Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
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SECTION
11.5
Replacement Study Over a Specified Study Period
option or many options; the longer the study period, the more complex this
analysis becomes.
The
AW D and AWe values are used to build the equiva-
lent cash flow series for each option.
2. Selection
of
the best option. The PW or A W for each option
is
calculated
over the study period. Select the option with the lowest cost, or h.ighest
in-
come if revenues are estimated. (As before, the best option will have the
numerically largest PW or AW value.)
The following three examples use this procedure and illustrate the importance
of
making cost estimates for the defender alternative when its remaining life is less
than the study period.
EXAMPLE
11.5 .
Claudia works with Lockheed-Martin (LMCO)
in
the aircraft maintenance division.
She is preparing for what she and her boss, the division chief, hope
to
be
a new 10-year
defense contract with the
U.S. Air Force on C-5A cargo aircraft. A key piece
of
equip-
ment for maintenance operations
is
an
avionics circuit diagnostics system. The current
system was purchased 7 years ago on
an
earlier contract. It has no capital recovery costs
remaining, and the following are reliable estimates: current market value
= $70,000, re-
maining life
of3
more years, no salvage value, and AOC = $30,000 per year. The only
options for this system are to replace it now or retain it for the full 3 additional years.
Claudia has found that there
is
only one good challenger system. Its cost estimates
are: first cost
of
$750,000, life
of
10
years, S = 0, and AOC = $50,000 per year.
Realizing the importance
of
accurate defender alternative cost estimates, Claudia
asked the division chief what system would
be
a logical follow-on to the current one
3 years hence,
if
LMCO wins the contract. The chief predicted LMCO would purchase
the very system she had identified
as
the challenger, because it is the best on the market.
The company would keep
it
for the entire
to
additional years for use on
an
extension
of
this contract or some other application that could recover the remaining 3 years
of
in-
vested capital. Claudia interpreted the response to mean that the last 3 years would
also be capital recovery years, but on some project other than this one. Claudia's estimate
of
the first cost
of
this same system 3 years from now is $900,000. Additionally, the
$50,000 per year
AOe
is
the best estimate at this time.
The division chief mentioned any study had
to
be conducted using the interest rate
of
10%,
as
mandated by the U.S. Office
of
Management and Budget (OMB). Perform a
replacement study for the fixed contract period
of
10
years.
Solution
The study period
is
fixed at 10 years, so the intent
of
the replacement study assumptions
is
not present. This means the defender foHow-on estimates are very important
to
the
analysis. Further, any analyses to determine the ESL values are unnecessary and incor-
rect since alternative lives are already set and
no
projected annual market values are
available. The first step
of
the replacement study procedure
is
to define the options.
Since the defender will be replaced now or
in
3 years, there are only two options:
I.
Challengerfor all 10 years.
2. Defender for 3 years, followed
by
challenger for 7 years.
405
406
CHAPTER
II
Replacement and Retention Decisions
The A W values for C and
0 are calculated. For option
1,
the challenger
is
used for all
10
years. Equation
[11
.
3J
is applied using the following estimates:
Challenger:
P = $750,000
n =
10
years
AOC = $50,000
S=O
AWc =
-750,000(A
/
P,1O
%,10)
-50
,000 =
$-172,063
The second option has more complex cost estimates. The
AW
for the in-place system
is
calculated over the first 3 years. Added to this is the capital recovery
for
the defender
follow-onfor
the next 7 years. However in this case, the
CR
amount
is
determined over
itsfull lO-year
Life
. (It
is
not unusual for the recovery
of
invested capital to be moved be-
tween projects, especially for contract work.) Refer
to
the
AW
components
as
AW
DC
(subscript DC for defender-current) and
AW
DF (subscript
OF
for defender follow-on).
The
final
cash flow diagrams are shown
in
Figure 11-6.
Defender current: market value = $70,000 AOC = $30,000
n = 3 years
s = O
AW
DC
=
[-70,000
- 30,000(P/A,10%,3)J(A/P,1O%,JO) =
$-23,534
0
2
3
4
5 6
7
8 9
10
Year
1
I
I I
I
I
I
I I I
I
T
AOC = $50,000
End
of
P = $750,000
study period
(a) Challenger (option
I)
0 2 3
4
5
6
7
8
9
10
AOC = $30,000
MV =$70,000
I
Defender-
-
---
current
CR=$146
,475
~
I----Defender-follow-on
--I
(b) Defender alternative (option 2)
Figure
11-6
II
12
l3 Year
I
I
I
I
I
I
I
~
CR for another
project
Cash
flow
diagrams for a 10-year study period replacement study, Example 11.5.
