Worked examples · Sheet 2

Equivalent rates, nominal rates and force of interest

Theory used: Equivalent rates Nominal rate convertible k times Continuous compounding

Exercise 1

Given the annual rate $i = 5\%$ , compute the equivalent semiannual, quarterly and monthly rates.

i_2 = 1.05^{\frac{1}{2}} - 1 \approx 0.024695 = 2.4695\%

i_4 = 1.05^{\frac{1}{4}} - 1 \approx 0.012272 = 1.2272\%

i_{12} = 1.05^{\frac{1}{12}} - 1 \approx 0.004074 = 0.4074\%

The periodic rate $i_k$ is equivalent to the annual rate if compounding $k$ times in a year at the rate $i_k$ gives the same accumulated value as annual compounding at $5\%$ , that is, if $(1 + i_k)^{k} = 1.05$ . Solving for $i_k$ one takes the $k$ -th root, which is written as a power with exponent $\dfrac{1}{k}$ , and subtracts $1$ .

Each equivalent rate is a little less than the annual rate divided by $k$ ; for example $i_2 \approx 2.47\%$ is less than $2.5\%$ . Indeed in the second half-year the interest is computed also on the interest of the first, so a lower semiannual rate is enough to obtain the same accumulated value. As a check, $1.024695^{2} \approx 1.05$ .

Exercise 2

Loan A has a nominal annual rate (APR) of $9\%$ with monthly installments; loan B has an APR of $9.2\%$ with semiannual installments. Compute the effective annual rates (EAR) and determine which loan costs less.

\text{A:} \quad i_{12} = \frac{0.09}{12} = 0.0075, \qquad i = 1.0075^{12} - 1 \approx 9.38\%

\text{B:} \quad i_{2} = \frac{0.092}{2} = 0.046, \qquad i = 1.046^{2} - 1 = 0.094116 \approx 9.41\%

The nominal rate convertible $k$ times is the periodic rate multiplied by $k$ , so the periodic rate is obtained by dividing the APR by $k$ ; this gives $0.75\%$ monthly for A and $4.6\%$ semiannually for B. The effective rate is the annual rate equivalent to the periodic rate, that is, $(1 + i_k)^{k} - 1$ .

The two nominal rates cannot be compared with each other, because they refer to different periods; the effective rates, instead, are both annual. Loan A has the lower EAR, about $9.38\%$ against $9.41\%$ , therefore A costs less. The EAR of A exceeds its APR by $0.38$ points, that of B by about $0.21$ points, because there are $12$ compoundings in a year for A and $2$ for B.

Exercise 3

A deposit earns interest under continuous compounding with force of interest $\delta = 4\%$ . Compute the equivalent annual compound rate, the accumulated value of $5000$ euros after $3$ years and $6$ months, and the time needed for a principal to double.

i = e^{0.04} - 1 \approx 0.040811 = 4.0811\%

M = 5000\,e^{0.04 \cdot 3.5} = 5000\,e^{0.14} \approx 5751.37 \text{ euros}

e^{0.04\,t} = 2 \iff 0.04\,t = \ln 2 \iff t = \frac{\ln 2}{0.04} \approx 17.33 \text{ years}

Under continuous compounding a principal is multiplied by $e^{\delta}$ after one year, so the equivalent annual compound rate is the one for which $1 + i = e^{\delta}$ . One obtains $i \approx 4.08\%$ , a little more than the force of interest, as happens for the effective rate compared with the nominal rate.

The time of $3$ years and $6$ months corresponds to $t = 3.5$ , and the accumulated value is $C\,e^{\delta t}$ . The same result is obtained under compound interest at the equivalent rate, because $(1 + i)^{3.5} = e^{0.04 \cdot 3.5}$ .

The principal doubles when the accumulation factor equals $2$ . The unknown is in the exponent, therefore one takes the natural logarithm of both sides, and the doubling time is about $17.33$ years, that is, $17$ years and about $4$ months; it does not depend on the initial principal.