Engineering Economic Analysis

446 INFLATION AND PRICECHANGE I

RB= 1950$ in 2005~ 3,357,000l Slrippmg

.

Out

55 yearsof

A$in 2005 = 82,701,600 6% Inflation

t

2005

Time

FIGURE 14-3 Translation ofA$in20.0.5toR195o.-based dollars in 2005.

2. In this method we start by recognizingthat theactual dollars in 1950are exactlyequivalent
   toreal 1950-hased dollars fhatexistin 1950.By definition,dollars thafhavetodayas the
   purchasing power base are the same asactual dollars today.'Thus,actual.flollm-~in 20.0.4
   are the same as real 2o.04.,-baseddollars that occur in 20.0.4.So in this example, the $1.2
   million can also be said to bereaLJ950-baseddollars that occur in 1950..As such, letus
   translate thosereal'dollm-sfrom 1950to the year 20.0.5.'Since they arereal dollar$,we use
   thereal interest rate.

Real 195Q-baseddollars in 20.05=(Real 1950-based dollm-s iIl195o.)(P 1P,i', 55)

!'

..

·($1;20.0.,0.0.0)f (FjP, 1.887%, 55)

I .~ $3,355,0.00.
t
FIGUllE14.,4 Translation ofR1950-based
dollars in 1950.toR195o.-based dollars in20.0.5.

~ $3.355,qoo

$1,200,000

t

1950

')

I

1:887% 2005

" Time

(No~e:The difference between the answets to parts 1 and 2 is due t9 rounding themarkeOnterest

rate"off to 1.887% versus caI"fYingitouttomotesignificant digits. The amount of difference due

to this rounding is less than 1%.'If vJewete to ca:Tythe caIcula.tlonofi'to a sufficientnumlJer6f

digffs, the answers to the two parts would,be identical.)

From Example 14-2we can see the relationship between dollars of different purchasing

power bases, the selection of appropriateinterest rates to use for moving dollars in time, and

stripping out or adding in inflation. In that example the $1.2 million initially investedgrew

in the 55-year period to over $82.7 million, which becomes the amount available to pay for

construction of the new complex. Does this mean that the new stadium will be82.~/1.2or

about 70.times "better" than the one built in 195o.?The answer to that question is no, because

in the year 20.0.5the purchasingpower of a dollar is less than it was in the year 1950..Assum-

ing that construction costs increased at the rate of 6% per year given in the problem, then

the amount available for the projectin terms of 1950-based dollarsis almost $3.4 million.

This means that the new stadium will be about3.4/1.2or approximately 2.8 times "better"

than the original one usingreal dollars-not the 70.times ratio ifactual dollarsare used.