Chapter 14, Problem 1P

(a):

To determine

Calculate the sunk cost.

Explanation of Solution

Time period is denoted by n and the interest rate is denoted by i. Sunk cost can be calculated as follows.

$\begin{array}{c}\text{Sunk cost}=\text{Initial cost}-\text{Current value}\\ =12,000-2,000\\ =10,000\end{array}$

Sunk cost is $9,000.

(b):

To determine

Calculate the opportunity cost.

Explanation of Solution

Opportunity cost is equal to the market value of that particular year. Thus, opportunity cost is $2,000. Present worth (PW) can be calculated as follows.

$\begin{array}{c}\text{PW}=\left(\text{Opportunity cost}+\text{Maintenance cost}\left(\frac{{\left(1+\text{i}\right)}^1-1}{\text{i}{\left(1+\text{i}\right)}^1}\right)\right)\\ =\left(\text{2,000}+10,000\left(\frac{{\left(1+0.15\right)}^3-1}{\text{0}\text{.15}{\left(1+0.15\right)}^3}\right)\right)\\ =24,832\end{array}$

Present worth is $24,832.

(c):

To determine

Calculate the present worth.

Explanation of Solution

Time period is denoted by n and the interest rate is denoted by i. Present worth (PW) can be calculated as follows.

$\begin{array}{c}\text{PW}=\left({\text{Price}}_{\text{Machine}}+\text{Operating cost}\left(\frac{{\left(1+\text{i}\right)}^1-1}{\text{i}{\left(1+\text{i}\right)}^1}\right)+\text{Salvage vlaue}\left(\frac{1}{{\left(1+\text{i}\right)}^2}\right)\right)\\ =\left(\text{15,000}+3,000\left(\frac{{\left(1+0.15\right)}^3-1}{\text{0}\text{.15}{\left(1+0.15\right)}^3}\right)+5,000\left(\frac{1}{{\left(1+0.15\right)}^3}\right)\right)\\ =18,562\end{array}$

Present worth is $18,562.