Solution manual engineering economic analysis 9th edition ch14 INflation and price change

Chapter 14: Inflation and Price Change
14-1
During times of inflation, the purchasing power of a monetary unit is reduced. In this way the
currency itself is less valuable on a per unit basis. In the USA, what this means is that during
inflationary times our dollars have less purchasing power, and thus we can purchase less
products, goods and services with the same $1, $10, or $100 dollar bill as we did in the
past.
14-2
Actual dollars are the cash dollars that we use to make transactions in our economy. These
are the dollars that we carry around in our wallets and purses, and have in our savings
accounts. Real dollars represent dollars that do not carry with them the effects of inflation,
these are sometimes called “inflation free” dollars. Real dollars are expressed as of
purchasing power base, such as Year-2000-based-dollars.
The inflation rate captures the loss in purchasing power of money in a percentage rate form.
The real interest rate captures the growth of purchasing power, it does not include the
effects of inflation is sometimes called the “inflation free” interest rate. The market interest
rate, also called the combined rate, combines the inflation and real rates into a single rate.
14-3
There are a number of mechanisms that cause prices to rise. In the chapter the authors talk
about how money supply, exchange rates, cost-push, and demand pull effects can
contribute to inflation.
14-4
Yes. Dollars, and interest rates, are used in engineering economic analyses to evaluate
projects. As such, the purchasing power of dollars, and the effects of inflation on interest
rates, are important.
The important principle in considering effects of inflation is not to mix-and-match dollars and
interest rates that include, or do not include, the effect of inflation. A constant dollar analysis
uses real dollars and a real interest rate, a then-current (or actual) dollar analysis uses
actual dollars and a market interest rate. In much of this book actual dollars (cash flows) are
used along with a market interest rate to evaluate projects  this is an example of the later
type of analysis.

14-5
The Consumer Price Index (CPI) is a composite price index that is managed by the US
Department of Labor Statistics. It measures the historical cost of a bundle of “consumer
goods” over time. The goods included in this index are those commonly purchased by
consumers in the US economy (e.g. food, clothing, entertainment, housing, etc.).


Composite indexes measure a collection of items that are related. The CPI and Producers
Price Index (PPI) are examples of composite indexes. The PPI measures the cost to
produce goods and services by companies in our economy (items in the PPI include
materials, wages, overhead, etc.). Commodity specific indexes track the costs of specific
and individual items, such as a labor cost index, a material cost index, a “football ticket”
index, etc.
Both commodity specific and composite indexes can be used in engineering economic
analyses. Their use depends on how the index is being used to measure (or predict) cash
flows. If, in the analysis, we are interested in estimating the labor costs of a new production
process, we would use a specific labor cost commodity index to develop the estimate. Much
along the same lines, if we wanted to know the cost of treated lumber 5 years from today,
we might use a commodity index that tracks costs of treated lumber. In the absence of
commodity indexes, or in cases where we are more interested in capturing aggregate effects
of inflation (such as with the CPI or PPI) one would use a composite index to
incorporate/estimate how purchasing power is affected.
14-6
The stable price assumption is really the same as analyzing a problem in Year 0 dollars,
where all the costs and benefits change at the same rate. Allowable depreciation charges
are based on the original equipment cost and do not increase. Thus the stable price
assumption may be suitable in some before-tax computations, but is not satisfactory where
depreciation affects the income tax computations.
14-7

F

= P (F/P, f%, 10 yrs) = $10 (F/P, 7%, 10) = $10 (1.967) = $19.67

14-8
iequivalent

= i’inflation corrected + f% + (i'inflation corrected) (f%)

In this problem:

iequivalent = 5%
f% = +2%
iinflation corrected
= unknown

0.05 = i’inflation corrected + 0.02 + (i'inflation corrected) (0.02)
i'inflation corrected = (0.05 – 0.02)/(1 + 0.02) = 0.02941

= 2.941%

That this is correct may be proved by the year-by-year computations.
Year

Cash Flow

0
1
2

-$1,000
+$50
+$50

(1 + f)-n
(P/F, f%, n)
0
0.9804
0.9612

Cash Flow in
Year 0 dollars
-$1,000.00
+$49.02
+$48.06

PW at 2.941%
-$1,000.00
+$47.62
+$45.35


3
4
5
6
7
8
9
10

11
12
13
14
15
16
17
18
19
20

+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$50
+$1,000

0.9423
0.9238
0.9057
0.8880
0.8706
0.8535
0.8368
0.8203
0.8043
0.7885
0.7730
0.7579
0.7430
0.7284
0.7142
0.7002
0.6864
0.6730

+$47.12
+$46.19
+$45.29
+$44.40
+$43.53
+$42.68
+$41.84
+$41.02
+$40.22
+$39.43
+$38.65

+$37.90
+$37.15
+$36.42
+$35.71
+$35.01
+$34.32
+$706.65

+$43.20
+$41.13
+$39.18
+$37.31
+$35.54
+$33.85
+$32.23
+$30.70
+$29.24
+$27.85
+$26.52
+$25.26
+$24.05
+$22.90
+$21.82
+$20.78
+$19.79
+$395.76
+$0.08

Therefore, iinflation corrected = 2.94%.

