Chapter 26, Problem 22PS

a.

Summary Introduction

To compute: The duration of securities when the rate of interest is 8%.

Explanation of Solution

The formula to calculate duration of securities is as follows:

$\text{Duration }=\left(\frac{1}{\text{V}}\right)\left\{\left[PV\left(C_1\right)\right]\left(1\right)+\left[PV\left(C_2\right)\right]\left(2\right)+\left[PV\left(C_3\right)\right]\left(3\right)\right\}$

$\begin{array}{c}\text{PV}=\text{Present value }\\ \text{C}=\text{Cash flow}\end{array}$

The calculation of duration of securities is as follows:

Security A:

$\begin{array}{c}\text{V}=40\left(\left(\frac{1}{.08}\right)–\left\{\frac{1}{\left[.08{\left(1.08\right)}^3\right]}\right\}\right)\\ =103.08\end{array}$

$\begin{array}{c}{\text{Duration}}_{\text{A}}=\frac{1}{103.08}\left\{\left[\left(\frac{40}{1.08}\right)\left(1\right)\right]+\left[\left(\frac{40}{{1.08}^2}\right)\left(2\right)\right]+\left[\left(\frac{40}{{1.08}^3}\right)\left(3\right)\right]\right\}\\ =1.95\text{ Years}\text{.}\end{array}$

Security B:

${\text{V}}_{\text{B}}=\frac{120}{\begin{array}{c}1.08\\ =111.11\end{array}}$

$\begin{array}{c}{\text{Duration}}_{\text{B}}=\frac{1}{111.11}\left[\left(\frac{120}{1.08}\right)\left(1\right)\right]\\ =1.00\text{ Year}\end{array}$

Security C:

$\begin{array}{c}{\text{V}}_{\text{C}}=10\left(\left(\frac{1}{.08}\right)–\left\{\frac{1}{\left[.08{\left(1.08\right)}^3\right]}\right\}+\frac{100}{{1.08}^3}\right)\\ =105.15\end{array}$

$\begin{array}{c}{\text{Duration}}_{\text{C}}=\frac{1}{105.15}\left\{\left[\left(\frac{10}{1.08}\right)\left(1\right)\right]+\left[\left(\frac{10}{{1.08}^2}\right)\left(2\right)\right]+\left[\left(\frac{110}{{1.08}^3}\right)\left(3\right)\right]\right\}\\ =2.74\text{ Years}\text{.}\end{array}$

b.

Summary Introduction

To compute: The mixture of B and C will hedge this investment against changes in interest rates.

Explanation of Solution

The formula to calculate mixture of B and C is as follows:

${\text{Duration}}_{\text{A}}=\left[x\left({\text{Duration}}_{\text{B}}\right)\right]+[(1–x)\left({\text{Duration}}_{\text{C}}\right)]$

The computation is as follows:

$\begin{array}{c}1.95=1.0x+[(1–x)\left(2.74\right)]\\ x=.4555\\ \left(1-x\right)=.5445\end{array}$

So that the following positions will protect the investment:

$\begin{array}{l}\text{Investment B }=.4555\left(\$10\text{m}\right)=\$4.555\text{ million }\left(\text{Short}\right)\\ \text{Investment C }=.5445\left(\$10\text{m}\right)=\$5.445\text{ million }\left(\text{Short}\right)\end{array}$

c.

Summary Introduction

To discuss: The manner in which the person X will hedge.

Explanation of Solution

$\begin{array}{c}{\text{Duration}}_{\text{B}}=\left[x\left({\text{Duration}}_{\text{A}}\right)\right]+\left[(1–x)\left({\text{Duration}}_{\text{C}}\right)\right]\\ 1.00=1.95x+\left[(1–x)\left(2.74\right)\right]\\ x=2.1955\\ \left(1-x\right)=-1.1955\end{array}$

So that the succeeding positions will protect the investment.

$\begin{array}{l}\text{Investment A }=2.1955\left(\$10\text{m}\right)=\$21.955\text{ million }\left(\text{Short}\right)\\ \text{Investment C }=–1.1955\left(\$10\text{m}\right)=–\$11.955\text{ million }\left(\text{Long}\right)\end{array}$