Chapter 6, Problem 57P

To determine

The alternative which should be selected.

Answer to Problem 57P

The net present worth of alternative B is greater than net present worth of alternative A therefore, alternative B should be selected.

The equivalent uniform annual cost of alternative B is less than equivalent uniform annual cost of alternative A therefore, alternative B should be selected.

The capitalized cost of alternative B is less than equivalent uniform annual cost of alternative A therefore, alternative B should be selected.

Explanation of Solution

Given:

The cost of machine A is $\$12500$.

The salvage value after $3$ years is $\$2000$.

The maintenance cost is $\$5000$.

The cost of machine B is $\$15000$.

The salvage value after $4$ years is $\$1500$.

The maintenance cost is $\$4000$.

The rate of interest is $5\%$ per year.

Concept used:

Write the expression for net present worth method.

$NPW=-\left[\begin{array}{l}P-A\left[\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}\right]+S\left[\frac{1}{{\left(1+i\right)}^n}\right]\\ -\left(P-S\right)\left[\left(\frac{1}{{\left(1+i\right)}^n}\right)+\left(\frac{1}{{\left(1+i\right)}^n}\right)+\left(\frac{1}{{\left(1+i\right)}^n}\right)\right]\end{array}\right]$ ...... (I)

Here, the net present worth is NPW, initial cost is $P$, uniform annual cost is $A$, salvage value is $S$, rate of interest is $i$ and time period is $n$.

Write the expression for Equivalent Uniform Annual Cost.

$EUAC=A+P\left[\frac{i{\left(i+1\right)}^n}{{\left(i+1\right)}^n-1}\right]-S\left[\frac{i}{{\left(i+1\right)}^n-1}\right]$ ...... (II)

Here, the Equivalent Uniform Annual Cost is EUAC.

Write the expression for capitalized cost.

$CC=P+\frac{A}{i}+\frac{S\left[\frac{i}{{\left(1+i\right)}^n-1}\right]}{i}$ ...... (III)

Here, the capitalized cost is CC.

Calculations:

Use the present worth method.

Calculate the net present worth for alternative A.

Substitute $\$12500$ for $P$, $\$5000$ for $A$, $\$2000$ for $S$ and $5\%$ for $i$ in Equation (I).

$\begin{array}{c}NP{W}_A=\left[\begin{array}{l}-\$12500-\$5000\left[\frac{{\left(1+0.05\right)}^{12}-1}{0.05{\left(1+0.05\right)}^{12}}\right]+\$2000\left[\frac{1}{{\left(1+0.05\right)}^{12}}\right]\\ -\left(\$12500-\$2000\right)\left[\left(\frac{1}{{\left(1+0.05\right)}^3}\right)+\left(\frac{1}{{\left(1+0.05\right)}^6}\right)+\left(\frac{1}{{\left(1+0.05\right)}^9}\right)\right]\end{array}\right]\\ =\left[\begin{array}{l}-\$12500-\$5000\left(8.863\right)+\$2000\left(0.5568\right)\\ -\left(\$10500\right)\left(0.863+0.7462+0.6446\right)\end{array}\right]\\ =-\$12500-\$44315+\$1113.6-\$23637.30\\ =-\$80374.70\end{array}$

Calculate the net present worth method. for alternative B.

Substitute $\$15000$ for $P$, $\$4000$ for $A$, $\$1500$ for $S$ and $5\%$ for $i$ in Equation (I).

$\begin{array}{c}NP{W}_B=\left[\begin{array}{l}-\$15000-\$4000\left[\frac{{\left(1+0.05\right)}^{12}-1}{0.05{\left(1+0.05\right)}^{12}}\right]+\$1500\left[\frac{1}{{\left(1+0.05\right)}^{12}}\right]\\ -\left(\$15000-\$1500\right)\left[\left(\frac{1}{{\left(1+0.05\right)}^4}\right)+\left(\frac{1}{{\left(1+0.05\right)}^8}\right)\right]\end{array}\right]\\ =\left[\begin{array}{c}-\$15000-\$4000\left(8.863\right)+\$1500\left(0.5568\right)\\ -\left(\$13500\right)\left(0.8227+0.06768\right)\end{array}\right]\\ =-\$15000-\$35452+\$835.20-\$20243.25\\ =-\$69860.05\end{array}$

The net present worth of alternative B is greater than net present worth of alternative A therefore, alternative B should be selected.

Use the annual worth method.

Calculate the equivalent uniform annual cost for alternative A.

Substitute $\$12500$ for $P$, $\$5000$ for $A$, $\$2000$ for $S$, $5\%$ for $i$ and $3$ for $n$ in Equation (II).

$\begin{array}{c}EUA{C}_A=\$5000+\$12500\left[\frac{0.05{\left(1+0.05\right)}^3}{{\left(1+0.05\right)}^3-1}\right]-\$2000\left[\frac{0.05}{{\left(1+0.05\right)}^3-1}\right]\\ =\$5000+\$12500\left(0.3672\right)-\$2000\left(0.3172\right)\\ =\$5000+\$4590-\$634.40\\ =\$85955.60\end{array}$

Calculate equivalent uniform annual cost for alternative B.

Substitute $\$15000$ for $P$, $\$4000$ for $A$, $\$1500$ for $S$, $5\%$ for $i$ and $4$ for $n$ in Equation (II).

$\begin{array}{c}EUA{C}_B=\$4000+\$15000\left[\frac{0.05{\left(1+0.05\right)}^4}{{\left(1+0.05\right)}^4-1}\right]-\$1500\left[\frac{0.05}{{\left(1+0.05\right)}^4-1}\right]\\ =\$4000+\$15000\left(0.2820\right)-\$1500\left(0.2320\right)\\ =\$4000+\$4230-\$348\\ =\$7882\end{array}$

The equivalent uniform annual cost of alternative B is less than equivalent uniform annual cost of alternative A therefore, alternative B should be selected.

Use the capitalized cost method.

Calculate the capitalized cost for alternative A.

Substitute $\$12500$ for $P$, $\$5000$ for $A$, $\$2000$ for $S$, $5\%$ for $i$ and $3$ for $n$ in Equation (III).

$\begin{array}{c}C{C}_A=\$12500+\frac{\$5000}{0.05}+\frac{\$2000\left[\frac{0.05}{{\left(1+0.05\right)}^3-1}\right]}{0.05}\\ =\$12500+\$100000+\frac{\$2000\left(0.2880\right)}{0.05}\\ =\$112500+\$11520\\ =\$124020\end{array}$

Calculate the capitalized cost for alternative B.

Substitute $\$15000$ for $P$, $\$4000$ for $A$, $\$1500$ for $S$, $5\%$ for $i$ and $4$ for $n$ in Equation (III).

$\begin{array}{c}C{C}_B=\$15000+\frac{\$4000}{0.05}+\frac{\$1500\left[\frac{0.05}{{\left(1+0.05\right)}^4-1}\right]}{0.05}\\ =\$15000+\$80000+\frac{\$1500\left(0.2003\right)}{0.05}\\ =\$95000+\$6009\\ =\$101009\end{array}$

The capitalized cost of alternative B is less than equivalent uniform annual cost of alternative A therefore, alternative B should be selected.

Conclusion:

The net present worth of alternative B is greater than net present worth of alternative A therefore, alternative B should be selected.

The equivalent uniform annual cost of alternative B is less than equivalent uniform annual cost of alternative A therefore, alternative B should be selected.

The capitalized cost of alternative B is less than equivalent uniform annual cost of alternative A therefore, alternative B should be selected.