Blank L., Tarquin A. Engineering Economy (McGraw-Hill Series in Industrial Engineering and Management)
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4.
15
An
interest rate
of
12% per year, com-
pounded monthly, is equivalent to what
nominal and effective interest rates per
6 months?
4.
16 (0)
An
interest rate
of
6.8% per semi-
annual period, compounded weekly,
is
equivalent to what weekJy interest
rate?
(b) Is the weekly rate a nominal or ef-
fective rate? Assume 26 weeks per
6 months.
Payment
and
Compounding Periods
4. 17 Deposits
of
$100 per week are made into a
savings account that pays interest
of
6%
per year, compounded quarterly. Identify
the payment and compounding periods.
4. 18 A certain national bank advertises quar-
terly compounding for business
checking
accounts. What payment and compound-
ing periods are associated with deposits
of
daily receipts?
4. 19 Determine the
F / P factor for 3 years at an
interest
rate'
of
8% per year, compounded
quarterly.
4.20 Determine the P / G factor for 5 years at an
effective interest rate
of
6% per year, com-
pounded semiannually.
Equivalence for Single Amounts
and
Series
4.2 1 A company that specializes
in
online secu-
rity software development wants to have
$85 million available in 3 years to pay
stock dividends. How much money must
the company set aside
now
in
an account
that earns interest at a rate
of
8% per year,
compounded quarterly?
4.22 Because testing
of
nuclear bombs was
halted in 1992, the
U.S. Department
of
Energy has been developing a laser project
that will allow engineers to simulate (in a
PROBLEMS
155
laboratory) conditions
in
a thermonuclear
reaction.
Due
to soaring cost overruns, a
congressional committee undertook an
investigation and discovered that the esti-
mated development cost
of
the project
increased
at
an average rate
of
3% per
month over a 5-year period.
If
the original
cost was estimated to be $2.7 billion
5 years ago, what is the expected cost
today?
4.23 A present sum
of
$5000 at an interest rate
of
8% per year, compounded semiannu-
ally,
is
equivalent to how much money
8 years ago?
4.24 In
an
effort to ensure the safety
of
cell
phone users, the Federal Communications
Commission (FCC) requires cell phones
to have a specific absorbed radiation
(SAR) number
of
1.6 watts per kilogram
(W
Ikg)
of
tissue or less. A new cell phone
company estimates that by advertising its
favorable 1.2 SAR number, it will increase
sales by $1.2 million 3 months from now
when its phones go on sale. At an interest
rate
of
20% per year, compounded quar-
terly, what is the maximum amount the
company can afford to spend
now
for
advertising in order to break even?
4.25 Radio Frequency Identification (RFlD)
is
technology that is used by drivers with
speed passes at toll booths and ranchers
who track livestock from farm to fork.
Wal-Ma.rt expects to begin using the tech-
nology to track products within its stores.
If
RFlD-tagged products will result in better
inventory control that will save the com-
pany $1.3 million per month beginning
3 months from now, how much could the
company afford to spend now to implement
the technology at an interest rate
of
12
%
per year, compounded monthly, if it wants
to recover its investment in
2
~
years?
1
56
CHAPTER 4 Nominal and Effective Interest Rates
4.26 The patriot missile, developed by Lock-
heed Martin for the
U.S. Army, is designed
to
shoot down aircraft and other missiles.
The
Patriot Advanced Capability-3 was
originally promised
to
cost $3.9 billion,
but due to extra time needed
to
write
computer code and scrapped tests (due to
high winds) at White Sands Missile
Range, the actual cost was much higher.
If
the total project development time was
10
years and costs increased at a rate
of
0.5% per month, what was the final cost
of
the project?
4.27 Video cards based on Nvidia's highly
praised GeForce2 GTS processor typi-
cally cost $250. Nvidia released a light
version
of
the chip that costs $150.
If
a
certain video game maker was purchasing
3000 chips per quarter, what was the pre-
sent worth
of
the savings associated with
the cheaper chip over a 2-year period at an
interest rate
of
16% per year, compounded
quarterly?
