW(+CFs)
PW(-CFs) - PW(+CFs) = 0
AW definition of ROR
AW(-CFs) = AW(+CFs)
AW(-CFs) - AW(+CFs) = 0
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7.2 ROR using Present Worth
Finding the ROR for most cash flows is a
trial and error effort.
The interest rate, i*, is the unknown
Solution is generally an approximation
effort
May require numerical analysis
approaches
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7.2 ROR using Present Worth
See Figure 7.2
0 1 2 3 4
5
-
$1,000
+$500
+$1,500
Assume you invest $1,000 at t = 0: Receive$500 @ t=3 and $1500 at t = 5. What is theROR of this project?
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7.2 ROR using Present Worth
Write a present worth expression, set equal to 0
and solve for the interest rate that satisfies the
formulation.
1000 = 500(P/F, i*,3) +1500(P/F, i*,5)
Can you solve this directly for the value ofi*?
NO!
Resort to trial and error approaches
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7.2 ROR using Present Worth
1000 = 500(P/F, i*,3) +1500(P/F, i*,5)
Guess at a rate and try it
Adjust accordingly
Bracket
Interpolate
i* approximately 16.9% per year onthe unrecovered investmentbalances.
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7.2 Trial and Error Approach
Iterative procedures require an initial
guess for i*
One makes an educated first guess and
calculates the resultant PV at the guess
rate.
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7.2 Trial and Error Approach
If the NPV is not = 0 then another i* value is
evaluated. Until NPV close to 0
The objective is to obtain a negative PV foran i* guess value then.
Adjust the i* value to obtain a positive PV
given the adjusted i* value
Then interpolate between the two i* values
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7.2 Example 7.3 In Excel
Excel Setup for ROR
MARR= 10.00%
Life 10 Year Cash Flow
0 -$500,000
1 $10,000
2 $10,000
3 $10,0004 $10,000
5 $10,000
6 $10,000
7 $10,000
8 $10,000
9 $10,000
10 $710,000$300,000
ROR-Guess 0%
ROR 5.16%
NPV= -$168,674.03
=IRR(D6:D16,D19)
D19 = Guess Value
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7.2 Example 7.2 InvestmentBalances
Investment balances
at the i* rate
The time t = 10balance = 0 at i*
As it should!
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CHAPTER VII
McGra
wHill
ENGINEERING ECONOMYFifth Edition
Blank andTarquin
Section 7.3Cautions when using the
ROR Method
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Section 7.3Cautions when using the
ROR Method
Important Cautions toremember when using the ROR
method
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7.3 Cautions when using theROR Method No.1
Many real-world cash flows may possess
multiple i* values
More than one i* value that will satisfy thedefinitions of ROR
If multiple i*s exist, which one, if any, is the
correct i*???
au ons w en us ng e
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. au ons w en us ng eROR Method. 2. ReinvestmentAssumptions
Reinvestment assumption for the ROR
method is not the same as the reinvestment
assumption for PW and AW
PW and AW assume reinvestment at the
MARR rate
ROR assumes reinvestment at the i* rate
Can get conflicting rankings with ROR vs.
PV and AW
. au ons w en us ng e
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. au ons w en us ng eROR Method: 3. ComputationalDifficulties
ROR method is computationally more
difficult than PW/AW
Can become a numerical analysis problem
and the result is an approximation
Conceptually more difficult to understand
7 3 Cautions when using the
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7.3 Cautions when using theROR Method: 4. Special Procedurefor Multiple Alternatives
For analysis of two or more alternatives
using ROR one must resort to a different
analysis approach as opposed to the PW/AW
methods
For ROR analysis of multiple alternatives
must apply an incremental analysisapproach
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7.3 Cautions when using theROR Method: ROR is more difficult!
ROR is computationally more difficult
But is a popular method with financial
managersROR is used internally by a substantial
number of firms
Suggest using PW/AW methods where
possible
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7.3 Valid Ranges for usable i*rates
Mathematically, i* rates must be:
*100% i < +
If an i*
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Section 7.4Multiple Rates of Return
A class of ROR problems exist that will
possess multiple i* values
Capability to predict the potential for
multiple i* values
Two tests can be applied
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7.4 Tests for Multiple i* values
1. Cash Flow Rule of Signs
2. Cumulative Cash Flow Rule of Signs
test
Example follows
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7.4 Cash Flow Rule of Signs Test
The total number of real values i*s is
always less than or equal to the number of
sign changes in the original cash flow series.
