Chapter 20, Problem 17P

To determine

Find the equivalent annual worth of the cash flow diagram.

Answer to Problem 17P

The equivalent annual worth of the cash flow diagram is $\$3868.37$.

Explanation of Solution

Given data:

The normal interest rate $\left(i\right)$ is $8\%$.

Formula used:

Formula to calculate the annual worth for the given cash flow diagram is,

$AW=\left[\begin{array}{l}\$5000\left(\frac{P}{A},i,n\right)\left(\frac{A}{P},i,n\right)-\$2000\left(\frac{P}{A},i,n\right)\left(\frac{P}{F},i,n\right)\left(\frac{A}{P},i,n\right)\\ +\$8000\left(\frac{P}{F},i,n\right)\left(\frac{A}{P},i,n\right)+\$5000\left(\frac{A}{F},i,n\right)\end{array}\right]$ (1)

Here,

$F$ is the future cost,

$P$ is the present cost,

$A$ is the uniform series payment,

$i$ is the normal interest rate, and

$n$ is the number of years.

Refer the Problem 20.17 Figure in the textbook, the equation (1) is rewritten with respect to the given number of years is,

$AW=\left[\begin{array}{l}\$5000\left(\frac{P}{A},i,6\right)\left(\frac{A}{P},i,10\right)-\$2000\left(\frac{P}{A},i,3\right)\left(\frac{P}{F},i,4\right)\left(\frac{A}{P},i,10\right)\\ +\$8000\left(\frac{P}{F},i,8\right)\left(\frac{A}{P},i,10\right)+\$5000\left(\frac{A}{F},i,10\right)\end{array}\right]$ (2)

Formula to calculate the present cost for the given uniform series payment is,

$\left(\frac{P}{A},i,n\right)=\left[\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right]$ (3)

Formula to calculate the uniform series payment for the given present cost is,

$\left(\frac{A}{P},i,n\right)=\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]$ (4)

Formula to calculate the present cost for the given future cost is,

$\left(\frac{P}{F},i,n\right)=\left[\frac{1}{{\left(1+i\right)}^n}\right]$ (5)

Formula to calculate the uniform series payment for the given future cost is,

$\left(\frac{A}{F},i,n\right)=\left[\frac{i}{{\left(1+i\right)}^n-1}\right]$ (6)

Calculation:

To calculate the equivalent annual worth of the cash flow diagram:

Substitute $8\%$ for $i$, and $6$ for $n$ in equation (3) to find $\left(\frac{P}{A},i,6\right)$.

$\begin{array}{c}\left(\frac{P}{A},i,6\right)=\left[\frac{{\left(1+\left(8\%\right)\right)}^6-1}{\left(8\%\right){\left(1+\left(8\%\right)\right)}^6}\right]\\ =\left[\frac{{\left(1+0.08\right)}^6-1}{0.08{\left(1+0.08\right)}^6}\right]\\ =4.623\end{array}$

Substitute $8\%$ for $i$, and $10$ for $n$ in equation (4) to find $\left(\frac{A}{P},i,10\right)$.

$\begin{array}{c}\left(\frac{A}{P},i,10\right)=\left[\frac{\left(8\%\right){\left(1+\left(8\%\right)\right)}^{10}}{{\left(1+\left(8\%\right)\right)}^{10}-1}\right]\\ =\left[\frac{0.08{\left(1+0.08\right)}^{10}}{{\left(1+0.08\right)}^{10}-1}\right]\\ =0.149\end{array}$

Substitute $8\%$ for $i$, and $3$ for $n$ in equation (3) to find $\left(\frac{P}{A},i,3\right)$.

$\left(\frac{P}{A},i,3\right)=\left[\frac{{\left(1+\left(8\%\right)\right)}^3-1}{\left(8\%\right){\left(1+\left(8\%\right)\right)}^3}\right]$

$\begin{array}{c}\left(\frac{P}{A},i,3\right)=\left[\frac{{\left(1+0.08\right)}^3-1}{0.08{\left(1+0.08\right)}^3}\right]\\ =2.577\end{array}$

Substitute $8\%$ for $i$, and $4$ for $n$ in equation (5) to find $\left(\frac{P}{F},i,4\right)$.

$\begin{array}{c}\left(\frac{P}{F},i,4\right)=\left[\frac{1}{{\left(1+\left(8\%\right)\right)}^4}\right]\\ =\left[\frac{1}{{\left(1+0.08\right)}^4}\right]\\ =0.735\end{array}$

Substitute $8\%$ for $i$, and $8$ for $n$ in equation (5) to find $\left(\frac{P}{F},i,8\right)$.

$\begin{array}{c}\left(\frac{P}{F},i,8\right)=\left[\frac{1}{{\left(1+\left(8\%\right)\right)}^8}\right]\\ =\left[\frac{1}{{\left(1+0.08\right)}^8}\right]\\ =0.540\end{array}$

Substitute $8\%$ for $i$, and $10$ for $n$ in equation (6) to find $\left(\frac{A}{F},i,10\right)$.

$\begin{array}{c}\left(\frac{A}{F},i,10\right)=\left[\frac{\left(8\%\right)}{{\left(1+\left(8\%\right)\right)}^{10}-1}\right]\\ =\left[\frac{0.08}{{\left(1+0.08\right)}^{10}-1}\right]\\ =0.069\end{array}$

Substitute $4.623$ for $\left(\frac{P}{A},i,6\right)$, $0.149$ for $\left(\frac{A}{P},i,10\right)$, $2.577$ for $\left(\frac{P}{A},i,3\right)$, $0.735$ for $\left(\frac{P}{F},i,4\right)$, $0.540$ for $\left(\frac{P}{F},i,8\right)$, and $0.069$ for $\left(\frac{A}{F},i,10\right)$ in equation (2) to find $AW$.

$\begin{array}{c}AW=\left[\begin{array}{l}\$5000\left(4.623\right)\left(0.149\right)-\$2000\left(2.577\right)\left(0.735\right)\left(0.149\right)+\$8000\left(0.540\right)\left(0.149\right)\\ +\$5000\left(0.069\right)\end{array}\right]\\ =\$3868.37\end{array}$

Therefore, the equivalent annual worth of the cash flow diagram is $\$3868.37$.

Conclusion:

Thus, the equivalent annual worth of the cash flow diagram is $\$3868.37$.