Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
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1.43
Use the rule
of
72
to
estimate the interest
rate that would be required for
$5000 to
accumulate to
$10,000
in
4 years.
1.44
If
you now have $62,500
in
your retire-
ment account and you want
to
retire when
FE
REVIEW PROBLEMS
1.45
An
example
of
an
intangible factor is
(a) Taxes
(b) Cost
of
materials
(c) Morale
(d)
Rent
l.46 The time it would take for money to dou-
ble at a simple interest rate
of
5% per year
is
closest
to
(a)
10
years
(b)
12
years
(c)
15
years
(d)
20 years
1.
47 At a compound interest rate
of
10% per
year,
$10,000 one year ago is now equiva-
lent to
(a) $8264
(b) $9091
(c) $11,000
(d)
$12,100
1.48
An
investment
of
$10,000 nine years ago
has accumulated
to
$20,000
now.
The
EXTENDED EXERCISE
EFFECTS OF COMPOUND INTEREST
EXTENDED EXERCISE
45
the account is worth $2 million, estimate
the rate
of
return that the account must
earn if you want to retire in
20 years
without adding any more money to the
account.
compound rate
of
return earned on the in-
vestment
is
closest
to
(a) 6%
(b)
8%
(c)
10%
(d)
12%
1.49
In
most engineering economy studies, the
best alternative is the one that
(a) Will last the longest time
(b) Is easiest to implement
(c) Costs the least
(d)
Is most politically correct
l.
50 The cost
of
tuition at a certain public uni-
versity was
$160 per credit-hour 5 years
ago. The cost today (exactly 5 years later)
is
$235. The annual rate
of
increase
is
closest to
(a) 4%
(b) 6%
(c)
8%
(d)
10%
In
an
effort
to
maintain compliance with noise emission standards on the pro-
cessing floor, National Semiconductors requires the use
of
noise-measuring in-
struments. The company plans
to
purchase new portable systems at the end
of
next year at a cost
of
$9000 each. National estimates the maintenance cost to be
$500 per year for 3 years, after which they will be salvaged for $2000 each.
46
CHAPTER I
Foundations
of
Engineering Economy
Questions
~
>
.
~
E-Solve
1.
2.
Construct the cash flow diagram. For a compound interest rate
of
8%
per
year, find the equivalent
F value after 4 years, using calculations by hand.
Find the
F value
in
question
1,
using a spreadsheet.
3.
Find the F value if the maintenance costs are $300, $500, and
$lOOO
for
each
of
the 3 years. By how much has the F value changed?
4.
Find the F value in question 1 in terms
of
dollars needed
in
the future with
an
adjustment for inflation
of
4% per year. This increases the interest rate
from 8% to 12.32% per year.
CASE
STUDY
DESCRIBING ALTERNATIVES
FOR
PRODUCING REFRIGERATOR SHELLS
Background
Large refrigerator manufacturers like Whirlpool, Gen-
eral Electric, Frigidaire, and others may subcontract
the molding
of
their plastic liners and door panels. One
prime national subcontractor
is
Innovations Plastics. It
is
expected that
in
about 2 years improvements
in
me-
chanical properties will allow the molded plastic to
sustain increased vertical and horizontal loading, thus
significantly reducing the need for attached metal
an-
chors for some shelving. However, improved molding
equipment will be needed
to
enter tbis market.
The
company president wants a recommendation on
whether Innovations should plan on offering the new
technology
to
the major manufacturers and an estimate
of
the necessary capital investment to enter this market
early.
You
work as an engineer for Innovations. At this
stage, you are not expected to perform a complete en-
gineering economic analysis, for not enough informa-
tion
is
available.
You
are asked to formulate reason-
able alternatives, determine what data and estimates
are needed for each one, and ascertain what criteria
(economic and noneconomic) should be utilized to
make the
final
decision.
Information
Some information useful
at
this time is as follows:
•
The
technology and equipment are expected to last
about
10 years before new methods are developed.
• Inflation and income taxes will not be considered
in
the analysis.
•
The
expected returns
on
capital investment used
for the last three new technology projects were
compound rates
of
15, 5, and 18%. The 5% rate
was the criterion for enhancing an employee-
safety system on an existing chemical-mixing
process.
• Equity capital financing beyond $5 million
is
not
possible.
The
amount
of
debt financing and its cost
are unknown.
• Annual operating costs have been averaging 8%
of
first cost for major equipment.
• Increased annual training costs and salary require-
ments for handling the new plastics and operat-
ing new equipment can range from $800,000 to
$1.2 million.
There are two manufacturers working on the
new-
generation equipment.
You
label these options as
alternatives A and B.
Case Study Exercises
1.
