Chapter 8, Problem 23P

(a):

To determine

Calculate the rate of return.

Explanation of Solution

Alternative X: First cost (F) is $84,000. Annual maintenance cost (AC) is $31,000 per year. Salvage value (SV) is $40,000. Annual revenue (AR) is $96,000per year.

Alternative Y: First cost (F) is $146,000. Annual maintenance cost (AC) is $28,000 per year. Salvage value (SV) is $47,000. Annual revenue (AR) is $119,000per year.

MARR is 15%. Time period (n) is 3.

Rate of return of alternative X (i) can be calculated as follows:

$\begin{array}{c}\text{FX}=\left(\text{ARX}-\text{ACX}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SVX}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ \text{84,000}=\left(96,000-\text{31,000}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{40,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\\ \text{84,000}=65,000\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{40,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\end{array}$

Substitute the rate of return as 65% by trial-and-error method in the above equation.

$\begin{array}{l}\text{84,000}=65,000\left(\frac{{\left(1+\text{0}\text{.65}\right)}^{\text{3}}-1}{\text{0}\text{.65}{\left(1+\text{0}\text{.65}\right)}^{\text{3}}}\right)+\frac{\text{40,000}}{{\left(1+\text{0}\text{.65}\right)}^{\text{3}}}\\ \text{84,000}=65,000\left(\frac{4.492125-1}{\text{0}\text{.65}\left(4.492125\right)}\right)+\frac{\text{40,000}}{4.492125}\\ \text{84,000}=65,000\left(\frac{3.492125}{2.91988125}\right)+8,904.47\\ \text{84,000}=65,000\left(1.19598\right)+8,904.47\\ \text{84,000}=77,738.7+8,904.47\\ \text{84,000}<86,643.17\end{array}$

The calculated value is greater than the present value of the first cost. Thus, increase the rate of return to 67.85%.

$\begin{array}{l}\text{84,000}=65,000\left(\frac{{\left(1+\text{0}\text{.6785}\right)}^{\text{3}}-1}{\text{0}\text{.6785}{\left(1+\text{0}\text{.6785}\right)}^{\text{3}}}\right)+\frac{\text{40,000}}{{\left(1+\text{0}\text{.6785}\right)}^{\text{3}}}\\ \text{84,000}=65,000\left(\frac{4.728943-1}{\text{0}\text{.6785}\left(4.728943\right)}\right)+\frac{\text{40,000}}{4.728943}\\ \text{84,000}=65,000\left(\frac{3.728943}{3.208588}\right)+\frac{\text{40,000}}{4.728943}\\ \text{84,000}=65,000\left(1.162176\right)+8,458.55\\ \text{84,000}=75,541.44+8,458.55\\ \text{84,000}\approx 83,999.99\end{array}$

The calculated value is nearly equal to the present value of the first cost. Thus, it is confirmed that the rate of return is 67.85%.

Rate of return of alternative Y (i) can be calculated as follows:

$\begin{array}{c}\text{FY}=\left(\text{ARY}-\text{ACY}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SVY}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ \text{146,000}=\left(119,000-\text{28,000}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{47,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\\ \text{146,000}=91,000\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{47,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\end{array}$

Substitute the rate of return as 47% by trial-and-error method in the above equation.

$\begin{array}{l}\text{146,000}=91,000\left(\frac{{\left(1+\text{0}\text{.47}\right)}^{\text{3}}-1}{\text{0}\text{.47}{\left(1+\text{0}\text{.47}\right)}^{\text{3}}}\right)+\frac{\text{47,000}}{{\left(1+\text{0}\text{.47}\right)}^{\text{3}}}\\ \text{146,000}=91,000\left(\frac{3.176523-1}{\text{0}\text{.47}\left(3.176523\right)}\right)+\frac{\text{47,000}}{3.176523}\\ \text{146,000}=91,000\left(\frac{2.176523}{1.49296581}\right)+\frac{\text{47,000}}{3.176523}\\ \text{146,000}=91,000\left(1.457852\right)+14,796.05\\ \text{146,000}=132,664.53+14,796.05\\ \text{146,000}<147,460.58\end{array}$

The calculated value is greater than the present value of the first cost. Thus, increase the rate of return to 47.78%.

$\begin{array}{l}\text{146,000}=91,000\left(\frac{{\left(1+\text{0}\text{.4778}\right)}^{\text{3}}-1}{\text{0}\text{.4778}{\left(1+\text{0}\text{.4778}\right)}^{\text{3}}}\right)+\frac{\text{47,000}}{{\left(1+\text{0}\text{.4778}\right)}^{\text{3}}}\\ \text{146,000}=91,000\left(\frac{3.227357-1}{\text{0}\text{.4778}\left(3.227357\right)}\right)+\frac{\text{47,000}}{3.227357}\\ \text{146,000}=91,000\left(\frac{2.227357}{1.542031}\right)+\frac{\text{47,000}}{3.227357}\\ \text{146,000}=91,000\left(1.444431\right)+14,796.05\\ \text{146,000}=131,443.22+14,563\\ \text{146,000}\approx 146,006.22\end{array}$

The calculated value is nearly equal to the present value of the first cost. Thus, it is confirmed that the rate of return is 47.78%.

Both the rates of returns are greater than MARR. Since the rate of return for X is greater, alternate X should be selected.

(b):

To determine

Calculate incremental rate of return.

