Chapter D, Problem 132P

To determine

The maximum amount that the firm should be willing to pay to an outside vendor for the subcomponent.

Answer to Problem 132P

The correct option is $\left(c\right)\$22\text{per}\text{unit}$.

Explanation of Solution

Given:

First cost is $\$350,000$.

Cost of the operation is $\$45,000$.

Salvage value is $\$15,000$.

Project life is $5\text{years}$.

Interest rate is $10\%$.

Material cost per unit is $\$8.50$.

Annual production is $10,000\text{units}$.

Calculation:

Calculate the total annual material cost.

$\begin{array}{c}\text{Annual}\text{material}\text{cost}=\$8.50\times 10,000\\ =\$85,000\end{array}$

Write the expression to calculate the present worth of making the part.

$P{W}_M=\left[\begin{array}{l}-\$350,000-\$45,000\left(P/A,i,n\right)\\ -\$85,000\left(P/A,i,n\right)+\$15,000\left(P/F,i,n\right)\end{array}\right]$ ...... (I)

Here, the present worth is P, the annual cost is A, the interest rate is i and the timer period is n.

Substitute $10\%$ for $i$ and $5$ for $n$ in Equation (I).

$\begin{array}{c}P{W}_{\text{M}}=\left[\begin{array}{l}-\$350,000-\$45,000\left(\frac{{\left(1+0.1\right)}^5-1}{0.1{\left(1+0.1\right)}^5}\right)\\ -\$85,000\left(\frac{{\left(1+0.1\right)}^5-1}{0.1{\left(1+0.1\right)}^5}\right)+\$15,000\left(\frac{1}{{\left(1+0.1\right)}^5}\right)\end{array}\right]\\ =\left[-\$350,000-\$45,000\left(3.7908\right)-\$85,000\left(3.7908\right)+\$15,000\left(0.621\right)\right]\\ =\left[-\$350,000-\$170,586-\$322,218+\$9,315\right]\end{array}$

$P{W}_{\text{M}}=-\$833489$ ....... (II)

Write the expression to calculate the present worth of buying the parts.

$\begin{array}{c}P{W}_{\text{B}}=-A\left(P/A,i,n\right)\\ =-A\left(\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right)\end{array}$ ...... (III)

Substitute $10\%$ for $i$ and $5$ for $n$ in Equation (III).

$\begin{array}{c}P{W}_{\text{B}}=-A\left(\frac{{\left(1+0.1\right)}^5-1}{0.1{\left(1+0.1\right)}^5}\right)\\ P{W}_{\text{B}}=-A\left(3.7908\right)\end{array}$ ....... (IV)

Equate Equation (II) and Equation (IV) and calculate the maximum annual amount.

$\begin{array}{l}P{W}_M=P{W}_B\\ -\$833489=-A\left(3.7908\right)\\ A=\frac{\$833489}{3.7908}\\ A=\$219,871.53\end{array}$

Calculate the annual amount per unit.

$\begin{array}{c}{A}_{\text{per}\text{unit}}=\frac{\$219,871.53}{10,000}\\ =\$21.9\\ \simeq \$22\end{array}$.

Conclusion:

Therefore, the correct option is $\left(c\right)\$22\text{per}\text{unit}$.