Chapter 13, Problem 1P

To determine

The most economic life.

Answer to Problem 1P

The most economic life is 5 years.

Explanation of Solution

Given:

Machine first cost is $1,050,000 and the salvage value for machine is $225,000.

Maintenance and operating cost is $235,000.

Maintenance and operating gradient is $75,000.

MARR is 10%.

Concept Used:

The number of years at which value of EUAB - EUAC is lowest, can be termed as most economic life.

Here,

EUAB is Equivalent uniform annual benefit.

EUAC is Equivalent uniform annual cost.

EUAB can be calculated with the help of below expression:

  $\begin{array}{l}\text{EUAB}=-\text{First}\text{cost}\left(\text{A/P,10\%,n}\right)+\text{Salvage}\text{value}\left(\text{A/F,10\%,n}\right)\\ \text{EUAB}=-\text{First}\text{cost}\times \left(\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right)+\text{Salvage}\text{value}\times \left(\frac{i}{{\left(1+i\right)}^n-1}\right)\end{array}$

EUAC can be calculated with the help of below expression:

  $\begin{array}{c}\text{EUAC}=\text{Maintenance}\text{cost}+\text{Maintenance}\text{and}\text{operating}\text{gradient}\left(\text{A/G},10\%,n\right)\\ =\text{Maintenance}\text{cost}+\text{Maintenance}\text{and}\text{operating}\text{gradient}\times \left(\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right)\end{array}$

Calculation:

As per the given problem

First cost = $1,050,000, Salvage value = $225,000, i = 10%.

Substitute these values in the formula $\text{EUAB}-\text{EUAC}$ ,

When $n=4$

  $\begin{array}{c}\text{EUAB}-\text{EUAC}=\left[\begin{array}{l}\left(-\$1,050,000\times \left(\frac{0.10\times {\left(1+0.10\right)}^4}{{\left(1+0.10\right)}^4-1}\right)+\$225,000\times \left(\frac{0.10}{{\left(1+0.10\right)}^4-1}\right)\right)\\ -\left(\$235,000+\$75,000\times \left(\frac{1}{0.10}-\frac{4}{{\left(1+0.10\right)}^4-1}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(-\$1,050,000\times 0.31547\right)+\left(\$225,000\times 0.21547\right)\\ -\left(\$235,000+\left(\$75,000\times \left(10-8.6188\right)\right)\right)\end{array}\right]\\ =-\$331,243.5+\$48,480.75-\$338,590\\ =-\$621,352.75\end{array}$

When $n=5$

  $\begin{array}{c}\text{EUAB}-\text{EUAC}=\left[\begin{array}{l}\left(-\$1,050,000\times \left(\frac{0.10\times {\left(1+0.10\right)}^5}{{\left(1+0.10\right)}^5-1}\right)+\$225,000\times \left(\frac{0.10}{{\left(1+0.10\right)}^5-1}\right)\right)\\ -\left(\$235,000+\$75,000\times \left(\frac{1}{0.10}-\frac{5}{{\left(1+0.10\right)}^5-1}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(-\$1,050,000\times 0.2638\right)+\left(\$225,000\times 0.1638\right)\\ -\left(\$235,000+\left(\$75,000\times \left(10-8.1898\right)\right)\right)\end{array}\right]\\ =-\$276,990+\$36,855-\$370,765\\ =-\$610,900\end{array}$

When $n=6$

  $\begin{array}{c}\text{EUAB}-\text{EUAC}=\left[\begin{array}{l}\left(-\$1,050,000\times \left(\frac{0.10\times {\left(1+0.10\right)}^6}{{\left(1+0.10\right)}^6-1}\right)+\$225,000\times \left(\frac{0.10}{{\left(1+0.10\right)}^6-1}\right)\right)\\ -\left(\$235,000+\$75,000\times \left(\frac{1}{0.10}-\frac{6}{{\left(1+0.10\right)}^6-1}\right)\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(-\$1,050,000\times 0.2296\right)+\left(\$225,000\times 0.1296\right)\\ -\left(\$235,000+\left(\$75,000\times \left(10-7.7764\right)\right)\right)\end{array}\right]\\ =-\$241,080+\$29,160-\$401,770\\ =-\$613,690\end{array}$