14-9
$20,000 in Year 0 dollars

n = 14 yrs
P = Lump Sum
deposit

Actual Dollars 14 years hence

= $20,000 (1 + f%)n
= $20,000 (1 + 0.08)14
= $58,744

At 5% interest:
P = F (1 + i)-n

= $58,744 (1 + 0.05)-14

= $29,670

Since the inflation rate (8%) exceeds the interest rate (5%), the money is annual losing
purchasing power.
Deposit $29,670.
14-10
To buy $1 worth of goods today will require:
F
= P (F/P, f%, n)
n years hence.
5
F

= $1 (1 + 0.05)
= $1.47
5 years hence.


For the subsequent 5 years the amount required will increase to:
$1.47 (F/P, f%, n)
= $1.47 (1 + 0.06)5
= $1.97
Thus for the ten year period $1 must be increased to $1.97. The average price change per
year is:
($1.97 - $1.00)/10 yrs = 9.7% per year
14-11
(1 + f)5
(1 + f)
f

= 1.50
= 1.501/5
= 0.845

= 1.0845
= 8.45%

14-12
Number of dollars required five years hence to have the buying power of one dollar today =
$1 (F/P, 7%, 5)
= $1.403
Number of cruzados required five years hence to have the buying power of 15 cruzados
today

= 15 (F/P, 25%, 5)
= 45.78 cruzados.
Combining: $1.403 = 45.78 cruzados
$1 = 32.6 cruzados

(Brazil uses cruzados.)

14-13
Price increase
= (1 + 0.12)8
Therefore, required fuel rating

= 2.476 x present price
= 10 x 2.476 = 24.76 km/liter

14-14
P= 1.00

F = 1.80

n = 10

1.80 = 1.00 (F/P, f%, 10)
(F/P, f%, 10) = 1.80
From tables, f is slightly greater than 6%. (f = 6.05% exactly).
14-15
i = i' + f + (i') (f)
0.15 = i' + 0.12 + 0.12 (i')
1.12 i' = 0.03
i' = 0.03/1.12

= 0.027 = 2.7%

f = ?


14-16
Compute equivalent interest/3 mo.

=x

= (1 + x)n – 1
= (1 + x)4 – 1
= 1.19250.25
= 1.045
= 0.045
= 4.5%/3 mo.

ieff
0.1925
(1 + x)
x

$3.00
n=?
i = 4.5%
$2.50

$2.50
= $3.00 (P/F, 4.5%, n)
(P/F, 4.5%, n)

= $2.50/$3.00 = 0.833
n is slightly greater than 4.
So purchase pads of paper- one for immediate use plus 4 extra pads.
14-17
f
i'
i

= 0.06
= 0.10
= 0.10 + 0.06 + (0.10) (0.06) = 16.6%

14-18
(a)
$109.6
1981

1986
n=5

$90.9

$109.6
(F/P, f%, 5)
f%

= $90.9 (F/P, f%, 5)
= $109.6/$90.9 = 1.2057
= 3.81%

(b)
CPI
1996
n=9
f = 3.81%
$113.6

CPI1996

= $113.6 (F/P, 3.81%, 9)
= $113.6 (1 + 0.0381)9 = $159.0

14-19
F

= $20,000 (F/P, 4%, 10) = $29,600

14-20
Compute an equivalent i:
iequivalent

= i' + f + (i') (f)
= 0.05 + 0.06 + (0.05) (0.06)
= 0.113
= 11.3%

Compute the PW of Benefits of the annuity:
PW of Benefits
= $2,500 (P/A, 11.3%, 10)

= $2,500 [((1.113)10 – 1)/(0.113 (1.113)10)]
= $14,540
Since the cost is $15,000, the benefits are less than the cost computed at a 5% real rate of
return. Thus the actual real rate of return is less than 5% and the annuity should not be
purchased.
14-21
1
log (1/0.20)

= 0.20 (1.06)n
= n log (1.06)

n

= 27.62 years

14-22
Use $97,000 (1 + f%)n, where f%=7% and n=15
$97,000 (1 + 0.07)15

= $97,000 (F/P, 7%, 15)
= $97,000 (2.759)


