Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
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SECTION
1.11
Rule of 72: Estimating Doubling Time and Interest Rate
1.11
RULE
OF 72: ESTIMATING DOUBLING TIME
AND
INTEREST
RATE
Sometimes it is helpful to estimate the number
of
years n
or
the rate
of
return i
required for a single cash flow amount to double in size.
The
rule
of
72 for com-
pound interest rates can be used to estimate
i
or
n,
given the other value. The
estimation is simple; the time required for an initial single amount to double in
size with compound interest is approximately equal to
72
divided by the rate
of
return in percent.
E
·
d 72
stImate n
=
-.
I
[1.9]
For
example, at a rate
of
5%
per
year, it would take approximately
72/5
=
14.4 years for a current amount to double.
(The
actual time required is 14.3
years, as will be shown in Chapter 2.) Table
1-4
compares the times estimated
from the rule
of
72 to the actual times required for doubling at several com-
pounded rates. As you can see, very good estimates are obtained.
Alternatively, the
compound
rate i in percent required for money to double in
a specified period
of
time n can
be
estimated by dividing
72
by the specified n
value.
72
Estimated i = -
n
[1.10]
In order for money to double in a time period
of
12 years, for example, a com-
pound rate
of
return
of
approximately
72/12
= 6%
per
year would be required.
The exact answer
is
5.946% per year.
If
the interest is simple, a rule
of
100 may be used in the same way. In this case
the answers obtained will always
be
exactly correct. As illustrations, money dou-
bles in exactly 12 years at
100/12
= 8.33% simple interest. Or, at 5% simple in-
terest it takes exactly
100/5
= 20 years to double.
TABLE
1-4
Doubling Time Estimates Using the Rule
of
72 and the Actual
Time
Using Compound Interest Calculations
Doubling Time, Years
Rate
of
Return,
Rule-of-72 Actual
%
per
Year Estimate
Years
1
I'
72 70
2 36
35.3
5
14.4 14.3
10 7.2 7.5
20 .
3.6 3.9
40
1.8
2.0
35
36
~
E-Solve
CHAPTER I Foundations
of
Engineering Economy
1.12
SPREADSHEET
APPLICATION-SIMPLE
AND
COMPOUND
INTEREST,
AND
CHANGING
CASH
FLOW
ESTIMATES
The example below demonstrates how an Excel spreadsheet can be used to
obtain equivalent future values. A key feature is the use
of
mathematical relations
developed in the cells to perform sensitivity analysis for changing cash flow
estimates and the interest rate.
To
answer these basic questions using hand solu-
tion can be time-consuming; the spreadsheet makes it much easier.
EXAMPLE
1.18
'Co
A Japan-based architectural
firm
has asked a United States-based software engineering
group to infuse
GIS (geographical information system) sensing capability via satellite
into monitoring software for high-rise structures
in
order to detect greater-than-expected
horizontal movements. This software could be very beneficial
as
an
advance warning
of
serious tremors
in
earthquake-prone areas
in
Japan and the United States. The inclusion
of
accurate GIS data
is
estimated to increase annual revenue over that for the
CUITent
soft-
ware system by
$200,000 for each
of
the next 2 years, and by $300,000 for each
of
years
3 and 4. The planning horizon is only 4 years due to the rapid advances made interna-
tionally
in
building-monitoring software. Develop spreadsheets to answer the questions
below.
(a) Determine the equivalent future value
in
year 4
of
the increased cash flows, using
an
8% per year rate
of
return. Obtain answers for both simple and compound
interest.
(b) Rework part (a) if the cash flow estimates
in
years 3 and 4 increase from $300,000
to
$600,000.
(c) The financial manager
of
the U.S. company wants
to
consider the effects
of
4% per
year inflation
in
the analysis
of
part (a). As mentioned in Section 1.4, inflation re-
duces the real rate
of
return. For the 8% rate
of
return, an inflation rate
of
4% per
year compounded each year reduces the return to 3.85% per year.
Solution
by
Computer
Refer to Figure 1-12a
to
c for the solutions. All three spreadsheets contain the same infor-
mation, but the cell values are altered as required
by
the question. (Actually, all the ques-
tions posed here can be answered on one spreadsheet by simply changing the numbers.
Three spreadsheets are shown here for explanation purposes only.)
