Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
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SECTION 6.3 Evaluating Alternatives
by
Annual Worth Analysis
Solution by Computer
The spreadsheet layout (Figure
6--4) shows the cash flows for the investment, the operating
costs, and annual income in separate columns. The functions use global variable format for
faster sensitivity analysis.
(a
and b) The capital recovery value
of
$5822 is displayed in cell B7, which is deter-
mined
by
a
PMT
function with an embedded NPV function. Cells C7 and
D7
also
use the
PMT function to find the annual equivalent for costs and incomes, again
with
an
embedded NPV function.
Cell
Pll
displays the final answer
of
AW
=
$-5362,
which is the sum
of
all
three
AW
components
in
row 7.
(c)
To
llnd the income (column D) necessary to justify the project, a value
of
AW
= $0
must be displayed
in
cell
Pll.
Other estimates remain the same. Because all the an-
nual incomes
in
column D receive their value from cell B4, change the entry
in
B4
until
PI1 displays
'$0
'. This occurs at $6562. (These amounts are not shown in B4
a
nd
FI!
of
Figure 6--4.) The owner
of
PizzaRush would have to increase the esti-
mate
of
extra income for the new system from $1200 to $6562 per year to make a
10
% return. This
is
a substantial increase.
Year
A.Wva
lue
0
1
2
3
4
Investment
$
$
$
$
$
$
income
1,200
1,200
1,
200
1,
200
1,200
~
E-Solve
5
$
300
(800)
(850)
$
,,\
2
0
0
~1~
;B~4
1
··_
.J
...
..
=.."",..=,....,.,
"
'-
+-
..
19<
11
Figure
6-4
Spre
ad
sheet solution, Example 6.3(a) and (b).
225
226
CHAPTER
6 Annual Worth Analysis
EXAMPLE
6.4
~
In
Example 5.12,
PW
analysis was performed (a) over the
LCM
of
24 years and (b) over a
study period
of
6 years. Compare the two plans for Southeastern Cement, under the same
conditions, using the
AW
method.
The
MARR
is
15%. Solve by hand and by computer.
Solution
by Hand
(a) Even though the two components
of
plan A, movers and pads, have different lives,
the
AW
analysis is conducted for only one life cycle
of
each component. Each AW
is comprised
of
CR
plus the annual operating cost.
Use
Equation [6.3] to find the
CR
amount.
AW
A
=
CRm
overs
+
CR
pad
+
AOCm
overs
+ AOC
pad
CRm
ov
ers
=
-90
,000(A/
P,15
%,
8)
+
1O,000(A
/ F,15
%,
8)
= $-
19
,328
CR
pad
=
-28
,000(A/P,15%,12) +
2000(A/F
,
15
%,
J2)
=
$-5096
Total AOC
A
=
$-J2000
- 300 =
$-12,300
The total AW for each plan is
AW
A
=
-19,328
- 5096 - 12,300 =
$-36,724
AW
B
=
CR
co
n
veyo
r + AOC
co
nv
eyo
r
=
-175
,000(A/P,15%,24) +
1O
,000(A/F,15%,24) - 2500 =
$-29
,646
Select plan B, the same decision as that for
PW
analysis.
(b)
For
the study period, perform the same analysis with n = 6 in all factors, after up-
dating the salvage values to the residual values.
CRmovers
= -90,000(A/P,15%,6) + 40,000(A/F,15%,6) = $- 19,212
CR
pad
= - 28,000(A/P,15%,6) + 2000(A/F,15%,6) = $- 7170
AW
A
=
-19,212
-7170
-
12
,300 = $- 38,682
AW B =
CR
co
n
vey
or
+ AOC
co
n
veyo
r
= -175,000(A/ P,
15
%,
6) +
25
,000(A/F,15%,6) - 2500
=
$-45
,886
Now, select plan A for its lower AW
of
costs.
