Chapter 16, Problem 36P

To determine

The value of MARR to be chosen for selecting plan B and the reason for doing so.

Answer to Problem 36P

The value of MARR is $4.76\%$.

Use this MARR as at $4.76\%$ the net present worth is zero which means neither profit nor loss condition.

Hence in order to select project $B$, use a MARR of $4.76\%$.

Explanation of Solution

Calculations:

Calculate the difference between cash flows $A$ and cash flow $B$.

Year	Cash flow $A$	Cash flows $B$	Incremental analysis $\left(B-A\right)$
0	$-\$300$	$-\$450$	$-\$150$
$1-15\left(A_1\right)$	$-\$75$	$-\$100$	$-\$25$
$15\left(F_1\right)$	$-\$250$	$-\$50$	$\$200$
$16-30\left(A_2\right)$	$-\$125$	$-\$125$	$\$0$
$30\left(F_2\right)$	$-\$300$	$\$0$	$\$300$
$31-40\left(A_3\right)$	$-\$250$	$-\$125$	$\$125$
$40\left(F_3\right)$	$\$0$	$\$150$	$\$150$

Table (1)

Write the expression to calculate the present worth.

$PW=\left(\begin{array}{l}P+A_1\left(\frac{{\left(1+i\right)}^{15}-1}{i{\left(1+i\right)}^{15}}\right)+\frac{F_1}{{\left(1+i\right)}^{15}}+A_2\left(\frac{{\left(1+i\right)}^{15}-1}{i{\left(1+i\right)}^{15}}\right)\left(\frac{1}{{\left(1+i\right)}^{15}}\right)\\ +\frac{F_2}{{\left(1+i\right)}^{30}}+A_3\left(\frac{{\left(1+i\right)}^{10}-1}{i{\left(1+i\right)}^{110}}\right)\left(\frac{1}{{\left(1+i\right)}^{30}}\right)+\frac{F_3}{{\left(1+i\right)}^{40}}\end{array}\right)$

Here, the initial cost is $P$, the annual amount for year $1-15$, the future price for year $15$ is $F_1$, the annual price for $year16-30$ is $A_2$, the future price for year $30$ is $F_2$, the annual price for year $31-40$ is $A_3$ and the future price for year $40$ is $F_3$.

Substitute $0$ for $PW$

$\$150$ for $P$, for $A_1$, $\$200$ for $F_1$, $\$0$ for $A_2$, $\$300$ for $F_2$, $\$125$ for $A_3$ and $\$150$ for $F_3$ in Equation (I).

$\begin{array}{l}PW=\left(\begin{array}{l}-\$150-\$25\left(\frac{{\left(1+i\right)}^{15}-1}{i{\left(1+i\right)}^{15}}\right)+\frac{\$200}{{\left(1+i\right)}^{15}}+\$0\left(\frac{{\left(1+i\right)}^{15}-1}{i{\left(1+i\right)}^{15}}\right)\left(\frac{1}{{\left(1+i\right)}^{15}}\right)\\ +\frac{\$300}{{\left(1+i\right)}^{30}}+\$125\left(\frac{{\left(1+i\right)}^{10}-1}{i{\left(1+i\right)}^{110}}\right)\left(\frac{1}{{\left(1+i\right)}^{30}}\right)+\frac{\$150}{{\left(1+i\right)}^{40}}\end{array}\right)\\ 0=\left(\begin{array}{l}-\$150-\$25\left(\frac{{\left(1+i\right)}^{15}-1}{i{\left(1+i\right)}^{15}}\right)+\frac{\$200}{{\left(1+i\right)}^{15}}+\frac{\$300}{{\left(1+i\right)}^{30}}\\ +\$125\left(\frac{{\left(1+i\right)}^{10}-1}{i{\left(1+i\right)}^{110}}\right)\left(\frac{1}{{\left(1+i\right)}^{30}}\right)+\frac{\$150}{{\left(1+i\right)}^{40}}\end{array}\right)\end{array}$

On solving for value of $i$ we get

$\begin{array}{c}i=0.0476\\ =\left(0.0476\times 100\right)\%\\ =4.76\%\end{array}$

Use this MARR as at $4.76\%$ the net present worth is zero which means neither profit nor loss condition.

Hence in order to select project $B$, use MARR of $4.76\%$.

Conclusion:

Thus, the value of MARR is $4.76\%$.

Use this MARR as at $4.76\%$ the net present worth is zero which means neither profit nor loss condition.

Hence in order to select project $B$, use a MARR of $4.76\%$.