Chapter 3, Problem 53P

To determine

Calculate the present value.

Explanation of Solution

The initial cost (C) is 1,800,000 for the first 4 years. Time period 1 (n1) is 3. From year 6, the cost decreases (d) by $30,000 till 10^th year. Time period (n) is 10. Interest rate (i) is 12%.

The present value (A) can be calculated as follows:

$\begin{array}{c}\text{P}=\text{C}\left(\frac{{\left(1+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n1}}}\right)+\left(\text{C}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}-\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}-\text{n1}}}\right)-\text{d}\times \frac{1}{\text{i}}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}-\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}-\text{n1}}}-\frac{\text{n}-\text{n1}}{{\left(1+\text{i}\right)}^{\text{n}-\text{n1}}}\right)\right)\left(\frac{1}{{\left(1+\text{i}\right)}^{\text{n1}}}\right)\\ =\text{1,800,000}\left(\frac{{\left(1+\text{0}\text{.12}\right)}^{\text{3}}-1}{\text{0}\text{.12}{\left(1+\text{0}\text{.12}\right)}^{\text{3}}}\right)+\left(\begin{array}{l}\text{1,800,000}\left(\frac{{\left(1+\text{0}\text{.12}\right)}^{\text{10}-\text{3}}-1}{\text{0}\text{.12}{\left(1+\text{0}\text{.12}\right)}^{\text{10}-\text{3}}}\right)\\ -\text{30,000}\times \frac{1}{\text{0}\text{.12}}\left(\begin{array}{l}\frac{{\left(1+\text{0}\text{.12}\right)}^{\text{10}-\text{3}}-1}{\text{0}\text{.12}{\left(1+\text{0}\text{.12}\right)}^{\text{10}-\text{3}}}\\ -\frac{10-3}{{\left(1+\text{0}\text{.12}\right)}^{\text{10}-\text{3}}}\end{array}\right)\end{array}\right)\left(\frac{1}{{\left(1+\text{0}\text{.12}\right)}^{\text{3}}}\right)\\ =\text{1,800,000}\left(\frac{1.404928-1}{\text{0}\text{.12}\left(1.404928\right)}\right)+\left(\begin{array}{l}\text{1,800,000}\left(\frac{2.210681-1}{\text{0}\text{.12}\left(2.210681\right)}\right)\\ -\text{30,000}\times 8.3333\left(\begin{array}{l}\frac{2.210681-1}{\text{0}\text{.12}\left(2.210681\right)}\\ -\frac{7}{2.210681}\end{array}\right)\end{array}\right)\left(\frac{1}{1.404928}\right)\\ =\text{1,800,000}\left(\frac{0.404928}{0.168591}\right)+\left(\begin{array}{l}\text{1,800,000}\left(\frac{1.210681}{0.265282}\right)\\ -\text{30,000}\times 8.3333\left(\frac{1.210681}{0.265282}-3.1664451\right)\end{array}\right)\left(0.71178\right)\\ =\text{1,800,000}\left(2.401836\right)\left(\begin{array}{l}+\text{1,800,000}\left(4.563751\right)\\ -\text{30,000}\times 8.3333\left(4.563751-3.1664451\right)\end{array}\right)\left(0.71178\right)\\ =4,323,304.8+\left(8,214,751.8-\text{30,000}\times 8.3333\left(1.397306\right)\right)\left(0.71178\right)\\ =4,323,304.8+\left(8,214,751.8-349,325.1027\right)\left(0.71178\right)\\ =4,323,304.8+5,598,453.4146\\ =9,921,758.21\end{array}$

The present value is $9,921,758.21.