Comment
There is a fundamental relation between the
PW
and
AW
values
of
part (a). As stated by
Equation
[6.1],
if
you have the
PW
of
a given plan, determine the AW by calculatingAW =
PW(A/ P,i,n); or
if
you have theAW, then
PW
=
AW(P
/ A,i,n). To obtain the correct value,
the
LCM
must be used for all n values, because the
PW
method
of
evaluation must take
place over an equal time period for each alternative to ensure an equal-service comparison.
The
PW
values, with round-off considered, are the same as determined
in
Example
5.
J 2,
Figure
5-9.
PW
A
= AW
A
(P/A,15%,24) = $- 236,275
PW
s
=
AW
B
(P/A,15%,24) = $- 190,736
SECTION 6.3 Evaluating Alternatives by Annual Worth Analysis
Solution by
Computer
(a) See Figure
6-5a.
This
is
exactly the same format as that used for the
PWevaluation
over the LCM
of
24 years (Figure
5-9),
except only the cash flo
wsfor
one life cycle
are shown here, and the NPV functions
at
the head
of
each column are now PMT
functions with the NPV function embedded. The cell tags detail two
of
the
PMT
functions, where the initial minus sign ensures the result is a
cost
amount in the total
AW
for each plan (cells H19 and H22). (The bottom portion
of
the spreadsheet is not
shown. Plan B continues through its entire life with the
$10,000 salvage value in
year 24, and the annual cost
of
$2500 continues through year 24.)
The
resulting
CR
and AW values obtained here are the same as those for the so-
lution by hand. Plan B is selected.
Plan A:
Plan B:
CRrn
overs = $- 19,328 (in B8)
AW
A
= $- 36,725 (in H19)
CR
co
nv
c
yor
= $- 27,146 (in F8)
CR
pad
=
$-5097
(in D8)
AWs
=
$-29,646
(in H22)
(b)
In Figure 6-5b, the lives are shortened to the study period
of
6 years. The esti-
mated residual values in year 6 are entered (row 16 cells), and all AOC amounts
(9
0,000)
$
$
(28,000)
$
$
(12,000)
$ $
(300)
(2,500)
$ (
12,000
)
$ $
(300) (2;500)
$ (12,000)
$ $
(300)
"(2,500)
$ (12,0
00
)
$ $
(300) (2,50
$
(1
2,000)
$ $
(300) (2,50
$ (12,000)
$
I :
(300)
"!
Bo
$ (12,000)
$
(300) (2,50
10
,0
00
$ (12,000)
$
I $
(300) (2,50
$ $
(300)
- $
""
(2,50
10
$ $ $
(
2,
11
$ $
"
$
"(
2,
12
$
2,
000 $ $
(
2,
$ (
2,
$
$
(a)
Figure
6-5
Spreadshe
et
solution
using
AW
comparison of
two
alternatives:
Ca)
one
life
cycle, (b)
study
period of 6
year
s, Example
6.4.
227
~
is-Solve
228
Re
ad
y
Figure
6-5
(Continued).
CHAP
TER
6 Annual Worth Analysis
(b)
beyond 6 years are deleted. When the n value in each
PMT
function is adjusted
from 8,
12,
or 24 years to 6
in
every case, new
CR
values are displayed, and cells
D20
and D21 display the new
AW
values. Now plan A is selected since it has a
lower
AW
of
costs. This
is
the same result as for the
PW
analysis
in
Figure 5- 10
for ExampJe 5.12h.
If
the projects are independent, the AW at the
MARR
is
calculated. All proj-
ects with
AW
:::::
0 are acceptable.
6.4
AW
OF A PERMANENT INVESTMENT
This section discusses the annual worth equivalent
of
the capitalized cost. Eval-
uation
of
public sector projects, such as flood control dams, irrigation canals,
bridges, or other large-scale projects, requires the comparison
of
alternatives that
have such long lives that they may be considered infinite in economic analysis
terms. For this type
of
analysis, the annual worth
of
the initial investment is the
perpetual annual interest earned on the initial investment, that
is,
A = Pi. This is
Equation
[5
.3]; however, the A value is also the capital recovery amount. (This
same
rel
ation will be used again when benefit/cost ratios are discussed.)