SECTION
11.5
Replacement Study Over a Specified Study Period
Defender follow-on: P = $900,000, n = 10 years for capital recovery calculation
only,
AOC = $50,000 for years 4 through
10
, S =
O.
The CR and A W for all
10
years are
CRop
=
-90
0
,O
O
O(AI
P,
10
%
,l0)
= $- 146,475 [11.6]
AW
OF = (- 146,475 -
50,OOO)(F
I A,lO%,7)(AI F,lO%,10) =
$-116,966
Total
AW
0 for the defender is the sum
of
the two annual worth values above. This is the
A W for option
2.
AWo
=
AW
oc
+
AW
OF
=
-23,534
-
116
,966 =
$-140,500
Option 2 has a lower cost
($-140,500
versus
$-172,063).
Retain the defender now
and expect
to
purchase the follow-on system 3 years hence.
Comment
The capital recovery cost for the defender follow-on will be borne
by
some yet-to-be-
identified project for years
11
through
13.
If this assumption were not made, its capital
recovery cost would be calculated over 7 years, not
10,
in
Equation [11.6], increasing
CR
to
$- 184,869. This raises the annual worth
to
AWo
= $- 163,357. The defender
alternative (option
1)
is
still selected.
EXAMPLE
11.6
Three years ago Chicago's O'hare Airport purchased a new fire truck. Because
of
flight
increases, new fire-fighting capacity is needed once again.
An
additional truck
of
the
same capacity can be purchased now, or a double-capacity truck can replace the current
fire
truck. Estimates are presented bel
ow.
Compare the options at 12% per year using
(a) a 12-year study period and
(b)
a 9-year study period.
First cost
P,
$
AOC, $
Market value, $
Salvage value, $
Life, years
Solution
Presently
Owned
- 151,000
(3
years ago)
- 1,500
70,000
10%
of
P
12
New
Pur
ch
a
se
Double Capacity
-175,000
- 1,500
12%
of
P
12
-190,000
-2,500
10
%
ofP
12
Identify option 1
as
retention
of
the presently owned truck and augmentation with a new
same-capacity vehicle. Define option 2 as replacement with the double-capacity truck.
Opt
ion 1
Opt
ion 2
Presently
Owned
Augmentation
Double Capacity
P, $
- 70,000 - 175,000
-190,000
AOC,$
-1,500
- 1,500
-2,500
S,$
15
,100 21,000 19,000
n,
years
9
12 12
407
408
CHAPTER
II
Replacement and Retention Decisions
(a) For a full-life 12-year study period
of
option 1,
J
AW I = (AW
of
presently owned) + (AW
of
augmentation)
= [
-7
0,000(A/ P,
12
%,9) + 15,100(A/ F,
12
%,9) - 1500]
+ [- 175,000(A/P,12%,12) + 21,000(A/F,12%,12) - 1500]
= -
13
,6
16
- 28,882
=
$-42,498
This computation assumes the equivalent services provided by the current fire
truck can be purchased at
$-13,616
per year for years
10
through
12.
AW
2
= - 190,000(A/P,
12%
,
12)
+ 19,000(A/F,12%,
12)
- 2500
=
$-32,386
Replace now with the double-capacity truck (option 2) at
an
advantage of$1 0,
112
per year.
(b) The analysis for
an
abbreviated 9-year study period is identical, except that
n = 9
in
each factor; that
is
, 3 fewer years are allowed for the augmentation and
double-capacity trucks to recover the capital investment plus a
12
% per year
return.
The
salvage values remain the same since they are quoted
as
a percentage
of
P for all years.
AWl =
$-46
,539
AW
2
=
$-36,873
Option 2
is
again selected; however, now the economic advantage
is
smaller.
If
the
st
udy period were abbreviated more severely, at some point the decision
should reverse.
If
this example were solved by computer, the n values
in
the PMT
functions cou
ld
be decreased to determine
if
and when the decision reverses from
option 2 to
1.
If
there are several options for the number
of
years that the defender may
bt-
retained before replacement with the challenger, the first step
of
the replacement
study
-s
uccession options and
AW
values-must
include all the viable options.
For example, if the study period is
5 years, and the defender will remain
in
ser-
v
ic
e 1 year, or 2 years, or 3 years, cost estimates must be made
to
determine A W
values for each defender retention period.
In
this case, there are four options; call
them
W,
X, Y, and
Z.
