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.

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.

## Solution manual accounting information systems 12th edition by romney and steinbart CH01

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