SECTION 6.4
AW
of
a Permanent Investment
Cash flows recurring at regular
or
irregular intervals are handled exactly as in
conventional
AW
computations; they are converted to equivalent
uniform
annual
amounts
A for
one
cycle.
This
automatically annualizes
them
for
each
succeed-
ing life cycle, as discussed in Section 6.1.
Add
all the A values to the
CR
amount
to find total
AW,
as in Equation [6.2].
EXAMPLE
6.5
c"'.
The U.S. Bureau
of
Reclamation
is
considering three proposals for increasing the
capacity
of
the main drainage canal in
an
agricultural region
of
Nebraska. Proposal A re-
quires dredging the canal
in
order to remove sediment and weeds which have accumu-
lated during previous years' operation. The capacity
of
the canal will have to be main-
tained
in
the future near its design peak flow because
of
increased water demand.
The
Bureau
is
planning
to
purchase the dredging equipment and accessories for $6S0,000.
The equipment
is
expected to have a lO-year life with a $17,000 salvage value. The an-
nual operating costs are estimated
to
total
$SO,OOO
. To control weeds in the canal itself
and along the banks, environmentally safe herbicides will
be
sprayed during the irriga-
tion season. The yearly cost
of
the weed control program is expected to be $120,000.
Proposal
B
is
to line the canal with concrete at
an
initial cost
of
$4 million. The lining
is
assumed to be permanent, but minor maintenance will be required every year at a cost
of
$SOOO.
In addition, lining repairs will have to be made every S years at a cost
of
$30,000.
Proposal
C
is
to
construct a new pipeline along a different route. Estimates are: an ini-
tial cos
tof$6
million, annual maintenanceof$3000forright-of-way, and a life
of
SO
years.
Compare the alternatives on the basis
of
annual worth, using an interest rate
of
S%
per year.
Solution
Since this
is
an investment for a permanent project, compute the
AW
for one cycle
of
all
recurring costs. For proposals A and C, the CR values are found using Equation [6.3],
with
n
A
= 10 and nc =
SO,
respectively. For proposal B, the CR
is
simply P(i).
Proposal A
CR
of
dredging equipment:
-
6S0
,000(A/P,S%,10) + 17,OOO(A/F,S%,10)
Annual cost
of
dredging
Annual cost
of
weed control
Proposal B
CR
of
initial investment: - 4,000,000(0.OS)
Annual maintenance cost
Lining repair cost:
- 30,OOO(A/F,S%,S)
Proposal C
CR
of
pipeline: - 6,000,000(A/ P,
S%,SO)
Annual maintenance cost
Proposal B
is
selected due to its lowest
AW
of
costs.
$ - 82,824
-SO,OOO
-120,000
$-2S2,824
$-200,000
-S,OOO
-S,429
-210,429
$-328,680
-3,000
$-331,680
229
230
CHAPTER 6 Annual Worth Analysis
Comment
Note the use
of
the
AI
F factor for the lining repair cost in proposal
B.
The AI F factor
is
used
in
stead
of
Al
P,
because the lining repa
ir
cost begins in year 5, not year 0, and
continues
ind
efinitely at 5-year intervals.
If
the 50-year life
of
proposal C is considered infinite, CR = P(i) =
$-300
,000,
in-
stead of
$-328,680
for n = 50. This
is
a sma
ll
economic difference. How long lives
of
40 or more years are treated economically
is
a matter
of
"local" practice.
EXAMPLE
6.6
.,
Q-
So
lv
~
E-Solve
An engineer with Becker Consulting has just received a bonus
of
$10,000.
If
she deposits
it
now at
an
interest rate
of
8% per year, how many years must the money accumulate before
she can withdraw
$2000
per
year forever? Use a computer to find the answer.
Solution by
Computer
Figure
6-6
presents the cash flow diagram. The first step is to
find
the total amount
of
money, call it P
",
that must be accumulated
in
year n, just 1 year prior
to
the first with-
drawal
of
the perpetual A = $2000 per year serie
s.
