Chapter 5, Problem 30P

(a):

To determine

Compare the project based on future worth.

Explanation of Solution

Table-1 shows the cash flow of different projects.

Table -1

Projects	A	B	C
First cost (C)	40,000	80,000	130,000
Annual cost (AC) per year	9,000	6,000	4,000
Salvage value (SV)			10% to C
Time period (n)	2	4	8

Interest rate (i) is 10%.

The time period for project A should be equated with project time period C. Thus, all the cash flows are repeated for other 6 years. The time period 1 (n) is 8 years, time period 2 (n1) is 6 years, time period 3 (n2) is 4 years, and time period 4 (n3) is 2 years.

Future value project A (FWA) can be calculated as follows:

$\begin{array}{c}\text{FWA}=-\text{C}\left({\left(1+\text{i}\right)}^{\text{n}}+{\left(1+\text{i}\right)}^{\text{n1}}+{\left(1+\text{i}\right)}^{\text{n2}}+{\left(1+\text{i}\right)}^{\text{n4}}\right)-\text{AC}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}}\right)\\ =-\text{40,000}\left(\begin{array}{l}{\left(1+\text{0}\text{.1}\right)}^{\text{8}}+{\left(1+\text{0}\text{.1}\right)}^{\text{6}}\\ +{\left(1+\text{0}\text{.1}\right)}^{\text{4}}+{\left(1+\text{0}\text{.1}\right)}^{\text{2}}\end{array}\right)-\text{9,000}\left(\frac{{\left(1+\text{0}\text{.1}\right)}^{\text{8}}-1}{\text{0}\text{.1}}\right)\\ =-\text{40,000}\left(\begin{array}{l}2.143589+1.771561\\ +1.4641+1.21\end{array}\right)-\text{9,000}\left(\frac{2.143589-1}{\text{0}\text{.1}}\right)\\ =-\text{40,000}\left(6.58925\right)-\text{9,000}\left(\frac{1.143589}{\text{0}\text{.1}}\right)\\ =-263,570-\text{9,000}\left(11.43589\right)\\ =-240,335.7-102,923.01\\ =-366,493.01\end{array}$

The future worth of project A is -$366,493.01.

The time period for project B should be equated with project time period C. Thus, all the cash flows are repeated for other 4 years. The time period 1 (n) is 8 years and time period 2 (n1) is 4 years.

Future value project B (FWB) can be calculated as follows:

$\begin{array}{c}\text{FWB}=-\text{C}\left({\left(1+\text{i}\right)}^{\text{n}}+{\left(1+\text{i}\right)}^{\text{n1}}\right)-\text{AC}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}}\right)\\ =-80,000\left({\left(1+\text{0}\text{.1}\right)}^8+{\left(1+\text{0}\text{.1}\right)}^4\right)-\text{6,000}\left(\frac{{\left(1+\text{0}\text{.1}\right)}^8-1}{\text{0}\text{.1}}\right)\\ =-80,000\left(2.143589+1.4641\right)-\text{6,000}\left(\frac{2.143589-1}{\text{0}\text{.1}}\right)\\ =-80,000\left(3.607689\right)-\text{6,000}\left(\frac{1.143589}{\text{0}\text{.1}}\right)\\ =-288,615.12-\text{6,000}\left(11.43589\right)\\ =-288,615.12-68,615.34\\ =-357,230.46\end{array}$

The future worth of project B is -$357,230.46.

Future value project C (FWC) can be calculated as follows:

$\begin{array}{c}\text{FWB}=-\text{C}{\left(1+\text{i}\right)}^{\text{n}}-\text{AC}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}}\right)+\text{C}\times \text{0}\text{.1}\\ =-130,000{\left(1+\text{0}\text{.1}\right)}^8-\text{4,000}\left(\frac{{\left(1+\text{0}\text{.1}\right)}^8-1}{\text{0}\text{.1}}\right)+130,000\times \text{0}\text{.1}\\ =-130,000\left(2.143589\right)-\text{4,000}\left(\frac{2.143589-1}{\text{0}\text{.1}}\right)+13,000\\ =-278,666.57-4\text{,000}\left(\frac{1.143589}{\text{0}\text{.1}}\right)+13,000\\ =-278,666.57-4\text{,000}\left(11.43589\right)+13,000\\ =-278,666.57-45,743.56+13,000\\ =-311,410.13\end{array}$

The future worth of project C is -$311,410.13. Since the future worth of project C is greater than the other available projects, select project C.

(b):

To determine

Compare the projects based on the present worth.

Explanation of Solution

Present worth of project A (PWA) can be calculated as follows:

$\begin{array}{c}\text{PWA}=\frac{\text{FWA}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ =\frac{-366,493.01}{{\left(1+\text{0}\text{.1}\right)}^{\text{8}}}\\ =\frac{-366,493.01}{2.143589}\\ =-170,971.68\end{array}$

Present worth of project A is -$170,971.68.

Present worth of project B (PWB) can be calculated as follows:

$\begin{array}{c}\text{PWB}=\frac{\text{FWA}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ =\frac{-357,230.46}{{\left(1+\text{0}\text{.1}\right)}^{\text{8}}}\\ =\frac{-357,230.46}{2.143589}\\ =-166,650.63\end{array}$

Present worth of project B is -$166,650.63.

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

$\begin{array}{c}\text{PWC}=\frac{\text{FWA}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ =\frac{-311,410.13}{{\left(1+\text{0}\text{.1}\right)}^{\text{8}}}\\ =\frac{-311,410.13}{2.143589}\\ =-145,275.11\end{array}$

Present worth of project C is -$145,275.11. Since the present worth of project C is greater than the other two projects, select project C.