Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
Подождите немного. Документ загружается.
There are usually no costs for the first
3 years, but thereafter maintenance is re-
quired for restriping, weed control,
light
replacement, shoulder repairs, etc.
For
one
section
of
a particular highway, these costs
are projected to be
$6000 in year 3, $7000
in
year 4, and amounts increasing by $1000
per year through the highway's expected
30-year life. Assuming it
is
replaced with
a similar roadway, what is its perpetual
equivalent annual worth (in
years]
through
infinity) at an interest rate
of8
% per year?
6.21 A phi lanthropist working to set up a per-
manent endowment wants to deposit
money each year, starting
now
and making
10
more (i.e.,
11)
deposits, so that money
will be available for research related to
planetary colonization. If the size
of
the
first deposit is $] million and each suc-
ceeding one is $]
00,000 larger than the
previous one, how much will be available
forever beginning in year
11
, if the fund
earns interest at a rate
of
10
% per year?
FE
REVIEW PROBLEMS
Note:
The
sign convention on the
FE
exam may
be opposite
of
that used here.
That
is, on the
FE exam, costs may be positive and receipts
negative.
6.24
For
the mutually exclusive alternatives
shown below, determine which one(s)
should be selected.
Alternative
A
B
C
D
(a) Only A
(b) Only D
(c) Only A and B
(d)
Only C and D
Annual
Worth,
$/yr
- 25,000
-
12
,000
10,000
15
,000
FE
REVIEW PROBLEMS
235
6.22 For the cash flow sequence shown below
(in thousands
of
dollars), determine the
amount
of
money that can be withdrawn
annually for an infinite period
of
time, if
the first withdrawal is to be made
in
year 10
and the interest rate is 12% per year.
Year 0 1 2 3 4 5 6
Deposit
amount,
$ 100 90 80 70 60 50 40
6.23 A company that manufactures magnetic
membrane switches is investigating three
production options that have the estimated
cash flows below.
(a) Determine which op-
tion
is
preferable at an interest rate
of
15
%
per year.
(b)
If
the options are independent,
determine which are economically accept-
able. (All dollar values are in millions.)
In-house
License
Contract
First cost, $
-30
- 2
0
Annual cost, $/year
-5
- 0.2
-2
Annual income, $/year
14
1.5
2
.5
Salvage value, $ 7
Life, years
10
5
6.25
The
annual worth (in years 1 through in-
finity)
of
$50,000 now,
$10
,000 per year
in
years 1 through 15, and $20,000 per
year
in
years 16 through infinity at 10%
per year is closest to
(a) Less than $16,900
(b)
$16,958
(c)
$l7,394
(d)
$19,573
6.26 An alumnus
of
West Virginia University
wishes to start an endowment that will pro-
vide scholarship money
of
$40,000 per
year beginning in year 5 and continuing in-
definitely.The donor plans to give money
now
and for each
of
the next 2 years.
If
the
size
of
each donation is exactly the same,
the amount that must be donated each year
at
i = 8% per year is closest to
236
CHAPTER
6 Annual Worth Analysis
(a) $
190
,8
20
(b) $122,280
(c)
$127,460
(d) $132,040
6.27.
How
much
must
you deposit
in
your
retire-
ment
account
each
year
for
10 years starting
now (i.e., years 0 through 9)
if
you
want
to
be
able
to withdraw
$50,000
per
year
for-
ever
beginning
30
years
from
now?
Assume
the
account
earns
interest at 10%
per
year.
(a) $4239
(b) $4662
(c) $4974
(d)
$5471
6.28
Assume
that a grateful engineering econ-
omy
graduate
starts an
endowment
at
UTEP
by donating
$100,000
now.
The
condit
i
ons
of
the donation are that scholarshi ps total ing
$10,000
per
year
are to be given to engineer-
ing
eco
nomy
students beginning now and
continuing through
year
5. After that (i.e.,
year
6), scholarships are to be given
in
an
amount
equal to the interest
thatis
generated
on the investment.
If
the investment earns an
effective rate
of
10
%
per
year,
compounded
continuously, how much money will be
available for scholarships from
year
6
on?