= $268,000
If there is 7% inflation per year, a $97,000 house today is equivalent to $268,000 15 years
hence. But will one have “profited” from the inflation?
Whether one will profit from owning the house depends somewhat on an examination of
the alternate use of the money. Only the differences between alternatives are relevant. If
the alterate is a 5% savings account, neglecting income taxes, the profit from owning the

house, rather than the savings account, would be:
$268,000 - $97,000 (F/P, 5%, 15)
= $66,300.
On the other hand, compared to an alternative investment at 7%, the profit is $0. And if
the alternative investment is at 9% there is a loss. If “profit” means an enrichment, or being
better off, then multiplying the price of everything does no enrich one in real terms.
14-23
Let x = selling price
Then long-term capital gain
Tax
After-Tax cash flow in year 10
Year
0
10

= x- $18,000
= 0.15 (x - $18,000)
= x – 0.15 (x - $18,000) = 0.85x + $2,700

ATCF
-$18,000
+0.85x + $2,700

Multiply by
1
1.06-10

Year 0 $ ATCF
-$18,000
0.4743x + $1,508

For a 10% rate of return:
$18,000 = (0.4746x + $1,508) (P/F, 10%, 10)
= 0.1830 x + $581
x = $95,186
Alternate Solution using an equivalent interest rate
iequiv
= i' + f + (i') (f) = 0.10 + 0.06 + (0.10) (0.06) = 0.166
So $18,000 (1 + 0.166)10 = 0.85x + $2,700
$83,610
= 0.85x + $2,700
Selling price of the lot

=x

= ($83,610 - $2,700)/0.85

14-24
Cash Flow:
Year $500 Kit $900 Kit
0
-$500
-$900
5
-$500
$0
(a) PW$500 kit
PW$900 kit

= $500 + $500 (P/F, 10%, 5) = $810

= $900

To minimize PW of Cost, choose $500 kit.

= $95,188


(b) Replacement cost of $500 kit, five years hence
= $500 (F/P, 7%, 5) = $701.5
PW$500 kit
PW$900 kit

= $500 + $701.5 (P/F, 10%, 5)
= $900

= $935.60

To minimize PW of Cost, choose $900 kit.
14-25
If one assumes the 5-year hence cost of the Filterco unit is:
$7,000 (F/P, 8%, 5) = $10,283
in Actual Dollars and $7,000 in Yr 0 dollars, the year 0 $ cash flows are:
Year
0
5

Filterco Duro
Duro – Filterco
-$7,000 -$10,000 -$3,000
-$7,000 $0

+$7,000

∆ROR
= 18.5%
Therefore, buy Filterco.
14-26
Year Cost to City (Year 0 $)
0
-$50,000
1- 10 -$5,000/yr

Benefits to City Description of Benefits

10

+$50,000

i

+A

Fixed annual sum in
thencurrent dollars
In then-current dollars

= i' + f + i'f
= 0.03 + 0.07 + 0.03 (0.07)
= 0.1021
= 10.21%
PW of Cost

$50,000 + $5,000 (P/A, 3%, 10)
$50,000 + $5,000 (8.530)
$92,650

A = ($92,650 - $18,915)/6.0895
= $12,109
* Computed on hand calculator

= PW of Benefits
= A(P/A, 10.21%, 10)
+$50,000 (P/F,10.21%,10)
= A (6.0895*) + $50,000 (0.3783*)
= 6.0895A + $18,915


14-27
Month
0
1- 36
36

BTCF
$0
-$1,000
+$40,365

$1,000 (F/A, i%, 36 mo)
(F/A, i%, 36)

= $40,365

= 40.365

Performing linear interpolation:
(F/A, i%, 36)) i
41.153
¾
%
39.336
½
%
i

= 0.50% + 0.25% [(40.365 – 39.336)/(41.153 – 39.336)]
= 0.6416% per month

Equivalent annual interest rate
i per year = (1 + 0.006416)12 – 1 = 0.080 = 8%
So, we know that i = 8% and f = 8%. Find i'.
i
= i' + f + (i') (f)
0.08 = i' + 0.08 + (i') (0.08)
i'
= 0%
Thus, before-Tax Rate of Return = 0%
14-28
Actual Dollars:

F= $10,000 (F/P, 10%, 15)

= $41,770

Real Dollars:
Year
1- 5
6- 10
11- 15
R$ in today’s base

Inflation
3%
5%
8%

= $41,770 (P/F, 8%, 5) (P/F, 5%, 5) (P/F, 3%, 5)
= $18,968

Thus, the real growth in purchasing power has been:
$18,968 = $10,000 (1 + i*)15
i* = 4.36%