Use the first four steps
of
the decision-making
process to generally describe the alternatives and
identify what economic-related estimates you
will need to complete
an
engineeling economy
analysis for the president.
2. Identify any noneconomic factors and criteria to
be considered
in
making the alternative selection.
3. During your inquiries about alternative B from
its manufacturer, you learn that this company has
already produced a prototype molding machine
and has sold it
to
a company
in
Germany for
CASE STUDY
$3 minion
(U.S. dollars). Upon inquiry, you fur-
ther discover that the German company already
has unused capacity on the equipment for manu-
facturing plastic shells. The company is willing
to sell time on the equipment to Innovations
immediately to produce its own shells for
U.S.
delivery. This could allow
an
earlier market entry
into the
United States. Consider this as alterna-
tive C, and develop the estimates necessary to
evaluate C at the same time
as
alternatives A
and B.
47
2
w
I-
«
I
u
Factors:
How
Time and
Interest Affect
Money
In
the
previous
chapter
we
learned
the
basic
concepts
of
engineering
econ-
omy
and
their
role in
decision
making.
The
cash
flow
is
fundamental
to
every
economic
study. Cash
flows
occur
in
many
configurations
and
amounts-
isolated
single
values, series
that
are
uniform,
and
series
that
increase
or
decrease
by
constant
amounts
or
constant
percentages.
This
chapter
devel-
ops
derivations
for
all
the
commonly
used
engineering
economy
factors
that
take
the
time
value
of
money
into
account.
The
app
li
cation
of
factors
is
illustrated
using
their
mathematical
forms
and
a
standard
notation
format.
Spreadsheet
functions
are
introduced
in
order
to
rapidly
work
with
cash
flow
series
and
to
perform
sensitivity
analysis.
The
case
study
focuses
on
the
significant
impacts
that
compound
interest
and
time
make
on
the
value
and
amount
of
money.
LEARNING OBJECTIVES
Purpose:
Derive
and
use
the
engineer
i
ng
economy
factors
to
account
for
the
time
value
of
money
.
F/
Pand
P/
Ffactors
P / A
and
A/ P
factors
F/ A
and
A/
Ffactors
Interpolate
factor
values
PI G
and
A/ G
factors
Geometric
gr
adient
Calculate;
Calculate
n
Spr
eadsheets
This
chapter
will
help
you
:
1. Derive
and
use
the
compound
amount
factor
and
present
worth
factor
for
single
payments.
2.
Derive
and
use
the
uniform
series pr
esent
worth
and
capital
recovery factors.
3.
Der
ive
and
use
the
uni
form
series
compound
amount
and
sin
ki
ng
f
und
facto
r
s.
4.
Li
nea
rl
y i
nt
er
po
late to
dete
rmine a factor
va
lu
e.
5. Derive and use the
arithmetic
grad
ient
present
wort
h
and
uniform
series factors.
6. Derive
and
use
the
geometric
gradient
series
formulas
.
7.
Determine
the
interest rate (rate
of
return)
for
a
sequence
of
cash flows.
8.
Determine
the
number
of
years
required
for
equivalence in a
cash
flow
series.
9.
Deve
l
op
a
spreadsheet
to
perform
basic sensiti
vity
analysis
using
spreadsheet
functions.
50
CHAPTER 2
Factors: How Time and Interest Affect Money
2.1
SINGLE-PAYMENT
FACTORS
(F
/ P
AND
P/
F)
The
most fundamental factor in engineering economy is the one that determines
the amount
of
money F accumulated after n years (or periods) from a single pre-
sent worth
P,
with interest compounded one time
per
year (or period). Recall that
compound interest refers to interest paid on top
of
interest. Therefore, if an
amount
P
is
invested at time t = 0, the amount
FI
accumulated 1 year hence at
an interest rate
of
i percent
per
year will be
FI
= P +
Pi
=
PO
+ i)
where the interest rate
is
expressed in decimal form.
At
the end
of
the second
year, the amount accumulated
F2
is the amount after year 1 plus the interest from
the end
of
year 1 to the end
of
year 2 on the entire Fl.
F2
=
FI
+ Fli
=
P(l
+ i) + P(1 + i)i
[2.
1]
This
is
the logic used in Chapter 1 for compound interest, specifically in Exam-
ples
1.8
and 1.18.
The
amount
F2
can be expressed as
F2 =
P(l
+ i + i +
p)
=
P(l
+ 2i +
p)
=
P(l
+
i)2
Similarly, the amount
of
money accumulated at
th
e end
of
year 3, using Equa-
tion
[2.1], will be
F3 = F2 +
F2i
Substituting
P(l
+
i)
2 for F2 and simp
li
fying, we get
F3 =
P(l
+
i)
3
From the preceding values, it
is
evident by mathematical induction that the for-
mula can be generalized for
n years to
F = P(l + i)"
[2
.
