Chapter 5, Problem 27QP

a)

Summary Introduction

To determine: The yield to maturity of the bond

Explanation of Solution

Compute the yield to maturity of the bond as follows:

$\begin{array}{c}\text{Bond value}=C\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^t}\right]}{r}+\frac{F}{{\left(1+r\right)}^t}\\ \$865=\$55\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^{21}}\right]}{r}+\frac{\$1,000}{{\left(1+r\right)}^{21}}\end{array}$…Equation (1)

Finding “r” in Equation (1) would give the yield to maturity. However, it is difficult to simplify the above the equation. Hence, the only method to solve for “r” is the trial and error method.

The attempt under the trial and error method using 6.70 percent as “r”:

$\begin{array}{c}\text{Bond value}=C\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^t}\right]}{r}+\frac{F}{{\left(1+r\right)}^t}\\ =\$55\times \frac{\left[1-\frac{1}{{\left(1+0.0670\right)}^{21}}\right]}{0.0670}+\frac{\$1,000}{{\left(1+0.0670\right)}^{21}}\\ =\$610.5975+\$256.1812\\ =\$866.78\end{array}$

The attempt under the trial and error method using 6.72 percent as “r”:

$\begin{array}{c}\text{Bond value}=C\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^t}\right]}{r}+\frac{F}{{\left(1+r\right)}^t}\\ =\$55\times \frac{\left[1-\frac{1}{{\left(1+0.0672\right)}^{21}}\right]}{0.0672}+\frac{\$1,000}{{\left(1+0.0672\right)}^{21}}\\ =\$609.6039+\$255.1748\\ =\$864.78\end{array}$

Hence, the YTM is 6.72%.

b)

Summary Introduction

To determine: The selling price of the bond after two years, the holding period yield, and the reason why the holding period yield is different from the yield to maturity given in Part A of the solution.

Explanation of Solution

Compute the selling price of the bond after two years as follows:

$\begin{array}{c}\text{Bond value}=C\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^t}\right]}{r}+\frac{F}{{\left(1+r\right)}^t}\\ =\$55\times \frac{\left[1-\frac{1}{{\left(1+0.0572\right)}^{19}}\right]}{0.0572}+\frac{\$1,000}{{\left(1+0.0572\right)}^{19}}\\ =\$975.15\end{array}$

Compute the holding period yield:

$\begin{array}{c}\text{Bond value}=C\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^t}\right]}{r}+\frac{F}{{\left(1+r\right)}^t}\\ \$865=\$55\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^2}\right]}{r}+\frac{\$975.15}{{\left(1+r\right)}^r}\end{array}$

Finding “r” in Equation (1) would give the yield to maturity. However, it is difficult to simplify the above the equation. Hence, the only method to solve for “r” is the trial and error method.

The attempt under the trial and error method using 12.36 percent as “r”:

$\begin{array}{c}\text{Bond value}=C\times \frac{\left[1-\frac{1}{{\left(1+r\right)}^t}\right]}{r}+\frac{F}{{\left(1+r\right)}^t}\\ =\$55\times \frac{\left[1-\frac{1}{{\left(1+0.1236\right)}^2}\right]}{0.1236}+\frac{\$975.15}{{\left(1+0.1236\right)}^2}\\ =\$92.5149+\$772.4101\\ =\$864.925\end{array}$