Follows from the analysis of a n-th degree
polynomial
A 0 value does not count as a sign change
Example follows
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7.4 C.F. Rule of Signs example
Consider Example 7.4
Year Cash Flow
0 $2,0001 -$5002 -$8,1003 $6,800
+--
+
Result: 2 sign changes in the CashFlow
l C l f Si
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7.4 Results: CF Rule of SignsTest
Two sign changes in this example
Means we can have a maximum of 2
real potential i* values for this problem
Beware: This test is fairly weak and
the second test must also be performed
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7.4 Norstroms Test
Norstroms Test1972 works with the
accumulated cash flow
For the example problem theaccumulated cash flow (ACF) is.
7 4 A l t d CF Si T t
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7.4 Accumulated CF Sign Test(ACF)
For the problem form the accumulated cash
flow from the original cash flow.
Year Cash Flow Accum.
C.F0 $2,000 $2,0001 -$500 $1,500
2 -$8,100 -$6,600
3 $6,800 $200
Count sign changeshere
7 4 A l t d CF i
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7.4 Accumulated CF signs -Example
Year Cash Flow Accum.C.F
0 $2,000 $2,0001 -$500 $1,500
2 -$8,100 -$6,6003 $6,800 $200
+
+-+
2 sign changes in the ACF.
A.C.F
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7.4 ACF Sign Test States:
A sufficient but not necessary condition
for a single positive i* value is:
The ACF value at year N is > 0
There is exactly one sign change in
the ACF
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7.4 ACF Test - continued
If the value of the ACF for year N is
0 then an i* of 0% exists
If the value of ACF for year N is > 0,this suggests an i* > 0
If ACF for year N is < 0 there mayexist one or more negative i* values
but not always
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7.4 ACF Test - continued
Alternatively:
If the ACF in year 0 < 0
And.
One sign change in the ACF series then
Have a unique i* value!
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7.4 ACF Test - continued
If the number of sign changes in the
ACF is 2 or greater this strongly suggest
that multiple rates of return exist.
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7.4 Example 7.4 ACF Sign Test
Year Cash Flow Accum.C.F
0 $2,000 $2,0001 -$500 $1,500
2 -$8,100 -$6,6003 $6,800 $200
2 Sign Changeshere
Strong evidence that we have multiple i*
values
ACF(t=3) = $200 > 0 suggests positive i* (s)
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7.4 Excel Analysis of Ex. 7.4
We find two i* values:
{7.47%,41.35%}
Year Cash Flow
0 $2,000
1 -$500
2 -$8,100
3 $6,800Sum $200
ROR-Guess 0%
ROR 7.47%
NPV= $200.00
ROR-Guess 30%
ROR 41.35%
First i* using aguess value of 0%
Second i* valueusing guess value
of 30%
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7.4 Investment (Project) Balance
Examine the Investment or Project
balances at each i* rateInv Bal
i-rate 7.47%
Year Project Balances@ i* Rate
0 $2,000.00
1 $1,649.362 -$6,327.473 $0.00
Inv Bal
i-rate 41.35%
Year Project Balances@ i* Rate
0 $2,000.00
1 $2,327.042 -$4,810.693 $0.00
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7 4 Investment Balances at both
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7.4 Investment Balances at bothi*s
Important Observations
The IBs for the terminal year (3) bothequal 0
Means that the two i* values are validRORs for this problem
Note: The IB amounts are not all the
same for the two i* values.IB amounts are a function of theinterest rate used to calculate theinvestment balances.
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7.4 PV Plot of 7.4
-150.000
-100.000
-50.000
0.000
50.000
100.000
150.000
200.000
250.000
0.00 0.20 0.40 0.60InterestRate
PV-
$$
i* = 7.47%
i*=41.35%
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7.4PV Plot - continued
-150.000
-100.000
-50.000
0.000
50.000
100.000
150.000
200.000
250.000
0.00 0.20 0.40 0.60
PV < 0If the
MARR isbetween
the two i*values thisinvestment would berejected!