The Excel functions are constructed with reference to the cells, not the values them-
selves, so that sensitivity analysis can be perfOlmed without function changes. This ap-
proach treats the value
in
a cell as a global variable for the spreadsheet. For example, the
8% (simple or compound interest) rate
in
cell B4 will be referenced
in
all functions as B4,
not 8%. Thus, a change
in
the rate requires only one alteration in the cell B4 entry, not
in
every spreadsheet relation and function where 8% is used. Key Excel relations are detailed
in
the cell tags.
(a) 8% simple interest. Refer to Figure 1-12a, columns C and D, for the answers.
Simple interest earned each year (column
C)
incorporates Equation [1.5] one year
SECTION 1.
12
Spreadsheet Appli
cation-Simple
and Compound Interest
:x
I.hcrosoft
beel
• E .ample 1.18Ia·.,)
••••
• •
'..
.....
••
II1iU3
Example 1.1 B (Contains 3 worksheets)
Part (a) - Find F
in
year 4
Ca)
(b)
37
38 CHAPTER I Foundations
of
Engineering Economy
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,IL
-;
1
2
(el Find F with 4% per year inflation
3
4
Rate of return
385%
~
6
7
Simple interest Compound interest
8
Cumulative Cumulative
-
9
End of year Interest earned equivalent
In
terest earned equivalent
10
(EOYl
Cash flow dunng year EOY cash flow during year
EO
Y cash flow
11
0
$
-
12 1
$
200,000 $
-
$
200_000
$
$
200,000
13
2 $ 200 ,000 $
7)00
$ 407,700 $
7
)00
$ 407,700
14
3 $ 300,000 $ 15,400 $ 723,100 $ 15,696 $ 723,396
15 4
$
300,000
$
26
,950
$ 1,050,050 $
27,851
$ 1,051,247
16
Total
$
1,
000,000 $ 50,050
$ 51,247
17
18
1Q
I
I
••
~
.1
\
Ex
1.18(0)
"J(
Ex
1.18(b)
~.
18(C)
.,.{
Sheet4 I SheetS I Sheet6 { Sheet'
I_
J
II
'Ir-
Ready
NUM
1
(el
Figure
1-12
Spreadsheet solution including sensitivity analysis, Example J .18(a)-(c).
at a time into the interest relation
by
using only the end-of-year (EOY) cash
flow
amounts ($200,000 or $300,000)
to
determine interest for the next year. This inter-
est
is
added to the interest from all previous years.
In
$1000 units,
Ye
ar
2:
C
13
= B 12*B4 = $200(0.08) = $16 (see the cell tag)
Year
3:
C14 =
CI3
+ B13*B4 = $
16
+ 200(0.08) = $32
Year
4:
CI5
=
Cl4
+ B14*B4 = $32 + 300(0.0
8)
= $56 (see the cell tag)
Remember,
an
= sign must precede each relation
in
the spreadsheet. Cell C
16
con-
tains the function SUM(Cl2:C15)
to
display the total simple interest
of
$104,000
over the 4 years. The future value
is
in
DIS.
It
is
F = $1, I 04,000 which includes the
cumulative amount
of
all cash flows and all simple interest. In $1000 units, exam-
ple functions are
Year
2:
DI3
= SUM(B13:C13) +
DI2
= ($200 +
16)
+ 200 = $416
Year 4:
DIS = SUM(BI5:CI5) +
014
= ($300 + 56) + 748 = $1104
8% compound inrerest. See Figure 1-12a, columns E and
F.
The spreadsheet
structure
is
the same, except that Equation [1.6] is incorporated into the compound
interest values
in
column E, thus adding interest
on
top
of
earned interest.
SECTION
1.12
Spreadsheet
Application-Simple
and Compound Interest
Interest at
80/0
is
based on the accumulated cash flow at the end
of
the previous year.
In
$1000 units,
Year 2 interest: E13
= F12*B4 = $200(0.08) = $16
Cumulative cash
flow:
F13 = B13 + E13 + F12 = $200 +
16
+ 200 = $416
Year 4 interest:
E15
= F14*B4 = $749.28(0.08) = $59.942
(see the cell tag)
Cumulative cash
flow:
Fl5
= B15 + E15 + F14
= $300 + 59.942 + 749.280 = $1109.222
The equivalent future value is
in
cell
FI5,
where F = $1,109,222 is shown.
The cash flows are equivalent to
$1,104,000 at a simple
80/0
interest rate, and
$1,109,222 at a compound
80/0
interest rate. Using a compound interest rate in-
creases the
F value by $5222.