(a) $7380
(b) $8389
(c) $
10
,
000
(d)
$11,611
Problem
s 6.29 through 6.31
are
based
on
the fol-
lowing cash flows and an interest rate
of
10
%
per
year,
compounded
semiannually.
CASE STUDY
Alternative
X
Alternative
Y
Fi
rst cost, $
Annual cost, $/year
Salvage value,
$
Life, years
- 200,000
- 60,000
20,000
5
-8
00
,000
-
10
,000
150,000
6.
29
In
comparing
the alternati ves by the an-
nual worth method, the
annual
worth
of
alternative
X is
represented
by
(a)
-2
00
,
000(0.1025
) -
60,000
+
20,000(0.1025)
(b)
-200
,000(Aj P,
10
%,5) -
60
,
000
+
20,000(AI F, 1 0%,5)
(c)
-2
00
,
000(Aj
P,5
%,
10) -
60,000
+
20,000(Aj F,5%, I 0)
(d)
-200,000(Aj
P, 10.25%,
5)
-
60
,
000
+
20
,000(A j F,1O.25%,5)
6.
30
The
annual
worth
of
perpetual service
for
alternative X is
repre
se
nted by
(a)
-200,000(0.1025)
-
60,000
+
20,000(0.1025)
(b)
-2
00,000(A j P,
1O
%,5) -
60,000
+
20,
000(Aj
F,
10
%,5)
(c)
-2
00,000(0.10)
-
60,000
+
20
,
000(0
.10)
(d) - 200,000(Aj
P,
10.25%,5)
-
60
,
000
+
20
,000(A j F,
10
.25%,
5)
6.31
The
annual
worth
of
alternative
Y is
closest
to
(a)
$-50
,000
(b)
$-76
,
625
(c)
$-90,000
(d) $- 92,000
THE CHANGING SCENE OF AN ANNUAL WORTH ANALYSIS
Harry,
owner
of
an automobile battery distributorship
in
Atlanta, Georgia, performed an economic analysis
3 years ago when he decided to place surge protectors
in-line for all his major piec
es
of
testing equipment.
The
estimates used and the annual worth analysis at
MAAR
=
15
% are summarized here. Two different
manufacturers' protectors were compared.
CASE STUDY
237
~
.,
~
105% -1
11)
ll
--
<l>!
-
A.
-
Ao
A
2
3
MARR
= 15%
4
5
PowrUp Lloyd's
6
Imestment Annual Repair
Investment
Annual Repair AWPowrUp
7
Year
and
salva e maint malnt. savin s
$
17,558
8 AW values
$
(6,642) $
$
(300) $ 35,000
9
AW Lloyd's
10 0
$
(26,000) $
$ $
(36,000) $ $ $
27
,
675
J
11
$
$
(800) $ 25,000
$ $
(300) $ 35,000
12
2 $ $ (800) $ 25,000 $ $ (300) $ 35,000
J~
3
$ $ (800 $
25
,000
$ $ 300 $ 35,000
14 4 $ $ (800) $
25
,000
$ $
(300) $
35
,0
00
15
5
$ $ (800) $ 25,000 $ $ (300) $ 35,000
16 6
$
2,000 $ (800) $ 25,000
$ $
(300) $ 35,000
17 7 $ $ (300) $ 35,000
18
B
$ $ (300) $ 35,000
19 9
$ $
(300) $
35,000
20
10
$ 3,000 $ (300) $ 35,000
21
,
..
.J
J
:&
I
~ ~
••
1
Sheet?
Sh~et8
5he~I'
1
Ready
NUM
Figure
6-8
AW analysis of two sur
ge
protector prop
osa
l
s,
case stud
y,
Chapter 6.
Cost and
in
sta
ll
a
ti
on
Annual maintenance cost
Salvage va
lu
e
Equipment repa
ir
sav
in
gs
Useful
li
fe, years
PowrUp
$-
26,000
- 800
2000
25,000
6
Lloyd's
$-36,
000
-3
00
3000
35,000
10
The spreadsheet
in
Figure
6-8
is
th
e one Harry used to
make
th
e decisio
n.
