Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
Подождите немного. Документ загружается.
SECTION
5.3
Present Worth Analysis
of
Different-Life Alternatives
The assumptions of a
PW
analysis
of
different-life alternatives for
the
LCM
method
are
as follows:
1.
The
service provided by the alternatives will be needed for
the
LCM
of years
or
more.
2. The selected alternative will be repeated over each life cycle
of
the
LCM
in exactly the same manner.
3. The cash flow estimates will be
the
same in every life cycle.
As will be shown in Chapter
14
, the third assumption is valid only when the cash
flows are expected to change by exactly the inflation (or deflation) rate that is ap-
plicable through the LCM time period.
If
the cash flows are expected to change
by any other rate, then the
PW analysis must be conducted using constant-value
dollars, which considers inflation (Chapter 14). A study period analysis is neces-
sary if the first assumption about the length
of
time the alternatives are needed
cannot be made. A present worth analysis over the
LCM
requires that the esti-
mated salvage values be included in each life cycle.
For the study period approach, a time horizon
is
chosen over which the eco-
nomic analysis
is
conducted, and only those cash flows which occur during that
time period are considered relevant to the analysis. All cash flows occurring
beyond the study period are ignored. An estimated market value at the end
of
the
study period must be made. The time horizon chosen might
be
relatively short,
especially when short-term business goals are very important. The study period
approach
is
often used in replacement analysis. It is also useful when the
LCM
of
alternatives yields an unrealistic evaluation period, for example, 5 and 9 years.
Example 5.2 includes evaluations based on the
LCM
and study period
approaches. Also, Example 5.12 in Section 5.9 illustrates the use
of
spreadsheets
in
PW
analysis for both different lives and a study period.
EXAMPLE 5.2 '
A project engineer with EnvironCare is assigned to start
up
a new office
in
a city where
a 6-year contract has been finalized to take and
to analyze ozone-level readings. Two
lease options are available, each with a first cost, annual lease cost, and deposit-return
estimates shown below.
First cost,
$
Annual lease cost, $ per year
Deposit return, $
Lease term, years
Location
A
-15,000
-3,500
1,000
6
Location
B
-18,000
-3,100
2,000
9
(a) Determine which lease option should be selected on the basis
of
a present worth
comparison, if the MARR
is
15% per year.
(b) EnvironCare has a standard practice
of
evaluating all projects over
as-year
period.
If
a study period
of
5 years
is
used and the deposit returns are not
expected to change, which location should be selected?
175
176 CHAPTER 5 Present Worth A
nal
ys
is
(c) Which location should be selected over a 6-year study period if the deposit
return at location B
is
estimated to be $6000 after 6 years?
Solution
(a) Since the leases have different terms (lives), compare them over the LCM
of
IS years. For life cycles after the first, the first cost
is
repeated
in
year 0 of each
new cycle, which
is
the last year
of
the previous cycle. These are years 6 and
12
for location A a
nd
year 9 for B. The cash
flow
diagram is
in
Figure 5-
2.
Calcu-
late
PW at
15
% over IS years.
$
15
,000
PW/J= ?
$
18
,000
PW
A
= -
15
,000 -
15
,000(P/ F,15%,6) + 1000(P/ F,
15
%,6)
-
15
,000(P/ F,
15
%
,l2
) +
LOOO
(P/ F,
l5
%,
12
) + 1000(P/ F,15%
,I
S)
- 3500(P / A,
15
%,
IS)
= $- 45,036
PW
s
= - IS,
OOO
-
IS
,
000(P/F,15%,9)
+
2000(P/F,15
%,9)
+
2000(P/F
,
15
%, lS) - 3100(P/A,
15
%,
IS)
= $- 41,3S4
Location B is selected, since it costs less
in
PW terms; that
is,
the PW B value is
numerically larger than
PW A '
2
2
$1000
6
~
$
15
,000
$3500
Locat.ionA
$2000
9
$18,000
Location B
$1000
$1000
12
16
J7
18
~
$
15
,000
$2000
16
17 18
Figure
5-2
Cash flow diagram for different-life alternatives, Example 5.2(a).
