Chapter 5, Problem 21P

(a):

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	B	C
First cost (C)	203,000	396,000
Operating cost (MO) per year	85,000	119,000
Replacement cost (O) per year	5,500
Sludge hauling cost (S) per year less		37,000
Salvage value (SV)	10% at C	10% at C
Time period (n)	5	10

MARR (i) is 6%.

The time period for project B should equate with project time period C. 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.

The present worth of project B (PWA) can be calculated as follows:

$\begin{array}{c}\text{PWB}=-\text{C}-\left(\text{MO}+\text{O}\right)\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n1}}}\right)+\frac{-\text{C}+\left(\text{C}\times \text{0}\text{.1}\right)}{{\left(\text{1}+\text{i}\right)}^{\text{n2}}}+\frac{\left(\text{C}\times \text{0}\text{.1}\right)}{{\left(\text{1}+\text{i}\right)}^{\text{n1}}}\\ =\left(\begin{array}{l}-203,000-\left(\begin{array}{l}85,000\\ +5,500\end{array}\right)\left(\frac{{\left(\text{1}+0.06\right)}^{10}-1}{\text{0}\text{.06}{\left(\text{1}+0.06\right)}^{10}}\right)\\ +\frac{-\text{203,000}+\left(\text{203,000}\times \text{0}\text{.1}\right)}{{\left(\text{1}+0.\text{06}\right)}^{\text{5}}}+\frac{\left(\text{203,000}\times \text{0}\text{.1}\right)}{{\left(\text{1}+0.\text{06}\right)}^{\text{10}}}\end{array}\right)\\ =\left(\begin{array}{l}-203,000-\left(90,500\right)\left(\frac{1.790848-1}{\text{0}\text{.06}\left(1.790848\right)}\right)\\ +\frac{-\text{203,000}+\text{20,300}}{1.338226}+\frac{\text{20,300}}{1.790848}\end{array}\right)\\ =-203,000-\left(90,500\right)\left(\frac{0.790848}{0.107451}\right)-136,524.03+11,335.42\\ =-203,000-\left(90,500\right)\left(7.36008\right)-136,524.03+11,335.42\\ =-203,000-666,087.24-136,524.03+11,335.42\\ =-994,275.85\end{array}$

The present worth of project B is -$994,275.85.

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

$\begin{array}{c}\text{PWC}=-\text{C}-\left(\text{MO}-\text{S}\right)\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)+\frac{\left(\text{C}\times \text{0}\text{.1}\right)}{{\left(\text{1}+\text{i}\right)}^{\text{n}}}\\ =\left(-396,000-\left(\begin{array}{l}119,000\\ -37,000\end{array}\right)\left(\frac{{\left(\text{1}+0.06\right)}^{10}-1}{\text{0}\text{.06}{\left(\text{1}+0.06\right)}^{10}}\right)+\frac{\left(\text{396,000}\times \text{0}\text{.1}\right)}{{\left(\text{1}+0.06\right)}^{\text{10}}}\right)\\ =\left(-396,000-82,000\left(\frac{1.790848-1}{\text{0}\text{.06}\left(1.790848\right)}\right)+\frac{\text{39,600}}{1.790848}\right)\\ =-396,000-82,000\left(\frac{0.790848}{0.107451}\right)+22,112.43\\ =-396,000-82,000\left(7.36008\right)+22,112.43\\ =-396,000-603,526.56+22,112.43\\ =-977,414.13\end{array}$

The present worth of project C is -$977,141.13. Since the present worth of project C is greater than project B, select project C.

(b):

To determine

Compare the projects based on the present worth.

Explanation of Solution

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

The present worth of the project B (PWA) can be calculated as follows:

$\begin{array}{c}\text{PWB}=-\text{C}-\left(\text{MO}+\text{O}\right)\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n1}}}\right)+\frac{-\text{C}+\left(\text{C}\times \text{0}\text{.1}\right)}{{\left(\text{1}+\text{i}\right)}^{\text{n2}}}+\frac{\left(\text{C}\times \text{0}\text{.1}\right)}{{\left(\text{1}+\text{i}\right)}^{\text{n1}}}\\ =\left(\begin{array}{l}-203,000-\left(\begin{array}{l}85,000\\ +5,500\end{array}\right)\left(\frac{{\left(\text{1}+0.06\right)}^8-1}{\text{0}\text{.06}{\left(\text{1}+0.06\right)}^8}\right)\\ +\frac{-\text{203,000}+\left(\text{203,000}\times \text{0}\text{.1}\right)}{{\left(\text{1}+0.\text{06}\right)}^{\text{5}}}+\frac{\left(\text{203,000}\times \text{0}\text{.1}\right)}{{\left(\text{1}+0.\text{06}\right)}^{\text{8}}}\end{array}\right)\\ =\left(\begin{array}{l}-203,000-\left(90,500\right)\left(\frac{1.593848-1}{\text{0}\text{.06}\left(1.593848\right)}\right)\\ +\frac{-\text{203,000}+\text{20,300}}{1.338226}+\frac{\text{20,300}}{1.593848}\end{array}\right)\\ =-203,000-\left(90,500\right)\left(\frac{0.593848}{0.095631}\right)-136,524.03+12,736.4717\\ =-203,000-\left(90,500\right)\left(6.209786\right)-136,524.03+12,736.4717\\ =-203,000-561,985.633-136,524.03+12,736.4717\\ =-888,773.25\end{array}$

The present worth of project B is -$888,773.2513.

Time period (n) is eight years. The present worth of the project C (PWC) can be calculated as follows:

$\begin{array}{c}\text{PWC}=-\text{C}-\left(\text{MO}-\text{S}\right)\left(\frac{{\left(\text{1}+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(\text{1}+\text{i}\right)}^{\text{n}}}\right)+\frac{\left(\text{C}\times \text{0}\text{.1}\right)}{{\left(\text{1}+\text{i}\right)}^{\text{n}}}\\ =\left(-396,000-\left(\begin{array}{l}119,000\\ -37,000\end{array}\right)\left(\frac{{\left(\text{1}+0.06\right)}^8-1}{\text{0}\text{.06}{\left(\text{1}+0.06\right)}^8}\right)+\frac{\left(\text{396,000}\times \text{0}\text{.1}\right)}{{\left(\text{1}+0.\text{06}\right)}^{\text{8}}}\right)\\ =\left(-396,000-82,000\left(\frac{1.593848-1}{\text{0}\text{.06}\left(1.593848\right)}\right)+\frac{\text{39,600}}{1.593848}\right)\\ =-396,000-82,000\left(\frac{1.593848-1}{\text{0}\text{.06}\left(1.593848\right)}\right)+24,845.5311\\ =-396,000-82,000\left(6.209786\right)+24,845.5311\\ =-396,000-509,202.452+24,845.5311\\ =-880,356.92\end{array}$

The present worth of project C is -$880.356.92. Since the present worth of project C is greater than project B, select project C.