Chapter 8, Problem 26P

To determine

Calculate incremental rate of return.

Explanation of Solution

Table-1 shows the cash flow (Values are in 1,000 units).

Table -1

Alternative	T	I
First cost (F)	-5,000	-6,500
AOC (AC) per year	-1,000	-650
Expected revenue (R) per year	2,500	2,500
Salvage value (SV)	100	200
Time period (n)	5	5

MARR is 25%.

Incremental rate of return of alternatives T and Do nothing can be calculated as follows:

$\begin{array}{l}\text{F}=\left(\text{R}+\text{AC}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SV}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ \text{5,000}=\left(\text{2,500}+\left(-1,000\right)\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)+\frac{\text{100}}{{\left(1+\text{i}\right)}^{\text{5}}}\\ \text{5,000}=1,500\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)+\frac{\text{100}}{{\left(1+\text{i}\right)}^{\text{5}}}\end{array}$

Substitute the incremental rate of return as 15% by trial-and-error method in the above equation.

$\begin{array}{l}\text{5,000}=1,500\left(\frac{{\left(1+\text{0}\text{.15}\right)}^{\text{5}}-1}{\text{0}\text{.15}{\left(1+\text{0}\text{.15}\right)}^{\text{5}}}\right)+\frac{\text{100}}{{\left(1+\text{0}\text{.15}\right)}^{\text{5}}}\\ \text{5,000}=1,500\left(\frac{2.011357-1}{\text{0}\text{.15}\left(2.011357\right)}\right)+\frac{\text{100}}{2.011357}\\ \text{5,000}=1,500\left(\frac{1.011357}{0.301704}\right)+\frac{\text{100}}{2.011357}\\ \text{5,000}=1,500\left(3.35215\right)+49.72\\ \text{5,000}=5,028.23+49.72\\ \text{5,000}=5,077.95\end{array}$

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 15.65%.

$\begin{array}{l}\text{5,000}=1,500\left(\frac{{\left(1+\text{0}\text{.1565}\right)}^{\text{5}}-1}{\text{0}\text{.1565}{\left(1+\text{0}\text{.1565}\right)}^{\text{5}}}\right)+\frac{\text{100}}{{\left(1+\text{0}\text{.1565}\right)}^{\text{5}}}\\ \text{5,000}=1,500\left(\frac{2.068846-1}{\text{0}\text{.1565}\left(2.068846\right)}\right)+\frac{\text{100}}{2.068846}\\ \text{5,000}=1,500\left(\frac{1.068846}{0.323774}\right)+\frac{\text{100}}{2.068846}\\ \text{5,000}=1,500\left(3.30121\right)+48.34\\ \text{5,000}=4,951.82+49.72\\ \text{5,000}\approx 5,001.54\end{array}$

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is ~15.65%. Since the incremental rate of return is less than MARR, reject the alternative T and select the do nothing.

Incremental rate of return of alternatives I and Do nothing can be calculated as follows:

$\begin{array}{l}\text{F}=\left(\text{R}+\text{AC}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SV}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ \text{6,500}=\left(\text{2,500}+\left(-650\right)\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)+\frac{\text{200}}{{\left(1+\text{i}\right)}^{\text{5}}}\\ \text{6,500}=1,850\left(\frac{{\left(1+\text{i}\right)}^{\text{5}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{5}}}\right)+\frac{\text{200}}{{\left(1+\text{i}\right)}^{\text{5}}}\end{array}$

Substitute the incremental rate of return as 13% by trial-and-error method in the above equation.

$\begin{array}{l}\text{6,500}=1,850\left(\frac{{\left(1+\text{0}\text{.13}\right)}^{\text{5}}-1}{\text{0}\text{.13}{\left(1+\text{0}\text{.13}\right)}^{\text{5}}}\right)+\frac{\text{200}}{{\left(1+\text{0}\text{.13}\right)}^{\text{5}}}\\ \text{6,500}=1,850\left(\frac{1.842435-1}{\text{0}\text{.13}\left(1.842435\right)}\right)+\frac{\text{200}}{1.842435}\\ \text{6,500}=1,850\left(\frac{0.842435}{0.239517}\right)+\frac{\text{200}}{1.842435}\\ \text{6,500}=1,850\left(3.517224\right)+108.55\\ \text{6,500}=6,506.86+108.55\\ \text{6,500}<6,615.41\end{array}$

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 13.72%.

$\begin{array}{l}\text{6,500}=1,850\left(\frac{{\left(1+\text{0}\text{.1372}\right)}^{\text{5}}-1}{\text{0}\text{.1372}{\left(1+\text{0}\text{.1372}\right)}^{\text{5}}}\right)+\frac{\text{200}}{{\left(1+\text{0}\text{.1372}\right)}^{\text{5}}}\\ \text{6,500}=1,850\left(\frac{1.901885-1}{\text{0}\text{.1372}\left(1.901885\right)}\right)+\frac{\text{200}}{1.901885}\\ \text{6,500}=1,850\left(\frac{0.901885}{0.260939}\right)+\frac{\text{200}}{1.901885}\\ \text{6,500}=1,850\left(3.456306\right)+105.16\\ \text{6,500}=6,394.17+108.55\\ \text{6,500}\approx 6,502.72\end{array}$

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is 13.72%. Since the incremental rate of return is less than MARR, reject the alternative I and select do nothing.