Chapter 8, Problem 25P

(a):

To determine

Calculate the incremental rate of return.

Explanation of Solution

Table-1 shows the cash flow.

Table -1

Alternative	P	Q
First cost (F)	-18,000	-35,000
AOC (AC) per year	-4,000	-3,600
Salvage value (SV)	1,000	2,700
Time period (n)	3	6

MARR is 10%.

Incremental rate of return of alternatives Y and X can be calculated as follows: Since alternative 1 has 2-year life time, which is less than alternative 2, its cash flow has to be repeated for two years.

$\begin{array}{c}-\left(\text{FQ}-\text{FP}\right)=\left(\begin{array}{l}\text{ACQ}\\ -\text{ACP}\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{nQ}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{nQ}}}\right)-\frac{\left(\text{SVP}+\text{FP}\right)}{{\left(1+\text{i}\right)}^{\text{nP}}}+\frac{\text{SVQ}-\text{SVP}}{{\left(1+\text{i}\right)}^{\text{nQ}}}\\ -\left(\begin{array}{l}-35,000\\ -\left(-18,000\right)\end{array}\right)=\left(\begin{array}{l}-\text{3,600}\\ -\left(-4,000\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{6}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{6}}}\right)-\frac{\left(\begin{array}{l}\text{1,000}\\ +\left(-18,000\right)\end{array}\right)}{{\left(1+\text{i}\right)}^{\text{3}}}+\frac{\left(\begin{array}{l}\text{2,700}\\ -\text{1,000}\end{array}\right)}{{\left(1+\text{i}\right)}^{\text{6}}}\\ 17,000=400\left(\frac{{\left(1+\text{i}\right)}^{\text{6}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{6}}}\right)+\frac{17,000}{{\left(1+\text{i}\right)}^{\text{3}}}+\frac{1,700}{{\left(1+\text{i}\right)}^{\text{6}}}\end{array}$

Substitute the incremental rate of return as 6% by trial-and-error method in the above equation.

$\begin{array}{l}17,000=400\left(\frac{{\left(1+\text{0}\text{.06}\right)}^{\text{6}}-1}{\text{0}\text{.06}{\left(1+\text{0}\text{.06}\right)}^{\text{6}}}\right)+\frac{17,000}{{\left(1+\text{0}\text{.06}\right)}^{\text{3}}}+\frac{1,700}{{\left(1+\text{0}\text{.06}\right)}^{\text{6}}}\\ 17,000=400\left(\frac{1.4185-1}{\text{0}\text{.06}\left(1.4185\right)}\right)+\frac{17,000}{1.191}+\frac{1,700}{\left(1.4185\right)}\\ 17,000=400\left(\frac{0.4185}{0.0851}\right)+\frac{17,000}{1.191}+\frac{1,700}{\left(1.4185\right)}\\ 17,000=400\left(4.9177\right)+14,273.72+1,198.45\\ 17,000=1,967.08+14,273.72+1,198.45\\ 17,000<17,439.25\end{array}$

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 6.84%.

$\begin{array}{l}17,000=400\left(\frac{{\left(1+\text{0}\text{.0684}\right)}^{\text{6}}-1}{\text{0}\text{.0684}{\left(1+\text{0}\text{.0684}\right)}^{\text{6}}}\right)+\frac{17,000}{{\left(1+\text{0}\text{.0684}\right)}^{\text{3}}}+\frac{1,700}{{\left(1+\text{0}\text{.0684}\right)}^{\text{6}}}\\ 17,000=400\left(\frac{1.4873-1}{\text{0}\text{.0684}\left(1.4873\right)}\right)+\frac{17,000}{1.2196}+\frac{1,700}{\left(1.4873\right)}\\ 17,000=400\left(\frac{0.4873}{0.1017}\right)+\frac{17,000}{1.2196}+\frac{1,700}{\left(1.4873\right)}\\ 17,000=400\left(4.7915\right)+13,938.99+1,143.01\\ 17,000=1,916.6+13,938.99+1,143.01\\ 17,000\approx 16,998.6\end{array}$

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is 0.0684%. Since the incremental rate of return is less than MARR, select the alternative P (Initial alternative).

(b):

To determine

Calculate the annual worth.

Explanation of Solution

Annual worth of project P (AW) can be calculated as follows:

$\begin{array}{c}\text{AW}=\text{F}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)+\text{AC}+\text{SV}\left(\frac{\text{i}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\\ =-\text{18,000}\left(\frac{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{3}}}{{\left(1+\text{0}\text{.1}\right)}^{\text{3}}-1}\right)-4\text{,000}+\text{1,000}\left(\frac{\text{0}\text{.1}}{{\left(1+\text{0}\text{.1}\right)}^{\text{3}}-1}\right)\\ =-\text{18,000}\left(\frac{\text{0}\text{.1}\left(1.331\right)}{1.331-1}\right)-4\text{,000}+\text{1,000}\left(\frac{\text{0}\text{.1}}{1.331-1}\right)\\ =-\text{18,000}\left(\frac{\text{0}\text{.1331}}{0.331}\right)-4\text{,000}+\text{1,000}\left(\frac{\text{0}\text{.1}}{0.331}\right)\\ =-\text{18,000}\left(0.402115\right)-4\text{,000}+\text{1,000}\left(0.302115\right)\\ =-7,238.07-4\text{,000}+302.12\\ =-10,935.96\end{array}$

Annual worth of alternative P is -$10,935.96.

Annual worth of project Q (AW) can be calculated as follows:

$\begin{array}{c}\text{AW}=\text{F}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)+\text{AC}+\text{SV}\left(\frac{\text{i}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\\ =-\text{35,000}\left(\frac{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{6}}}{{\left(1+\text{0}\text{.1}\right)}^{\text{6}}-1}\right)-3,6\text{00}+\text{2,700}\left(\frac{\text{0}\text{.1}}{{\left(1+\text{0}\text{.1}\right)}^{\text{6}}-1}\right)\\ =-\text{35,000}\left(\frac{\text{0}\text{.1}\left(1.771561\right)}{1.771561-1}\right)-3,6\text{00}+\text{2,700}\left(\frac{\text{0}\text{.1}}{1.771561-1}\right)\\ =-\text{35,000}\left(\frac{\text{0}\text{.177156}}{0.771561}\right)-3,6\text{00}+\text{2,700}\left(\frac{\text{0}\text{.1}}{0.771561}\right)\\ =-\text{35,000}\left(0.229607\right)-3,6\text{00}+\text{2,700}\left(0.129607\right)\\ =-8,036.25-3,6\text{00}+349.94\\ =-11,286.31\end{array}$

Annual worth of alternative Q is -$11,286.31. Since the annual worth of alternative P is greater than alternate Q, select alternate P.