Chapter 13, Problem 10P

To determine

The minimum EUAC for the machine and its economic life.

Answer to Problem 10P

The minimum EUAC is $\$17240$.

The economic life is $\text{4}\text{years}$.

Explanation of Solution

Given:

The first cost is $\$50000$.

Decline in its value per annum is $20\%$.

Operating and maintenance cost is $\$3500$ per year.

Climb in the operating and maintenance cost is $\$2000$ per year.

The MARR is $9\%$.

Calculation:

Write the expression for capital recovery.

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

Here, the present value is P, the annual expenditure is A, the rate of interest is i and the period is n.

Write the expression for uniform gradient series.

$\left(A/G\right)=\left(\frac{1}{i}-\frac{n}{{\left(1+i\right)}^n-1}\right)$ ...... (II)

Here, the gradient is G.

Calculate the Equivalent Uniform Annualised Cost of capital.

Year	Fixed Cost (P) $\left(a\right)$	Salvage percent $\left(b\right)=0.80$	Salvage Value $\left(c\right)=\left(c_{n-1}\times b\right)$	$\left(A/P,9\%,n\right)$ $\left(d\right)$	EUAC $\left(\left(a-c\right)\times d\right)+\left(c\times 0.09\right)$
$0$	$\$50000$		$\$50000$
$1$	$\$50000$	$0.80$	$\$40000$	$1.090$	$\$14500$
$2$	$\$50000$	$0.80$	$\$32000$	$0.5685$	$\$13113$
$3$	$\$50000$	$0.80$	$\$25600$	$0.3951$	$\$11944.44$
$4$	$\$50000$	$0.80$	$\$20480$	$0.3087$	$\$10956.024$
$5$	$\$50000$	$0.80$	$\$16384$	$0.2571$	$\$10117.024$
$6$	$\$50000$	$0.80$	$\$13107.2$	$0.2229$	$\$9403.05312$
$7$	$\$50000$	$0.80$	$\$10485.76$	$0.1987$	$\$8795.19788$
$8$	$\$50000$	$0.80$	$\$8388.608$	$0.1807$	$\$8274.153254$
$9$	$\$50000$	$0.80$	$\$6710.8864$	$0.1668$	$\$7824.603924$
$10$	$\$50000$	$0.80$	$\$5368.70912$	$0.1558$	$\$7437.73894$

Now the annual cost is and the climb in cost is given.

This means that the provided cash flow will be of the uniform gradient series where $\$2000$ will act as $G$ and $\$3500$ as $A$.

Calculate the equal uniform annualised cost of machine.

Year	$A$ $\left(a\right)$	$G$ $\left(b\right)$	$\left(A/G,9\%,n\right)$ $\left(c\right)$	EUAC $\left(a+\left(b\times c\right)\right)$
$0$
$1$	$\$3500$	$\$2000$	$0$	$\$3500$
$2$	$\$3500$	$\$2000$	$0.478$	$\$4456$
$3$	$\$3500$	$\$2000$	$0.943$	$\$5386$
$4$	$\$3500$	$\$2000$	$1.393$	$\$6286$
$5$	$\$3500$	$\$2000$	$1.828$	$\$7156$
$6$	$\$3500$	$\$2000$	$2.25$	$\$8000$
$7$	$\$3500$	$\$2000$	$2.657$	$\$8814$
$8$	$\$3500$	$\$2000$	$3.051$	$\$9602$
$9$	$\$3500$	$\$2000$	$3.431$	$\$10362$
$10$	$\$3500$	$\$2000$	$3.798$	$\$11096$

Calculate the total equivalent uniform annualized cost.

Year	EUAC of capital $\left(a\right)$	EUAC of machine $\left(b\right)$	Total EUAC $c=\left(a+b\right)$
$1$	$\$14500$	$\$3500$	$\$18000$
$2$	$\$13113$	$\$4456$	$\$17569$
$3$	$\$11944.44$	$\$5386$	$\$17330.44$
$4$	$\$10956.024$	$\$6286$	$\$17242.024$
$5$	$\$10117.024$	$\$7156$	$\$17273.233$
$6$	$\$9403.05312$	$\$8000$	$\$17403.053120$
$7$	$\$8795.19788$	$\$8814$	$\$17609.197888$
$8$	$\$8274.153254$	$\$9602$	$\$17876.153254$
$9$	$\$7824.603924$	$\$10362$	$\$18186.603924$
$10$	$\$7437.73894$	$\$11096$	$\$18532.738940$

From the table it is clearly known that the minimum EUAC is $\$17242.024\approx \$17240$ and the economic life is $\text{4}\text{years}$.

Conclusion:

The minimum EUAC is $\$17240$.

The economic life is $\text{4}\text{years}$.