The b F (
fv] command computes
the future value of an investment. It takes three arguments
from the stack: ‘fv(rate, n, payment)’.
If you give payments of payment every year for n
years, and the money you have paid earns interest at rate per
year, then this function tells you what your investment would be
worth at the end of the period. (The actual interval doesn’t
have to be years, as long as n and rate are expressed
in terms of the same intervals.) This function assumes payments
occur at the end of each interval.
The I b F [
fvb] command does the same computation,
but assuming your payments are at the beginning of each interval.
Suppose you plan to deposit $1000 per year in a savings account
earning 5.4% interest, starting right now. How much will be
in the account after five years?
fvb(5.4%, 5, 1000) = 5870.73.
Thus you will have earned $870 worth of interest over the years.
Using the stack, this calculation would have been
5.4 M-% 5 RET 1000 I b F. Note that the rate is expressed
as a number between 0 and 1, not as a percentage.
The H b F [
fvl] command computes the future value
of an initial lump sum investment. Suppose you could deposit
those five thousand dollars in the bank right now; how much would
they be worth in five years?
fvl(5.4%, 5, 5000) = 6503.89.
The algebraic functions
fvb accept an optional
fourth argument, which is used as an initial lump sum in the sense
fvl. In other words,
payment, initial) = fv(rate, n, payment)
+ fvl(rate, n, initial).
To illustrate the relationships between these functions, we could
fvb calculation “by hand” using
final balance will be the sum of the contributions of our five
deposits at various times. The first deposit earns interest for
fvl(5.4%, 5, 1000) = 1300.78. The second
deposit only earns interest for four years:
fvl(5.4%, 4, 1000) =
1234.13. And so on down to the last deposit, which earns one
fvl(5.4%, 1, 1000) = 1054.00. The sum of
these five values is, sure enough, $5870.73, just as was computed
fv(5.4%, 5, 1000) = 5569.96 mean? The payments
are now at the ends of the periods. The end of one year is the same
as the beginning of the next, so what this really means is that we’ve
lost the payment at year zero (which contributed $1300.78), but we’re
now counting the payment at year five (which, since it didn’t have
a chance to earn interest, counts as $1000). Indeed, ‘5569.96 =
5870.73 - 1300.78 + 1000’ (give or take a bit of roundoff error).