Chapter B, Problem 1P

To determine

The present worth of the machinery.

Answer to Problem 1P

The present worth of the machinery is $\$10,330.5$.

Explanation of Solution

Given:

Cost of machinery is $\$70,000$.

Savings are $\$18,000$ annually.

Salvage value is $\$12,500$.

Interest rate is $12\%$.

Calculation:

Write the formula to calculate the present worth.

$PW=-P+A\left(\frac{P}{A},i,n\right)+F\left(\frac{P}{F},i,n\right)$         ...... (I)

Here, initial cost of machine is P, amount of annual savings is A, salvage value is F, interest rate is i and time period is n.

Substitute $\$70,000$ for P, $\$18,000$ for A, $\$12,500$ for F and $12\%$ for i in Equation (I).

$\begin{array}{c}PW=-\$70,000+\$18,000\left(\frac{P}{A},12\%,6\right)+\$12,500\left(\frac{P}{F},12\%,6\right)\\ =-\$70,000+\$18,000\left(\frac{{\left(1+0.12\right)}^6-1}{0.12{\left(1+0.12\right)}^6}\right)+\$12,500\left(\frac{1}{{\left(1+0.12\right)}^6}\right)\\ =-\$70,000+\$73,998+\$6,332.5\\ =\$10,330.5\end{array}$.

Conclusion:

Thus, the present worth of the machinery is $\$10,330.5$.