14-29
(a) F = $2,500 (1.10)50 = $293,477

in A$ today


(b) R$ today in (-50) purchasing power = $293,477 (P/F, 4%, 50)
= $41,296
14-30
(a) PW

= $2,000 (P/A, ic, 8)
icombined = ireal + f + (ireal) (f)
= 0.0815
PW
(b) PW

= 0.03 + 0.05 + (0.03) (0.05)

= $2,000 (P/A, 8.15%, 9)

= $11,428

= $2,000 (P/A, 3%, 8)

= $14,040

14-31
Find PW of each plan over the next 5-year period.
ir

= (ic – f)/(1 + f) = (0.08 – 0.06)/1.06

PW(A)
PW(B)
PW(C)

= 1.19%

= $50,000 (P/A, 11.5%, 5)
= $236,359

= $45,000 (P/A, 8%, 5) + $2,500 (P/G, 8%, 5)
= $65,000 (P/A, 1.19, 5) (P/F, 6%, 5) = $229,612

= $198,115

Here we choose Company A’s salary to maximize PW.
14-32
(a) R today $ in year 15 = $10,000 (P/F, ir%, 15)
ir = (0.15 – 0.08)/1.08 = 6.5%
R today $ in year 15 = $10,000 (1.065)15

= $25,718

(b) ic = 15%f = 8%
F = $10,000 (1.15)15 = $81,371
14-33
No Inflation Situation
Alternative A:
Alternative B:

PW of Cost
PW of Cost

Alternative C:

PW of Cost

= $6,000
= $4,500 + $2,500 (P/F, 8%, 8)
= $4,500 + $2,500 (0.5403)

= $5,851
= $2,500 + $2,500 (P/F, 8%, 4)
+ $2,500 (P/F, 8%, 8)
= $2,500 ( 1 + 0.7350 + 0.5403)
= $5,688

To minimize PW of Cost, choose Alternative C.


For f = +5% (Inflation)
Alternative A:
Alternative B:

PW of Cost
PW of Cost

Alternative C:

PW of Cost

= $6,000
= $4,500 + $2,500 (F/P, 5%, 8) (P/F, 8%, 8)
= $4,500 + $2,500 (1 + f%)8 (P/F, 8%, 8)
= $4,500 + $2,500 (1.477) (0.5403)
= $6,495
= $2,500 + $2,500 (F/P, 5%, 4) (P/F, 8%, 4)
+ $2,500 (F/P, 5%, 8) (P/F, 8%, 8)
= $2,500 + $2,500 (1.216) (0.7350)
+ $2,500 (1.477) (0.5403)
= $6,729

To minimize PW of Cost in year 0 dollars, choose Alternative A.
This problem illustrates the fact that the prospect of future inflation encourages current
expenditures to be able to avoid higher future expenditures.
14-34
Alternative I: Continue to Rent the Duplex Home
Compute the Present Worth of renting and utility costs in Year 0 dollars.
Assuming end-of-year payments, the Year 1 payment is:
= ($750 + $139) (12) = $7,068
The equivalent Year 0 payment in Year 0 dollars is:
$7,068 (1 + 0.05)-1
= $6,713.40
Compute an equivalent i
iequivalent = i' + f + (i') (f)
Where i'
= interest rate without inflation
= 15.5%
f
= inflation rate
= 5%
iequivalent = 0.155 + 0.05 + (0.155) (0.05)
= 0.21275
= 21.275%
PW of 10 years of rent plus utilities:
= $6,731.40 (P/A, 21.275%, 10)
= $6,731.40 [(1 + 0.21275)(10-1))/(0.21275 (1 + 0.21275)10)]
= $6,731.40 (4.9246)
= $33,149
An Alternative computation, but a lot more work:
Compute the PW of the 10 years of inflation adjusted rent plus utilities using 15.5% interest.