SECTlON 5.4 Future Worth Analysis
(b) For a 5-year study period
no
cycle repeats are necessary. The PW analysis
is
PW A = - 15,000 - 3500(P/A,
J5
%,5) + 1000(P/ F,15%,5)
= $- 26,236
PW
B
= - 18,000 - 3100(P/A,15%,5) + 2000(P/F,15%,5)
= $- 27,397
Location A
is
now the better choice.
(c) For a 6-year study period, the deposit return for B
is
$6000 in year
6.
PW
A
= -
15
,000 - 3500(P/A,
15
%,6) + 1000(P/ F,
l5
%,6) =
$-27
,
813
PW
B
= -
18
,000 - 3100(P/A,
15
%,
6) + 6000(P/ F,
15
%,6) =
$-27
,138
Location B now has a small economic advantage. Noneconomic factors are
likely
to
enter into the final decision.
Comments
In
part (a) and Figure 5-
2,
the deposit return for each lease
is
recovered after each life
cycle, that
is
,
in
years
6,
12,
and
18
for A and
in
years 9 and
18
for
B.
In
part (c), the
increase
of
the deposit return from $2000
to
$6000 (one year later), switches the
selected location from A
to
B.
The project engineer should reexamine these estimates
before making a final decision.
5.4 FUTURE WORTH ANALYSIS
The future worth (FW)
of
an
alternative may be determined directly from the
cash flows
by
determining the future worth value, or by multiplying the
PW
value by
th
e F / P factor, at the established MARR. Therefore, it is an extension
of
present worth analysis.
The
n value in the F / P factor depends upon which
time period has been used to determine
PW-the
LCM
value or a specified study
period. Analysis
of
one alternative, or the comparison
of
two or more alterna-
tives, using FW values is especially applicable to large capital investment deci-
sions when a prime goal
is
to maximize the
futur
e wealth
of
a corporation's
stockholders.
Future worth analysis
is
often utilized if the asset (equipment, a corporation, a
building, etc.) might be sold or traded at some time after its start-
up
or
acquisi-
tion, but before the expected life is reached. An
FW
value at an intermediate year
estimates the alternative's worth at the time
of
sale
or
disposal. Suppose an en-
trepreneur is planning to buy a company and expects to trade it within 3 years.
FW
analysis is the best method to help with the decision to sell or keep
it
3 years
hence. Example 5.3 illustrates this use
of
FW
analysis. Another excellent appli-
cation
of
FW analysis is for projects that will not come online until the end
of
the
investment period. Alternatives such
as
electric generation facilities, toll roads,
hotels, and the like can be analyzed using the
FW
value
of
investment commit-
ments made during construction.
177
178
CHAPTER 5
Present Worth Analysis
Once the FW value
is
determined, the selection guidelines are the same
as
with PW analysis; FW
:::=:
0 means the MARR is met or exceeded (one alterna-
tive). For two (or more) mutually exclusive alternatives, select the one with the
numerically larger (largest) FW value.
EXAMPLE
5.3
A British food distribution conglomerate purchased a Canadian food store chain for $75
million
(U.S
.)
three years ago. There was a net loss
of
$10 million at
th
e e
nd
of
year 1
of
ownership. Net cash
flow
is increasing with
an
arithmetic gradient
of
$+
5 million
per year starting the second year, and this pattern
is
expected to continue for the fore-
seeable future. This means that breakeven net cash flow was achieved
thi
s year.
Because
of
the heavy debt
fin
ancing used to purchase
th
e Canadian chain, the interna-
tional board
of
directors expects a MARR
of
25% per year from any sale.