4.28 A 40-day strike at Boeing resulted
in
50
fewer deliveries
of
commercial jetliners at
the end
of
the first quarter
of
2000. At a cost
of
$20 million per plane, what was the
equivalent end-of-year cost
of
the strike
(i.e., end offourth quarter) at an interest rate
of
18
% per year, compounded monthly?
4.29 The optical products division ofPanasonic
is planning a $3.5 million building expan-
sion for manufacturing
its
powerful
Lumix DMC digital zoom camera.
If
the
company uses
an
interest rate
of
20% per
year, compounded quarterly, for all new
investments, what
is
the uniform amount
per quarter the company must make
to
re-
cover its investment in 3 years?
4.30 Thermal Systems, a company that special-
izes in odor control, made deposits
of
$10,000 now, $25,000 at the end
of
month
6,
and $30,000 at the end
of
month
9.
Determine the future worth (end
of
year])
of
the deposits at
an
interest rate
of
16% per year, compounded quarterly.
4.31 Lotus Development has a software rental
plan called SmartSuite that is available on
the World Wide Web. A number
of
pro-
grams are available at $2.99 for 48 hours.
If
a construction company uses the service
an average
of
48 hours per week, what is
the present worth
of
the rental costs for
10 months at an interest rate
of
1 % per
month, compounded weekly? (Assume
4 weeks per month.)
4.32 Northwest Iron and Steel is considering
getting involved in electronic commerce.
A modest e-cornrnerce package
is
avail-
able for $20,000.
If
the company wants
to
recover the cost
in
2 years, what is the
equivalent amount
of
new income that
must be realized every 6 months,
if
the in-
terest rate is 3% per quarter?
4.
33
Metro~litan
Water Utilities purchases
surface water from Elephant Butte Irriga-
tion District at a cost
of
$100,000 per
month in the months
of
February through
September. Instead
of
paying monthly, the
utility makes a single payment
of
$800,000
at the end
of
the year (i.e., end
of
Decem-
ber) for the water it used. The delayed pay-
ment essentially represents a subsidy
by
the irrigation district to the water utility. At
an
interest rate
of
0.25% per month, what is
the amount
of
the subsidy?
4.34 Scott Specialty Manufacturing
is
consid-
ering consolidating all its electronic ser-
vices with one company. By purchasing a
digital phone from AT&T Wireless, the
company can buy wireless e-mail and fax
services for $6.99 per month. For $14.99
per month, the company will get unlimited
Web access and personal organization
functions.
For
a 2-year contract period,
what is the present worth
of
the difference
between the services at an interest rate
of
12%
per
year, compounded monthly?
4.35 Magnetek Instrument and Controls, a
manufacturer
of
liquid-level sensors, ex-
pects sales for one
of
its models to increase
by
20% every 6 months into the foresee-
able future.
If
the sales 6 months from now
are expected to be
$150,000, determine the
equivalent semiannual worth
of
sales for a
5-year period at an interest rate
of
14% per
year, compounded semiannually.
4.36 Metalfab
Pump and Filter projects that the
cost
of
steel bodies for certain valves will
increase by $2 every 3 months.
If
the cost
for the first quarter is expected to be
$80,
what
is
the present worth
of
the costs for a
3-year period at an interest rate
of
3% per
quarter?
4.37 Fieldsaver Technologies, a manufacturer
of
precision laboratory equipment, bor-
rowed $2 million to renovate one
of
its
testing labs. The loan was repaid in 2 years
through quarterly payments that increased
by
$50,000 each time. At an interest rate
of
3% per quarter, what was the size
of
the
first quarterly payment?
4.38 For the cash flows shown below, deter-
mine the present worth (time
0), using an
interest rate
of
18% per year, compounded
monthly.
Month
o
1-12
13-28
Cash Flow,
$/Month
1000
2000
3000
PROBLEMS
157
4.39
The
cash flows (in thousands) associated
with
Fisher-Price's Touch learning system
are shown below. Determine the uniform
quarterly series
in
quarters 0 through 8
that would be equivalent to the cash flows
shown at an interest rate
of
16% per year,
compounded quarterly.