Defender
Challenger
Option
Retained
Serves
W
3 years
2 years
X
2
3
Y 1
4
Z 0 5
The respective
AW
values for defender retention a
nd
challenger use define thr
cash flows for each option. Example 11.7 illustrates the procedure.
SECTION
11.5
Replacement Study Over a Spec
ifi
ed Study Period
EXAMP
LE
11.7
Amoco Canada has oil field equipment placed into service 5 years ago for which a replacement
study has been requested. Due to its special purpose, it has been decided that the current equip-
ment will have to serve for either 2, 3,
or
4 more years before replacement. The equipment has
a current market value
of
$100,000, which is expected to decrease by $25,000 per year.
TheAOC
is constant now, and is expected to rema
in
so, at $25,000 per year.
The
replacement challenger is
a
fix
ed-price contract to provide
th
e sa
me
services at $60,000 per year for a minimum
of2
years
and a maximum
of
5 year
s.
Use MARR
of
l2
% per year
to
perform a replacement study over a
6-year period
to
determine when to sell the
CUiTent
equipment and purchase the contract services.
Solution
Since the defe
nd
er will be retained for 2,
3,
or4
years, there are three viable options (X, Y, and Z).
Defender
Challenger
Option
Retained Serves
X
Y
Z
2 years
3
4
4 years
3
2
The defe
nd
er annual worth values are identified with subscripts D2, D3, and D4 for the num-
ber
of
years retaine
d.
AW
D2
= -
100
,000(A/ P,
12
%,2) + 50,000(A/ F,12%,2) - 25,000 = $
-60
,585
AW
D3
= - 100,000(A/ P,
l2
%
,3
) + 25,000(A /F,12%,3) - 25,000 =
$-59,226
AW
D4
= - J
OO,OOO(A
/ P, I 2
%,
4) - 25,000 = $- 57,923
For a
ll
options,
th
e cha
ll
eng
er
has an annual worth
of
AW
e = $- 60,000
Ta
bl
e
11
-3
prese
nt
s
th
e cash
flo
ws and PW values for each option over the 6-year study
period. A sample
PW computation for option Y is
PWy = - 59,226(
P/A
,
12
%,3) - 60,000(F/A,
12
%
,3)(P/F,12
%,6) = $- 244,817
Option Z has
th
e lowest cost PW value ($- 240,369). Keep the defender all 4 years, then re-
pl
ace it. Obviously, the
sa
me answer will result
if
the annual worth, or future worth,
of
each
option is calculated at
th
e MARR.
TABLE
11-3
Equivalent
Cash
Flows and
PW
Va
lu
es for a 6-Year Study Period Replacement
Ana
lysis,
Examp
le
11
.7
Time
in
Service,
Years
Defen-
Challen-
AW
Cash
Flows for
Each
Option, $lYear
Option
Option
der
ger
1 2 3 4 5 6
PW, $
X 2
4
- 60,585 - 60,585 - 60,000
-60
,000
-60
,000
-60
,
000
- 247,666
Y 3 3
- 59,226 - 59,226 - 59,226
- 60,000
-60
,000
-60
,
000
-244
,817
Z 4 2
- 57,923 - 57,923 - 57,923 - 57,923
-60,000
-60,000
-240,369
409
410
PROBLEMS
CHAPTER
11
Replacement and Retention Decisions
Comment
If
the study period
is
long enough, it is possible that the ESL
of
the challenger should be
determined and its
AW
value used
in
developing the options and cash flow series. An option
may include more than one life cycle
of
the challenger for its ESL period. Partial life cycles
of
the challenger can be included. Regardless, any years beyond the study period must be
disregarded for the replacement study, or treated explicitly,
in
order to ensure that equal-
service comparison
is
maintained, especially if
PW
is used to select the best option.
CHAPTER SUMMARY
It
is important
in
a replacement study to compare the best challenger with the
defender. Best (economic) challenger is described as the one with the lowest an-
nual worth (AW)
of
costs
for
some period
of
years .
If
the expected remaining life
of
the defender and the estimated life
of
the challenger are specified, the A W values
over these years are determined and the replacement study proceeds. However,
if
reasonable estimates
of
the expected market value (MV) and
AOe
for each year
of
ownership can be made, these year-by-year (marginal) costs help determine the
best challenger.
The economic service life (ESL) analysis is designed to determine the best
challenger's years
of
service and the resulting lowest total
AW
of
costs. The re-
sulting nc and AWe values are used in the replacement study procedure. The same
analysis can be performed for the ESL
of
the defender.