(a)
The
British conglomerate has just been offered $159.5 million (U.S.)
by
a
French compa
ny
wishing to get a foothold in Canada. Use FW analysis
to
deter-
mine if the
MARR
will be realized at this selling price.
(b)
If
th
e British conglomerate continues
to
own the chain, what selling price
mu
st
be obtained at the end
of
5 years
of
ownership to make
th
e MARR?
Solution
(a) Set up the future worth
rel
ation
in
year 3 (FW
3
)
at i = 25% per year a
nd
an
offer
price
of
$159.5 million. Figure 5
-3a
prese
nt
s the cash flow diagram
in
million $
unit
s.
o
$75
FW
3
=
-75(F/
P,25%,3) -
IO
(F/
P
,2
5%,2) -
5(F/P,25%,I)
+ 159.5
= - 168.36 + 159.5 =
$-8.86
million
No,
th
e MARR
of
25% will not
be
rea
li
zed if the $159
.5
million offer is
accepted.
$159.5
FW=?
i
~
25
'70
?
i=25
%
2
3
o
2
3
$75
(a)
(b)
Figure
5-3
Cash
fl
ow diagrams for Example 5.3. (a) Is MARR = 25% realized? (b) What
is
FW
in
year
5?
Amounts are
in
million $ units.
SECTION S.S Capitalized Cost Calculation a
nd
Analysis
(b) Determine the future worth S years from now at 2S% per year. Figure S
-3
b pre-
sents the cash flow diagram. The
A/G
and
F/A
factors are applied
to
the arith-
metic gradient.
FW
s = - 7S
(F
/ P,2S%,S) -
1O
(F/A,2S%,S) + S(A/ G,2S%,S)(F
/A,2
S%,S)
=
$-2
46.81 miUion
The offer must be for
at
least $246.81 million
to
make the MARR. This
i.s
approximately 3.3 times the purchase price only S years earlier,
in
large part
based on the required MARR
of
2S%.
Comment
If the 'rule of
72'
in
Equation
[l.9]
is applied at 2S% per year, the sales price must
double approximately eve
ry
72
/ 2S% = 2.9 years. This does not consider any annual
net positive or negative cash
flows during the years
of
ownership.
5.5 CAPITALIZED COST CALCULATION
AND
ANALYSIS
Capitaliz
ed
cost (CC) is the present worth
of
an alternative
th
at will last "for-
ever." Public
sector projects such as bridges, dams, irrigation systems, and rail-
roads fall into this category. In addition, permanent and charitable orgartization
endowments are evaluated using the capitalized cost methods.
The formula to calculate CC
is
derived from the relation P =
A(P
/ A,i,n),
where
n =
00
.
The
equation for P using the P
/;1
factor formula is
P = Al(1 + it -
1]
i(1 + i)"
Divide the numerator and denominator by
(1
+ i)".
r
1 -
(1
~
i)" l
P
=A
i
As n approaches
00
, the bracketed term becomes 1/
i,
and the symbol
CC
replaces
PW and
P.
cc=
~
[5.1]
l
If
the A value
is
an annual worth
(A
W) determined through equivalence calcula-
tions
of
cash flows over n years, the CC value
is
cc=
A~
[5.2]
t
The
validity
of
Equation [5.1] can be illustrated by considering the time value
of
money.
If
$10,000
ea
rns 20% per year, compounded annually, the maximum
179
180
CHAPTER 5
Present Worth Analysis
amount
of
money that can be withdrawn at the end
of
every year for eternity
is
$2000, or the interest accumulated each year. This leaves the original $10,000 to
earn interest so that another
$2000 will be accumulated the next year. Mathe-
matically, the amount
A
of
new money generated each consecutive interest
period for an infinite number
of
periods is
A =
Pi
=
CC(i)
[5.3]
The capitalized cost calculation
in
Equation [5.1]
is
Equation [5.3] solved for
P and renamed
Cc.
