Chapter 14, Problem 56P

(a):

To determine

Calculate annual worth in constant dollar.

Explanation of Solution

Machine A: First cost (FC) is $150,000. Maintenance and operating cost (MO) is $70,000. Salvage value (SV) is $40,000. Time period (n) is 5.

Machine B: First cost (FC) is $1,025,000. Maintenance and operating cost (MO) is $5,000. Salvage value (SV) is $2,00,000. Time period (n) is infinitive.

Real return (i) is 12%.

The annual worth of Machine A (AWA) can be calculated as follows:

$\begin{array}{c}\text{AWA}=-\text{FC}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)-\text{MO}+\text{SV}\left(\frac{\text{i}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\\ =-\text{150,000}\left(\frac{\text{0}\text{.12}{\left(1+\text{0}\text{.12}\right)}^{\text{5}}}{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}-1}\right)-\text{70,000}+\text{40,000}\left(\frac{\text{0}\text{.12}}{{\left(1+\text{0}\text{.12}\right)}^{\text{5}}-1}\right)\\ =-\text{150,000}\left(0.27741\right)-\text{70,000}+\text{40,000}\left(0.15741\right)\\ =-41,611.5-70,000+6,296.4\\ =-105,315\end{array}$

The annual worth is -$105,315.

The annual worth of Machine B (AWBA) can be calculated as follows:

$\begin{array}{c}\text{AWB}=-\text{FC}\times \left(\text{i}\right)-\text{MO}\\ =-1,025,000\left(0.12\right)-5,000\\ =-123,000-5,000\\ =-128,000\end{array}$

The annual worth is -$128,000. Since the annual worth of the cost is less for Machine A, select Machine A.

(b):

To determine

Calculate annual worth in current dollars.

Explanation of Solution

Inflation adjusted interest rate (if) can be calculated as follows:

$\begin{array}{c}\text{if}=\text{i}+\text{f}+\text{i}\times \text{f}\\ =\text{0}\text{.12}+\text{0}\text{.07}+\text{0}\text{.12}\times \text{0}\text{.07}\\ =0.19+0.0084\\ =0.1984\end{array}$

Inflation adjusted interest rate is 19.84%.

The annual worth of machine A (AWA) can be calculated as follows:

$\begin{array}{c}\text{AWA}=-\text{FC}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)-\text{MO}+\text{SV}\left(\frac{\text{i}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\\ =-\text{150,000}\left(\frac{\text{0}\text{.1984}{\left(1+\text{0}\text{.1984}\right)}^{\text{5}}}{{\left(1+\text{0}\text{.1984}\right)}^{\text{5}}-1}\right)-\text{70,000}+\text{40,000}\left(\frac{\text{0}\text{.12}}{{\left(1+\text{0}\text{.1984}\right)}^{\text{5}}-1}\right)\\ =-\text{150,000}\left(0.3332\right)-\text{70,000}+\text{40,000}\left(0.1348\right)\\ =-49,980-70,000+5,392\\ =-114,588\end{array}$

The annual worth is -$114,588.

The annual worth of Machine B (AWBA) can be calculated as follows:

$\begin{array}{c}\text{AWB}=-\text{FC}\times \left(\text{i}\right)-\text{MO}\\ =-1,025,000\left(0.1984\right)-5,000\\ =-203,360-5,000\\ =-208,360\end{array}$

The annual worth is -$208,360. Since the annual worth of the cost is less for Machine A, select Machine A.

Annual worth can be calculated using the spreadsheet as follows: