Chapter 13, Problem 14P

To determine

The preference of one alternative out of the three based on economic criteria by using the present worth, equivalent annual cost, benefit-cost ratio and rate of return methods.

Answer to Problem 14P

Alternative 3- Intersection widening

Explanation of Solution

Given:

Analysis period = 20 years

Annual interest rate = 15 percent

Calculation:

Determine the present worth of alternative 1 (Traffic signals):

  $\begin{array}{l}P{W}_1=-340,000-\left(10,000+450,000\right)\left(P/A-15-20\right)+25,000\left(P/F-15-20\right)\\ \left(P/A-15-20\right)=\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}=\frac{{\left(1+0.15\right)}^{20}-1}{0.15{\left(1+0.15\right)}^{20}}=6.26\\ \left(P/F-15-20\right)=\frac{1}{{\left(1+i\right)}^n}=\frac{1}{{\left(1+0.15\right)}^{20}}=0.0611\\ P{W}_1=-340,000-\left(10,000+450,000\right)\left(6.26\right)+25,000\left(0.0611\right)\\ =\$-3218072.50\end{array}$

Determine the present worth of alternative 2 (Intersection widening):

  $\begin{array}{l}P{W}_2=-850,000-\left(5,000+300,000\right)\left(P/A-15-20\right)+12,000\left(P/F-15-20\right)\\ \left(P/A-15-20\right)=\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}=\frac{{\left(1+0.15\right)}^{20}-1}{0.15{\left(1+0.15\right)}^{20}}=6.26\\ \left(P/F-15-20\right)=\frac{1}{{\left(1+i\right)}^n}=\frac{1}{{\left(1+0.15\right)}^{20}}=0.0611\\ P{W}_2=-850,000-\left(5,000+300,000\right)\left(6.26\right)+12,000\left(0.0611\right)\\ =\$-2758566.80\end{array}$

Determine the present worth of alternative 3 (Grade separation):

  $\begin{array}{l}P{W}_3=-2,120,000-\left(5,000+225,000\right)\left(P/A-15-20\right)\\ \left(P/A-15-20\right)=\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}=\frac{{\left(1+0.15\right)}^{20}-1}{0.15{\left(1+0.15\right)}^{20}}=6.26\\ P{W}_3=-2,120,000-\left(5,000+225,000\right)\left(6.26\right)\\ =\$-3559800\end{array}$

The project with the highest present worth is alternative 2.

Determine the equivalent annual cost of alternative 1 (Traffic signals):

  $\begin{array}{l}EUAW=NPW\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]\\ EUA{W}_1=-3218072.50\left[\frac{0.15{\left(1+0.15\right)}^{20}}{{\left(1+0.15\right)}^{20}-1}\right]\\ EUA{W}_1=\$-514891.60\end{array}$

Determine the equivalent annual cost of alternative 2 (Intersection widening):

  $\begin{array}{l}EUAW=NPW\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]\\ EUA{W}_2=-2758566.80\left[\frac{0.15{\left(1+0.15\right)}^{20}}{{\left(1+0.15\right)}^{20}-1}\right]\\ EUA{W}_2=\$-441370.688\end{array}$

Determine the equivalent annual cost of alternative 3 (Grade separation):

  $\begin{array}{l}EUAW=NPW\left[\frac{i{\left(1+i\right)}^n}{{\left(1+i\right)}^n-1}\right]\\ EUA{W}_3=-3559800\left[\frac{0.15{\left(1+0.15\right)}^{20}}{{\left(1+0.15\right)}^{20}-1}\right]\\ EUA{W}_3=\$-569568\end{array}$

The project with the highest equivalent annual costis alternative 2.