Lloyd's was the clear choice due
to
its s
ub
sta
nti
a
lly
large
AW
va
lu
e.
The Lloyd's protec-
tors were
in
sta
ll
e
d.
During a quick review
thi
s l
as
t year (year 3
of
op-
eration), it was obvious
th
e maintenance costs a
nd
re-
pa
ir
sav
in
gs have n
ot
fo
ll
owed (a
nd
w
ill
not fo
ll
ow)
th
e estimates made 3 years ago. ]n fact,
th
e mainte-
nance contract cost (w
hi
ch includes
qu
arte
rl
y in-
sp
ect
ion) is go
in
g from $300 to $1200 per
yea
r next
year a
nd
w
ill
th
en
in
crease
10
% per year for
th
e next
10 years. Also,
th
e repair savings for the last 3 years
were
$35,000, $32,000, and $28,000, as best
as
Harry
can determine. He believes savings will decrease by
$2000 per year hereafter. Finally, these 3-year-old pro-
tectors are worth nothing on the market now, so
th
e
salvage
in
7 years is zero, not $3000.
Case Study Exercises
1.
Plot a graph of the newly estimated maintenance
cos
ts
and repa
ir
savings projections, assuming
the protectors last for 7 more years.
2.
With these new estimates, what is the recalcu-
lated
AW
for
th
e Lloyd's protectors? Use the o
ld
first cost a
nd
maintenance cost estimates for
th
e
first 3 years.
If
th
ese estimates had been made
3 years ago, would Lloy
d'
s sti
ll
have been
th
e
economic choice?
3.
How has
th
e capital recovery amount changed
fo
r
the Lloyd's protectors with these new estimates?
7
w
I-
«
I
u
Rate
of
Return Analysis:
Single
Alternative
Although
the
most
commonly
quoted
measure
of
economic
worth
for
a
proj-
ect
or
alternative
is
the
rate
of
return (ROR), its
meaning
is
easily misinter-
preted,
and
the
methods
to
determine
ROR
are
often
applied
incorrectly.
In
this chapter,
the
procedures
to
correctly
interpret
and
calculate
the
ROR
of
a
cash
flow
series are
explained,
based
on a PW
or
AW
equation.
The
ROR
is
known
by
several
other
names: internal rate
of
return
(lRR),
return
on
invest-
ment
(ROI), and
profitability
index
(PI),
to
name
three.
The
determination
of
ROR
is
accomplished
using a manual
trial-and-error
process or,
more
rapidly,
using
spreadsheet
functions.
In
some
cases,
more
than
one
ROR
value may satisfy
the
PW
or
AW
equa-
tion
. This
chapter
describes
how
to
recognize this
possibility
and an ap-
proach
to
find
the
multiple
values. Alternatively,
one
unique
ROR
value can
be
obtained
by
using a
reinvestment
rate
that
is
established
independently
of
the
project
cash flows.
Only
one
alternative
is
considered
here;
the
next
chapter
applies
these
same
principles
to
multiple
alternatives. Finally,
the
rate
of
return
for
a
bond
investment
is
discussed here.
The
case
study
focuses
on
a cash
flow
series
that
has
multiple
rates
of
return.
LEARNING OBJECTIVES
Purpose: Understand
the
meaning
of
rate
of
return (ROR) and
perform
ROR calculations
for
one alternative.
Definition
of
ROR
ROR using
PW
and
AW
Cautions about
ROR
Multiple
RORs
Composite
ROR
ROR
of
bonds
This ch
ap
ter
will he
lp
you:
1.
State
the
meaning
of
rate
of
re
turn
.
2.
Calculate
the
rate
of
return using a
present
worth
or
annual
worth
equation.
3.
Understand
the
diff
iculties
of
us
i
ng
the
ROR
method,
relative
to
PW
and
AW
methods
.
4. De
termin
e
the
maximum
number
of
possible
ROR
va
lu
es
and
their
va
lu
es
for
a specific cash fl
ow
series.
5.
Calcul
ate
the
composite
rate
of
return using a
stated
reinvestment
rate.
6.
Calculate
the
nominal
and
effective
interest rate
for
a
bond
investment.