Note that
it
is
not possible to use the FV function
in
this case because the A val-
ues
are not the same for all 4 years.
We
wilt learn how to use all the basic functions
in
more versatile ways
in
the next few chapters.
(b) Refer to Figure 1-12b.
In
order to initialize the spreadsheet with the two increased
cash flow estimates, replace the
$300,000 values in B
14
and B I 5 with $600,000. All
spreadsheet relations are identical, and the new interest and accumulated cash flow
values are shown immediately. The equivalent fourth-year
F values have increased
for both the
80/0
simple and compound interest rates
(015
and F15, respectivel
y).
(c) Figure 1-12c is identical to the spreadsheet
in
Figure 1-12a, except the cell B4 now
contains the rate
of
3.850/0
. The corresponding F value for compound interest in F15
has decreased to $1,051,247 from the $1,109,222 at
80/0.
This represents
an
effect
of
inflation
of
$57,975
in
only four years. It is no surprise that governments, corpora-
tions, engineers, and all individuals are concerned when inflation rises and the cur-
rency is worth less over time.
Comment
When working with an Excel spreadsheet, it is possible to display all
of
the entries and
functions on the screen
by
simultaneously touching the
<Ctrl>
and
<'
> keys, which may
be
in
the upper left
of
the keyboard
on
the key with
<->.
Additionally, it may be neces-
sary to widen some columns
in
order to display the entire function statement.
ADDITIONAL EXAMPLES
CASH FLOW DIAGRAMS
A rental company spent $2500
on
a new air compressor 7 years ago. The annual rental
income from the compressor has been $750. Additionally, the
$100 spent on mainte-
nance during the first year has increased each year by $25. The company plans to sell
the compressor at the end of next year for $150. Construct the cash flow diagram from
the company's perspective.
39
40
CHAPTER I Foundations
of
Engineering Economy
Solution
Use now as time t =
O.
The incomes and costs for years
-7
through I (next year) are
tabulated below with net cash
flow
computed using Equation [1.8]. The net cash flows
(one negative, eight positive) are diagrammed
in
Figure 1-13.
End
of
year
Income
Cost
-7
$
0
$2500
-6
750
100
-5
750 125
-4
750 150
-3
750 175
-2
750
200
-1
750
225
0
750 250
750
+ 150
275
$650
$625
$600
-7
-6
-5 -4
-3
-2
-1
P = $2500
Figure
1-13
Cash
flow
diagram, Example 1.19.
CASH FLOW DIAGRAMS
Net
Cash
Flow
$-2500
650
625
600
575
550
525
500
625
$625
o
J Year
An
electrical engineer wants to deposit an amount P now such that she can withdraw an
equal annual amount
of
A I = $2000 per year for the first 5 years starting 1 year after the
deposit, and a different annual withdrawal
of
A2
= $3000 per year for the following
3 years. How would the cash
flow
diagram appear if i = 8.5% per year?
CHAPTER SUMMARY
Solution
The cash flows are shown
in
Figure 1-14. The negative cash outflow P occurs now. The
first withdrawal (positive cash inflow) for the
AI series occurs at the end
of
year
1,
and
A2
occurs
in
years 6 through
8.
A2
= $3000
AI = $2000
o 2 3
4
5 6 7
8 Year
i = 8.5%
p=?
Figure
1-14
Cash
flow
diagram with two different A series, Example 1.20.
CHAPTER SUMMARY
Engineering economy is the application
of
economic factors and criteria to eval-
uate alternatives, considering the time value
of
money. The engineering econ-
omy study involves computing a specific economic measure
of
worth for esti-
mated cash flows over a specific period
of
time.
The concept
of
equivalence helps in understanding how different sums
of
money at different times are equal
in
economic terms. The differences between
simple interest (based on principal only) and compound interest (based on prin-
cipal and interest upon interest) have been described in formulas, tables, and
graphs. This power
of
compounding is very noticeable, especially over long pe-
riods
of
time,
as
is the effect
of
inflation, introduced here.
The MARR
is
a reasonable rate
of
return established
as
a hurdle rate
to
deter-
mine if an alternative
is
economically viable. The MARR is always higher than
the return from a safe investment.
Also, we learned about cash flows:
Difficulties with their estimation.
Difference between estimated and actual value.
End-of-year convention for cash flow location.
Net cash flow computation.
Different perspectives in determining the cash flow sign.
Construction
of
a cash flow diagram.
