Number of Periods

Study guide: Exam FM-2

The number of periods n (or time t) is the length of time, counted in the same unit as the interest rate, over which an amount is accumulated or discounted; in a time-value-of-money problem it is the unknown found when the rate, the present value and the future value are given.

n = \frac{\ln(FV/PV)}{\ln(1+i)}

t = \frac{FV/PV - 1}{i}

• The first block solves FV = PV(1+i)^n under compound interest; the second solves FV = PV(1 + it) under simple interest.
• Units follow the rate. With a nominal rate i^{(12)} the natural rate is i^{(12)}/12 per month, so n comes out in months. At 6\% convertible monthly, money doubles in \ln 2 / \ln 1.005 = 138.98 months, or 11.58 years.
• Since \ln(1+i) = \delta, the force of interest, the compound answer is also n = \ln(FV/PV)/\delta. Under a variable force \delta(t), solve \int_0^n \delta(t)\,dt = \ln(FV/PV) for n.
• Under compound interest n need not be a whole number — (1+i)^t is defined for every real t. For an annuity or loan, the number of payments is the Term of Annuity or Term of Loan, and a non-integer answer means a smaller final Drop Payment or a larger Balloon Payment.
• It is one of the four quantities in a time-value-of-money equation: given any three of rate, time, present value and future value, the Equation of Value gives the fourth.

Time for a Claims Fund to Reach Its Target An insurer sets aside $2,000,000 to meet a settlement it expects to cost $2,500,000. How long must the fund be invested at 4% per year to reach the target under (a) compound interest and (b) simple interest?

Answer The ratio needed is FV/PV = 2{,}500{,}000/2{,}000{,}000 = 1.25.

\begin{align*}
n_{\text{compound}} &= \frac{\ln 1.25}{\ln 1.04} \\
&= \frac{0.22314}{0.03922} \\
&= 5.69 \text{ years}
\end{align*}

\begin{align*}
t_{\text{simple}} &= \frac{1.25 - 1}{0.04} \\
&= 6.25 \text{ years}
\end{align*}

Compound interest reaches the target about seven months sooner, because interest earned in the early years itself earns interest.

Solving for Time Under a Variable Force of Interest A fund grows under the force of interest \delta(t) = 0.02 + 0.004t. How long does it take a deposit to grow to 150% of its original amount?

Answer Set the accumulated force equal to \ln 1.5:

\begin{align*}
\int_0^n (0.02 + 0.004t)\,dt &= \ln 1.5 \\
0.02n + 0.002n^2 &= 0.405465
\end{align*}

Solve the quadratic 0.002n^2 + 0.02n - 0.405465 = 0, keeping the positive root:

\begin{align*}
n &= \frac{-0.02 + \sqrt{0.02^2 + 4(0.002)(0.405465)}}{2(0.002)} \\
&= \frac{-0.02 + 0.060363}{0.004} \\
&= 10.09 \text{ years}
\end{align*}

The deposit reaches 150% of its value after about 10.1 years.