Engineering Economic Analysis

Analysis in Constant Dollars Versus Then-Current Dollars 449

SOLUTION..

The costs for each of the two alternativesare as follows:

Year

1

2

3

4

5

Then-Current Costs

Stated by Alpha

$150,000 x (1.05)° = $150,000

150,000 x (1.05)1= 157,500

150,000 x (1.05)2- 165,375

150,000 x (1.05)3- 173,644

150,000 x (1.05)4 = 182,326

Constant Dollar Costs

Stated by Beta

$150,000

150,000

150,000

150,000

150,000

We inflate (or escalate) the stated yearlYocost given by Company Alpha by 5% per year to obtain'

the then-current (actual) dollars each year. Company Beta's costs are given in terms of today-based

constant dollars.

:I

Using a Constant Dollar Analysis

Here we must convert the then-currentcosts given by Company Alpha to constant today-based

dollars. We do this by stripping the appropriate number of years of general inflation from each

year's cost using(P/F, f,n)or (l + f)-n.

- - - - ---.------

Constant Dollar Costs

Stated by Beta

$150,000

150,000

150,000

150,000

150,000

We use thereal interest rate (i')calculated from Equation 14-1tqcalculate the presentwor$ of

costs for each alternative:

i'=(i- f)/(1 +f) =(0.25- 0.035)/(1+ 0.035)=0.208
t I I i r I

PW of cost (Alpha) =$144,928(P /F,20.8%, 1) +$147,028(P I F,20.8%, 2)

+$149, 159(P /F,20.8%, 3) +$151,321(P /F,20.8%, 4)

+ $153,514(P /F, 20.8%,5) -- $436,000

I?Wof cost (Beta)=$150,000(P/A,20.8%,5) = $150,000(2.9387)-- $441,000

_.Using a'Then-Current Dollar Amilysis __ a .... ;II

1fere w~ must convert the constant dollar costs of Company Beta to then-current dollars. We

do this by using(F IP, f, n)or (1 + f)n to "add in" the appropriate number of years of gen,eral

. =' lll~tlOq. fl<>1.'t0 e~c.hy~ilt~s' GP§t..: ~ ~I:' '=Ii;;;;;:=; e: == =='" ~ ==11I:II:= = ~== !II= 11_

Year

1

2

3

4

5

Constant Dollar Costs

Stated by 1\1pha

$150,000 x (1.035)-1 =$144,928

157,500 x (1.035)-2 = 147,02?

165,375 x (1.035)-3 = 149,159

173,644 x (1.035)-4 151,321

182,326 x (1.035)-5= 153,514

t