41
42
CHAPTER I
Foundations
of
Engineering Economy
PROBLEMS
Basic Concepts
1.1 What
is
meant by the term time value
of
money?
1.2 List three intangible factors.
1
.3
(a) What is meant by evaluation criterion?
(b) What is the primary evaluation crite-
rion used in economic analysis?
1.4 List three evaluation criteria besides the eco-
nomic one for selecting the best restaurant.
1.5
Discuss the importance
of
ident{fying
alternatives in the engineering economic
process.
1.6 What is the difference between simple and
compound interest?
1.7
What
is
meant by minimum attractive rate
of
return?
1
.8
What
is
the difference between debt and
equity financing? Give an example
of
each.
Interest Rate and Rate
of
Return
1.9 Trucking giant Yellow
Corp
agreed to pur-
chase rival Roadway for $966 million in
order to reduce so-called back-office costs
(e.g., payroll and insurance) by $45 mil-
lion per year.
If
the savings were realized
as planned, what would be the rate
of
return on the investment?
1.10
If
Ford
Motor
Company's
profits in-
creased from 22 cents per share to 29 cents
per
share
in
the
April-June
quarter
com
-
pared to the previous quarter, what was the
rate
of
increase
in
profits for that quarter?
1.
11
A broadband service company borrowed
$2 million for new equipment and repaid
the principal
of
the loan plus $275,000
\J
(V
,,17\)
interest after 1 year. What was the interest
rate on the loan?
1.1
2 A design-build engineering firm com-
pleted a pipeline project wherein the com-
pany realized a profit
of
$2.3 million
in
1 year.
If
the amount
of
money the com-
pany had invested was
$6
million, what
was the rate
of
return on the investment?
1.
13
US Filter received a contract for a small
water desalting plant whereby the com-
pany expected to make a 28% rate
of
return
on its investment.
If
the company invested
$8 million
in
equipment the first year, what
was the amount
of
the profit in that year?
1.1
4 A publicly traded construction company
reported that it
just
paid
off
a loan that it
received 1 year earlier.
If
the total
amount
of
money the company paid was $1.6 mil-
lion and the interest rate on the loan was
10% per year, how much money did the
company borrow 1 year ago?
1.15 A start-up chemical company has estab-
lished a goal
of
making at least a 35% per
year rate
of
return on its investment.
If
the
company acquired
$50 million in venture
capital, how much did it have to earn in
the first year?
Equivalence
1.1
6 At an interest rate
of
8%
per
year, $10,000
today is equivalent to how much (a) 1 year
from now and
(b) 1 year ago?
1.17 A medium-size consulting engineering
firm is trying to decide whether it should
replace its office furniture now
or
wait and
do it 1 year from now.
If
it waits 1 year, the
cost is expected
to
be $16,000. At an inter-
est rate
of
10%
per
year, what would be
the equivalent cost now?
1.18
An
investment
of
$40,000 one year ago
and
$50,000 now are equivalent at what
interest rate?
1.19 At what interest rate would
$100,000 now
be equivalent to
$80,000 one year ago?
Simple and Compound Interest
1.20 Certain certificates
of
deposit accumulate
interest at 10%
per
year simple interest.
If
a company invests $240,000 now
in
these
certificates for the purchase
of
a new ma-
chine 3 years from now, how much will
the company have at the end
of
the 3-year
period?
1.21 A local bank is offering to pay compound
interest
of
7% per year on new savings
accounts. An e-bank
is
offering 7.5%
per
year simple interest on a 5-year certificate
of
deposit. Which offer
is
more attractive
to a company that wants to set aside
$1,000,000 now for a plant expansion
5 years from now?
1.22 Badger
Pump Company invested
$500,000 five years ago
in
a new product
line that is now worth
$1,000,000. What
rate
of
return did the company earn (a) on
a simple interest basis and
(b) on a com-
pound interest basis?
1.23 How long will it take for an investment to
double at 5% per year
(a) simple interest
and
(b) compound interest?
1.24 A company that manufactures regenera-
tive thermal oxidizers made an investment
10 years ago that is now worth $1,300,000.
How much was the initial investment at
an interest rate
of
15% per year (a) simple
interest and
(b) compound interest?
