Chapter 8, Problem 38P

(a):

To determine

Calculate the missing value.

Explanation of Solution

Table -1 shows the cash flow.

Table -1

Alternative	1	2	3	4
Initial cost (F)	-	-60,000	-72,000	-98,000
Annual cost (AC)	-70,000	-64,000	-61,000	-58,000
Annual saving (AS)	80,000	80,000	80,000	80,000
Overall rate of return (i)	18.6%	-	23.1%	20.8%
Comparing alternative		2 to 1	3 to 2	4 to 3
Incremental investment (II)		-16,000	-	-26,000
Incremental cash flow per year (IC)		6,000	3,00	-
Incremental rate of return (ir)		35.7%	-	-
Time period (n)	10	10	10	10

Initial cost of machine 1 can be calculated as follows.

$\begin{array}{c}\text{Initial cost 1}=\text{Inital cost 2}-\text{Incremental investment 2}\\ =-60,000-\left(-16,000\right)\\ =-44,000\end{array}$

Thus, the initial cost of alternate 1 is -44,000.

Overall rate of return for alternate 2 can be calculated as follows.

$\begin{array}{c}-\text{First cost}=\left(\begin{array}{l}\text{Annual saving}\\ -\text{Annual cost}\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)\\ -\left(-60,000\right)=\left(-\text{64,000}-\left(-80,000\right)\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)\\ 60,000=16,000\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)\\ \left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{10}}}\right)=\frac{60,000}{16,000}\\ \left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{10}}}\right)=3.75\end{array}$

Substitute the incremental rate of return as 23% by trial and error method in the above equation.

$\begin{array}{c}\left(\frac{{\left(1+\text{0}\text{.23}\right)}^{\text{10}}-1}{\text{0}\text{.23}{\left(1+\text{0}\text{.23}\right)}^{\text{10}}}\right)=3.75\\ \left(\frac{7.925946-1}{\text{0}\text{.23}\left(7.925946\right)}\right)=3.75\\ \left(\frac{6.925946}{1.822968}\right)=3.75\\ 3.799>3.75\end{array}$

The calculated value is greater than the present value factor to the annual value. Thus, increase the incremental rate of return to 23.41%.

$\begin{array}{c}\left(\frac{{\left(1+\text{0}\text{.2341}\right)}^{\text{10}}-1}{\text{0}\text{.2341}{\left(1+\text{0}\text{.2341}\right)}^{\text{10}}}\right)=3.75\\ \left(\frac{8.194143-1}{\text{0}\text{.2341}\left(8.194143\right)}\right)=3.75\\ \left(\frac{7.194143}{1.918249}\right)=3.75\\ 3.75=3.75\end{array}$

The calculated value is equal to the present value factor to the annual value. Thus, it is confirmed that the incremental rate of return is 23.41%.

Incremental investment between alternate 3 and 2 can be calculated as follows.

$\begin{array}{c}\text{Increemntal cash flow 3 to 2}=\text{Inital investment 2}-\text{Inital investment 3}\\ =-72,000-\left(-60,000\right)\\ =-72,000+60,000\\ =-12,000\end{array}$

Incremental cash flow 3 to 2 is -$12,000.

Incremental investment cash flow per year between alternate 4 and 3 can be calculated as follows.

$\begin{array}{c}\text{Increemntal cash flow per year 4 to 3}=\left(\begin{array}{l}\text{Annual saving 4}\\ +\text{Annual cost 4}\end{array}\right)-\left(\begin{array}{l}\text{Annual saving 3}\\ +\text{Annual cost 3}\end{array}\right)\\ =\left(82,000+\left(-58,000\right)\right)-\left(80,000+\left(-61,000\right)\right)\\ =24,000-19,000\\ =5,000\end{array}$

Incremental cash flow per year 4 to 3 is $5,000.

Incremental rate of return between alternate 3 to 2 can be calculated as follows.

$\begin{array}{c}-\left(\begin{array}{l}\text{First cost 3}\\ -\text{First cost 2}\end{array}\right)=\left(\left(\begin{array}{l}\text{Annual saving 3}\\ +\text{Annual cost 3}\end{array}\right)-\left(\begin{array}{l}\text{Annual saving 2}\\ +\text{Annual cost 2}\end{array}\right)\right)\left(\frac{{\left(1+\text{ir}\right)}^{\text{n}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{n}}}\right)\\ -\left(\begin{array}{l}-72,000\\ -\left(-60,000\right)\end{array}\right)=\left(\begin{array}{l}\left(-\text{61,000}+80,000\right)\\ -\left(-\text{64,000}+80,000\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{10}}}\right)\\ 12,000=\left(19,000-\left(16,000\right)\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)\\ 12,000=3,000\left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)\\ \left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)=\frac{12,000}{3,000}\\ \left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)=4\end{array}$

Substitute the incremental rate of return as 21% by trial and error method in the above equation.