PWyear 0
= 12[$589 (1 + 0.155)-1 + $619 (1 + 0.155)-2 + …
+ $914 (1 + 0.155)-10]
= 12 ($2,762.44)
= $33,149


Alternative II: Buying a House
$3,750 down payment plus about $750 in closing costs for a cash requirement of $4,500.
Mortgage interest rate per month = (1+I)6 = 1.04 I = 0.656%
n = 30 years x 12 = 360 payments
Monthly Payment:

A

= ($75,000 - $3,750) (A/P, 0.656%, 360)
= $71,250 [(0.00656 (1.00656)360)/((1.00656)360 – 1)]
= -$516.36

Mortgage Balance After the 10-year Comparison Period:
A’
= $523 (P/A, 0.656%, 240)
= $523 [((1.00656)240 – 1)/(0.00656 (1.00656)240)]
= $62,335
Thus:
$523 x 12 x 10
$71,250 - $62,504

= $62,760
= $8,746

= $54,014

total payments
principal repayments (12.28% of loan)
interest payments

Sale of the property at 6% appreciation per year in year 10:
F = $75,000 (1.06)10 = $134,314
Less 5% commission = -$6,716
Less mortgage balance
= -$62,335
Net Income from the sale
= $65,263
Assuming no capital gain tax is imposed, the Present Worth of Cost is:
PW= $4,500 [Down payment + closing costs in constant dollars]
+ $516.36 x 12 (P/A, 15.5%, 10) [actual dollar mortgage]
+ $160 x 12 (P/A, 10%, 10) [constant dollar utilities]
+ $50 x 12 (P/A, 10%, 10) [constant dollar insurance & maint.]
- $65,094 (P/F, 15.5%, 10) [actual dollar net income from sale]
PW= $4,500 + $516.36 x 12 (4.9246) + $160 x 12 (6.145)
+ $50 x 12 (6.145)- $65,263 (0.2367)
= $35,051
The PW of Cost of owning the house for 10 years = $35,051 in Year 0 dollars. Thus
$33,149 < $35,329 and so buying a house is the more attractive alternative.
14-35
Year

Cost- 1

Cost- 2

Cost- 3

Cost- 4

TOTAL

1
2
3
4
5

$4,500
$4,613
$4,728
$4,846
$4,967

$7,000
$7,700
$8,470
$9,317
$10,249

$10,000
$10,650
$11,342
$12,079
$12,865

$8,500
$8,288
$8,080
$7,878
$7,681

$30,000
$31,250
$32,620
$34,121
$35,762

PWTOTAL
$24,000
$20,000
$16,702
$13,976
$11,718


6
7
8
9
10

$5,091
$5,219
$5,349

$5,483
$5,620

$11,274
$12,401
$13,641
$15,005
$16,506

$13,701
$14,591
$15,540
$16,550
$17,626

$7,489
$7,302
$7,120
$6,942
$6,768

$37,555
$39,513
$41,649
$43,979
$46,519

$9,845
$8,286
$6,988

$5,903
$4,995

PW = -$60,000 – ($24,000 + $20,000 + $16,702 + … +$4,995)
+ $15,000 (P/F, 25%, 10)
= $180,802
14-36
(a) Unknown Quantities are calculated as follows:
% change
= [($100 - $89)/$89] x 100%
PSI
= 100 (1.04)
% change
= ($107 - $104)/$104
% change
= ($116 - $107)/$107
PSI
= 116 (1.0517)
(b)

= 12.36%
= 104
= 2.88%
= 8.41%
= 122

The base year is 1993. This is the year of which the index has a value of 100.

(c)
(i)

(ii)

PSI (1991)
PSI (1995)
h
i*
i*

= 82
= 107
= 4 years
=?
= (107/82)0.25 – 1

= 6.88%

PSI (1992)
PSI (1998)
n
i*
i*

= 89
= 132
= 6 years
=?
= (132/89)(1/6) – 1

= 6.79%

14-37
(a) LCI(-1970)
LCI(-1979)
n
i*
i*
(b) LCI(1980)
LCI(1989)
n
i*
i*

= 100
= 250
=9
=?
= (250/100)(1/9) – 1
= 250
= 417
=9
=?
= (417/250)(1/9) – 1

(c) LCI(1990)
LCI(1998)

= 417
= 550

= 10.7%

= 5.85%


n
i*
i*

=8
=?
= (550/417)(1/8) – 1

= 3.12%

14-38
(a) Overall LCI change
(b) Overall LCI change
(c) Overall LCI change

= [(250 – 100)/100] x 100%
= [(415 – 250)/250] x 100%
= [(650 – 417)/417] x 100%

= 150%
= 66.8%
= 31.9%

14-39
(a) CPI (1978)

CPI (1982)
n
i*
i*

= 43.6
= 65.3
=4
=?
= (65.3/43.6)(1/4) – 1

= 9.8%

(b) CPI (1980)
CPI (1989)
n
i*
i*

= 52.4
= 89.0
=9
=?
= (89.0/52.4)(1/9) – 1

= 6.1%

(c) CPI (1985)
CPI (1997)
n

i*
i*

= 75.0
= 107.6
= 12
=?
= (107.6/75.0)(1/12) – 1 = 3.1%

14-40
(a)
Year Brick Cost CBI
1970 2.10
442
1998 X
618
x/2.10 = 618/442
x
= $2.94
Total Material Cost
= 800 x $2.94 = $2,350
(b) Here we need f% of brick cost
CBI(1970)
= 442
CBI(1998)
= 618
n
= 18
i*
=?