240
EXAMPLE
7.1
CHAPTER 7 Rate
of
Return Analysis: Single Alternative
7.1 INTERPRETATION
OF
A
RATE
OF
RETURN VALUE
From
the perspective
of
someone
who
has borrowed money, the interest rate
is
applied to the unpaid balance so that the total loan
amount
and interest are paid
in
full exactly with the last loan payment.
From
the perspective
of
the lender,
there is an
unrecovered balance at each time period.
The
interest rate is the re-
turn on this unrecovered balance so that
the
total
amount
lent and interest are
recovered exactly with the last receipt.
Rate
of
return describes both
of
these
perspecti ves.
Rate
of
return
(ROR) is the
rate
paid on
the
unpaid
balance
of
borrowed
money,
or
the
rate
earned
on the unrecovered balance of
an
investment,
so
that
the final
payment
or
receipt
brings
the balance to exactly zero
with interest considered.
The
rate
of
return is expressed as a percent
per
period, for example, i = 10% per
year. It is stated as a positive percentage; the fact that interest paid on a loan
is
actually a negative rate
of
return
from
the
borrower's
perspective
is
not
considered.
The
numerical value
of
i can range from
-100
% to infinity, that is,
- 100% < i <
00
. In terms
of
an
investment
, a return
of
i = - 100% means the
entire
amount
is
lost.
The
definition
above
does not state that the rate
of
return is
on
the initial
amount
of
the investment, rather it
is
on the unrecovered balance, which
changes
each time period.
The
example
below illustrates this difference.
Wells Fargo Bank lent a newly graduated engineer $1000 at i = 10% per year for 4 years
to buy home office equipment. From the bank's perspective (the lender), the investment
in
this young engineer
is
expected to produce an equi valent net cash flow
of
$315.47 for each
of
4 years.
A = $1000(A/ P,10%,4) = $315.47
This represents a 10% per year rate
of
return on the bank's unrecovered balance. Compute
the amount
of
the unrecovered investment for each
of
the 4 years using
(a)
the rate
of
re-
turn on the unrecovered balance (the correct basis) and
(b) the return
on
the initi
al
$1000
investment. (c) Explain why all the initial $1000 amount
is
not recovered by the final pay-
ment
in
part (b).
Solution
(a) Table 7-1 shows the unrecovered balance
at
the end
of
each year
in
column 6 using
the
10
% rate on the unrecovered balance
at
the beginning
of
the yew: After 4 years
the total
$1000
is
recovered, and the balance
in
column 6 is exactly zero.
(b) Table
7-2
shows the unrecovered balance
if
the 10% return
is
always figured
011
the initial $1000. Column 6
in
year 4 shows a remaining unrecovered amount
of
$138.12, because only $861.88
is
recovered
in
the 4 years (column 5).
SECTION
7.1
Interpretation
of
a Rate
of
Return
Va
lue
TABLE
7-1
Unrecovered Balances Using a Rate
of
Return
of
lOC
k on the Unrecovered
Balance
(1
) (2)
(3)
= 0.
10
x (2)
(4) (5)
= (4) - (3)
Beginning
Interest
on
Unrecovered
Unrecovered
Cash
Recovered
Year Balance Balance
Flow
Amount
0
$- 1,000.00
$-1,000.00
$100.00 +315.47 $215.47
2
- 784.53
78.45 + 315.47 237.02
3
- 547.51
54.
75
+3
15.47 260.72
4
-286
.79 28.68
+3
15.47 286.79
---
$261.88
$1,000.00
TABLE
7-2
Unrecovered Balances Using a
10(
k Return on the Initial
Amount
(1
) (2)
(3)
=
0.10
x (2)
(4)
(5) = (4) - (3)
Beginning
Unrecovered
Interest
on
Cash
Recovered
Year Balance
Initial
Amount
Flow
Amount
0
$-1,000.00
$-
1,000.00
$100
+315.47 $215.47
2 - 784.53
100
+3
15.47 215.47
3
- 569.06
100 + 315.47 215.47
4
-3
53.59
100
+3
15.47 215.47
$400 $861.88
(c) A total
of
$400
in
interest must
be
earned if the
10
% return each year is based on
the initial amount
of
$1000. However, only $261.88 in interest must be earned
if
a
10
% return
on
the unrecovered balance is used. There is more
of
the annual cash
flow available to reduce the remaining loan when the rate is applied to
th
e unrecov-
ered balance as
in
part (a) a
nd
Table 7-
1.