That is,
A
P" =
2000
0.08
= $25,000
$2000
o 2
n-J
II=
? t t
I--+--+--j
f---1--
____
,.---+--+--+---1
Jf-
I
--jf---+--i~~
oo
$JO
,
OOO
Figure 6-6
P"
=?
Diagram to determine n for a perpetual withdrawal, Example 6.6.
Use the NPER function
in
one cell
to
determine when the initial $10,000 deposit will ac-
cumulate to
$25,000 (Figure
6-7
, cell B4). The answer
is
11
.
91
years.
If
the engineer
leaves the money
in
for 12 years, and if 8%
is
earned every year, forever, the $2000 per
year is ensured.
Figure
6-7
also presents a more general spreadsheet solution
in
cells B7
d1J'ough
B
lJ.
Cell B I 0 determines the amount to accumulate
in
order to receive any amount (ce
ll
B9)
forever at 8% (cell B7), and
BII
includes the NPER function developed
in
cell reference
format for any interest
rate, deposit, and accumulated amount.
I
Ready
Interest rate
Amount deposited today
Amount to withdraw forever
Time required
Figure
6-7
CHAPTER SUMMARY
8%
$10,000
$~;~~
~
~R9
~
7
1
\ 11.91 years
Two spreadsheet solutions to find an n value using the NPER function, Example 6.6.
S
'iD'
( ; :
CHAPTER
SUMMARY
The
annual worth method
of
comparing alternatives is often preferred to the
present worth method, because the AW comparison is performed for only one
life cycle. This is a distinct advantage when comparing different-life alterna-
tives. A W for the first life cycle is the AW for the second, third, and all suc-
ceeding life cycles, under certain assumptions. When a study period is specified,
the
AW
calculation
is
determined for that time period, regardless
of
the lives
of
the alternatives. As in the present worth method, the remaining value
of
an
alternative at the end
of
the study period is recognized by estimating a
market
value.
For
infinite-life alternatives, the initial cost is annualized simply by multiply-
ing
P by
i.
For
finite-life alternatives, the AW through one life cycle is equal to
the perpetual equivalent annual worth.
231
232
CHAPTER 6 AnnuaJ Worth Analysis
PROBLEMS
6
.1
Assume that an alternative has a 3-year life
and that you calculated its annual worth
over its 3-year life cycle.
If
you were told to
provide the annual worth
of
that alternative
for a 4-year study period, would the annual
worth value you calculated from the alter-
native's 3-year life cycle be a valid esti-
mate
of
the annual worth over the 4-year
study period? Why or why not?
6.2 Machine A has a 3-year life with no sal-
vage value. Assume that you were told
that the service provided
by
these ma-
chines would be needed for only 5 years.
Alternative A would have to be repur-
chased and kept for only 2 years. What
would its salvage value have to be after
the 2 years in order to make its annual
worth the same as it is for its 3-year life
cycle at
an
interest rate
of
10% per year?
Year
Alternative
A,
$
Alternative
B,
$
0
-
10
,000
- 20,000
- 7,000
- 5,000
2 - 7,000 - 5,000
3 - 7,000 - 5,000
4
- 5,000
5 - 5,000
Alternatives Comparison
6.3 A consulting engineering firm is consider-
ing two models
of
SUVs for the company
principals. A
GM
model will have a first
cost
of
$26,000, an operating cost
of
$2000, and a salvage value
of
$12,000
after 3 years. A Ford model will have a
first cost
of
$29,000, an operating cost
of
$1200, and a $15,000 resale value after
3 years. At an interest rate
of
15% per year,
which model should the consulting firm
buy? Conduct an annual worth analysis.
6.4 A large textile company is trying to decide
which sludge dewatering process it should
use ahead
of
its
sludge drying operation. The
costs associated with centrifuge and belt
press systems are shown below. Compare
them on the basis
of
their annual worths,
using an interest rate
of
10% per year.