i*
= (618/442)(1/18) – 1 = 1.9%
We assume the past average inflation rate continues for 10 more years.
Brick Unit Cost in 2008

= 2.94 (F/P, 1.9%, 10) = $3.54


Total Material Cost

= 800 x $3.54

= $2.833

14-41
EAT(today) = $330 (F/P, 12$, 10) = $1,025
14-42
Item
Structural
Roofing
Heat etc.
Insulating
Labor
Total

Year 1
$125,160
$14,280
$35,560
$9,522

$89,250
$273,772

Year 2
$129,165
$14,637
$36,306
$10,093
$93,266
$283,467

Year 3
$137,690
$15,076
$37,614
$10,850
$97,463
$298,693

(a) $89,250; $93,266; $97,463
(b) PW

= $9,522 (P/F, 25%, 1) + $10,093 (P/F, 25%, 2)
+ $10,850 (P/F, 25%, 3)
= $19,632

(c) FW

= ($9,522 + $89,250) (F/P, 25%, 2) + ($10,093
+ $93,266) (F/P, 25%, 1) + ($10,850 + $97,463)

= $391,843

(d) PW

= $273,772 (P/F, 25%, 1) + $283,467 (P/F, 25%, 2)
+ $298,693 (P/F, 25%, 3)
= $553,367

14-43
The total cost of the bike 10 years from today would be $2,770

Item
Frame
Wheels
Gearing
Braking
Saddle
Finishes
Sum=

Current
Cost
800
350
200
150
70
125
1695

Inflation
2.0%
10.0%
5.0%
3.0%
2.5%
8.0%
Sum=

Future
Cost
975.2
907.8
325.8
201.6
89.6
269.9
2769.8


14-44
To minimize purchase price Mary Clare should select the vehicle from company X.

Car
X
Y
Z

Current
Price

27500
30000
25000

Inflation
4.0%
1.5%
8.0%
Min=

Future
Price
30933.8
31370.4
31492.8
30933.8

14-45
FYEAR 5 = $100 (F/A, 12/4=3%, 5 x 4=20) = $2,687
FYEAR 10 = $2,687 (F/P, 4%, 20) + $100 (F/A, 4%, 20) = $8,865
FYEAR 15 (TODAY) = $8,865 (F/P, 2%, 20) + $100 (F/A, 2%, 20) = $15,603

14-46
To pay off the loan Andrew will need to write a check for $ 18,116

Year
1
2
3

Amt due
Begin yr
15000
15750.0
16773.8

Inflation
5.0%
6.5%
8.0%
Due=

Amt due
End yr
15750.0
16773.8
18115.7
18115.7

14-47
See the table below for (a) through (e)

Year
5 years ago
4 years ago
3 years ago
2 years ago
last year
This year

Ave
Price
165000.0
167000.0
172000.0
180000.0
183000.0
190000.0

Inflation
for year
(a) = 1.2%
(b) = 3.0%
(c) = 4.7%
(d) = 1.7%
(e) = 3.8%
(f) see below


One could predict the inflation (appreciation) in the home prices this year using a number of
approaches. One simple rule might involve using the average of the last 5 years inflation
rates. This rate would be (1.2+3+4.7+1.7+3.8)/5 = 2.9%.
14-48
Depreciation charges that a firm makes in its accounting records allow a profitable firm to
have that amount of money available for replacement equipment without any deduction for
income taxes.
If the money available from depreciation charges is inadequate to purchase needed
replacement equipment, then the firm may need also to use after-tax profit for this purpose.
Depreciation charges produce a tax-free source of money; profit has been subjected to
income taxes. Thus substantial inflation forces a firm to increasingly finance replacement

equipment out of (costly) after-tax profit.

14-49
(a)
Year

BTCF

0
1
2
3
4
5

-$85,000
$8,000
$8,000
$8,000
$8,000
$8,000
$77,500

Sum

SL
Deprec
.

TI

34%
Income
Taxes

$1,500
$1,500
$1,500
$1,500
$1,500

$6,500
$6,500
$6,500
$6,500
$6,500
$0

-$2,210
-$2,210
-$2,210
-$2,210
-$2,210

ATCF
-$85,000
$5,790
$5,790
$5,790
$5,790

$83,290

$7,500

SL Depreciation
= ($67,500 - $0)/45 = $1,500
Book Value at end of 5 years = $85,000 – 5 ($1,500) = $77,500
After-Tax Rate of Return

= 5.2%

(b)
Year

BTCF

0
1
2
3
4
5

-$85,000
$8,560
$9,159
$9,800
$10,486
$11,220
$136,935*

Sum

SL
Deprec
.

