Chapter 6, Problem 50P

To determine

The best alternative from the given alternatives.

Answer to Problem 50P

Alternative C is the best.

Explanation of Solution

Given for alternative A :

Cost is $\$10000$.

Annual benefit is $\$1000$.

Useful life is infinite years.

Calculation:

Calculate the EUAW for A.

${\left(EUAW\right)}_A={\left(EUAB\right)}_A-{\left(EUAC\right)}_A$ ....... (I)

Write the Equation for EUAC for A.

${\left(EUAC\right)}_A=\$10000\left(\frac{A}{P},8\%,\infty \right)$ ....... (II).

Calculate the factor $\left(\frac{A}{P},8\%,\infty \right)$.

$\begin{array}{c}\left(\frac{A}{P},8\%,\infty \right)=i\\ =0.08\end{array}$

Substitute $0.08$ for $\left(\frac{A}{P},8\%,\infty \right)$ in Equation (II).

$\begin{array}{c}{\left(EUAC\right)}_A=\$10000\times 0.08\\ =\$800\end{array}$

Substitute $\$1000$ for ${\left(EUAB\right)}_A$, and $\$800$ for ${\left(EUAC\right)}_A$ in Equation (I).

$\begin{array}{c}{\left(EUAW\right)}_A=\$1000-\$800\\ =\$200\end{array}$

Given for alternative B :

Cost is $\$15000$.

Annual benefit is $\$1762$.

Useful life is $20$ year.

Calculations:

Calculate the EUAW for B.

${\left(EUAW\right)}_B={\left(EUAB\right)}_B-{\left(EUAC\right)}_B$ ....... (III)

Write the Equation for EUAC for B.

${\left(EUAC\right)}_B=\$10000\left(\frac{A}{P},8\%,20\right)$ ....... (IV)

Write the equation for factor $\left(\frac{A}{P},8\%,20\right)$.

$\left(\frac{A}{P},8\%,20\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$

Substitute $8\%$ for i, and $20\text{years}$ for n.

$\begin{array}{c}\left(\frac{A}{P},8\%,20\right)=\left[\frac{0.08{\left(1+0.08\right)}^{20}}{{\left(1+0.08\right)}^{20}-1}\right]\\ =0.1019\end{array}$

Substitute $0.1019$ for $\left(\frac{A}{P},8\%,20\right)$ in Equation (IV).

$\begin{array}{c}{\left(EUAC\right)}_B=\$15000\times 0.1019\\ =\$1529\end{array}$

Substitute $\$1762$ for ${\left(EUAB\right)}_B$, and $\$1529$ for ${\left(EUAC\right)}_B$ in Equation (III).

$\begin{array}{c}{\left(EUAW\right)}_B=\$1762-\$1529\\ =\$234\end{array}$

Given for alternative C :

Cost is $\$20000$.

Annual benefit is $\$5548$.

Useful life is $5$ year.

Calculations:

Calculate the EUAW for C.

${\left(EUAW\right)}_C={\left(EUAB\right)}_C-{\left(EUAC\right)}_C$ ....... (V)

Write the equation for EUAC for C.

${\left(EUAC\right)}_C=\$10000\left(\frac{A}{P},8\%,5\right)$ ....... (VI)

Write the equation for factor $\left(\frac{A}{P},8\%,5\right)$.

$\left(\frac{A}{P},8\%,5\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$

Substitute $8\%$ for i, and $20\text{years}$ for n.

$\begin{array}{c}\left(\frac{A}{P},8\%,20\right)=\left[\frac{0.08{\left(1+0.08\right)}^5}{{\left(1+0.08\right)}^5-1}\right]\\ =0.2505\end{array}$

Substitute $0.2505$ for $\left(\frac{A}{P},8\%,5\right)$ in Equation (VI).

$\begin{array}{c}{\left(EUAC\right)}_C=\$20000\times 0.2505\\ =\$5010\end{array}$

Substitute $\$5548$ for ${\left(EUAB\right)}_C$, and $\$5010$ for ${\left(EUAC\right)}_C$ in Equation (V).

$\begin{array}{c}{\left(EUAW\right)}_C=\$5548-\$5010\\ =\$538\end{array}$.

Conclusion:

Here, EUAW for the alternative C is largest. Hence, alternative C is the best alternative.