1.25 Companies frequently borrow money
under an arrangement that requires them
to make periodic payments
of
only interest
and then pay the principal
of
the loan all at
PROBLEMS
43
once. A company that manufactures odor
control chemicals borrowed $400,000 for
3 years at 10% per year compound interest
under such an arrangement. What
is
the
difference in the
total amount
paid
between
this arrangement (identified as plan
1)
and
plan 2, in which the company makes no in-
terest payments until the loan is due and
then pays it off in one lump sum?
1.26 A company that manufactures in-line mix-
ers for bulk manufacturing
is
considering
borrowing $1.75 million to update a pro-
duction line.
If
it bon·ows the money now, it
can do so at an interest rate
of7.5%
per year
simple interest for 5 years.
If
it borrows next
year, the interest rate will be 8% per year
compound interest, but it will be for only
4 years.
(a) How much interest (total) will
be paid under each scenario, and
(b) should
the company borrow now or 1 year from
now? Assume the total amount due will be
paid when the loan is due
in
either case.
Symbols and Spreadsheets
1.27 Define the symbols involved when a con-
struction company wants to know how
much money it can spend 3 years from
now in lieu
of
spending $50,000 now to
purchase a new truck, when the compound
interest rate is
15
% per year.
1.28 State the purpose for each
of
the following
built-in Excel functions:
(a) FV(i%,n,A,P)
(b) IRR(firsccell:last_cell)
(c) PMT(i%,n,
P,F)
(d) PV(i%,n,A,F)
1.29 What are the values
of
the engineering
economy symbols
P,
F, A, i, and n in the
following Excel functions?
Use a ? for the
symbol that is to be determined.
(a) FV(7%,1O,2000,9000)
(b) PMT(11 %,20,14000)
(c) PV(8%, 15, 1000,800)
44
CHAPTER I
Foundations
of
Engineering Economy
1.30 Write the engineering economy symbol
that corresponds
to
each
of
the following
Excel functions.
(a)
PV
(b)
PMT
(c) NPER
(d)
IRR
(e)
FV
1.31
In a built-in Excel function, if a certain
parameter does not apply, under what
circumstances can it be left blank? When
must a comma be entered
in
its place?
MARR
and
Cost
of
Capital
1.32 Identify each
of
the following
as
either a
safe investment or a risky one.
(a) New restaurant business
(b) Savings account
in
a bank
(c) Certificate
of
deposit
(d)
Government bond
(e) Relative's "get-rich-quick" idea
1.33 Identify each
of
the following
as
either
equity or debt financing.
(a) Money from savings
(b) Money from a certificate
of
deposit
(c) Money from a relative who is a part-
ner
in
the business
(d)
Bank loan
(e)
Credit card
1.34 Rank the following from highest to lowest
rate
of
return or interest rate: government
bond, corporate bond, credit card, bank
loan
to
new business, interest on checking
account.
1.35 Rank the following from highest
to
lowest
interest rate: cost
of
capital, acceptable
rate
of
return on a risky investment, mini-
mum attractive rate
of
return, rate
of
return
on
a safe investment, interest on checking
account, interest on savings account.
1.36 Five separate projects have calculated
rates
of
return
of
8,
11,
12.4, 14, and 19%
per year. An engineer wants to know
which projects to accept on the basis
of
rate
of
return. She learns from the finance
department that company funds, which
have a cost
of
capital
of
18% per year, are
commonly used
to
fund 25%
of
all capital
projects. Later, she is told that borrowed
money
is
currently costing 10% per year.
If
the MARR is established at exactly the
weighted average cost
of
capital, which
projects should she accept?
Cash
Flows
1.37 What is meant by the end-of-period con-
vention?
1.38 Identify the following
as
cash inflows or
outflows to Daimler-Chrysler: income
taxes, loan interest, salvage value, rebates
to dealers, sales revenues, accounting ser-
vices, cost reductions.
1.39 Construct a cash flow diagram for the fol-
lowing cash flows:
$10,000 outflow at
time zero,
$3000 per year outflow in years
I through 3 and
$9000 inflow in years 4
through 8 at an interest rate
of
10% per
year, and an unknown future amount
in
year 8.
1.40 Construct a cash flow diagram
to
find
the
present worth
of
a future outflow
of
$40,000 in year 5 at an interest rate
of
15
%
per year.
Doubling
the
Value
1.41 Use the rule
of
72
to
estimate the time it
would take for
an
initial investment
of
$10,000
to
accumulate
to
$20,000 at a
compound rate
of
8% per year.
1.42 Estimate the time it would take (according
to
the rule
of
72) for money
to
quadruple
in value at a compound interest rate
of
9% per year.