TI

34%
Income
Taxes

$1,500
$1,500
$1,500
$1,500
$1,500

$7,060
$7,659
$8,300
$8,986
$9,720

-$2,400
-$2,604
-$2,822
-$3,055
-$3,305

-$16,242**

Actual
Dollars ATCF
-$85,000
$6,160
$6,555
$6,978
$7,431
$131,913

$7,500

*Selling Price = $85,000 (F/P, 10%, 5) = $85,000 (1.611) = $136,935
** On disposal, there are capital gains and depreciation recapture
Capital Gain = $136,935 - $85,00 = $51,935


Tax on Cap. Gain = (20%) ($51,935) = $10,387
Recaptured Depr. = $85,000 - $77,500 = $7,500
Tax on Recap. Depr. = (34%)($7,500) = $2,550
Total Tax on Disposal = $10,387 + $2,550 = $12,937
After Tax IRR = 14.9%
After-Tax Rate of Return in Year 0 Dollars
Year
0
1
2
3
4

5
Sum

Actual Dollars
ATCF
-$85,000
$6,160
$6,555
$6,978
$7,431
$131,913

Multiply by
1
1.07-1
1.07-2
1.07-3
1.07-4
1.07-5

In year 0 dollars, After-Tax Rate of Return

Year 0
$ ATCF
-$85,000
$5,757
$5,725
$5,696
$5,669
$94,052

= 7.4%

14-50
Year

BTCF

TI

42%
Income
Taxes

0
1
2
3
4
5

-$10,000
$1,200
$1,200
$1,200
$1,200
$1,200
$10,000

$1,200
$1,200

$1,200
$1,200
$1,200

-$504
-$504
-$504
-$504
-$504

ATCF

Multiply by Year 0 $ ATCF

-$10,000
$696
$696
$696
$696
$10,696

1
1.07-1
1.07-2
1.07-3
1.07-4
1.07-5

-$10,000
$650

$608
$568
$531
$7,626

Sum

-$17

(a) Before-Tax Rate of Return ignoring inflation
Since the $10,000 principal is returned unchanged,
i = A/P = $1,200/$10,000 = 12%
If this is not observed, then the rate of return may be computed by conventional means.
$10,000 = $1,200 (P/A, i%, 5) + $10,000 (P/F, i%, 5)
Rate of Return = 12%
(b) After-Tax Rate of Return ignoring inflation
Solved in the same manner as Part (a):
i = A/P = $696/$10,000 = 6.96%
(c) After-Tax Rate of Return after accounting for inflation


An examination of the Year 0 dollars after-tax cash flow shows the algebraic sum of the
cash flow is -$17. Stated in Year 0 dollars, the total receipts are less than the cost,
hence there is no positive rate of return.
14-51
Now:
Taxable Income
Income Taxes
After-Tax Income

= $60,000
= $35,000x0.16+25,000x0.22 = $11,100
= $60,000 - $11,100 = $48,900

Twenty Years Hence: To have some buying power, need:
After-Tax Income
= $48,900(1.07)20
= $189,227.60
= Taxable Income – Income Taxes
Income Taxes

= $24,689.04+0.29(Taxable Income - $113,804)

Taxable Income

= After-Tax Income + Income Taxes
= $189,227.60+$24,689.04+0.22(TI - $113,804)
= $242,153.50

14-52
P=
CCA=
t=
S=

$10,000
30%
50%
0

f=

7%

Year
0
1
2
3
4
5
6
7

Actual $s
Received
-$10,000
$2,000
$3,000
$4,000
$5,000
$6,000
$7,000
$8,000

Actual
CCA

Actual $s
Tax

Net
Salvage

-$1,500
-$2,550
-$1,785
-$1,250
-$875
-$612
-$429

$250
$225
$1,108
$1,875
$2,563
$3,194
$3,786

$500

Actual $s
ATCF
-$10,000
$1,750
$2,775
$2,893
$3,125
$3,437

$3,806
$4,714

IRR=

Real $s
ATCF
-$10,000
$1,636
$2,424
$2,361
$2,384
$2,451
$2,536
$2,936

13.71%

Net Salvage Calculation from Equation 12-7, 12-8, & 11-8
UCC7 = $1,000

=$B$1*(1-$B$2/2)*(1-$B$2)^(7-1)