Replacement studies in which no study period (planning horizon) is specified
utilize the annual worth method
of
comparing two unequal-life alternatives. The
better A W value determines how long the defender is retained before replacement.
When a study period is specified for the replacement study, it is vital that mar-
ket value and cost estimates for the defender be as accurate as possible. When the
defender's remaining life is shorter than the study period, it is critical that the cost
for continuing service
be
estimated carefully. All the viable options for using the
defender and challenger are enumerated, and their
AW
equivalent cash flows are
determined. For each option, the
PW
or A W value is used to select the best option.
This option determines how long the defender is retained before replacement.
Foundations
of
Replacement
11.2 In a replacement analysis, what numeri-
cal value should be used as the first cost
for the defender? How is this value best
obtained?
11.1
Identify the basic assumptions made
specifically about the challenger alter-
native when a replacement study is
performed.
11.3 Why is it important to take a consultant's
viewpoint in a replacement analysis?
11.4 Chris is tired
of
driving the old used car
she bought 2 years ago for
$18,000.
She
estimates it is worth about $8000 now. A
car salesman gave her this deal:
"Look,
I'll give you $10,000
in
trade for this
year's model. This is
$2000 more than
you expect, and it is
$3000 more than the
current official Kelly Blue Book value for
your car.
Our
sales price for your new
car
is
only $28,000, which is $6000 less
than the manufacturer's sticker price
of
$34,000. Considering the extra $3000 on
the trade-in and the
$6000 reduction
from sticker, you are paying
$9000 less
for the new car.
So
, I am giving you a
great deal, and you get
$2000
more for
your old clunker than you estimated it
was worth. So, let's trade now.
Okay?"
If
Chris were to perform a replacement
study at this moment, what
is
the correct
first cost for
(a) the defender and (b) the
challenger?
11.5 New microelectronics testing equipment
was purchased 2 years ago by Mytesmall
Industries at a cost
of
$600,000. At that
time,
it
was expected to be used for 5 years
and then traded
or
sold for its salvage
value
of
$75,000. Expanded business in
newly developed international markets
is forcing the decision to trade now for a
new unit at a cost
of
$800,000.
The
cur-
rent equipment could be retained, if nec-
essary, for another 2 years, at which time
it would have a
$5000 estimated market
value.
The
current unit
is
appraised at
$350,000 on the international market,
and if it
is
used for another 2 years, it
will have
M&O
costs (exclusive
of
oper-
ator costs)
of
$125,000 per year. Deter-
mine the values
of
P,
n,
S,
and
AOC
for
this defender if a replacement analysis
were performed today.
.
PROBLEMS
411
11.6 Buffett Enterprises installed a new
fire monitoring and control system for
its manufacturing process lines
in
California exactly 2 years ago for
$450,000 with an expected life
of
5 years.
The
market value was described then
by the relation
$400,000 - 50,000kI4,
where k was the years from time
of
pur-
chase. Previous experience with fire
monitoring equipment indicated that its
annual operating costs follow the rela-
tion
10,
000
+ 100!?.
If
the relations are
correct over time, determine the val ues
of
P, S, and AOC for this defender if
a replacement analysis is performed
(a) now with a study period
of
3 years
specified and
(b) 2 years from now with
no study period identified.
11.7 A machine purchased 1 year ago for
$85,000 costs
more
to operate than an-
ticipated. When purchased, the machine
was expected to be used for
10 years
with annual maintenance costs
of
$22,000 and a
$10
,
000
salvage value.
However, last year, it cost the company
$35,000 to maintain it, and these costs
are expected to escalate to
$36,500 this
year and increase by
$1500 each year
thereafter.
The
salvage value is now esti-
mated to be
$85,000 - $10,000k, where
k is the number
of
years since the ma-
chine was purchased. It is now estimated
that this machine will be useful for a
maximum
of
5 more years. Determine
the values
of
P,
AOC,
n,
and S for a re-
placement study performed now.
Economic Service Life
11.8 Halcrow, Inc., expects to replace a down-
time tracking system currently installed
on
CNC
machines.
The
challenger sys-
tem has a first cost
of
$70,000, an esti-
mated annual operating cost
of
$20,000,
a
maximum
useful life
of
5 years, and
a
$10,000 salvage value anytime it
is
412
CHAPTER
11
Replacement and Retention Decisions
replaced. At an interest rate
of
10% per
year, determine its economic service life
and coooesponding
AW
value. Work this
problem using a hand calculator.