2]
The
factor
(l
+
i)"
is
called the single-payment
compound
amount
factor
(SPCAF)
, but it
is
usually referred to as the F / P factor. This is the conversion
factor that, when multiplied by
P,
yields the future amount F
of
an initi
al
amount
P after n years at interest rate
i.
The
cash flow diagram is seen
in
Figure
2-1a.
Reverse the situation to determine the P value for a stated amount F that
occurs
n periods
in
the future. Simply solve Equation [2.2] for
P.
P-F
l
1 ]
(1 + i)"
[2.3]
SECTION
2.1
Single-Payment Factors (F I P and PI F)
P = gi
ve
n
p= ?
i = given
i = given
01-
--
-1-
-
-+
2
-----1
fl-
n
-+
_-
2
--
I1
--1
--
I
---
t
o
2 n - 2
F=
?
(a)
(b)
Figure 2-1
Cash
flow
di
ag
rams for singl
e-
paymenl
fa
c
lor
s: (a) find F
and
(b) find
P.
The expression
is
brackets
is
known as the single-payrnent present worth
factor
(SPPWF),
or
the P / F factor. This expression determines the present worth P
of
a given future amount F after n years at interest rate
i.
The cash flow diagram
is
shown
in
Figure
2-lh.
Note that the two factors derived here are for single payments; that is, they are
used to find the present
or
future amount when only
one
payment
or
receipt is
involved.
A standard notation has been adopted for all factors.
The
notation includes
two cash flow symbols, the interest rate, and the number
of
periods.
It
is always
in the general form
(X/ Y,i,n).
The
letter X represents what is sought, while the
letter
Y represents what is give
n.
For
example, F / P means find F when given
P.
The i is the interest rate in percent, and n represents the number
of
periods in-
volved. Thus,
(F
/ P,6%,20) represents the factor that is used to calculate the
future amount
F accumulated in 20 periods
if
the interest rate is 6% per period.
The
P
is
given.
The
standard notation, simpler to use than formulas and factor
names, will be used hereafter.
Table
2-
1 summarizes the standard notation and equations for the F / P and
P / F factors. This information is also included inside the front cover.
To simplify routine engineering economy calculations, tables
of
factor values
have been prepared for interest rates from 0.25 to 50% and time periods from 1
to large n values, depending on the
i value. These tables, found at the rear
of
the
TABLE
2-1 F / P and P / F Factors: Notation and Equations
Factor
Standard
Notation
Equation
n - I
Notation
Name
Find/Given
Equation
with
Factor
Formula
(FIP,i,n) Single-payment
FI P
d~::JJ
F = P(FIP,i,n)
F =
P(1
+ 0"
compo
und
amount
y r J
(PIF,i,n)
Single-payment
PI F
trf
=t£
P = F(P I F,i,n)
P = F[I /
(l
+ i)")
present worth
Ire
II'>
51
n
F = given
Excel
Functions
FV(i
%,
n"P)
PV(i%
,n
"F)
52
Q-
S
ol
v
CH
A
PTER
2 Factors: How Time and Interest Affect Money
book,
are
arranged with factors across the top
and
the
number
of
periods n down
the left.
The
word discrete in the title
of
each table emphasizes that these tables
utilize the end-of-period convention and that interest is
compounded
once
each
interest period.
For
a given factor, interest rate, and time, the correct factor value
is found at the intersection
of
the factor
name
and
n.
For
example, the value
of
the fa
ctor
(P / F,5%, 1 0) is found in the P / F
column
of
Table 10 at period 10 as
0.6139. This value is
determined
by using Equation [2.3].
1
(P/F
,5%,
1O)
=
(1
+ i)"
(1.05)10
1.6~89
=
0.6139
For
solution
by
computer, the F value is calculated by the
FV
function using the
format
FV(i
%
,n"P)
An = sign must
precede
the function when it is
en
tered.
The
amount
P is deter-
mined using the
PV
function with the format
PV(i
%
,n"F)
These
functions are included in Table 2-1.
Refer
to Appendix A
or
Excel online
help for
more
information on the
FV
and
PV
functions.
Examples
2.1 and 2.2
illustrate solutions by
computer
using both
of
these functions.
An
industrial engineer received a bonus
of
$12,000 that he will invest now.
He
wants
to
calculate the equivalent value after 24 years, when he plans
to
use all the resulting money
as the down payment on
an
island vacation home. Assume a rate
of
return
of
8% per year
for each
of
the 24 years. (a) Find the amount he can pay down, using both the standard
notation and the factor formula.
(b) Use a computer to find the amount he can pay down.