For a public sector alternative with an infinite or very long life, the A value
determined by Equation [5.3] is used when the benefit/cost (B/C) ratio is the
comparison basis for public projects. This method
is
covered in Chapter 9.
The cash flows (costs or receipts) in a capitalized cost calculation are usually
of
two types: recurring, also called periodic, and nonrecurring. An annual oper-
ating cost
of
$50,000 and a rework cost estimated at $40,000 every 12 years are
examples
of
recurring cash flows. Examples
of
nonrecurring cash flows are the
initial investment amount in year
0 and one-time cash flow estimates at future
times, for example,
$500,000
in
royalty fees 2 years hence.
The
following pro-
cedure assists
in
calculating the CC for an infinite sequence
of
cash flows.
I. Draw a cash flow diagram showing all nonrecurring (one-time) cash flows
and at least two cycles
of
all recurring (periodic) cash flows.
2.
Find the present worth
of
all nonrecurring amounts. This is their CC value.
3. Find the equivalent uniform annual worth
(A
value) through one
lif
e cy
cl
e
of
all recurring amounts. This is the same value
in
all succeeding life
cycles, as explained in Chapter
6.
Add this to all other uniform amounts
occurring in years 1 through infinity and the result
is
the total equivalent
uniform annual worth (AW).
4. Divide the
AW
obtained in step 3 by the interest rate i to obtain a CC value.
This is an application
of
Equation [5.2].
5.
Add the CC values obtained in steps 2 and 4.
Drawing the cash flow diagram (step
1)
is
more important in CC calculations
than elsewhere, because it helps separate nonrecurring and recurring amounts.
In
step 5 the present worths
of
all component cash flows have been obtained; the
total capitalized cost
is
simply their sum.
EXAMPLE
5.4
: .
The
property appraisal district for Marin Coun
ty
has
just
installed new software to track
residential market values for property tax computations.
The
manager wants
to
know the
total equivalent cost
of
a
ll
future costs incurred when the three county judges agreed to pur-
chase the software system.
If
the new system will be used for
d1e
indefinite future, find the
equivalent va
lu
e (a) now and (b) for each year hereafter.
The
system has an installed cost
of
$150,000 and an additional cost
of
$50,000 after
10
years.
The
annual software maintenance contract cost is $5000 for the first 4 years and
SECTION 5.5 Capitalized Cost Calculation and Analysis
$8000 thereafter.
In
addition, there
is
expected to be a recurring major upgrade cost
of
$15,000 every
13
years. Assume that i = 5% per year for county funds.
Solution
(a) The five-step procedure
is
applied.
1. Draw a cash flow diagram for two cycles (Figure
5-4)
.
2.
Find the present worth
of
the nonrecurring costs
of
$150,000 now and $50,000
in
year
10
at i = 5%. Label this CCI'
CC
I
= - 150,000 - 50,000(P/F,5%,1O) =
$-180,695
3.
Convert the recurring cost
of
$15,000 every
13
years into
an
annual worth
AI
for the first
13
years.
AI = -
15
,000(A/
F,5
%,
13)
=
$-847
The same value,
AI
= $- 847, applies
to
all the other 13-year periods as well.
4.
The capitalized cost for the two annual maintenance cost selies may be
determined
in
either
of
two ways:
(I)
consider a series
of
$- 5000 from now to
infinity and
find
the present worth
of
-$8000
-
($-5000)
=
$-3000
from
year
5 on;
or
(2) find the CC
of
$-5000
for 4 years and the present worth
of
$-8000
from year 5 to infinity. Using the first method, the annual cost (Az)
is
$- 5000 forever. The capitalized cost CC
2
of
$- 3000 from year 5 to infinity
is
found using Equation [5.1] times the P / F factor.
-3000
CC
2
= - - (P/F,5%,4) =
$-49,362
0.05
The two annual cost series are converted into a capitalized cost CC
3
.
CC
=A
I
+A
2
=
-847+(-5000)=$
_ 116940
3 i 0.05 '
5.