Quarter
2-3
5-8
Cash
Flow,
$/Quarter
1000
2000
3000
Equivalence When PP < cp
4.40 An engineer deposits $300 per month into
a savings account that pays interest at a
rate
of
6% per year, compounded semian-
nually. How much will be in the account at
the end
of
15
years? Assume no interpe-
riod compounding.
4.41
Attimet
= 0,
an
engineerdeposited
$1
0,000
into an account that pays interest at 8% per
year, compounded semiannually.
If
she
withdrew
$1000
in
months 2,
11
, and 23,
what was the total value
of
the account at
the end
of
3 years? Assume no interperiod
compounding.
4.42 For the transactions shown below, deter-
mine the amount
of
money in the account
at the end
of
year 3 if the interest rate
is
8% per year, compounded semiannually.
Assume no interperiod compounding.
End
of
Quarter
1
2-4
7
II
Amount
of
Deposit,
$/Quarter
900
700
1000
Amount
of
Withdrawal,
$/Quarter
2600
1000
158
CHAPTER 4 Nominal and Effective Interest Rates
4.43
The
New Mexico State Police and Public
Safety
Department owns a helicopter that
it uses to provide transportation and logis-
tical support for high-level state officials.
The
$495 hourly rate covers operating ex-
penses and the pilot's salary.
If
the gover-
nor uses the helicopter an average
of
2 days
per month for 6 hours each day, what is the
equivalent future worth
of
the costs for
1 year at an interest rate
of
6% per year,
compounded quarterly (treat the costs as
deposits)?
Continuous Compounding
4.44 What effective interest rate per year, com-
pounded continuously,
is
equivalent to a
nominal rate
of
J 3% per year?
4.45 What effective interest rate per 6 months
is
equal to a nominal 2% per month, com-
pounded continuously?
4.46 What nominal rate per quarter is equiva-
lent to an effective rate
of
12.7% per year,
compounded continuously?
4.47 Corrosion problems and manufacturing
defects rendered a gasoline pipeline be-
tween
EI
Paso and Phoenix subject to lon-
gitudinal weld seam failures. Therefore,
pressure was reduced to
80%
of
the de-
sign value.
If
the reduced pressure results
in
delivery
of
$100,000 per month less
product, what will be the value
of
the lost
revenue after a 2-year period at an inter-
est rate
of
15% per year, compounded
continuously?
4.48 Because
of
a chronic water shortage
in
Santa Fe, new athletic fields must use arti-
ficial turf
or
xeriscape landscaping.
If
the
value
of
the water saved each month is
$6000, how much can a private developer
afford to spend on artifical turf
if
he wants
to recover his investment in 5 years at an
interest rate
of
18% per year, compounded
continuously?
4.49 A Taiwan-based chemical company had to
file for bankruptcy because
of
a nation-
wide phase-out
of
methyl tertiary butyl
ether (MTBE).
If
the company reorganizes
and invests
$50 million
in
a new ethanol
production facility, how much money
must it
make
each month
if
it wants to re-
cover its investment in 3 years at an inter-
est rate
of
2% per month, compounded
continuously?
4.50 In order to have $85,000 four years from
now for equipment replacement, a con-
struction company plans to set aside money
today in government-insured bonds.
If
the
bonds earn interest at a rate
of
6% per
year, compounded continuously, how much
money must the company invest?
4.51 How long would it take for a lump-sum in-
vestment to double in value at
an
interest
rate
of
1.5% per month, compounded con-
tinuously?
4.52 What effective interest rate per month,
compounded continuously, would be re-
quired for a single deposit to triple
in
value in 5 years?
Varying Interest Rates
4.53 How much money could the maker
of
fluidized-bed scrubbers afford to spend
now instead
of
spending $150,000 in
year 5
if
the interest rate
is
10%
in
years
1 through 3 and 12% in years 4 and 5?
4.54
What
is the future worth in year 8
of
a
present sum
of
$50,000
if
the interest rate
is 10%
per
year in years 1 through 4 and
1 % per month in years 5 through 8?
4.55 For the cash flows shown below, deter-
mine
(a) the future worth in year 5 and
(b) the equivalent A value for years 0
through 5.
