Chapter 11.3, Problem 11E

a.

To determine

Complete the table by finding the effective annual yield for each account.

Answer to Problem 11E

  $\begin{array}{ccc}InterestCompounded& Interest\left(\$\right)& EffectiveAnnualYield\\ Annually& 80& 8\%\\ Semi-annually& 81.6& 8.16\%\\ Quarterly& 82.4316& 8.24\%\\ monthly& 83& 8.3\%\\ Daily& 83.28& 8.328\%\\ Continuously& 83.29& 8.329\%\end{array}$

Explanation of Solution

Given information:

A bank earns interest that is compounded more than one time per year, $E$ is the interest rate that would give the same amount of interest when compounded once per year.

The principal is $\$1000$ and annual interest rate is $8\%$ the term is one year.

  $\begin{array}{ccc}InterestCompounded& Interest& EffectiveAnnualYield\\ Annually& & \\ Semi-annually& & \\ Quarterly& & \\ monthly& & \\ Daily& & \\ Continuously& & \end{array}$

Calculation:

Consider,

  $\begin{array}{l}\mathrm{Pr}incipal=\$1000\\ \mathrm{int}erestrate=8\%\\ Amount=p{\left(1+\frac{r}{n}\right)}^{nt}\\ I=Interest=amountaftercompound\mathrm{in}g-principal\end{array}$

  $\begin{array}{l}p=pricipal\\ r=\mathrm{int}erestrate\\ n=numberoftimes\mathrm{int}erestpaid\end{array}$

Now calculate interest and effective annual yield for each period,

Annually.

  $\begin{array}{l}I=1000{\left(1+\frac{0.08}{1}\right)}^1-1000\\ =\$80\\ Effectiveyield=\frac{I}{1000}\\ =\frac{80}{1000}\\ =8\%\end{array}$

Semi-Annually,

  $\begin{array}{l}I=1000{\left(1+\frac{0.08}{2}\right)}^2-1000\\ =\$81.6\\ Effectiveyield=\frac{I}{1000}\\ =\frac{81.6}{1000}\\ =8.16\%\end{array}$

Quarterly,

  $\begin{array}{l}I=1000{\left(1+\frac{0.08}{4}\right)}^4-1000\\ =\$82.4316\\ Effectiveyield=\frac{I}{1000}\\ =\frac{82.4316}{1000}\\ =8.24\%\end{array}$

Monthly,

  $\begin{array}{l}I=1000{\left(1+\frac{0.08}{12}\right)}^{12}-1000\\ =\$83\\ Effectiveyield=\frac{I}{1000}\\ =\frac{83}{1000}\\ =8.3\%\end{array}$

Daily,

  $\begin{array}{l}I=1000{\left(1+\frac{0.08}{365}\right)}^{365}-1000\\ =\$83.28\\ Effectiveyield=\frac{I}{1000}\\ =\frac{83.28}{1000}\\ =8.328\%\end{array}$

Continuously,

  $\begin{array}{l}principal\times c^{rt}\\ I=1000c^{0.08\times 1}-1000\\ =\$83.29\\ Effectiveyield=\frac{I}{1000}\\ =\frac{83.29}{1000}\\ =8.329\%\end{array}$

Hence, the table will be,

  $\begin{array}{ccc}InterestCompounded& Interest\left(\$\right)& EffectiveAnnualYield\\ Annually& 80& 8\%\\ Semi-annually& 81.6& 8.16\%\\ Quarterly& 82.4316& 8.24\%\\ monthly& 83& 8.3\%\\ Daily& 83.28& 8.328\%\\ Continuously& 83.29& 8.329\%\end{array}$

b.

To determine

Find the maximum effective annual yield account.

Answer to Problem 11E

Continuously.

Explanation of Solution

Given information:

A bank earns interest that is compounded more than one time per year, $E$ is the interest rate that would give the same amount of interest when compounded once per year.

The principal is $\$1000$ and annual interest rate is $8\%$ the term is one year.

Calculation:

Consider the table,

  $\begin{array}{ccc}InterestCompounded& Interest\left(\$\right)& EffectiveAnnualYield\\ Annually& 80& 8\%\\ Semi-annually& 81.6& 8.16\%\\ Quarterly& 82.4316& 8.24\%\\ monthly& 83& 8.3\%\\ Daily& 83.28& 8.328\%\\ Continuously& 83.29& 8.329\%\end{array}$

Hence, $8.329\%$ is the maximum effective annual yield by interest compounded continuously.

c.

To determine

Find the formula for effective annual yield with interest compounded $n$ times per year.

Answer to Problem 11E

  $E={\left(1+\frac{r}{n}\right)}^n-1$

Explanation of Solution

Given information:

A bank earns interest that is compounded more than one time per year, $E$ is the interest rate that would give the same amount of interest when compounded once per year.

The principal is $\$1000$ and annual interest rate is $8\%$ the term is one year.

  $P$ represents the principal and $A$ is the total value of the investment, $A=P\left(1+E\right)$ .

Calculation:

The total value of investment $A$ for $1$ year is,

  $\begin{array}{l}A=p{\left(1+\frac{r}{n}\right)}^n\\ A=p\left(1+{\left(1+\frac{r}{n}\right)}^n-1\right)\\ nowwehave\\ A=p\left(1+E\right)\\ compareboth\exp ressions\\ A=p\left(1+{\left(1+\frac{r}{n}\right)}^n-1\right)=A=p\left(1+E\right)\\ so,\\ E={\left(1+\frac{r}{n}\right)}^n-1\end{array}$

Hence, $E={\left(1+\frac{r}{n}\right)}^n-1$ .

  $8.329\%$ is the maximum effective annual yield by interest compounded continuously.

d.

To determine

Find the formula for effective annual yield with interest compounded continuously.

Answer to Problem 11E

  $E=e^r-1$

Explanation of Solution

Given information:

A bank earns interest that is compounded more than one time per year, $E$ is the interest rate that would give the same amount of interest when compounded once per year.

The principal is $\$1000$ and annual interest rate is $8\%$ the term is one year.

  $P$ represents the principal and $A$ is the total value of the investment, $A=P\left(1+E\right)$ .

Calculation:

The total value of investment $A$ for $1$ year is,

  $\begin{array}{l}A=p\times e^r\\ A=p\times \left(1+e^r-1\right)\\ nowwehave\\ A=p\left(1+E\right)\\ compareboth\exp ressions\\ A=p\left(1+e^r-1\right)=A=p\left(1+E\right)\\ so,\\ E=e^r-1\end{array}$

Hence, $E=e^r-1$ .

  $8.329\%$ is the maximum effective annual yield by interest compounded continuously.