The total capitalized cost CC
T
is
obtained
by
adding the three CC values.
CC
T
= - 180,695 - 49,362 - 116,940 = $- 346,997
i = 5% per year
o
2 4
6 8 10
12 14
TTl
$5000
$150,000
Figure
5-4
$8000
j
$50,000
t
$15,000
Cash flows for two cycles
of
recutTing costs and
all
nonrecurring amount
s,
Example 5.4.
t
$
15
,000
181
182
CHAPTER 5 Present Worth Analysis
(b) Equation [5.3] determines the A value forever.
A =
Pi
= CCT(i) = $346,997(0.05) = $17,350
Correctly interpreted, this means Marin County officials have committed the
equivalent
of
$17,350 forever
to
operate and maintain the property appraisal
software.
Comment
The CC
2
value
is
calculated using
n.
= 4
in
the P / F factor because the present worth
of
the
annual
$3000 cost
is
located
in
year 4, since P is always one period ahead
of
the first A.
Rework the problem using the second method suggested for calculating CC
2
.
For the comparison
of
two or more alternatives on the basis
of
capitalized
cost, use the procedure above
to
find CC
T
for each alternative. Since the capital-
ized cost represents the total present worth
of
financing and maintaining a given
alternative forever, the alternatives will automatically be compared for the
same number
of
years (i.e., infinity). The alternative with the smaller capitalized
cost will represent the more economical one. This evaluation is illustrated in
Example
5.5.
As
in
present worth analysis, it is only the differences in cash
flow
between
the alternatives that must
be
considered for comparative purposes. Therefore,
whenever possible, the calculations should be simplified by eliminating the
elements
of
cash flow which are common to both alternatives. On the other hand,
if true capitalized cost values are needed to reflect actual financial obligations,
actual cash flows should be used.
EXAMPLE
5.5
..
~;
Two sites are currently under consideration for a blidge to cross a river
in
New York.
The north site, which connects a major state highway with
an
interstate loop around the
city, would alleviate much
of
the local through traffic.
The
disadvantages
of
this site are
that the bridge would do little to ease local traffic congestion during rush hours, and the
bridge would have
to
stretch from one hill to another to span the widest part
of
the river,
railroad tracks, and local highways below. This bridge would therefore be a suspension
bridge. The south site would require a much shorter span, allowing for
consu·uction
of
a truss bridge, but
it
would require new road construction.
The suspension bridge will cost
$50 million with annual inspection and maintenance
costs
of
$35,000.
In
addition, the concrete deck would have to
be
resurfaced every 10
years at a cost
of
$100,000. The truss bridge and approach roads are expected to cost
$25 million and have annual maintenance costs
of
$20,000. The bridge would have to
SECTION 5.5 Capitalized Cost Calculation and Analysis
be painted every
3 years at a cost
of
$40,000.
In
addition, the bridge would have to be
sandblasted every
10
years at a cost
of
$190,000. The cost
of
purchasing right-of-way
is
expected to be $2 million for the suspension bridge and $15 million for the truss
bridge. Compare the alternatives on the basis
of
their capitalized cost if the interest rate
is
6% per year.
Solution
Construct the cash
flow
diagrams over two cycles (20 years).
Capitalized cost
of
suspension bridge (CC
s
):
CC
I
= capitalized cost
of
initial cost
= - 50.0 - 2.0 =
$-52.0
milljon
The recurring operating cost
is
A I =
$-35,
000, and the annual equivalent
of
the resur-
face cost
is
A2
= - JOO,000(A/F,6%,
IO)
=
$-7587
CC
2
= capitalized cost
of
recurring costs = A I + A2
I
=
-35,000
+
(-
7587) =
$-709
,
783
0.06
The total capitalized cost
is
CC
s
= CC
I
+ CC
2
=
$-52.71
million
Capitalized cost
of
truss bridge (CC
r
):
CC
I
= - 25.0 +
(-15.0)
=
$-40.0
million
AI =
$-20,000
A2
= annual cost
of
painting = - 40,00
0(A/
F,6%,3) =
$-12,564
A3 = annual cost
of
sandblasting = - 190,000(A/ F,6%,1
0)
= $-
14
,415
CC = AI + A2 + A3 =
$-46,979
=
$-782983
2 i 0.06 '
CC
r
= CC
I
+ CC
2
=
$-40.78
million
Conclusion: Build the truss
bJidge, since its capitalized cost is lower.
