Chapter 4.7, Problem 5MRPS

Solution

a.

To find: The effective interest rate that has a nominal interest rate of 5%.

The effective interest rate is $e^{0.05}-1$ .

Given:

  $N=\ln \left(E+1\right)$ , where $N$ is the nominal interest rate and $E$ is the effective interest rate.

Calculation:

Here,

  $\begin{array}{c}N=\frac{5}{100}\\ =0.05\end{array}$

Now put $N=0.05$ in $N=\ln \left(E+1\right)$ , it follows that

  $\begin{array}{c}0.05=\ln \left(E+1\right)\\ E+1=e^{0.05}\\ E=e^{0.05}-1\end{array}$

Hence, the effective interest rate is $e^{0.05}-1$ .

b.

To find: The effective interest rate that has a nominal interest rate of 10%.

The effective interest rate is $e^{0.1}-1$ .

Given:

  $N=\ln \left(E+1\right)$ , where $N$ is the nominal interest rate and $E$ is the effective interest rate.

Calculation:

Here,

  $\begin{array}{c}N=\frac{10}{100}\\ =0.1\end{array}$

Now put $N=0.1$ in $N=\ln \left(E+1\right)$ , it follows that

  $\begin{array}{c}0.1=\ln \left(E+1\right)\\ E+1=e^{0.1}\\ E=e^{0.1}-1\end{array}$

Hence, the effective interest rate is $e^{0.1}-1$ .

c.

To compare: The effective interests in part (a) and part (b).

The effective interest rate in part (b) is $\frac{e^{0.1}-1}{e^{0.05}-1}$ times greater than the effective interest rate in part (a).

Given:

The effective interest rate in part (a) is $e^{0.05}-1$ , and the effective interest rate in part (b) is $e^{0.1}-1$ .

Calculation:

The ratio of the interest rates in part (a) and part (b) is,

  $\frac{e^{0.1}-1}{e^{0.05}-1}$

So, the effective interest rate in part (b) is $\frac{e^{0.1}-1}{e^{0.05}-1}$ times greater than the effective interest rate in part (a).

d.

To solve: The ratio in part (c).

  $\frac{e^{0.1}-1}{e^{0.05}-1}=e^{0.05}+1$

Given:

The effective interest rate in part (a) is $e^{0.05}-1$ , and the effective interest rate in part (b) is $e^{0.1}-1$ .

Calculation:

Consider the ratio,

  $\frac{e^{0.1}-1}{e^{0.05}-1}$

On simplification, it follows that

  $\begin{array}{c}\frac{e^{0.1}-1}{e^{0.05}-1}=\frac{{\left(e^{0.05}\right)}^2-1^2}{e^{0.05}-1}\\ =\frac{\left(e^{0.05}-1\right)\left(e^{0.05}+1\right)}{e^{0.05}-1}\left[\because a^2-b^2=\left(a-b\right)\left(a+b\right)\right]\\ =e^{0.05}+1\end{array}$