Explanation of Solution

Incremental rate of return of alternatives Y and X can be calculated as follows:

$\begin{array}{c}\left(\text{FY}-\text{FX}\right)=\left(\begin{array}{l}\left(\text{ARY}-\text{ARX}\right)\\ +\left(\text{ACY}-\text{ACX}\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SVY}-\text{SVX}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ \left(\begin{array}{l}\text{146,000}\\ -\text{84,000}\end{array}\right)=\left(\begin{array}{l}\left(119,000-96,000\right)\\ -\left(\text{28,000}-\text{31,000}\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{47,000}-\text{40,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\\ 62,000=\left(\begin{array}{l}23,000\\ -\left(-3,000.\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\\ 62,000=26,000\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\end{array}$

Substitute the incremental rate of return as 16% by trial-and-error method in the above equation.

$\begin{array}{l}62,000=26,000\left(\frac{{\left(1+\text{0}\text{.16}\right)}^{\text{3}}-1}{\text{0}\text{.16}{\left(1+\text{0}\text{.16}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{0}\text{.16}\right)}^{\text{3}}}\\ 62,000=26,000\left(\frac{1.560896-1}{\text{0}\text{.16}\left(1.560896\right)}\right)+\frac{\text{7,000}}{1.560896}\\ 62,000=26,000\left(\frac{0.560896}{0.24974336}\right)+\frac{\text{7,000}}{1.560896}\\ 62,000=26,000\left(2.24589\right)+4,484.6\\ 62,000=58,393.14+4,484.6\\ 62,000<62,877.74\end{array}$

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 16.83%.

$\begin{array}{l}62,000=26,000\left(\frac{{\left(1+\text{0}\text{.1683}\right)}^{\text{3}}-1}{\text{0}\text{.1683}{\left(1+\text{0}\text{.1683}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{0}\text{.1683}\right)}^{\text{3}}}\\ 62,000=26,000\left(\frac{1.594642-1}{\text{0}\text{.1683}\left(1.594642\right)}\right)+\frac{\text{7,000}}{1.594642}\\ 62,000=26,000\left(\frac{0.594642}{0.268378}\right)+\frac{\text{7,000}}{1.594642}\\ 62,000=26,000\left(2.215688\right)+4,389.7\\ 62,000=57,607.89+4,389.7\\ 62,000\approx 61,997.59\end{array}$

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is 16.83%. Since the incremental rate of return is greater than MARR, alternate Y must be selected.

(c):

To determine

Calculate incremental rate of return.

Explanation of Solution

The procedure used in subpart (b) is correct (Incremental rate of return). Since the procedure used in subpart (a) is incorrect, Project X should not be selected.

Incremental rate of return of alternatives Y and X can be calculated as follows:

$\begin{array}{c}\left(\text{FY}-\text{FX}\right)=\left(\begin{array}{l}\left(\text{ARY}-\text{ARX}\right)\\ +\left(\text{ACY}-\text{ACX}\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\frac{\text{SVY}-\text{SVX}}{{\left(1+\text{i}\right)}^{\text{n}}}\\ \left(\begin{array}{l}\text{146,000}\\ -\text{84,000}\end{array}\right)=\left(\begin{array}{l}\left(119,000-96,000\right)\\ -\left(\text{28,000}-\text{31,000}\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{47,000}-\text{40,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\\ 62,000=\left(\begin{array}{l}23,000\\ -\left(-3,000.\right)\end{array}\right)\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\\ 62,000=26,000\left(\frac{{\left(1+\text{i}\right)}^{\text{3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{i}\right)}^{\text{3}}}\end{array}$

Substitute the incremental rate of return as 16% by trial-and-error method in the above equation.

$\begin{array}{l}62,000=26,000\left(\frac{{\left(1+\text{0}\text{.16}\right)}^{\text{3}}-1}{\text{0}\text{.16}{\left(1+\text{0}\text{.16}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{0}\text{.16}\right)}^{\text{3}}}\\ 62,000=26,000\left(\frac{1.560896-1}{\text{0}\text{.16}\left(1.560896\right)}\right)+\frac{\text{7,000}}{1.560896}\\ 62,000=26,000\left(\frac{0.560896}{0.24974336}\right)+\frac{\text{7,000}}{1.560896}\\ 62,000=26,000\left(2.24589\right)+4,484.6\\ 62,000=58,393.14+4,484.6\\ 62,000<62,877.74\end{array}$

The calculated value is greater than the present value of the incremental first cost. Thus, increase the incremental rate of return to 16.83%.

$\begin{array}{l}62,000=26,000\left(\frac{{\left(1+\text{0}\text{.1683}\right)}^{\text{3}}-1}{\text{0}\text{.1683}{\left(1+\text{0}\text{.1683}\right)}^{\text{3}}}\right)+\frac{\text{7,000}}{{\left(1+\text{0}\text{.1683}\right)}^{\text{3}}}\\ 62,000=26,000\left(\frac{1.594642-1}{\text{0}\text{.1683}\left(1.594642\right)}\right)+\frac{\text{7,000}}{1.594642}\\ 62,000=26,000\left(\frac{0.594642}{0.268378}\right)+\frac{\text{7,000}}{1.594642}\\ 62,000=26,000\left(2.215688\right)+4,389.7\\ 62,000=57,607.89+4,389.7\\ 62,000\approx 61,997.59\end{array}$

The calculated value is nearly equal to the incremental present value. Thus, it is confirmed that the incremental rate of return is 16.83%. Since the incremental rate of return is greater than MARR, alternate Y must be selected.