Chapter 4, Problem 5QP

Summary Introduction

To determine: The number of periods of investment

Introduction:

The future value of money refers to the amount of dollars that an investment grows over a definite period at a particular rate of interest rate. The value increases based on the period of investment. The value of the investment will be higher if the duration of investment is longer.

Answer to Problem 5QP

The number of periods is as follows:

Particulars	Present value	Years	Interest Rate	Future value
Investment A	$195	20.13	9%	$1,105
Investment B	$2,105	8.34	7%	$3,700
Investment C	$47,800	18.45	12%	$387,120
Investment D	$38,650	9.39	19%	$198,212

Explanation of Solution

Given information:

Investment A has a present value of $195, future value of $1,105, and an interest rate of 9 percent. Investment B has a present value of $2,105, future value of $3,700, and an interest rate of 7 percent.

Investment C has a present value of $47,800, future value of $387,120, and an interest rate of 12 percent. Investment D has a present value of $38,650, future value of $198,212, and an interest rate of 19 percent.

Formula:

Derive the formula to calculate the number of periods from the present value equation as follows:

$\begin{array}{c}PV=\frac{FV}{{\left(1+r\right)}^t}\\ {\left(1+r\right)}^t=\frac{FV}{PV}\\ \ln {\left(1+r\right)}^t=\ln \left(\frac{FV}{PV}\right)\end{array}$

$\begin{array}{l}\text{The power rule of natural}\text{logarithm states that}\left[\ln \left(x^y\right)= y\times \ln \left(x\right)\right]\\ \text{Hence,}\\ t\times \ln \left(1+r\right)=\ln \left(\frac{FV}{PV}\right)\\ \text{Divide the right hand side and left hand side by}\ln \left(1+r\right)\end{array}$

$\begin{array}{c}\frac{t\times \ln \left(1+r\right)}{\ln \left(1+r\right)}=\frac{\ln \left(\frac{FV}{PV}\right)}{\ln \left(1+r\right)}\\ t=\frac{\ln \left(\frac{FV}{PV}\right)}{\ln \left(1+r\right)}\end{array}$

The formula to calculate the number of periods:

$t=\frac{\ln \left(\frac{\text{FV}}{\text{PV}}\right)}{\ln \left(1+r\right)}$

Where,

“t” refers to the number of years or periods of investment,

“ln” refers to the log value,

 “FV” refers to the future value,

“PV” refers to the present value,

“r” refers to the simple rate of interest.

Compute the number of periods of Investment A:

$\begin{array}{c}t=\frac{\ln \left(\frac{\text{FV}}{\text{PV}}\right)}{\ln \left(1+r\right)}\\ =\frac{\ln \left(\frac{\text{\$1,105}}{\$195}\right)}{\ln \left(1+0.09\right)}\\ =\frac{\ln \left(5.66\right)}{\ln \left(1.09\right)}\\ =20.13\text{ years}\end{array}$

Hence, the number of periods of Investment A is 20.13 years.

Compute the number of periods of Investment B:

$\begin{array}{c}t=\frac{\ln \left(\frac{\text{FV}}{\text{PV}}\right)}{\ln \left(1+r\right)}\\ =\frac{\ln \left(\frac{\text{\$3,700}}{\$2,105}\right)}{\ln \left(1+0.07\right)}\\ =\frac{\ln \left(1.76\right)}{\ln \left(1.07\right)}\\ =8.34\text{ years}\end{array}$

Hence, the number of periods of Investment B is 8.34 years.

Compute the number of periods of Investment C:

$\begin{array}{c}t=\frac{\ln \left(\frac{\text{FV}}{\text{PV}}\right)}{\ln \left(1+r\right)}\\ =\frac{\ln \left(\frac{\text{\$387,120}}{\$47,800}\right)}{\ln \left(1+0.12\right)}\\ =\frac{\ln \left(8.099\right)}{\ln \left(1.12\right)}\\ =18.45\text{ years}\end{array}$

Hence, the number of periods of Investment C is 18.45 years.

Compute the number of periods of Investment D:

$\begin{array}{c}t=\frac{\ln \left(\frac{\text{FV}}{\text{PV}}\right)}{\ln \left(1+r\right)}\\ =\frac{\ln \left(\frac{\text{\$198,212}}{\$38,650}\right)}{\ln \left(1+0.19\right)}\\ =\frac{\ln \left(5.12\right)}{\ln \left(1.19\right)}\\ =9.39\text{ years}\end{array}$

Hence, the number of periods of Investment D is 9.39 years.