Chapter 5, Problem 23P

To determine

Compare the projects based on the present worth.

Explanation of Solution

Table -1 shows the cash flow of different projects.

Table -1

Projects	C	D
First cost (C)	-40,000	-32,000
Operating cost (MO) per year	-7,000	-3,000
Increasing operating cost (IO) per year	-1,000	-
Salvage value (SV)	9,000	500
Time period (n)	10	5

MARR (i) is 10%.

The present worth of project C (PWC) can be calculated as follows:

$\begin{array}{c}\text{PW}=\text{C}+\left(\text{MO}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\text{IO}\frac{1}{\text{i}}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}-\frac{\text{n}}{{\left(1+\text{i}\right)}^{\text{n}}}\right)\right)+\frac{\text{SV}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ =-\text{40,000}-\left(\begin{array}{l}7,000\left(\frac{{\left(1+\text{0}\text{.1}\right)}^{\text{10}}-1}{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{10}}}\right)\\ +\left(-1,000\right)\frac{1}{\text{0}\text{.1}}\left(\frac{{\left(1+\text{0}\text{.1}\right)}^{\text{10}}-1}{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{10}}}-\frac{\text{10}}{{\left(1+\text{0}\text{.1}\right)}^{\text{10}}}\right)\end{array}\right)+\frac{\text{9,000}}{{\left(1+\text{0}\text{.1}\right)}^{\text{10}}}\\ =-\text{40,000}-\left(\begin{array}{l}7,000\left(\frac{2.593742-1}{\text{0}\text{.1}\left(2.593742\right)}\right)\\ +\left(1,000\right)10\left(\begin{array}{l}\frac{2.593742-1}{\text{0}\text{.1}\left(2.593742\right)}\\ -\frac{\text{10}}{2.593742}\end{array}\right)\end{array}\right)+\frac{\text{9,000}}{2.593742}\\ =-\text{40,000}-\left(\begin{array}{l}7,000\left(\frac{1.593742}{\text{0}.259374}\right)\\ +10,000\left(\frac{1.593742}{\text{0}.259374}-3.855434\right)\end{array}\right)+3,469.8902\\ =-\text{36,530}\text{.1098}-\left(7,000\left(6.144571\right)+10,000\left(6.144571-3.855434\right)\right)\\ =-\text{36,530}\text{.1098}-\left(43,011.997+10,000\left(2.289137\right)\right)\\ =-\text{36,530}\text{.1098}-\left(43,011.997+22,891.37\right)\\ =-102,433.48\end{array}$

The present worth of project C is -$102,433.48.

The time period for project C should equate with project time period B. Thus, all the cash flows are repeated for other five years. The time period (n1) is 10 years and time period 2 (n2) is five years.

Present worth of project D (PWD) can be calculated as follows:

$\begin{array}{c}\text{PWD}=\text{C}+\text{MO}\left(\frac{{\left(1+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n1}}}\right)+\frac{\text{C}+\text{SV}}{{\left(1+\text{i}\right)}^{\text{n2}}}+\frac{\text{SV}}{{\left(1+\text{i}\right)}^{\text{n1}}}\\ =-\text{32,000}-3,000\left(\frac{{\left(1+\text{0}\text{.1}\right)}^{\text{10}}-1}{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{10}}}\right)+\frac{-\text{32,000}+\text{500}}{{\left(1+\text{0}\text{.1}\right)}^{\text{5}}}+\frac{\text{500}}{{\left(1+\text{0}\text{.1}\right)}^{\text{10}}}\\ =-\text{32,000}-3,000\left(\frac{2.593742-1}{\text{0}\text{.1}\left(2.593742\right)}\right)-\frac{\text{31,500}}{1.61051}+\frac{\text{500}}{2.593742}\\ =-\text{32,000}-3,000\left(\frac{1.593742}{\text{0}.259374}\right)-19,559.2017+192.7717\\ =-\text{51,366}\text{.43}-3,000\left(6.144571\right)\\ =-\text{51,366}\text{.43}-18,433.71\\ =-69,800.14\end{array}$

Present worth of project D is -69,800.14. Since the present worth of project D is greater than project C, select project D.