Chapter 6, Problem 23P

(a)

To determine

Calculate the equivalent annual worth.

Explanation of Solution

Table -1 shows the cash flow of different models.

Table -1

Models	1	2
Price (P)	-30,000	-2,000
AOC (MO) per year	-1,000	-150
Salvage value (SV)	8,000
Number of pallet (PA)	500	800
Cost per pallet (CP)	-10	-10
Wage (W)	-32,000	-55,000
Time period (n)	12	4
Interest  (i)	8%	8%

Pallets are purchased every 2 years once.

The equivalent annual worth of project 1 (AW1) can be calculated as follows:

$\begin{array}{c}\text{AW1}=\left(\begin{array}{l}\text{C}\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{W}\\ +\left(\text{PA}\times \text{CP}\right)\left(\begin{array}{l}\frac{1}{{\left(1+\text{i}\right)}^2}+\frac{1}{{\left(1+\text{i}\right)}^4}+\frac{1}{{\left(1+\text{i}\right)}^6}\\ +\frac{1}{{\left(1+\text{i}\right)}^8}+\frac{1}{{\left(1+\text{i}\right)}^{10}}+1\end{array}\right)\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\end{array}\right)\\ =\left(\begin{array}{l}-\text{30,000}\left(\frac{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{12}}}{{\left(1+\text{0}\text{.08}\right)}^{\text{12}}-1}\right)-1,000+\text{8,000}\left(\frac{\text{0}\text{.08}}{{\left(1+\text{0}\text{.08}\right)}^{\text{12}}-1}\right)-32,000\\ +\left(\text{500}\times -10\right)\left(\begin{array}{l}\frac{1}{{\left(1+\text{0}\text{.08}\right)}^2}+\frac{1}{{\left(1+\text{0}\text{.08}\right)}^4}+\frac{1}{{\left(1+\text{0}\text{.08}\right)}^6}\\ +\frac{1}{{\left(1+\text{0}\text{.08}\right)}^8}+\frac{1}{{\left(1+\text{0}\text{.08}\right)}^{10}}+1\end{array}\right)\left(\frac{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{12}}}{{\left(1+\text{0}\text{.08}\right)}^{\text{12}}-1}\right)\end{array}\right)\\ =\left(\begin{array}{l}-\text{30,000}\left(\frac{\text{0}\text{.08}\left(2.51817\right)}{2.51817-1}\right)+\text{8,000}\left(\frac{0.08}{2.51817-1}\right)-33,000\\ -5,000\left(\begin{array}{l}\frac{1}{1.1664}+\frac{1}{1.360489}+\frac{1}{1.586874}\\ +\frac{1}{1.85093}+\frac{1}{2.158925}+1\end{array}\right)\left(\frac{\text{0}\text{.08}\left(2.51817\right)}{2.51817-1}\right)\end{array}\right)\\ =\left(\begin{array}{l}-\text{30,000}\left(\frac{0.20145}{1.51817}\right)+\text{8,000}\left(\frac{0.08}{1.51817}\right)-33,000\\ -5,000\left(\begin{array}{l}0.857339+0.735029+0.630169\\ +0.540269+0.463193+1\end{array}\right)\left(\frac{0.20145}{1.51817}\right)\end{array}\right)\\ =\left(\begin{array}{l}-\text{30,000}\left(0.132693\right)+\text{8,000}\left(0.052695\right)-33,000\\ -5,000\left(4.225999\right)\left(0.132693\right)\end{array}\right)\\ =\left(-3,980.79+421.56-33,000-2,803.8\right)\\ =-39,363.03\end{array}$

The annual worth of the project 1 is -$39,363.03.

Number of electric pallet mover (N) is 2. The equivalent annual worth of project 2 (AW2) can be calculated as follows:

$\begin{array}{c}\text{AW2}=\left(\begin{array}{l}\left(\text{N}\right)\text{C}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)+\left(\text{N}\right)\text{MO}+\text{W}\\ +\left(\text{PA}\times \text{CP}\right)\left(1+\frac{1}{{\left(1+\text{i}\right)}^2}\right)\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\end{array}\right)\\ =\left(\begin{array}{l}2\left(-2\text{,000}\right)\left(\frac{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{4}}}{{\left(1+\text{0}\text{.08}\right)}^{\text{4}}-1}\right)+2\left(-150\right)-55,000\\ +\left(\text{800}\times -10\right)\left(1+\frac{1}{{\left(1+\text{0}\text{.08}\right)}^2}\right)\left(\frac{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{4}}}{{\left(1+\text{0}\text{.08}\right)}^{\text{4}}-1}\right)\end{array}\right)\\ =\left(\begin{array}{l}-\text{4,000}\left(\frac{\text{0}\text{.08}\left(1.360489\right)}{1.360489-1}\right)-300-55,000\\ -8,000\left(1+\frac{1}{1.1664}\right)\left(\frac{\text{0}\text{.08}\left(1.360489\right)}{1.360489-1}\right)\end{array}\right)\\ =\left(\begin{array}{l}-\text{4,000}\left(\frac{0.108839}{0.360489}\right)-55,300\\ -8,000\left(1+0.857339\right)\left(\frac{0.108839}{0.360489}\right)\end{array}\right)\\ =-\text{4,000}\left(0.30192\right)-55,300-14,858.712\left(0.30192\right)\\ =-1,207.68-55,300-4,486.14\\ =-60,993.82\end{array}$

The annual worth of the project 2 is -$38,699.56.

Since the annual worth of the project 1 is greater than the project 2, select the project 1.

(b)

To determine

Calculate the annual worth using spreadsheet.

Explanation of Solution

The annual worth of the project 1 and 2 is calculated by using spreadsheet as follows:

Since the annual worth of the project 1 (Forklift) is greater than the project 2 (Walkies), select the project 1.