Cash Flow,
Interest
Rate
Year $/Year
per
Year, %
0 5000
12
1--4
6000
12
5
9000
20
FE
REVIEW PROBLEMS
4.57 An interest rate
of
effective 14% per
month, compounded weekly, is
(a) An effective rate per year
(b) An effective rate per month
(c) A nominal rate
per
year
(d)
A nominal rate per month
4.58 An interest rate
of
2% per month is the
same as
(a) 24% per year, compounded monthly
(b) A nominal 24% per year, com-
pounded monthly
(c) An effective 24% per year, com-
pounded monthly
Cd)
Both (a) and (b)
4.59 An interest rate
of
12%
per
year, com-
pounded monthly,
is
nearest to
(a) 12.08% per year
(b)
12
.28% per year
(c) 12.48% per year
(d)
12.68% per year
4.
60
An interest rate
of
1.5% per month, com-
pounded continuously, is closest to an
effective
(a) 1.51% per quarter
(b) 4.5% per quarter
FE REVIEW PROBLEMS
159
4.56
For
the cash flow series shown below,
find the equivalent
A value in years L
through 5.
Cash Flow,
Interest
Rate
Year $/Year
per
Year, %
0 0
1-3 5000
4-5 7000
(c) 4.6% per quarter
(d)
9%
per 6 months
10
12
4.61 An interest rate
of
2%
per
quarter is the
same as
(a) Nominal 2% per quarter
(b) Nominal 6%
per
year, compounded
quarterly
(c) Effective 2% every 4 months
(d)
Effective 2% every 3 months
4.62 An interest rate
expre~sed
as an effective
12%
per
year, compounded monthly, is the
same as
(a) 12% per year
(b) 1 % per month
(c) 12.68% per year
(d)
Any
of
the above
4.63 A
I-year interest rate
of
20% per year,
compounded continuously, is closest to
the following interest rate:
(a) Simple 22% per year
(b) 21%
per
year, compounded quarterly
(c)
21
% per year, compounded monthly
(d)
22% per year, compounded semi-
annually
160
CHAPTER 4 Nominal and Effective Interest Rates
4.
64
For
an
interest rate
of
1 % per quarter,
compounded continuously, the effective
semiannual interest rate is closest to
(a) Less than 2.0%
(b) 2.02%
(c) 2.20%
(d)
Over
2.25%
4.65
The
only time you change the amount and
timing
of
the original cash flows
in
prob-
lems involving a uniform series is when
(a) Payment period is longer than the
compounding period.
(b) Payment period is equal to the
com
-
pounding period.
(c) Payment period is shorter than the
compounding period.
(d)
It
may be done
in
any
of
the cases
above, depending upon how the
ef
-
fective interest rate
is
calculated.
4.66 Exotic Faucets and Sinks, Ltd., guaran-
tees that its new
in
frared sensor faucet
will save any household that has two or
more children at least
$30 per month
in
water costs beginning 1 month after the
faucet
is
installed.
If
the faucet is under
full warranty for 5 years, the minimum
amount a family
cou
ld
afford to spend
now on such a faucet at an interest rate
of
6% per year, compounded monthly, is
closest to
(a) $149
(b) $1552
(c) $1787
(d)
$1890
4.67
The
multistate Powerball Lottery, worth
$182 million, was won by a single individ-
ual who had purchased five tickets at
$1
each.
The
individual was given two
choices: Receive 26 payments
of
$7 mil-
lion each, with the first payment to be
made
now
and the rest to be made at the
end
of
each
of
the next 25 years; or receive
a single lump-sum payment
now
that
wouid be equivalent to the 26 payments of
$7 million each.
If
the state uses an inter-
est rate
of
4% per year, the amount
of
th
e
lump-sum payment would be closest to
(a) Less than $109 million
(b) $109,355,000
(c) $116,355,000
(d)
Over $117 million
4.68 Royalties paid to holders
of
mineral
rights tend to decrease with time as re-
sources become depleted. In one particular
case, the light holder received a royalty
check
of
$18,000 six months after the
lease was signed. She continued to receive
checks at 6-month intervals, but the
amount decreased by
$2000 each time. At
an interest rate
of
6% per year, com-
pounded semiannually, the equivalent
uniform semiannual worth
of
the royalty
payments through the first 4 years is rep-
resented by
(a) A =
18
,000 - 2000(A/ G, 3%, 8)
(b) A =
18
,000 - 2000(A/ G, 6%, 4)
(c) A =
18
,000(AIP, 3%, 8) - 2000
(d)
A = 18,000 +
2000(A/
G,
3%, 8)
4.69
The
cost for increasing the production ca-
pacity
in
a certain manufacturing facility
is expected to increase by 7%
per
year
over the next 5-year period.
If
the cost at
the end
of
year 1 is $39,000 and the inter-
est rate is 10% per year, the present worth
of
the costs through the end
of
the 5-year
period is represented by
(a) P = 39,000{ 1 -
[(1
+ 0.07)6/(1 +
0.10)6])
/(0
.10 - 0.07)
(b) P = 39,000{ 1 -
[(1
+ 0.07)5/(1 +
0.10)5
]}
/ (0.1O + 0.07)
(c) P =
39,000{l
-
[(I
+ 0.07)4/(1 +
0.1O)
4
]}
/ (0.10 - 0.07)
(d)
P = 39,
000{l
-
[(1
+ 0.07)
5/
(1
+
0.1O)5
]}
/ (0.10 - 0.07)
4.
70
A plant manager wants to know the prese
nt
worth
of
the maintenance costs for a
certain assembly line. An industrial engi-
neer, who designed the system, estimates
that the maintenance costs are expected to
be zero for the first 3 years,
$2000
in
year 4,
$2500
in
year 5, and amounts increasing
by
$500 each year through year 10. At an
interest rate
of
8% per year, compounded
semiannually,
th
e value
of
n to use in the
P / G equation for this problem
is
(a) 7
(b)
8
(c)
10
(d)
14
4.71 A public relations company hired by the
city
of
EI
Paso to increase tourism to the
Sun City proposed that the city build
the world's only roller coaster that travels
through two different countries. The idea is
to
build the ride along the Rio Grande River
and have part
of
the tracks in the United
States and part in Mexico. The ride would
be built such that coaster cars could be
launched from either side
of
the border,
but riders would get
off
at the same point
where they got on. After the ride becomes
operational, tourism revenue is projected
to be
$1
million initially (i.e., at time 0),
$1.05
million after the first month,
$1.1025
mi
Ilion after the second month,
and amounts increasing by 5% each
month through the first year. At an interest
rate
of
12%
per year, compounded monthly,
the present worth (time
0)
of
the tourism
revenue generated
by
the ride
is
closest to
(a) $15.59 million
(b)
$16.59 million
(c) $17.59 million
(d)
Over $18 million
4.72 In problems that involve an arithmetic
gradient G wherein the payment period is
longer than the interest period, the interest
rate to use
in
the equations
FE REVIEW PROBLEMS
161
(a)
Can
be
any effective rate, as long as
the time units on
i and n are the same
(b)
Must be the interest rate that is ex-
actly as it is stated in the problem
(c) Must be an effective interest rate that
is expressed over a
I-year
period
(d)
Must be the effective interest rate
that is expressed over the time period
equal to the time where the first
change equal to G occurs
4.73 An engineer analyzing cost data dis-
covered that the information for the first
three years was missing. However, he
knew that the cost in year 4 was
$1250 and that it increased by 5% per
year thereafter.
If
the same trend applied
to the first three years, the cost in year 1
was closest to
(a) $1235.70
(b)
$119l.66
(c)
$1133.79
(d)
$1079.80
4.74 Encon Environmental Testing needs to
purchase
$40,000 worth
of
equipment
2 years from now. At an interest rate
of
20% per year, compounded quarterly, the
uniform quarterly worth
of
the equipment
(quarters 1 through 8)
is
closest to
(a) $3958
(b)
$4041
(c) $4189
(d)
Over $4200
4.75 Border Steel invested $800,000 in a new
shearing unit.
At
an interest rate
of
12%
per
year, compounded quarterly, the quar-
terly income required to recover the in-
vestment in 3 years is closest to
(a) $69,610
(b) $75,880
(c) $80,370
(d) $83,550
162
CHAPTER 4
Nom
in
al and Effec
ti
ve Interest Rates
CASE
STUDY
FINANCING A HOUSE
Introduction
When a person or a couple dec
id
e to purchase a house,
one
of
th
e most important considera
ti
ons is
th
e
fin
anc-
ing. There are many me
th
ods of
fin
anc
in
g the purchase
of
res
id
e
nti
al proper
ty,
each hav
in
g ad
va
ntages w
hi
ch
make it
th
e me
th
od
of
choice under a given set
of
cir-
c
um
stances. The selec
ti
on
of
one me
th
od from sev
er
al
fo
r a given set
of
conditions is
th
e to
pi
c of this case
stud
y.
Three methods
of
financing are described
in
d
et
ail.
Pl
ans A and B are evaluated; you are asked
to evaluate plan C and perfo
rm
some additional
analyses.
The criterion used here is:
Select
th
e
fin
anc
in
g plan
w
hi
ch has
th
e largest amount of money remaining at
th
e e
nd
of a
10
-year pe
ri
od. Therefore, calc
ul
ate
th
e
future worth of each plan, a
nd
select
th
e one with
th
e
largest future worth va
lu
e.
Plan
A
B
C
Description
30-year fixed rate
of
10% per year
in
terest, 5% do
wn
payment
30-year adjustable-rate mo
rt
gage
(A
RM
), 9%
fir
st 3 years, 9
)1%
in
year
4,
10
)4%
in
years 5 through 10
(assumed), 5% down payment
IS-year fixed rate
of
9
)1%
per year
in
terest, 5% down payment
Other
in
fo
rm
a
ti
o
n:
• Price of house is $150,000.
•
House will be sold
inl
0 y
ea
rs for $ I 70,000 (net
proceeds after selling expenses).
• Taxes a
nd
in
sura
nc
e (T&I) are $300 per month.
•
Am
ount availabl
e:
maximum of $
40
,000 for
down paymen
t,
$1600 per month, including
T&
l.
• New loan expenses:
or
ig
in
a
ti
on fee
of
1
%,
ap-
praisal fee $300, survey fee $200, atto
rn
ey's fee
$200, process
in
g fee $350, escrow fees $
150
,
other costs $300.
•
An
y money not spent on
th
e down payment or
monthly payments w
ill
ea
rn
tax-free interest at
)4%
per month.
Analysis
of
Financing Plans
Plan
A:
30-Year Fixed Rate
The amount
of
money required
up
front is
(a) Do
wn
payme
nt
(5
%
of
$150,000)
(b) Orig
in
ation fee ( I %
of
$142,500)
(c) Apprais
al
(d)
Survey
(e) Atto
rn
ey's fee
(f)
Process
in
g
(g) Escrow
(h) Other (recording, credit report, etc.)
Total
$7,500
1,425
300
200
200
350
150
300
$10,425
The amo
un
t
of
th
e loan is $142,500. The equiva-
lent monthly principal and interest (P&T) payment
is determined at
10
%/
12 per month
fo
r 30( 12) =
360 months.
A = 142,500
(A/
P,
1O
%/
12,360)
=
$1250.56
When T&T are added to
P&l
,
th
e total monthly pay-
ment PMT
A
is
PMT
A
= 1250.56 + 300
= $1550.56
We can now determine the future worth
of
pl
an A by
summing three future worth amollnt
s:
th
e remaining
funds not used for
th
e down payme
nt
a
nd
up-front fees
(F
IA
) and for monthly payments
(F
2A
),
a
nd
th
e
in
-
crease
in
th
e va
lu
e
of
the hO
ll
se
(F
3A
).
Since non-
expe
nd
ed money earns interest at
X%
per month, in
10 years
th
e
fir
st future worth is
F
IA
= (40,000 -
1O
,425
)(F
/ P,0.25%, 120)
= $39,907.1 3
The available money not spent on monthly payments
is $49.44
= $1600 - 1550.56.
It
s future worth after
10 years
is
F
2A
=
49.44(F
/A,0.25%, 120) = $6908.
