FM for Actuaries

26 CHAPTER 1
Solution: Over one quarter, the interest rate is 1.5%. With C 0 = $100, C j =0
for j>0, andF = $300, from (1.37) the equation of value is (n is in quarters)
300 = 100(1.015) n ,
so that
i.e., 18.45 years.
n =
ln(3)
ln(1.015)
=73.79 quarters,
Another problem is the solution of the rate of interest that will give rise to a
targeted present value or future value with given cash flows. The example below
illustrates this point.
Example 1.23: A student takes out a tuition loan of $15,000, and is required to
pay back with a step-up payment $7,000 in year 1 and $8,500 in year 2. What is
the effective rate of interest she is charged?
Solution: The equation of value is, from (1.36),
15,000 = 7,000v +8,500v 2 .
Solving for the quadratic equation (see Appendix A.3), we obtain, after dropping
thenegativeroot,
v = −7+√ 7 2 +4× 8.5 × 15
2 × 8.5
=0.979,
which implies
i = 1
0.979 − 1=2.14%.
Note that the solution of the above problem can be obtained analytically by
solving a quadratic equation, as there are only two payments. When there are more
payments, the solution will require the use of numerical methods, the details of
which can be found in Chapter 4.