Figure 7-1 illustrates the correct inter-
pretation
of
rate
of
return
in
Table 7-
1.
Ea
ch year the $315.47 receipt represents
10
% interest on the unrecovered balance in column 2 plus the recovered amount
in
column
5.
Because rate
of
return is the interest rate on the unrecovered balance, the computations
in
Table
7-
1
for
part (a)
pr
esent a co
rre
ct
interpretation
of
a 10% rate
of
return. Clearly, an
interest rate applied only
to
the principal represe
nt
s a higher rate than
is
stated. In practice,
a so-ca
ll
ed add-on interest rate
is
frequently based on principal only, as in part (b). This is
somet
im
es referred to as the install
l17
en
tfinan
cing problem.
(6) = (2) + (5)
Ending
Unrecovered
Balance
$-1,000.00
-784.53
-547.51
-286.79
0
(6) = (2) + (5)
Ending
Unrecovered
Balance
$-1,000.00
-784.53
-569.06
- 353.59
-
13
8.
12
241
242
CHAPTER 7
Rate
of
Return Analysis: Single Alternative
Loan balance
of
$1000
1000.00
_
....
'" '"
"'
lJ
$100.00
--
Interest
784.53
'"
.r
u
t:
'"
-;;
547.5 1
.0
@
0
....J
Loan balance
$215.47
--
reduction
___
__
_
........
'"
$78.45
I ,
: $237.02
: :
11
286.79
_____
l
______
i
_____
_ -U
~
:~
7:
I I I
$28.68
I I I
I I I
I I I $286.79
o
Figure 7- 1
I I I
I I I
2 3
Loan balance
of
$0
Plot
of
unrecovered balances and 10% per year rate
of
return on a $1000
amount, Table 7-
1.
Year
Installmentjinancing can be discovered in many forms in everyday finances.
One popular example
is
a
"no
-interest program" offered by retail stores on the
sale
of
major appliances, audio and video equipment, furniture, and other con-
sumer
items. Many variations are possible, but in most cases, if the purchase
is
not paid for
in
full by the time the promotion is over, usually 6 months to I year
later,
jinance charges are assessed from the original date
of
purchase. Further,
the program's fine print may stipulate that the purchaser use a credit card issued
by the retail company, which often has a higher interest rate than that
of
a regu-
lar credit card, for example, 24% per year compared to
18
% per year. In a
ll
these
types
of
programs, the one common theme is more interest paid over time by the
consumer. Usually, the correct definition
of
i as interest on the unpaid balance
does not apply directly;
i has often been manipulated to the financial disadvan-
tage
of
the purchaser.
7.2
RATE
OF
RETURN
CALCULATION USING A PW OR
AW
EQUATION
To determine the rate
of
return
of
a cash flow series, set up the ROR equation
using either
PW
or
AW
relatio
ns
.
The
present worth
of
costs
or
disbursements
PW D is equated to the present worth
of
incomes
or
receipts
PW
w Equivalently,
SECT10N 7.2
Rate
of
Return Calculation Using a PW or A W Equation
the two can be subtracted and set equal to zero. That is, solve for i using
PW
D
=
PW
R
0=
-PW
D
+
PW
R
[7.1]
The
annual worth approach utilizes the AW values
in
the same fashion to solve
for
i.
AW
D
=AW
R
0=
-AW
D
+AW
R
[7.2]
The
i value that makes these equations numerically correct is called i*.
It
is
the
root
of
the ROR relation. To determine if the alternative's cash flow series is
viable, compare
i* with the estab
li
shed MARR.
If
i*
;::::
MARR, accept the alternative as economically viable.
If
i*
< MARR, the alternative is not economically viable.