$\begin{array}{c}\left(\frac{{\left(1+\text{0}\text{.21}\right)}^{\text{10}}-1}{\text{0}\text{.21}{\left(1+\text{0}\text{.21}\right)}^{\text{10}}}\right)=4\\ \left(\frac{6.7275-1}{\text{0}\text{.21}\left(6.7275\right)}\right)=4\\ \left(\frac{5.7275}{1.4128}\right)=4\\ 4.05>4\end{array}$

The calculated value is greater than the present value factor to the annual value. Thus, increase the incremental rate of return to 21.41%.

$\begin{array}{c}\left(\frac{{\left(1+\text{0}\text{.2141}\right)}^{\text{10}}-1}{\text{0}\text{.2141}{\left(1+\text{0}\text{.21}\right)}^{\text{10}}}\right)=4\\ \left(\frac{6.958964-1}{\text{0}\text{.2141}\left(6.958964\right)}\right)=4\\ \left(\frac{5.958964}{1.489914}\right)=4\\ 4=4\end{array}$

The calculated value is equal to the present value factor to the annual value. Thus, it is confirmed that the incremental rate of return is 21.41%.

Incremental rate of return between alternate 4 to 3 can be calculated as follows.

$\begin{array}{c}-\left(\begin{array}{l}\text{First cost 4}\\ -\text{First cost 3}\end{array}\right)=\left(\left(\begin{array}{l}\text{Annual saving 4}\\ +\text{Annual cost 4}\end{array}\right)-\left(\begin{array}{l}\text{Annual saving 3}\\ +\text{Annual cost 3}\end{array}\right)\right)\left(\frac{{\left(1+\text{ir}\right)}^{\text{n}}-1}{\text{ir}{\left(1+\text{ir}\right)}^{\text{n}}}\right)\\ -\left(\begin{array}{l}-98,000\\ -\left(-72,000\right)\end{array}\right)=\left(\begin{array}{l}\left(-\text{58,000}+82,000\right)\\ -\left(-\text{61,000}+80,000\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{10}}}\right)\\ 26,000=\left(24,000-\left(19,000\right)\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)\\ 26,000=5,000\left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)\\ \left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)=\frac{26,000}{5,000}\\ \left(\frac{{\left(1+\text{i}\right)}^{\text{10}}-1}{\text{i}{\left(1+\text{i}\right)}^{10}}\right)=5.2\end{array}$

Substitute the incremental rate of return as 14% by trial and error method in the above equation.

$\begin{array}{c}\left(\frac{{\left(1+\text{0}\text{.14}\right)}^{\text{10}}-1}{\text{0}\text{.14}{\left(1+\text{0}\text{.14}\right)}^{\text{10}}}\right)=5.2\\ \left(\frac{3.707221-1}{\text{0}\text{.14}\left(3.707221\right)}\right)=5.2\\ \left(\frac{2.707221}{0.519011}\right)=5.2\\ 5.22>5.2\end{array}$

The calculated value is greater than the present value factor to the annual value. Thus, increase the incremental rate of return to 14.08%.

$\begin{array}{c}\left(\frac{{\left(1+\text{0}\text{.1408}\right)}^{\text{10}}-1}{\text{0}\text{.1408}{\left(1+\text{0}\text{.1408}\right)}^{\text{10}}}\right)=5.2\\ \left(\frac{3.733319-1}{\text{0}\text{.1408}\left(3.733319\right)}\right)=5.2\\ \left(\frac{2.733319}{0.525651}\right)=5.2\\ 5.2=5.2\end{array}$

The calculated value is equal to the present value factor to the annual value. Thus, it is confirmed that the incremental rate of return is 14.08%.

(b):

To determine

Selection of alternate.

Explanation of Solution

MARRR is 18%.

Since the alternatives are mutually exclusive, should select only one alternative.

The overall rate of return cans be considered as incremental rate of return from the particular alternative with the initial alternate Do nothing.

Incremental rate of return between initial alternate Do nothing and new alternate 1 is 18.6%. Since the incremental rate of return is greater than the MARR, select the alternate 1 and eliminate the alternate Do nothing.

Incremental rate of return between initial alternate 1 and new alternate 2 is 35.7%. Since the incremental rate of return is greater than the MARR, select the alternate 2 and eliminate the alternate 1.

Incremental rate of return between initial alternate 2 and new alternate 3 is 21.41%. Since the incremental rate of return is greater than the MARR, select the alternate 3 and eliminate the alternate 2.

Incremental rate of return between initial alternate 3 and new alternate 4 is 14.08%. Since the incremental rate of return is less than the MARR, select the alternate 3 and eliminate the alternate 4.

Thus, ultimately select the alternate 3.