Chapter 13, Problem 32P

To determine

(a)

The challenger’s economic life.

Answer to Problem 32P

The challenger’s economic life is $4$ years as the defender with lowest equivalent annual cost is $\$3305.54$.

Explanation of Solution

Given:

The interest rate is $10\%$.

Concept used:

The minimum cost of life is where the equivalent annual cost of ownership is minimized for the number of years held.

Write the expression for equivalent annual cost of year $1$.

$EUA{C}_1=\text{Defender cost year}1\left[\frac{i}{{\left(1+i\right)}^n-1}\right]$ ...... (I)

Here, rate of interest is $i$ and time period is $n$.

Write the expression for equivalent annual of year $2$.

$EUA{C}_2=\left[\begin{array}{l}\text{Defender cost year}1\left(\frac{1}{{\left(1+i\right)}^{n_1}}\right)\\ +\text{Defender cost year2}\left(\frac{1}{{\left(1+i\right)}^{n_2}}\right)\end{array}\right]\left(\frac{i{\left(1+i\right)}^{n_2}}{{\left(1+i\right)}^{n_2}-1}\right)$ ...... (II)

Write the expression for equivalent annual of year $3$.

$EUA{C}_3=\left[\begin{array}{l}\text{Defender cost year}1\left(\frac{1}{{\left(1+i\right)}^{n_1}}\right)\\ +\text{Defender cost year2}\left(\frac{1}{{\left(1+i\right)}^{n_2}}\right)\\ +\text{Defender cost year3}\left(\frac{1}{{\left(1+i\right)}^{n_3}}\right)\end{array}\right]\left(\frac{i{\left(1+i\right)}^{n_3}}{{\left(1+i\right)}^{n_3}-1}\right)$ ...... (III)

Write the expression for equivalent annual of year $4$.

$EUA{C}_4=\left[\begin{array}{l}\text{Defender cost year}1\left(\frac{1}{{\left(1+i\right)}^{n_1}}\right)\\ +\text{Defender cost year2}\left(\frac{1}{{\left(1+i\right)}^{n_2}}\right)\\ +\text{Defender cost year3}\left(\frac{1}{{\left(1+i\right)}^{n_3}}\right)\\ +\text{Defender cost year4}\left(\frac{1}{{\left(1+i\right)}^{n_4}}\right)\end{array}\right]\left(\frac{i{\left(1+i\right)}^{n_4}}{{\left(1+i\right)}^{n_4}-1}\right)$ ...... (IV)

Calculations:

Calculate the equivalent annual cost of year $1$.

Substitute $\$3000$ for defender cost for year $1$, $10\%$ for $i$ and $1$ for $n$ in Equation (I).

$\begin{array}{c}EUA{C}_1=\$3000\times \left[\frac{0.10}{{\left(1+0.10\right)}^1-1}\right]\\ =\$3000\times \left[\frac{0.10}{1.1-1}\right]\\ =\$3000\end{array}$

Calculate the equivalent annual cost of year $2$.

Substitute $\$3000$ for defender cost for year $1$, $\$3150$ for defender cost for year $2$, $10\%$ for $i$, $1$ for $n_1$ and $2$ for $n_2$ in Equation (II).

$\begin{array}{c}EUA{C}_2=\left[\begin{array}{l}\$3000\left(\frac{1}{{\left(1+0.1\right)}^1}\right)\\ +\$3150\left(\frac{1}{{\left(1+0.1\right)}^2}\right)\end{array}\right]\left(\frac{0.1{\left(1+0.1\right)}^2}{{\left(1+0.1\right)}^2-1}\right)\\ =\left[\$3000\left(\frac{1}{1.1}\right)+\$3150\left(\frac{1}{1.21}\right)\right]\left(\frac{0.1\times 1.21}{1.21-1}\right)\\ =\left[\$2727.27+\$2603.30\right]\left(0.576\right)\\ =\$3070.41\end{array}$

Calculate the equivalent annual cost of year $3$.

Substitute $\$3000$ for defender cost for year $1$, $\$3150$ for defender cost for year $2$, $\$3400$ for defender cost for year $3$, $10\%$ for $i$, $1$ for $n_1$, $2$ for $n_2$ and $3$ for $n_3$ in Equation (III).

$\begin{array}{c}EUA{C}_3=\left[\begin{array}{l}\$3000\left(\frac{1}{{\left(1+0.1\right)}^1}\right)\\ +\$3150\left(\frac{1}{{\left(1+0.1\right)}^2}\right)\\ +\$3400\left(\frac{1}{{\left(1+0.1\right)}^3}\right)\end{array}\right]\left(\frac{0.1{\left(1+0.1\right)}^3}{{\left(1+0.1\right)}^3-1}\right)\\ =\left[\$3000\left(\frac{1}{1.1}\right)+\$3150\left(\frac{1}{1.21}\right)+\$3400\left(\frac{1}{1.331}\right)\right]\left(\frac{0.1\times 1.331}{1.331-1}\right)\\ =\left[\$2727.27+\$2603.30+\$2554.47\right]\left(0.4021\right)\\ =\$3171.78\end{array}$

Calculate the equivalent annual cost of year $4$.

Substitute $\$3000$ for defender cost for year $1$, $\$3150$ for defender cost for year $2$, $\$3400$ for defender cost for year $3$, $\$3800$ for defender cost for year $4$, $10\%$ for $i$, $1$ for $n_1$, $2$ for $n_2$

$3$ for $n_3$ and $4$ for $n_4$ in Equation (IV).

$\begin{array}{c}EUA{C}_4=\left[\begin{array}{l}\$3000\left(\frac{1}{{\left(1+0.1\right)}^1}\right)\\ +\$3150\left(\frac{1}{{\left(1+0.1\right)}^2}\right)\\ +\$3400\left(\frac{1}{{\left(1+0.1\right)}^3}\right)\\ +\$3800\left(\frac{1}{{\left(1+0.1\right)}^4}\right)\end{array}\right]\left(\frac{0.1{\left(1+0.1\right)}^4}{{\left(1+0.1\right)}^4-1}\right)\\ =\left[\begin{array}{l}\$3000\left(\frac{1}{1.1}\right)+\$3150\left(\frac{1}{1.21}\right)\\ +\$3400\left(\frac{1}{1.331}\right)++\$3800\left(\frac{1}{1.4641}\right)\end{array}\right]\left(\frac{0.1\times 1.4641}{1.4641-1}\right)\\ =\left[\$2727.27+\$2603.30+\$2554.47+\$2595.45\right]\left(0.3154\right)\\ =\$3305.54\end{array}$

The minimum cost of life is where the equivalent annual cost of ownership is minimized for the number of years held. This would occur at $4$ years for the challenge where the equivalent annual cost is $\$3305.54$.

Conclusion:

The challenger’s economic life is $4$ years as the defender with lowest equivalent annual cost is $\$3305.54$.

To determine

(b)

The time after which the defender should be replaced with the challenger.

Answer to Problem 32P

Defender can be replaced with the challenger with 2 more years of defender.

Explanation of Solution

Calculation:

Using replacement analysis technique, the defender’s lowest equivalent annual cost value should keep the defender for 2 more years.

Conclusion:

Thus, the defender can be replaced with the challenger with 2 more years of defender.