In Chapter 2 the method for calculating the rate
of
return on an investment
was illustrated when only
one
engineering economy factor was involved. Here
the present worth equation is the basis for calculating the rate
of
return when sev-
eral factors are involved. Remember that the basis for engineering economy cal-
culations is
equivalence
in
PYI
,
FW
,
or
AW terms for a stated i
;::::
0%. In rate
of
return calculations, the objective
is
to find the interest rate i* at which the cash
flows are equivalent.
The
calculations are the reverse
of
those made in previous
chapters, where the interest rate was known.
For
example,
if
you deposit $1000
243
now and are promised payments
of
$SOO
three years from now and
$1
SOO
five
PW
and A W relations
years from now, the rate
of
return relation using
PW
factors is
1000 =
SOO(?
/ F,
i*
,3) + 1
SOO(P
/ F,i* ,S)
[7.3]
The
value
of
i* to make the equality correct
is
to be computed (see Figure
7-2).
If
the $1000
is
moved to the right side
of
Equation [7.3], we have
o
$1000
o = - 1000 +
SOO(P
/ F,i
*,
3) + 1
SOO(P
/ F,
i*
,S)
$
500
2
3
i
=?
4
$1500
Figure
7-2
5
Cash flow for which a
value
of
i is to be
determined.
[7.4]
244
CHAPTER
7
Rate
of
Return Analysis: Single Alternative
which
is
the general form
of
Equation [7.1].
The
equation
is
solved for i to obtain
i*
= 16.9% by hand using trial and error
or
by
computer
using spreadsheet func-
tions.
The
rate
of
return will always
be
greater than zero
if
the total
amount
of
receipts is greater than the total
amount
of
disbursements, when the time value
of
money is considered. Using
i':'
= 16.9%, a graph similar to
Figure
7-1
can be
constructed. It will show that the unrecovered balances
each
year, starting
with
$- 1000 in
year
I, are exactly recovered by the $500 and $1500 receipts
in
years 3 and
5.
It should be evident that rate
of
return relations are merely a rearrangement
of
a present worth equation.
That
is,
if
the
above
interest rate
is
known
to be 16.9%,
and it
is
used to find the present worth
of
$500 three years from now and $1500
five years from now, the
PW
relation
is
PW
=
SOO(P
IF,16.9%,3) + IS00(PIF,
16.9
%,
S)
= $1000
This
illustrates that rate
of
return and present worth equations are set up
in
ex-
actly the
same
fashion.
The
only differences are
what
is
given and what
is
sought.
There
are two ways to determine
i*
once
the
PW
relation is established: solu-
tion via trial
and
error by hand and solution by spreadsheet function.
The
second
is
faster; the first helps in understanding
how
ROR
computations work. We sum-
marize both
methods
here and in
Example
7.2.
i* Using Trial
and
Error by
Hand
The
general procedure
of
using a PW-based
equation
is
I.
Draw
a cash flow diagram.
2. Set up the rate
of
return equation in the form
of
Equation [7.1].
3. Select values
of
i by trial and error until the equation
is
balanced.
When
the trial-and-error
method
is applied to determine i*, it is advantageous
in
step 3 to get fairly close to the
correct
answer
on
the first trial.
If
the cash flows
are combined in such a
manner
that the
income
and disbursements can
be
repre-
sented by a
single
factor
such as
PI
For
PIA, it is possible to
look
up the inter-
est rate (in the tables) corresponding to the value
of
that factor for n years.
The
problem, then,
is
to
combine
the cash flows into the format
of
only
one
of
the fac-
tors.
This
may be
done
through the following procedure:
I.
Convert
all disbursements into either single amounts (P
or
F)
or
uniform
amounts
(A) by neglecting the time value
of
money.
For
example,
if
it
is
de-
sired to convert an A to an
F value, simply multiply the A by the
number
of
years n.
The
scheme
selected for
movement
of
cash flows should be the
one
which minimizes the error caused by neglecting the time value
of
money.
That
is,
if
most
of
the cash flow is an A and a small
amount
is an
F,
convert
the F to an A rather than the
other
way around.
2.
Convert
all receipts to either single
or
uniform values.
3.
Having
combined
the disbursements
and
receipts so that a PI F, PiA,
or
AI F format applies, use the interest tables to find the
approximate
interest