Centrifuge
Belt
Press
First cost, $ -250,000
- 170,000
Annual operating cos
t,
-3
1,000
-3
5,000
$/year
Overhaul in year 2, $
-26,0
00
Salvage value, $
40,000
10
,0
00
Life, years
6 4
6.5
A chemical engineer is considering two
styles
of
pipes for moving distillate from a
refinery to the tank farm. A small pipeline
will cost less to purchase (including valves
and other appurtenances) but will have a
high head loss and, therefore, a higher
pumping cost. The small pipeline will cost
$l.7
million installed and will have an
operating cost
of
$12,000 per month. A
larger-diameter pipeline will cost $2.1 mil-
lion installed, but its operating cost will be
only
$8000 per month. Which pipe size
is more economical at an interest rate
of
1 % per month on the basis
of
an annual
worth analysis? Assume the salvage value
is
10%
of
the first cost for each pipeline at
the end
of
the lO-year project period.
6.6
Polymer Molding, Inc., is considering two
processes for manufacturing storm drain
s.
Plan A involves conventional injection
molding that will require making a steel
mold at a cost
of
$2 million. The cost for
inspecting, maintaining, and cleaning the
molds is expected to be
$5000 per month.
Since the cost
of
materials for this plan is
expected to be the same as for the other
plan, this cost will not be included in the
comparison. The salvage value for plan A
is expected to be
10%
of
the first cost.
Plan B involves using an innovative process
known as virtual engineered composites
wherein a floating mold uses an operating
system that constantly adjusts the water
pressure around the mold and the chemi-
cals entering the process.
The
first cost to
tool the floating mold
is
only $25,000, but
because
of
the newness
of
the process,
personnel and product-reject costs are ex-
pected to be higher than those for a conven-
tional process.
The
company expects the
operating costs to be
$45,000 per month for
the first 8 months and then to decrease to
$10,000 per month thereafter. There will
be no salvage value with this plan. At an
interest rate
of
12
% per year, compounded
monthly, which process should the com-
pany select on the basis
of
an annual worth
analysis over a 3-year study period?
6.7 An industrial engineer
is
considering two
robots for purchase by a fiber-optic manu-
facturing company. Robot X will have a
first cost
of
$85,000, an annual mainte-
nance and operation
(M&O) cost
of
$30,000, and a $40,000 salvage value.
Robot Y will have a first cost
of
$97,000,
an annual
M&O
cost
of
$27,000, and a
$48,000 salvage value. Which should
be
selected on the basis
of
an annual worth
comparison at an interest rate
of
12
% per
year?
Use a 3-year study period.
6.8 Accurate airflow measurement requires
straight unobstructed pipe for a minimum
of
10 diameters upstream and 5 diameters
downstream
of
the measuring device.
In
one particular application, physical con-
straints compromised the pipe layout, so
the engineer was considering installing the
airflow probes
in
an
elbow, knowing that
flow measurement would be less accurate
but good enough for process control. This
was plan A, which would be acceptable for
only 2 years, after which a
more
accurate
flow measurement system with the
same
costs as plan A will be available. This plan
would have a first cost
of
$25,
000
with
annual maintenance estimated at
$4000.
Plan
B involved installation
of
a recently
PROBLEMS
233
designed submersible airflow probe.
The
stainless steel probe could be installed
in
a
drop pipe with the transmitter located
in
a
waterproof enclosure on the handrail. The
cost
of
this system would be $88,000, but
because it
is
accurate, it would not have to
be replaced for at least 6 years. Its mainte-
nance cost is estimated to be
$1400
per
year. Neither system will have a salvage
value. At an interest rate
of
12% per year,
which one should
be
selected on the basis
of
an annual worth comparison?
6.9 A mechanical engineer is considering
two types
of
pressure sensors for a low-
pressure steam line.
The
costs are shown
below. Which should be selected based on
an annual worth comparison at an interest
rate
of
12 % per year?
Type X Type Y
First cost, $
- 7,650 -
12
,900
Maintenance cost, $/year
-1,200
- 900
Salvage value, $
0 2,000
Life, years
2 4
6. 10 The machines shown below are under
consideration for an improvement to an
automated candy
bar
wrapping process.
Determine which should be selected on
the basis
of
an annual worth analysis using
an interest rate
of
15
% per year.