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7.4 Comments on ROR
Multiple i* values lead tointerpretation problems
If multiple i*s which one, if any is
the correct one to use in ananalysis?
Serves to illustrate the
computational difficulties associatedwith ROR analysis
Section 7.5 provides an alternativeROR approach Composite ROR (C-ROR
Section 7 5
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Section 7.5Composite ROR Approach
Consider the following investment
-$10,000
0 1 2 3 45
+$8,000
+$9,000
Determine the ROR as.
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7 5 Composite ROR Approach:
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7.5 Composite ROR Approach:IBs
The Investment Balances ar i* are: Year Cash Flow Project Balances
@ i* Rate
0 -$10,000 -$10,000.00
1 $0 -$11,681.592 $8,000 -$5,645.95
3 $0 -$6,595.36
4 $0 -$7,704.435 $9,000 $0.00
Sum $7,000
ROR-Guess 0%
ROR 16.82%
All IBs are
negative for t= 0 4: IB(5) =0
Conventional
(pure)investment
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7.5 Composite ROR Approach
i* = 16.82%
-$10,000
0 1 2 3 45
+$8,000
+$9,000
Question: Is it reasonable toassume that the +8,000 can beinvested forward at 16.82%?
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7.5 Composite ROR Approach
Remember .
ROR assumes reinvestment at
the calculated i* rate
What if it is not practical for the+$8,000 to be reinvested forward
one year at 16.82%?
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7.5 Reinvestment Rates
Most firms can reinvest surplusfunds at some conservative marketrate of interest in effect at the time
the surplus funds become available.
Often, the current market rate is lessthan a calculated ROR value
What then is the firm to do with the+$8,000 when it comes in to the firm?
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7.5 Reinvesting
Surplus funds must be put to gooduse by the firm
These funds belong to the owners not to the firm!
Owners expect such funds to be putto work for benefit of future wealth ofthe owners
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7.5 Composite ROR Approach
Consider the followingrepresentation.
TheFirm
Project
Invested Funds
Returns
back
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7.5 Composite ROR Approach
Or, put in another context.
TheFirm
Project
Project borrows from the
firm
Project Lends back to the
firm
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7.5 Composite ROR Approach
Or, put in another context.
TheFirm
Project
Project borrows from thefirm
At the i* rate (16.82%)
Project Lends back to the
firm
But can the firm reinvestthese funds at 16.82%?Probably not!
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7.5 Composite ROR Approach
So, we may have to consider areinvestment rate that is closer to thecurrent market rate for reinvestment
of the $8,000 for the next timeperiod(s)
Assume a reasonable market rate is
say, 8% per year.
Call this rate an external rate - c
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7.5 Composite ROR Approach
Finding i is a much more involvedprocess
Prior to digital computers, only verysmall (N
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7.5 Investment Balances are:
IB(7.468%)
Inv Bal
i-rate 7.468%
Year Project Balances@ i* Rate
0 $2,000.00
1 $1,649.36
2 -$6,327.473 $0.00
Inv Bal
i-rate 41.352%
Year Project Balances@ i* Rate
0 $2,000.00
1 $2,327.04
2 -$4,810.693 $0.00
IB(41.352%)
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7.5 Project Balances @ 41.35%
Project Balances
-$6,000
-$5,000
-$4,000
-$3,000
-$2,000
-$1,000
$0$1,000
$2,000
$3,000
0 1 2 3
Years
PB's
-$
Over-recovered
Under-recovered
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Project Balances
-$8,000
-$6,000
-$4,000
-$2,000
$0
$2,000
$4,000
0 1 2 3
Years
PB's
-$
7.5 Project Balances @ 7.468%%
Over-recovered
Under-recovered
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7.5 Assume you Reinvest at 20%
c = 20%
Positive IBs are reinvested at 20% -
Not at the computed i* rate
Now, what is the modified ROR i?
Must perform a recursive analysis
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7.5 Recursive IBs are.
