Chapter 18, Problem 1CS1

To determine

Calculate the present worth analysis.

Explanation of Solution

Time period is denoted by ‘n’ and MARR is denoted by ‘i’. Time period 1 (n) is 40. Time period 2 (n1) is 20.

The present worth (PW) of Plan A can be calculated using the following formula:

$\text{PW}=\text{First cost}+\text{AOC}\times \left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\text{Salvage value}\times \left(\frac{1}{{\left(1+\text{i}\right)}^{\text{n}}}\right)$        (1)

Substitute the respective values in Equation (1) to calculate the present worth of Plan A with the actual time period.

$\begin{array}{c}\text{PW}=-10,000-500\left(\frac{{\left(1+0.\text{1}\right)}^{\text{40}}-1}{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{40}}}\right)+\text{1,000}\times \left(\frac{1}{{\left(1+\text{0}\text{.1}\right)}^{\text{40}}}\right)\\ =-10,000-500\left(9.7791\right)+\text{1,000}\times \left(0.0221\right)\\ =-14,867\end{array}$

Thus, the present worth is -$14,867.

Location is denoted by ‘L’. Present worth (PW) of Plan B can be calculated using the following formula:

$\text{PW}=\left(\begin{array}{l}{\text{First cost}}_{\text{L 1}}+{\text{AOC}}_{\text{L 1}}\times \left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+{\text{Salvage value}}_{\text{L 1}}\times \left(\frac{1}{{\left(1+\text{i}\right)}^{\text{n}}}\right)\\ +{\text{First cost}}_{\text{L 2}}+{\text{AOC}}_{\text{L 2}}\times \left(\frac{1}{{\left(1+\text{i}\right)}^{\text{n1}}}\right)+{\text{First cost}}_{\text{L 2}}\left(\frac{1}{{\left(1+\text{i}\right)}^{\text{n1}}}\right)\\ +{\text{Salvage value}}_{\text{L 2}}\times \left(\frac{1}{{\left(1+\text{i}\right)}^{\text{n}}}\right)+{\text{AOC}}_{\text{L 2}}\times \left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)\end{array}\right)$        (2)

Substitute the respective values in Equation (1) to calculate the present worth of Plan B with the actual time period.

$\begin{array}{c}\text{PW}=\left(\begin{array}{l}-30,000-100\times \left(\frac{{\left(1+\text{0}\text{.1}\right)}^{\text{40}}-1}{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{40}}}\right)+5000\left(\frac{1}{{\left(1+\text{0}\text{.1}\right)}^{\text{40}}}\right)\\ -5,000-200\times \left(\frac{1}{{\left(1+\text{0}\text{.1}\right)}^{\text{20}}}\right)-5,000\left(\frac{1}{{\left(1+\text{0}\text{.1}\right)}^{\text{20}}}\right)\\ -200\times \left(\frac{1}{{\left(1+\text{0}\text{.1}\right)}^{\text{40}}}\right)-200\times \left(\frac{{\left(1+\text{0}\text{.1}\right)}^{\text{40}}-1}{\text{0}\text{.1}{\left(1+\text{0}\text{.1}\right)}^{\text{40}}}\right)\end{array}\right)\\ =\left(\begin{array}{l}-30,000-100\times \left(9.7791\right)+5000\left(0.0221\right)\\ -5,000-200\times \left(0.1486\right)-5,000\left(0.1486\right)\\ -200\times \left(0.0221\right)-200\times \left(9.7791\right)\end{array}\right)\\ =-38,600\end{array}$

Thus, the present worth is -$38,600.

Table 1 shows the value of present worth by changing the time period to different percentages that is obtained using Equations (1) and (2):

Table 1

MARR	PW A	PW B
0.1	-14,867	-38,600
0.07	-16,599	-40,023
0.04	-19,688	-42,311

Table 1 reveals that at different levels of MARR, the present worth of alternate A is greater than alternate B. Thus, there is no sensitivity. Plan A is selected for different levels of MARR.