First cost, $
Annual cost, $/year
Salvage value, $
Life, years
Machine
C
Machine
D
- 40,000
-10,000
12,000
3
-
65
,000
- 12,000
25,000
6
6.11 Two processes can
be
used for producing
a polymer that reduces friction loss
in
en-
gines. Process K will have a first cost
of
$160,000, an operating cost
of
$7000 per
month, and a salvage value
of
$40,000
after its 2-year life. Process L will have a
first cost
of
$21
0,000, an operating cost
of
$5000 per month, and a $26,000 salvage
value after its 4-year life. Which process
234
CHAPTER 6 Annual Worth Analysis
should be selected on the basis
of
an an-
nual worth analysis at
an
interest rate
of
12% per year, compounded monthly?
6. l2 Two mutually exclusive projects have the
estimated cash flows shown below.
Use an
annual worth analysis to determine which
should be selected at an interest rate
of
10
% per year.
Project
Q
Proje
ct
R
First cost, $
Annual cost,
$/year
Salvage value, $
Life, years
- 42,000
-6
,000
o
2
-8
0,000
-7,000
year I, increasing
by
$ I 000 per year
4,000
4
6.13 An environmental engineer is considering
three methods for disposing
of
a nonhaz-
ardous chemical sludge: land application,
fluidized-bed incineration, and private dis-
posal contract.
Th
e details
of
each method
are shown below. Det
er
mine which has
the least cost on the basis
of
an annual
worth comparison at
12
% per year.
Fi
rst cost, $
Annual cost, $/year
Salvage
va
lu
e, $
Life, years
Land Inciner-
Application
ation
Contract
-
110
,000
-95
,000
15
,000
3
-8
00,000 0
-6
0,000 -
190
,0
00
250,000 0
6 2
6.
14 A state highway department is trying to
decide whether it shou
ld
"hot-patch" a short
section
of
an
existing county road or resur-
face
it.
If
the hot-patch method is used, ap-
proximately
300 cubic meters
of
material
would be required at a cost
of
$700 per
cubic meter (in place). Additionally, the
shoulders will have to be improved at the
same time at a cost
of
$24,000. These im-
provements will last 2 years, at which time
they will have to be redone.
The
annual
cost
of
routine maintenance on the patched
up road would be
$5000. Alternatively, the
state can resurface the road at a cost
of
$850,000. This surface will last 10 years if
the road
is
maintained at a cost
of
$2000
per year beginning 3 years from now. No
matter which alternative
is
selected, the
road will be completely rebuilt in 10 years.
At an interest rate
of
8% per year, which
alternative should the state select on the
basis
of
an annual worth analysis?
Permanent
Investments
and
Projects
6.15 How much must you deposit in your re-
tirement account starting
now
and contin-
uing each year through year 9 (i.e., 10 de-
posits)
if
you want to be able to withdraw
$80,000 per year forever beginning 30 years
from now? Assume the account earns inter-
est at
10% per year.
6.16 What
is
the difference
in
annual worth
between an investment
of
$100,000 per
year for
100 years and an investment
of
$100,000 per year forever at an interest
rate
of
10% per year?
6.
17
A stockbroker claims she can consistently
earn
15
% per year on an investor's money.
If
she invests $20,000 now, $40,000 two
years from now, and
$10,000 per year
through year
11
starting 4 years from now,
how much money can the client withdraw
every year forever, beginning 12 years
from now,
if
the stockbroker delivers what
she said and the account earns 6% per year
from year 12 forward? Disregard taxes.
6.18 Determine the perpetual equivalent annual
worth (in years 1 through infinity)
of
an
investment
of
$50,000 at time 0 and
$50,000 per year thereafter (forever) at
an
interest rate
of
10%
per
year.
6.19
The
cash flow associated with landscaping
and maintaining a certain monument
in
Washington, D.C., is $100,000 now and
$50,000 every 5 years forever. Determine
its perpetual equivalent annual worth (in
years 1 through infinity) at an interest rate
of
8% per year.
6.20 The cost associated with maintaining rural
highways follows a predictable pattern.