(
a)
Solution
by
Hand
The symbols and their values are
P = $12,000
F=?
i = 8% per year
n = 24 years
The
cash flow diagram is the same as that
in
Figure 2-
1a
.
Standard notation: Determine
F,
using the F / P factor for 8% and 24 years. Table
13
provides the factor value.
F =
P(F
/P,i,n) =
12
,000(F/
P,8
%,
24)
= 12,000(6.3412)
= $76,094.40
SECTION
2. 1 Single-Payment
Factor
s
(F
I P and PI F )
Factor formula: Apply Equation [2.2) to calculate the future worth
F.
F =
P(l
+ i)" = 12,000(1 +
0.08i
4
=
12
,000(6.341181 )
= $76,
094
.17
The
slight difference in answers
is
due
to the round-off
error
introduced by the tabulated
factor values. An equi valence interpretation
of
this result is that $12,
000
today is worth
$76,094 after
24
years
of
growth
at
8% per
year
,
compounded
annually.
(b) Solution
by
Computer
To find the future value use the
FY
function that has the format
FY(i
%,
n,A,
P).
The
spreadsheet will look like the
one
in
Figure I-
Sa,
except
the cell entry is
FY
(8%,24" 12000).
Th
e F va
lue
displayed by Excel is ($76,094.17) in red to indicate a
cash outflow.
The
FY
function has performed the computation F =
P(l
+
i)
" =
12,000(1 + 0.08
?4
and
pre
sented the
answer
on the screen.
EXAMPLE 2.2
~
Hewlett-Packard has completed a study indicating that $50,000 in reduced maintenance
this year (i.e.,
year
zero) on one processing line resulted from improved integrated circuit
(IC) fabrication technology based on rapidly changing designs.
(a)
If
Hewlett-Packard considers these types
of
savings worth
20
%
per
year, find the
equivalent val
ue
of
this result after 5 years.
(b)
If
the $50,000 maintenance savings occurs now, find its
equiva
l
ent
value 3 years
earli
er
with interest
at
20%
per
year
.
(c)
Deve
lop a spreadsheet to
answer
the two parts
above
at
compound
rates
of
20% and
5% per year. Additionally develop an Excel column chart indicating the
equivalent
values
at
the three
diff
erent times for both rate-of-return values.
Solution
(a)
The
cash
fl
ow diagram appears as in
Figure
2-1 a.
The
symbols and their values are
P = $
50
,
000
F
=?
i = 20%
pe
f
year
n = 5
year
s
Use the F I P factor to determine F after 5 years.
F =
P(F
I P,i,n) = $50,000(F I P,20%,5)
=
50
,000(2.4883)
=
$124
,
415.00
The
function
FY(20
%
,5
,,50000) provides the
same
answer.
See
Figure 2-2a, cell
C4
.
(b)
In
this case, the cash flow diagram appears as in Figure
2-1
b with F placed
at
time
t = °
aDd
the P value placed 3 years earlier at t = - 3.
The
symbols and their values
are
P
=?
F = $50,000 i =
20
% per
year
n = 3 years
53
Q-Solv
Q-Solv
54
CHAPTER 2 Factors: How Time and Interest Affect Money
X
IoAICIO
SOft
Ex
cel E.
ampl
e 2 2
Itl
s I!lfilEf
H
PV(20
%,
3"SOOOO)
Sh
eet
I
Sheet2
Sheet3
Sheet'1
SheetS Sheet6 5heet7
SheetS
(a)
X:
MicrosoitE~cel
-
EKample
i2
-V~-
---------
---
-
~--
-
-
- -
--
- -
--
-
III!J
-
Ef
0
14
A
1
2
=
=F
V(D5
,5
,0,-50000)
BC
D
01
E
E
xa
mple
2.2(c) I
F
L
G
3
4
5
En
d of
De
termin
e P
and
F
at
varyin
g i
rate
s
r----------,
~=_c___,_______=-_,_______,_,__----=~
--.:.:
':L:::--I=
PV
(C5,3" - 50000)
ear
E
urva
l
ence
at
i
;::
20
%
~
------=
LI=
I
6
-3
7
-2
8
-1
9
0
--
10
1
11
2
12
3
13
4
5
Figure
2-2
P
,3
yea
rs
ago
$
28,93
5
$4
3,192
PV(D5,3" - 50000)
C
urrent
s
avings
$
50,000
$50,000
= FV(C5,5,0, - 50000)
Sh
eet2
Sheet3
(b)
$140,000
$1
20,000
$100,000
St
,
H
1
(a) Q-solv spreadsheet
fo
r Example 2.2(
a)
and (bl; (b) complete spreadsheet with graphic,
Ex
ample 2.2.