11.9
Use a spreadsheet to work Problem 11.8
and plot the total
AW
curve and its com-
ponents,
(a) using the estimates origi-
nally made and
(b)
using new, more pre-
cise estimates, namely, an expected
maximum life
of
10 years, an
AOe
that
will increase
by
15% per year from the
initial estimate
of
$20,000, and a salvage
value that
is
expected to decrease
by
$1000 per year from the $10,000 esti-
mated for the first year.
11.10
An
asset with a first cost
of
$250,000
is
expected to have a maximum useful
lif
e
of
10
years a
nd
a market value that decreases
$25,000 each year. The annual operating
cost is expected
to
be constant at $25,000
per year for 5 years and to increase at a
substantial
25
% per year thereafter.
The
interest rate
is
a low 4% per year, because
the company,
Public Services Corp., is
majority-owned
by
a municipality and re-
garded as a semiprivate corporation that
enjoys public project interest rates on its
loans.
(a) Verify that the
ESL
is 5 years. Is
the ESL sensitive to the changing market
value and
AOe
estimates? (b) The engi-
neer doing a replacement analysis deter-
mines that this asset should have an ESL
of
10
years when it
is
pitted against any
challenger.
If
the estimated
AOe
series
has proved to be correct, determine the
minimum market value that will make
ESL equal
10
years. Solve by hand or
spreadsheet as
in
structed.
11.11 A new gear grinding machine for com-
posite materials has a first cost
of
P =
$100,000 and can be used for a maxi-
mum
of
6 years. Its salvage value
is
estimated by the relation S = P(0.85)/,
11.12
where n is the number
of
years after
purchase.
The
operating cost will be
$75,000 the first year and will increase
by
$10,000 per year thereafter. Use i =
18% per year.
(a) Determine the economic service
life and corresponding A W
of
this
challenger.
(b) It was hoped that the machine
is
economically justified for retention
for all 6 years, but that is not the case
since the
ESL
in
part (a)
is
consider-
ably less than 6 years. Determine
the reduction in first cost that would
have to be negotiated to make the
equivalent annual cost for a
fu
ll
6 years
of
ownership numerically
equal to the A W esti mate deter-
mined for the calculated ESL. As-
sume all other estimates remain the
sa
me
, and neglect the fact that this
lower
P value will still not make a
newly calculated
ESLequal6
years.
(a)
Set up a general (ce
ll
reference for-
mat) spreadsheet that will indicate
the ESL and associated A W value
for any challenger asset that has a
maximum useful life
of
10 years.
The
relation for A W should be a
single-cell formula to calculate A W
for each year
of
ownership, using
all the necessary estimates.
(b) Use your spreadsheet to find the
ESL
and AW values for the esti-
mates tabulated. Assume
i = 10%
per year.
Estimated Estimated
Year
Market
Value, $
AOC,$
0
80,000
0
60,000 60,000
2 50,000 65,000
3
40,000 70,000
4 30,000 75,000
5 20,000
80,000
11.1
3 A piece
of
equipment has a first cost
of
$150,000, a maximum useful life
of
7 years, and a salvage value described by
S = 120,000 - 20,000k, where k
is
the
number
of
years since it was purchased.
The
salvage value does not go below
zero. The
AOe
series is estimated using
AOe
= 60,
000
+
1O
,
000k.
The interest
rate is
15
% per year. Determine the eco-
nomic service life
(a) by hand solution,
using regular AW computations, and
(b) by computer, using annual marginal
cost estimates.
11
.14 Determine the economic service life and
corresponding
AW
for a machine that
has the following cash flow estimates.
Use an interest rate
of
14% per year and
hand solution.
Salvage
Operating
Year
Value, $ Cost, $
0
100
,000
75,000
- 28,000
2 60,000
- 31,000
3
50,000
- 34,000
4 40,000
- 34,000
5
25,000
-3
4,000
6
15
,000
- 45,000
7 0
- 49,000
II
. I 5 Use the annual marginal costs to find the
economic service life for
Problem 11.14
on a spreadsheet. Assume the salvage
values are the best estimates
of
future
market value. Develop an Excel chart
of
annual marginal costs
(Me)
and A W
of
Me
over the 7 years.
Replacement Study
11
.16 During a 3-year period Shanna, a project
manager with Sherholme Medical De-
v
ic
es, performed replacement studies
on
microwave-based cancer detection equip-
ment used
in
the diagnostic labs. She tab-
ulated the ESL and
AW
values each year.