Net Salvage end of yr 7 = $500
=$B$4+$B$3*IF($B$4>$B$1,$C$18-$B$1,$C$18-$B$4)-1/2*$B$3*MAX(($B$4-$B$1),0)


14-53
P=
CCA=

t=
S=

$

$103,500
10%
35%
103,500

f=

0%
Actual $s
Received
-$103,500
$15,750
$15,750
$15,750
$15,750
$15,750

Year
0
1
2
3
4
5

Actual
CCA
-$5,175
-$9,833
-$8,849
-$7,964
-$7,168

Actual $s
Tax
$3,701
$2,071
$2,415
$2,725
$3,004

Net
Salvage
-$46,500

$136,354

Actual $s
ATCF
-$150,000
$12,049
$13,679
$13,335
$13,025
$149,100

Real $s
ATCF
-$150,000
$12,049
$13,679
$13,335
$13,025
$149,100

7.06%

=IRR

Net Salvage Calculation from Equation 12-7, 12-8, &
11-8
UCC5 =$64,511 =$B$1*(1-$B$2/2)*(1-$B$2)^(5-1)
Net Salvage end of yr 5 = $89,854
=$B$4+$B$3*IF($B$4>$B$1,$C$18-$B$1,$C$18-$B$4)-1/2*$B$3*MAX(($B$4-$B$1),0)

14-54
P=
CCA=
t=
S=

f=

Year
0

1
2
3
4
5

$

$103,500
10%
35%
103,500

10%

Actual $s
Received
-$103,500
$12,000
$13,440
$15,053
$16,859
$18,882

Actual
CCA

Actual $s Tax

Net Salvage

-$46,500

-$5,175
-$9,833
-$8,849
-$7,964
-$7,168

$2,389
$1,263
$2,171
$3,113
$4,100

$230,694

Actual $s
ATCF
-$150,000
$9,611
$12,177
$12,882
$13,746
$245,476

16.01%

Real $s
ATCF
-$150,000

$8,738
$10,064
$9,678
$9,389
$152,421

5.46%

=IRR


Net Salvage Calculation from Equation 12-7, 12-8, & 11-8
UCC5 = $64,511

=$B$1*(1-$B$2/2)*(1-$B$2)^(5-1)

Net Salvage end of yr 5 = $89,854
=$B$4+$B$3*IF($B$4>$B$1,$C$18-$B$1,$C$18-$B$4)-1/2*$B$3*MAX(($B$4-$B$1),0)

Market Value in 5 years = 150000 x (1+12%)^5
house value
land sale
capital gain's
tax
net land
salvage

$264,351.25
$103,500.00
$160,851.25

$ 20,011.47
$140,839.78

14-55
Alternative A
Year

Cash
Flow in
Year 0 $

Cash
Flow in
Actual $

SL
Deprec.

TI

25%
Income
Tax

0
1
2
3

-$420

-$420

$200
$200
$200

$210
$220.5
$231.5

$140
$140
$140

$70
$80.5
$91.5

-$17.5
-$20.1
-$22.9

SL
Deprec.

TI

25%
Income

Tax

$100
$100
$100

$57.5
$65.4
$73.6

-$14.4
-$16.4
-$18.4

ATCF in
Actual $

ATCF in
Year 0 $

-$420

-$420

$192.5
$200.4
$208.6

$183.3
$181.8

$180.2

ATCF in
Actual $

ATCF in
Year 0 $

-$300

-$300

$143.1
$149.0
$155.2

$136.3
$135.1
$134.1

Alternative B
Year

Cash
Flow in
Year 0 $

Cash
Flow in
Actual $

0
1
2
3

-$300

-$300

$150
$150
$150

$157.5
$165.4
$173.6

Quick Approximation of Rates of Return:
Alternative A:
$420 = $182 (P/A, i%, 3)
(P/A, i%, 3) = $420/$182 = 2.31
12% < ROR < 15%
(Actual ROR = 14.3%)
Alternative B:
$300 = $135 (P/A, i%, 3)
(P/A, i%, 3) = $300/$135 = 2.22
15% < ROR < 18%
(Actual ROR = 16.8%)

Incremental ROR Analysis for A- B
Year A
0
-$420
1
$183.3
2
$181.8
3
$180.2

B
-$300
$136.3
$135.1
$134.1

A- B
-$120
$47
$46.7
$46.1

Try i = 7%
NPW
= -$120 + $47 (P/F, 7%, 1) + $46.7 (P/F, 7%, 2)
+ $46.1 (P/F, 7%, 3)
= +$2.3
So the rate of return for the increment A- B is greater than 7% (actually 8.1%). Choose the

higher cost alternative: choose Alternative A.