Determine the benefit-cost ratioof alternative 1 (Traffic signals):

  $\begin{array}{l}BC{R}_1=\frac{-\left(10,000+450,000\right)\left(P/A-15-20\right)+25,000\left(P/F-15-20\right)}{-340,000}\\ \left(P/A-15-20\right)=\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}=\frac{{\left(1+0.15\right)}^{20}-1}{0.15{\left(1+0.15\right)}^{20}}=6.26\\ \left(P/F-15-20\right)=\frac{1}{{\left(1+i\right)}^n}=\frac{1}{{\left(1+0.15\right)}^{20}}=0.0611\\ BC{R}_1=\frac{-2878072.5}{-340,000}=8.465\end{array}$

Determine the benefit-cost ratio of alternative 2 (Intersection widening):

  $\begin{array}{l}BC{R}_2=\frac{-\left(5,000+300,000\right)\left(P/A-15-20\right)+12,000\left(P/F-15-20\right)}{-850,000}\\ \left(P/A-15-20\right)=\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}=\frac{{\left(1+0.15\right)}^{20}-1}{0.15{\left(1+0.15\right)}^{20}}=6.26\\ \left(P/F-15-20\right)=\frac{1}{{\left(1+i\right)}^n}=\frac{1}{{\left(1+0.15\right)}^{20}}=0.0611\\ BC{R}_2=\frac{-1908566.8}{-850,000}=2.245>1\end{array}$

Since BCR >1, we would select Alternative 2.

Determine the benefit-cost ratio of alternative 3 (Grade separation):

  $\begin{array}{l}BC{R}_3=\frac{-\left(5,000+225,000\right)\left(P/A-15-20\right)}{-2,120,000}\\ \left(P/A-15-20\right)=\frac{{\left(1+i\right)}^n-1}{i{\left(1+i\right)}^n}=\frac{{\left(1+0.15\right)}^{20}-1}{0.15{\left(1+0.15\right)}^{20}}=6.26\\ BC{R}_3=\frac{-1439800}{-2,120,000}=0.68<1\end{array}$

Since BCR is less than 1, we would not select Alternative 3

Determine the benefit-cost ratio of alternative 2 with respect to 1:

  $\begin{array}{l}BC{R}_{2-1}=\frac{1,908,566.8-2,878,072.5}{850,000-340,000}\\ =1.9\end{array}$

We reachthe same conclusion as previously, which is to select Alternative 2

Determine the rate of return (ROR)of alternative 1versus 2:

  $NPW=0=\left(850,000-340,000\right)-\left(5,000+150,000\right)\left(P/A-i-20\right)+\left(25,000-12,000\right)\left(P/F-i-20\right)$

It is very difficult to solve this explicitly for i. By trial and error, we can easily find the i that makes the right side of the equation equal to the left side.

  $510,000=-155,000\left(P/A-i-20\right)+13,000\left(P/F-i-20\right)$

Try $i=30\%$,

  $-155,000\left(P/A-30-20\right)+13,000\left(P/F-30-20\right)=513911.62$

For $i=31\%$

  $-155,000\left(P/A-31-20\right)+13,000\left(P/F-31-20\right)=497684.2512$

Then, we can try $i=30.5\%$ and by interpolation 30.237 % is the rate of return to make the left side to equal to the right side.

Since ROR is greater than 15 percent, we select Alternative 2

Determine the rate of return (ROR) of alternative 3 versus 1:

  $NPW=0=\left(2,120,000-340,000\right)-\left(5,000+225,000\right)\left(P/A-i-20\right)+\left(25,000-0\right)\left(P/F-i-20\right)$

By trial and error, we can easily find the i that makes the right side of the equation equal to the left side.

  $1,780,000=-230,000\left(P/A-i-20\right)+25,000\left(P/F-i-20\right)$

Try $i=11\%$,

  $-230,000\left(P/A-11-20\right)+25,000\left(P/F-11-20\right)=1828464.62$

For $i=12\%$

  $-230,000\left(P/A-12-20\right)+25,000\left(P/F-12-20\right)=1715277.50$

By interpolation 11.428 % is the rate of return to make the left side to equal to the right side.

Since the ROR is lower than 15 percent, we discard Alternative 3.

Conclusion:

The second alternative i.e. Intersection widening is selected by using the three economic methods.