Chapter 6, Problem 48P

To determine

The best alternative from the given alternatives.

Answer to Problem 48P

The route $105$ is the best alternative.

Explanation of Solution

Given for Route 105:

Cost is $\$500000$, annual saving is $\$100000$ and useful life is $6$ year.

Calculation:

Calculate the EUAW for Route 105.

${\left(EUAW\right)}_{105}={\left(EUAB\right)}_{105}-{\left(EUAC\right)}_{105}$ ....... (I)

Write the equation for EUAC for Route 105.

${\left(EUAC\right)}_{105}=\$500000\left(\frac{A}{P},4\%,6\right)$ ....... (II).

Calculate the factor $\left(\frac{A}{P},4\%,6\right)$.

$\left(\frac{A}{P},4\%,6\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$

Here, the present value is P, rate is i, and time is n.

Substitute $4\%$ for i, and $6\text{years}$ for n.

$\begin{array}{c}\left(\frac{A}{P},4\%,6\right)=\left[\frac{0.04{\left(1+0.04\right)}^6}{{\left(1+0.06\right)}^6-1}\right]\\ =0.1209\end{array}$

Substitute $0.1209$ for $\left(\frac{A}{P},4\%,6\right)$ in Equation (II).

$\begin{array}{c}{\left(EUAC\right)}_{105}=\$500000\times 0.1209\\ =\$60450\end{array}$

Substitute $\$100000$ for ${\left(EUAB\right)}_{105}$, and $\$604500$ for ${\left(EUAC\right)}_{105}$ in Equation (I).

$\begin{array}{c}{\left(EUAW\right)}_A=\$100000-\$60450\\ =\$39550\end{array}$

Given for Route 205:

Cost is $\$450000$, annual saving is $\$125000$ and useful life is $4$ year.

Calculations:

Calculate the EUAW for Route 205.

${\left(EUAW\right)}_{205}={\left(EUAB\right)}_{205}-{\left(EUAC\right)}_{205}$ ....... (III)

Write the Equation for EUAC for Route 105.

${\left(EUAC\right)}_{205}=\$500000\left(\frac{A}{P},4\%,4\right)$ ....... (IV).

Calculate the factor $\left(\frac{A}{P},4\%,4\right)$.

$\left(\frac{A}{P},4\%,4\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$

Substitute $4\%$ for i, and $4\text{years}$ for n.

$\begin{array}{c}\left(\frac{A}{P},4\%,4\right)=\left[\frac{0.04{\left(1+0.04\right)}^4}{{\left(1+0.04\right)}^4-1}\right]\\ =0.2755\end{array}$

Substitute $0.2755$ for $\left(\frac{A}{P},4\%,4\right)$ in Equation (IV).

$\begin{array}{c}{\left(EUAC\right)}_{205}=\$450000\times 0.2755\\ =\$123975\end{array}$

Substitute $\$125000$ for ${\left(EUAB\right)}_{205}$, and $\$123975$ for ${\left(EUAC\right)}_{205}$ in Equation (I).

$\begin{array}{c}{\left(EUAW\right)}_A=\$125000-\$123975\\ =\$1025\end{array}$

Given for Route 305:

Cost is $\$400000$, annual saving is $\$90000$ and useful life is $5$ year.

Calculations:

Calculate the EUAW for Route 305.

${\left(EUAW\right)}_{305}={\left(EUAB\right)}_{305}-{\left(EUAC\right)}_{305}$ ....... (V)

Write the equation for EUAC for Route 305.

${\left(EUAC\right)}_{305}=\$500000\left(\frac{A}{P},4\%,5\right)$ ....... (VI).

Calculate the factor $\left(\frac{A}{P},4\%,5\right)$.

$\left(\frac{A}{P},4\%,5\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$

Substitute $4\%$ for i, and $5\text{years}$ for n.

$\begin{array}{c}\left(\frac{A}{P},4\%,5\right)=\left[\frac{0.04{\left(1+0.04\right)}^5}{{\left(1+0.04\right)}^5-1}\right]\\ =0.2246\end{array}$

Substitute $0.2246$ for $\left(\frac{A}{P},4\%,5\right)$ in Equation (IV).

$\begin{array}{c}{\left(EUAC\right)}_{305}=\$400000\times 0.2246\\ =\$89840\end{array}$

Substitute $\$90000$ for ${\left(EUAB\right)}_{305}$, and $\$89840$ for ${\left(EUAC\right)}_{305}$ in Equation (V).

$\begin{array}{c}{\left(EUAW\right)}_A=\$90000-\$89840\\ =\$160\end{array}$.

Conclusion:

Here, EUAW for the alternative $105$ is largest. Hence, alternative $105$ is the best alternative.