If
a finite-life alternative (e.g., 5 years) is compared to one with an indefinite
or very long life, capitalized costs can be used for the evaluation. To determine
capitalized cost for the alternative with a finite life, calculate the equivalent
A
value for one
li
fe cycle and divide by the interest rate (Equation [5.1]). This pro-
cedure is illustrated
in
the next example.
183
184
CHAPTERS Present Worth Analys
is
EXAMPLE
5.6
'.~
~
E-Solve
APSco, a large electronics subcontractor for the Air Force, needs to immediately acquire
10
so
ld
er
in
g machines with specially prepared ji
gs
for assembling components onto
printed circuit
boards, More machines may
be
needed
in
the future. The lead production en-
gineer has outlined below two simplified, but viable, alternatives. The company's MARR
is
15
% per year.
Alternative LT (long-term). For $8 million now, a contractor will provide the necessary
number
of
machines (up
to
a maximum
of
20), now and
in
the future, for
as
long
as
APSco needs them. The annual contract fee is a total
of
$25,000 with no additional
per-machine annual cost. There
is
no time limit placed
on
the contract, and the cos
ts
do not escalate.
Alternative
ST
(short-term). APSco buys its own machines for $275,000 each and ex-
pends
an
est
im
ated $12,000 per machine
in
annual operating cost (AOC). The use-
fu
l
li
fe
of
a soldering system
is
5 years.
Perform a capitalized cost evaluation
by
hand and
by
computer. Once the eva
lu
ation
is
complete, use the spreadsheet for sensitivity anaJysis
to
determine the maximum number
of
so
ld
ering machines that can be purchased now and still have a capitalized cost less than
that
of
the long-term alternative.
Solution
by
Hand
For the
LT
alternative,
find
the CC
of
the AOC using Equation [5.1], CC =
A/i.
Add this
aJTIount
to
the initial contract fee, which is already a capitalized cost (present worth)
amount.
CCLT
= CC
of
contract fee + CC
of
AOC
= - 8 million - 25,000/ 0.15 =
$-8
,166,667
For the
ST
alternative, first calculate the equivalent annual amount for the purchase cost
over the 5-year life, and add the
AOC values for all
10
machines. Then determine the total
CC using Equation [5.2].
AW
sT
=
AW
for purchase + AOC
= - 2.75 million(A/ P,
15
%,
5) - 120,000 = $- 940,380
CC
ST
= - 940,380/0.15 = $- 6,269,200
The
ST alternative has a lower capitalized cost by approximately $1.9 million present
va
lu
e dollars.
Solution
by
Computer
Figure 5- 5 contains the solution for
10
machines
in
column
B.
Cell B8 uses the same rela-
tion as
in
the so
lu
tion by hand. Cell B
15
uses the PMT function to detelmine the equiva-
lent annual amount
A for the purchase
of
10
machines,
to
which the AOC
is
added. Cell
B
16
uses Equation
[5
.
2]
to
find
the total CC for tbe ST alternative. As expected, alternative
ST
is
selected. (Compare CC
ST
for the hand and computer solutions to note that the round-
off
eITor
using the tabulated interest factors gets larger for large P
va
lu
es.)
The type
of
sensitivity analysis requested here is easy to perform once a spreadsheet
is
developed. The PMT function
in
B
15
is
expressed generally in terms
of
cell B
12,
the