81
Net money available from the sale
of
the house is the
difference between the net selling price and the bal-
ance
of
the loan. The balance
of
the loan
is
Loan balance = 142,500(F/ P,
IO
%/
12,120)
- 1250.56(F/ A,
IO
%/
12,120)
= 385,753.40 - 256,170.92
=
$l29
,582.48
S
in
ce the net proceeds from the sale
of
the house
are
$]70,000,
F3A = ]70,000 - 129,582.48 = $40,417.52
The total future worth
of
plan A
is
FA =
FI
A + F
2A
+ F
3A
= 39,907.l3 + 6908.81 +
40,4]7
.5
2
= $87,233.46
Plan B: 30-Year Adjustable Rate Mortgage
Aclj
u table rate mortgages are tied to some index such
as U.S. Treasury bonds. For
thi
s example, we have
ass
um
ed that the rate is 9% for the
fir
st 3 years,
9
~%
in
year 4, and
10
1.%
in
years 5 through
10
. Since this
option also requires 5%
clown
, the up-front money
required will be the same as for plan
A,
that
is
, $10,425.
The monthly
P&l
amount for the
fir
st 3 years
is
ba
sed
on
9% per year for 30 years.
A =
142
,500(A / P,9
%/
12,360) = $1146
.5
8
The total monthly payment for the first 3 years
is
PMT B = $1146.58 + 300 = $1446.58
At the end
of
year 3, the interest rate changes to
9;,j%
per year. This new rate applies to the balance
of
the
loan at that time:
Lo
an
balance
at end
of
year 3 = 142,
500(F
/ P,0.75%,36)
- I I 46.58(F/A,0.75%,36)
=
$139,297.08
CASE
STUDY
163
The P&I monthly payment for year 4 is now
A = 139,297.08(A/ P,9.5
%/
12,324) = $1195.67
The total payment for year 4
is
PMT
B
= 1195.67 + 300 = $1495.67
At the end
of
year 4, the interest rate changes again,
this time
to
10
1.%
per year, and it stays at
thi
s rate for
the remainder
of
the 10-year
peI10cl.
The loan balance
at the end
of
year 4
is
Loan balance
at e
nd
of
year 4 = 139,297.08(F / P,9
.5
%/
12, 12)
- 1195.67(F/A,9.5
%/
12,12)
= $138,132.42
The new
P&I amount is
A = 138, I 32.42(A/P,1O.25
%/
12
,312) = $1269.22
The new total payment for years 5 through
lOis
PMT
B
= 1269.22 + 300 = $1569.22
The loan balance at the end
of
10
years is
Loan balance
after
10
years = 138,132.42(F/ P,1O.25
%/
l2
,72)
- 1269.22(F/
A,IO
.
25
%/
12
,72)
=
$129,296.16
The future worth
of
plan B can now be determined
using the same three fuulre worth values. The future
worth
of
the money not spent on a down payment is
the same as for plan
A.
FIB
= (40,000 -
1O
,
425)(F
/ P,0.25%,120)
= $39,907.13
The future worth
of
the money not spe
nt
on monthly
payments
is
more complex than
in
plan
A.
F2B = (1600 - 1446.58)(F/A,0.25%,36)
X
(F
/ P,0.25%,84) + (1600 -
1495
.67)
X
(F
/ A,0.
25
%,
l2
)(F
/ P,0.25%,
72
)
+ (1600 -
1569
.22)(F
/
A
,
0.25
~
,72
)
=
7ll8.61
+ 1519.31 + 2424.83
=
$1
1,062.75
164
CHAPTER 4 Nominal and Effective Interest Rates
The amount
of
money left from the sale
of
the
house
is
F3B = 170,000 - 129,296.16
==
$40,703.84
The total future worth
of
plan 8
is
FB
=
FIB
+ F
28
+ F 38 = $91,673.72
Case Study Exercises
1.
Evaluate plan C and select the best financing
method.
2. What
is
the total amount
of
interest paid
in
plan A
through the lO-year period?
3.
What is the total amount
of
interest paid in plan 8
through year 4?
4.
What
is the maximum amount
of
money avail-
able for a down payment under plan
A,
if
$40,000 is the total amount available?
5.
By
how much does the payment increase in plan
A for each 1 % increase
in
interest rate?
6.
If
you wanted to "buy down" the interest rate
from
10% to 9% in plan
A,
how much extra down
payment would you have
to
make?