Chapter D, Problem 80P

To determine

The interest rate that makes alternatives $A$ and $B$ equivalent is in what range.

Answer to Problem 80P

Therefore, the correct option is $\left(c\right)\text{Greater}\text{than}5\%$.

Explanation of Solution

Given:

Year	Alternative I	Alternative II
$0$	$-\$1000$	$-\$3000$
$1$	$100$	$500$
$2$	$100$	$550$
$3$	$100$	$600$
$4$	$200$	$650$
$5$	$200$	$700$

Concept used:

Write the expression to calculate the present worth.

$PW=\frac{F}{{\left(1+i\right)}^n}$

Here, future amount is $F$, number of years is $n$ and the interest rate is $i$.

Calculation:

Write the expression to calculate present worth of Alternative I.

$P{W}_{\text{A}}=\left[\begin{array}{c}-1000+100\left(\frac{1}{{\left(1+i\right)}^1}\right)+100\left(\frac{1}{{\left(1+i\right)}^2}\right)\\ +100\left(\frac{1}{{\left(1+i\right)}^3}\right)+200\left(\frac{1}{{\left(1+i\right)}^4}\right)+200\left(\frac{1}{{\left(1+i\right)}^5}\right)\end{array}\right]$ ...... (I)

Write the expression to calculate present worth of Alternative II.

$P{W}_{\text{B}}=\left[\begin{array}{c}-3000+500\left(\frac{1}{{\left(1+i\right)}^1}\right)+550\left(\frac{1}{{\left(1+i\right)}^2}\right)\\ +600\left(\frac{1}{{\left(1+i\right)}^3}\right)+650\left(\frac{1}{{\left(1+i\right)}^4}\right)+700\left(\frac{1}{{\left(1+i\right)}^5}\right)\end{array}\right]$ ...... (II)

Equate Equation (I) and Equation (II).

$\begin{array}{l}\left[\begin{array}{c}-1000+100\left(\frac{1}{{\left(1+i\right)}^1}\right)\\ +100\left(\frac{1}{{\left(1+i\right)}^2}\right)+100\left(\frac{1}{{\left(1+i\right)}^3}\right)\\ +200\left(\frac{1}{{\left(1+i\right)}^4}\right)+200\left(\frac{1}{{\left(1+i\right)}^5}\right)\end{array}\right]=\left[\begin{array}{c}-3000+500\left(\frac{1}{{\left(1+i\right)}^1}\right)\\ +550\left(\frac{1}{{\left(1+i\right)}^2}\right)+600\left(\frac{1}{{\left(1+i\right)}^3}\right)\\ +650\left(\frac{1}{{\left(1+i\right)}^4}\right)+700\left(\frac{1}{{\left(1+i\right)}^5}\right)\end{array}\right]\\ \frac{8}{{\left(1+i\right)}^1}+\frac{9}{{\left(1+i\right)}^2}+\frac{10}{{\left(1+i\right)}^3}+\frac{9}{{\left(1+i\right)}^4}+\frac{10}{{\left(1+i\right)}^5}-40=0\end{array}$ ...... (III)

Calculate the internal rate of return by hit and trial.

Substitute $4\%$ for $i$ in Equation (III) which is $V_1$.

$\begin{array}{c}{V}_1=\frac{8}{{\left(1+0.04\right)}^1}+\frac{9}{{\left(1+0.04\right)}^2}+\frac{10}{{\left(1+0.04\right)}^3}+\frac{9}{{\left(1+0.04\right)}^4}+\frac{10}{{\left(1+0.04\right)}^5}-40\\ =7.6923+8.321+8.89+7.6932+8.2193-40\\ =0.8158\end{array}$

Substitute $5\%$ for $i$ in Equation (III) which is $V_2$.

$\begin{array}{c}{V}_2=\frac{8}{{\left(1+0.05\right)}^1}+\frac{9}{{\left(1+0.05\right)}^2}+\frac{10}{{\left(1+0.05\right)}^3}+\frac{9}{{\left(1+0.05\right)}^4}+\frac{10}{{\left(1+0.05\right)}^5}-40\\ =7.6190+8.1633+8.6384+7.4043+7.8352-40\\ =-0.3398\end{array}$

Calculate the internal rate of return by interpolating $V_1$ and $V_2$.

$IRR=i_1+\left(i_2-i_1\right)\times \frac{V_1}{V_1-V_2}$ ...... (IV)

Substitute, $4\%$ for $i_1$, $5\%$ for $i_2$, $0.8158$ for $V_1$ and $-0.3398$ for $V_2$ in Equation (IV)

$\begin{array}{c}IRR=0.04+\left(0.05-0.04\right)\left(\frac{0.8158}{0.8158-0.3398}\right)\\ =0.057\\ =5.7\%\end{array}$.

Conclusion:

Therefore, the correct option is $\left(c\right)\text{Greater}\text{than}5\%$.