Question

Determine the effective annual yield for each investment. Then select the better investment. Assume 360 days in a year. 9.5% compounded monthly; 9.75% compounded annually. The effective annual yield for a 9.5% compounded monthly investment Round to two decimal places as needed. The effective annual yield for a 9.75% compounded annually investment Round to two decimal places as needed

Answer 1

In order to calculate the final value of the investment, we can use the following equation:

$P=P_0\cdot(1+\frac{r}{n})^{nt}$

Where P is the final value, P0 is the initial value, r is the annual rate, t is the amount of time and n is a factor relative to the period of compound.

For the first investment, the rate is 9.5 compounded monthly (a year has 12 months, so we use n = 12). So for one year, we have that:

$\begin{gathered} P=P_0(1+\frac{0.095}{12})^{12} \\ \frac{P}{P_0}=(1+0.007916667)^{12} \\ \frac{P}{P_0}=(1.007916667)^{12}=1.09925 \end{gathered}$

The increase in the initial investment is 0.0992 times, that is, 9.92%.

For the second investment, the rate is 9.75% compounded annually (so n = 1), so we have:

$\begin{gathered} P=P_0(1+\frac{0.0975}{1})^1 \\ \frac{P}{P_0}=1+\frac{0.0975}{1}=1.0975 \end{gathered}$

The increase in the initial investment is 0.0975 times, that is, 9.75%.

The first investment has a greater increase in one year, so the first investment is better.