When the university stadium was completed in 1960, the total cost was $1.2 million. At that time a wealthy alumnus made the university a gift of $1.2 million to be used for a future replacement. University administrators are now considering building the new facility in the year 2015. Assume

Question

When the university stadium was completed in 1960, the total cost was $1.2 million. At that time a wealthy alumnus made the university a gift of $1.2 million to be used for a future replacement. University administrators are now considering building the new facility in the year 2015. Assume the following:

•Inflation is 6.0% a year from 1960 to 2015.

• In 1955 the university invested the gift at a market interest rate of 8.0% per year.

(a) Define i, i',f, A$, and R$ from the problem.

(b) How many actual dollars in the year 2015 will the gift be worth?

(c) How much would the actual dollars in 2015 be in terms of 1960 purchasing power?

(d) How much better or worse should the new stadium be?

Accepted Answer

SOLUTION TO (a)

Since 6.0% is the inflation rate (f) and 8.0% is the market interest rate (i), we can write

i' = (0.08 -0.06)/(1 + 0.06) = 0.01887, or 1.887%

The cost of the building in 1960 was $1,200,000. These were the actual dollars (A$) spent in 1960. We will take the basis year for our real-dollar calculations to be 1960, so we can also say that $1,200,000 in R$ were spent in 1960.

SOLUTION TO (b)

From Figure 14-1 we are going from actual dollars at t, in 1960, to actual dollars at n, in 2015. To do so, we use the market interest rate and compound this amount forward 55 years, as illustrated in Figure 14-2.

Actual dollars in 2015 = (Actual dollars in 1960)(F/P, i, 55 years)

= $1,200,00(F/P, 8%, 55)
= $82,696,600

SOLUTION TO (c)
Now we want to find how many real 1960 dollars are equivalent in 2015 to the $82.7 million from the solution to part (b). Let us solve this problem two ways.
1. Translate the actual dollars in the year 2015 into real 1960 dollars in the year 2015. From Figure 14-1 we can use the inflation rate to strip 55 years of inflation from the actual dollars. We do this by using the P/F factor for 55 years at the inflation rate. This is illustrated in the following equation and Figure 14-3.
Real 1960-based dollars in 2015 = (Actual dollars in 2015)(P/F, f, 55)
= ($82,696,600)(P/F, 6%, 55)
= $3,357,000

2. Translate the real 1960 dollars in 1960 into real 1960 dollars in 2015. Since they are real dollars, we use the real interest rate.
Real 1960-based dollars in 2015 = (Real 1960-based dollars in 1960)(P/F, i', 55)
= ($1,200,000)(F/P, 1.887%, 55)
= $3,355,000

(Note: The answers differ because we rounded off the market interest rate to 1.887% rather than carry it out to more significant digits. The difference from this rounding is less than 0.1%. If i' and the factors had enough digits, the answers to the two parts would be identical.)
SOLUTION TO (d)
Assuming that construction costs increased at the rate of 6% a year, then the amount available for the project in terms of 1960-based dollars is almost $3.4 million. This means that the new stadium will be about 3.4/1.2, or approximately 2.8, times better than the original.

Question 14.2: When the university stadium was completed in 1960, the total...

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