Let Ibj = Fj
F0 = +2000 (> 0; invest at 20%)
F1 = 2000(1.20) 500 = +1900
+1900 > 0; invest at 20%
F2 = 1900(1.20) 8100 = -5820
-5820 < 0; invest at i rate
Under-recovered
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7.5 Recursive IBs are.
F2 = 1900(1.20) 8100 = -5820
-5820 < 0; invest at i rate
F3 =-5820(1+i) + 6800
Since N = 3, F3
= 0
Solve; -5820(1+i) + 6800 = 0
Repeated from the
previous slide forclarity
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7.5 Single conditional ROR value
-5820(1+i) + 6800 = 0
i = [6800/5820] 1
i = 16.84% given c = 20%
If MARR = 20% and i = 16.84% thisinvestment would be rejected
Reduced the problem to a singleconditional ROR value.
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7.5 What ifc = i*1 (7.47%)?
Here, we examine what happens IFwe assume the reinvestment rate, c,equals one of the computed i* values.
This approach assumes that thereinvestment rate, c, is one of the i*rates
Recursive calculations follow.
7.5 Recursive IBs are.
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(@7.47%)
Let IBj = Fj
F0 = +2000 (> 0; invest at 7.47%)
F1 = 2000(1.0747) 500 = +1649.40
>0: Over-recovered balance
F2 = +1649.40(1.074) 8100 =
-6327.39
Now, under-recovered balance
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7.5 Recursive IBs are.
F2 = +1649.40(1.074) 8100 =
-6327.39
F3 =-6327(1+i) + 6800
Since N = 3, F3must = 0 (terminal IB
condition)
Solve; -6327(1+i) + 6800 = 0
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7.5 Single conditional ROR value
Must solve:
-6327.39(1+i) + 6800 = 0
(1+i) = 6800/6327.39
i = 0.0747 = 7.47%
Given c = one of the i*s, the i willequal the i* used as the
reinvestment rate!
Section 7 6
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Section 7.6Rate of Return on a Bond
Investment
Review Chapter 5 Section on Bonds
Bond problems represent a classicalROR type problem
Bond problems represent aconventional investment with a
unique ROR
Example Follows:
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7.6 Example 7.8
Purchase Price: $800/bond
4%, $1000 bond
Life = 20 years
Interest paid semiannually
Question: If you pay $800 per bond
what is the ROR (yield) on thisinvestment?
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7.6 Ex. 7.8: Cash Flow Diagram
. ..
0 1 2 3 439 40
$800
F40 =
$1000
$I/6 mos = $1000(0.04/2) = +$20.00 (every 6months)
A = +$20/6 months
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7.6 Ex. 7-8: Closed Form Setup
Setup is:
0 = -$800 +20(P/A,i*,40 + $1000(P/F,i*,40)
Solve for i*
Manual or computer solution yields:i*=2.87%/6 months
Nominal ROR/year = (2.87%)(2) = 5.74%/yr,
C.S.A.
Effective ROR/year = (1.0287)2 1 =5.82%/yr
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Chapter 7 Summary
ROR is a popular analysis method
Understood in part by practitioners
Presents computational difficulties
No ROR may exist
Multiple RORs may exist
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Chapter 7 Summary cont.
For cash flows where multiple i*values exist, one must decide to:
Go with one of the i* rates or,
Calculate the composite rate, i
If an exact ROR is not required, it is
suggested to go with PW or AW
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Chapter 7 Summary cont.
For cash flows where multiple i*values exist, one must decide to:
Go with one of the i* rates or,
Calculate the composite rate, i
If an exact ROR is not required, it is
suggested to go with PW or AW
h
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Chapter 7 Summary cont.
To have a possible i* value with agiven cash flow, there must be atleast one negative CF amount in the
series
IROR can be demanding and is bestaccomplished using a computer
algorithmIROR is often misunderstood bypractitioners but is a popular methodof analysis
Ch 7 S
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Chapter 7 Summary cont.
In real world applications one mayencounter a cash flow that possess nousable i* value!
These types of problems are difficultto explain to uninformed decisionmakers.
ENGINEERING ECONOMYSixthEditi
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McGrawHill
EditionBlank andTarquin
END OF SLIDE SET
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