Chapter 3, Problem 2CS

To determine

Calculate the equivalent annual value.

Explanation of Solution

Table 1 shows the cash flow of two projects that described the case as follows:

Year	Land	Park
1	4,000,000
2	3,000,000
3	2,250,000
4	1,265,000	550,000
5	949,219	650,000
6		750,000

Interest rate is 7%. CFL_n indicates the cash flow of project land in the year n. CFL_P indicates the cash flow of Project Park in the year n. Time period (n) for this calculation is 3 years. Revenue collected from bond (B) is $3,000,000.

The equivalent annual value (A) for each in 2 years can be calculated as follows:

$\begin{array}{c}\text{A}=\left(\begin{array}{l}\text{CFL0}+\frac{\text{CFL1}}{{\left(1+\text{i}\right)}^1}+\frac{\text{CFL2}}{{\left(1+\text{i}\right)}^2}+\frac{\text{CFL3}}{{\left(1+\text{i}\right)}^3}\\ +\frac{\text{CFL4}}{{\left(1+\text{i}\right)}^4}+\frac{\text{CFL5}}{{\left(1+\text{i}\right)}^5}+\frac{\text{CFP4}}{{\left(1+\text{i}\right)}^4}\\ +\frac{\text{CFP5}}{{\left(1+\text{i}\right)}^5}+\frac{\text{CFP6}}{{\left(1+\text{i}\right)}^6}-\text{B}\end{array}\right){\left(1+\text{i}\right)}^{\text{n}}\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\\ =\left(\begin{array}{l}\text{4,000,000}+\frac{\text{3,000,000}}{{\left(1+\text{0}\text{.07}\right)}^1}+\frac{\text{2,250,000}}{{\left(1+\text{0}\text{.07}\right)}^2}\\ +\frac{\text{1,678,000}}{{\left(1+\text{0}\text{.07}\right)}^3}+\frac{\text{1,265,625}}{{\left(1+\text{0}\text{.07}\right)}^4}+\frac{\text{949,219}}{{\left(1+\text{0}\text{.07}\right)}^5}\\ +\frac{\text{550,000}}{{\left(1+\text{0}\text{.07}\right)}^4}+\frac{\text{650,000}}{{\left(1+\text{0}\text{.07}\right)}^5}+\frac{\text{750,000}}{{\left(1+\text{0}\text{.07}\right)}^6}\\ -3,000,000\end{array}\right){\left(1+\text{0}\text{.07}\right)}^3\left(\frac{\text{0}\text{.07}{\left(1+\text{0}\text{.07}\right)}^3}{{\left(1+\text{0}\text{.07}\right)}^3-1}\right)\end{array}$

$\begin{array}{c}=\left(\begin{array}{l}\text{4,000,000}+\frac{\text{3,000,000}}{1.07}+\frac{\text{2,250,000}}{1.1449}\\ +\frac{\text{1,678,000}}{1.225043}+\frac{\text{1,265,625}}{1.310796}+\frac{\text{949,219}}{1.402552}\\ +\frac{\text{550,000}}{1.310796}+\frac{\text{650,000}}{1.402552}+\frac{\text{750,000}}{1.50073}\\ -3,000,000\end{array}\right)\left(1.225043\right)\left(\frac{\text{0}\text{.07}\left(1.225043\right)}{1.225043-1}\right)\\ =\left(\begin{array}{l}\text{4,000,000}+2,803,738.3178\\ +1,965,237.1386+1,369,747.8374\\ +965,539.2601+676,779.8984\\ +419,592.3698+463,440.9277\\ +499,756.785-3,000,000\end{array}\right)\left(1.225043\right)\left(\frac{0.08575301}{0.225043}\right)\\ =10,163,832.534\left(1.225043\right)\left(0.381052\right)\\ =4,744,528.71\end{array}$

The equivalent annual value is $4,744,528.71.