Chapter 4, Problem 3P

a.

Summary Introduction

To determine: Payment for each period if interest is accrued monthly.

Explanation of Solution

The formula to calculate payment for each period is as follows:

$\text{Payment}=\frac{\text{Principle amount}}{\raisebox{1ex}{$\left\{\left[{\left(1+\text{r}\right)}^{360}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[\text{r}{\left(1+\text{r}\right)}^{360}\right]$}\right.}$

$\begin{array}{c}\text{Payment}=\frac{\$100,000}{\raisebox{1ex}{$\left\{\left[{\left(1+0.005\right)}^{360}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[0.005{\left(1+0.005\right)}^{360}\right]$}\right.}\\ =\frac{\$100,000}{\$166.7916}\\ =\$599.55\end{array}$

Hence, the payment will be of $599.55 if the interest is accrued monthly.

b.

Summary Introduction

To determine: Payment for each period if interest is accrued Quarterly.

Explanation of Solution

The formula to calculate payment for each period is as follows:

$\text{Payment}=\frac{\text{Principle amount}}{\raisebox{1ex}{$\left\{\left[{\left(1+\text{r}\right)}^{120}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[\text{r}{\left(1+\text{r}\right)}^{120}\right]$}\right.}$

$\begin{array}{c}\text{Payment}=\frac{\$100,000}{\raisebox{1ex}{$\left\{\left[{\left(1+0.015\right)}^{120}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[0.015{\left(1+0.015\right)}^{120}\right]$}\right.}\\ =\frac{\$100,000}{\$55.4985}\\ =\$1,801.85\end{array}$

Hence, the payment will be of $1,801.85 if the interest is accrued quarterly.

c.

Summary Introduction

To determine: Payment for each period if interest is accrued annually.

Explanation of Solution

The formula to calculate payment for each period is as follows:

$\text{Payment}=\frac{\text{Principle amount}}{\raisebox{1ex}{$\left\{\left[{\left(1+\text{r}\right)}^{30}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[\text{r}{\left(1+\text{r}\right)}^{30}\right]$}\right.}$

$\begin{array}{c}\text{Payment}=\frac{\$100,000}{\raisebox{1ex}{$\left\{\left[{\left(1+0.06\right)}^{30}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[0.06{\left(1+0.06\right)}^{30}\right]$}\right.}\\ =\frac{\$100,000}{\$13.76483}\\ =\$7,264.89\end{array}$

Hence, the payment will be of $7,264.89 if the interest is accrued annually.

d.

Summary Introduction

To determine: Payment for each period if interest is accrued annually.

Explanation of Solution

The formula to calculate payment for each period is as follows:

$\text{Payment}=\frac{\text{Principle amount}}{\raisebox{1ex}{$\left\{\left[{\left(1+\text{r}\right)}^{1,560}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[\text{r}{\left(1+\text{r}\right)}^{1,560}\right]$}\right.}$

$\begin{array}{c}\text{Payment}=\frac{\$100,000}{\raisebox{1ex}{$\left\{\left[{\left(1+0.0012\right)}^{1,560}\right]-1\right\}$}\!\left/ \!\raisebox{-1ex}{$\left[0.0012{\left(1+0.0012\right)}^{1,560}\right]$}\right.}\\ =\frac{\$100,000}{\$723.2749}\\ =\$138.26\end{array}$

Hence, the payment will be of $138.26 if the interest is accrued weekly.