Chapter 6, Problem 51P

To determine

The machine should be chosen.

Answer to Problem 51P

The machine B is the best choice.

Explanation of Solution

Given:

Cost & Details	Machine A	Machine B
Initial cost of machine	$\$15000$	$\$25000$
Annual maintenance cost	$\$1600$	$\$400$
Annual benefits	$\$8000$	$\$13000$
Salvage value	$\$3000$	$\$6000$
Life	$6$	$10$

Calculation:

Calculate the EUAC for Machine A.

$\text{EUAC}=\left[\begin{array}{l}P\left(\frac{A}{P},i,n\right)-\text{Operating}\text{and}\text{maintenance}\text{cost}+\\ \text{Annual}\text{benefits}+F\left(\frac{A}{F},i,n\right)\end{array}\right]$ ...... (I)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor $\left(\frac{A}{P},i,n\right)$.

$\left(\frac{A}{P},i,n\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$ ...... (II)

Substitute $12\%$ for interest and $6\text{years}$ in Equation (II).

$\begin{array}{c}\left(\frac{A}{P},i,n\right)=\left[\frac{0.12{\left(1+0.12\right)}^6}{{\left(1+0.12\right)}^6-1}\right]\\ =0.2432\end{array}$

Calculate the factor $\left(\frac{A}{F},i,n\right)$.

$\left(\frac{A}{F},i,n\right)=\left[\frac{i}{{\left(1+i\right)}^n-1}\right]$ ...... (III).

Substitute $12\%$ for interest and $6\text{years}$ in Equation (III).

$\begin{array}{c}\left(\frac{A}{F},i,n\right)=\left[\frac{0.12}{{\left(1+0.12\right)}^6-1}\right]\\ =0.1232\end{array}$

Substitute $-\$15000$ for P, $\$3000$ for F, $\$1600$ for operation and maintenance cost, $\$8000$ for annual benefits, $0.2432$ for $\left(\frac{A}{P},i,n\right)$, and $0.1232$ for $\left(\frac{A}{F},i,n\right)$ in Equation (I).

$\begin{array}{c}\text{EUAC}=\left[\begin{array}{l}-\left(\$15000\times 0.2432\right)-\text{\$1600}+\$8000\\ +\left(\$3000\times 0.1232\right)\end{array}\right]\\ =\$3121.6\end{array}$

Calculate the EUAC for Machine B.

$\text{EUAC}=\left[\begin{array}{l}P\left(\frac{A}{P},i,n\right)-\text{Operating}\text{and}\text{maintenance}\text{cost}+\\ \text{Annual}\text{benefits}+F\left(\frac{A}{F},i,n\right)\end{array}\right]$ ...... (IV)

Here, present worth is P, future value is F, rate is i, year is n, and annual cost is A.

Calculate the factor $\left(\frac{A}{P},i,n\right)$.

$\left(\frac{A}{P},i,n\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$ ...... (V)

Substitute $12\%$ for interest and $10\text{years}$ in Equation (V).

$\begin{array}{c}\left(\frac{A}{P},i,n\right)=\left[\frac{0.12{\left(1+0.12\right)}^{10}}{{\left(1+0.12\right)}^{10}-1}\right]\\ =0.1770\end{array}$

Calculate the factor $\left(\frac{A}{F},i,n\right)$.

$\left(\frac{A}{F},i,n\right)=\left[\frac{i}{{\left(1+i\right)}^n-1}\right]$ ...... (VI).

Substitute $12\%$ for interest and $10\text{years}$ in Equation (VI).

$\begin{array}{c}\left(\frac{A}{F},i,n\right)=\left[\frac{0.12}{{\left(1+0.12\right)}^{10}-1}\right]\\ =0.0570\end{array}$

Substitute $-\$25000$ for P, $\$6000$ for F, $\$400$ for operation and maintenance cost, $\$13000$ for annual benefits, $0.1770$ for $\left(\frac{A}{P},i,n\right)$, and $0.0570$ for $\left(\frac{A}{F},i,n\right)$ in Equation (IV).

$\begin{array}{c}\text{EUAC}=\left[\begin{array}{l}-\left(\$25000\times 0.1770\right)-\text{\$400}+\$13000\\ +\left(\$6000\times 0.0570\right)\end{array}\right]\\ =\$8517\end{array}$.

Conclusion:

Therefore the machine B is the best choice.