Defender
PROBLEMS
413
(a)
What
decision should be made each
year?
(b)
From the data, describe what
changes took place
in
the defender
and challenger over the 3 years.
Maximum
ESL,
AW,
Life, Years Years $/Year
First
Year
, 200X
3 3
-
10
,000
Challenger I
]0
5
-
15
,000
Second Year, 200X
+ I
Defender 2 I
-
14
,000
Challenger I
10
5
-
15
,000
Third
Year
, 200X
+2
Defender I I
-
14
,000
Challenger 2 5 3
- 9,000
11
.
17
A consulting aerospace engineer at
Aerospatial estimated A W values for a
presently
owned
, highly accurate steel
rivet inserter based on company records
of
similar equipment.
If
Retained This
Number
of
Years
2
3
4
5
AWValue,
$lYear
- 62,000
- 50,000
- 47,000
-
53
,000
- 70,000
A challenger has
ESL
= 2 years and
AWe
=
$-49,000
per year.
If
the con-
sultant must recommend a replace/retain
decision today, should the company pur-
chase the challenger?
The
MARR
is
15
% per year.
11.18
If
a replacement study is performed and
the defender is selected for retention for n
D
years, explain what should be done I year
later if a new challenger is identified.
11.19 BioHealth, a biodevice systems leasing
company, is considering a new equipment
414
CHAPTER II Replacement and Retention Decisions
purchase to replace a currently owned
asset that was purchased 2 years ago for
$250,000. It
is
appraised at a current
market value
of
only $50,000. An up-
grade is possible for
$200,000 now that
would be adequate for another 3 years
of
lease rights, after which the entire sys-
tem cou
ld
be so
ld
on the international
circuit for
an
estimated $40,000.
The
challenger can be purchased at a cost
of
$300,000, has an expected life
of
] 0 years, and has a $50,000 salvage value.
Determine whether the company should
upgrade
or
replace at a
MARR
of
12%
per year. Assume the
AOC
estimates are
the
same
for both alternatives.
1l.20
For
the estimates
in
Problem 11.19, use a
spreadsheet-based analysis to determine
the
maximum
first cost for the augmen-
tation
of
the cun'ent system that will
make the defender and challenger break
even. Is this a maximum
or
minimum for
the upgrade,
if
the current system is to be
retained?
1l.
21 A lumber company that cuts fine woods
for cabinetry
is
evaluating wheth
er
it
should retain the current bleaching system
or
replace it with a new one. The relevant
costs for each system are known
or
esti-
mated. Use
an
interest rate
of
10% per year
to
(a) perform the replacement analysis
and
(b) determine the minimum resale
price needed to make the challenger re-
placement choice now. Is this a reasonable
amount to expect for the current system?
Current
New
System System
First cost 7 years ago, $ - 450,000
First cost, $
- 700,000
Remaining life, years
5
10
Current market value, $
50.000
AOC,
$ per year -
160
,000
-
150
,000
Future salvage, $
0
50,000
11.22 Five years ago, the
Nuyork
Port
Authority purchased several container-
ized transport vehicles for
$350,000
each. Last year a replacement study was
performed with the decision to retain the
vehicles for 2 more years. However, this
year the situation has changed in that
each transport vehicle is estimated
to
have a value
of
only
$8000
now.
If
they
are kept in service, upgrading at a cost
of
$50,000 will make them useful for up to
2 more years. Operating cost is expected
to be
$10,000 the first year and $15,
000
the second year, with no salvage value
at all. Alternatively, the company can
purchase a new vehicle with an
ESL
of
7 years, no salvage value, and an equiva-
lent annual cost
of
$-55,540
per year.
The
MARR
is 10% per year.
If
the bud-
get to upgrade the current vehicles is
available this year, use these estimates to
determine
(a) when the company should
replace the upgraded vehicles and
(b) the
minimum future salvage value
of
a new
vehicle necessary to indicate that purchas-
ing now is economically advantageous
to upgrading.
11
.23 Annabelle went to work this month for
Caterpillar, a heavy equipment manufac-
turing company. When asked to verify
the results
of
a replacement study that
concluded in favor
of
the challenger, a
new piece
of
heavy-duty metal forming
equipment for the bulldozer processing
plant, at first she conculTed because
th
e
numerical results were in favor
of
this
challenger.
Life, years
A
W,
$ per year
Challenger
Defender
4
-
80
,000
6 more
-
130
,000
Curious about past decisions
of
this same
kind, she learned that similar replacement