Traffic and Highway Engineering
Traffic and Highway Engineering
5th Edition
ISBN: 9781305156241
Author: Garber, Nicholas J.
Publisher: Cengage Learning
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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.

Expert Solution & Answer
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Answer to Problem 14P

Alternative 3- Intersection widening

Explanation of Solution

Given:

Analysis period = 20 years

Annual interest rate = 15 percent

Traffic and Highway Engineering, Chapter 13, Problem 14P

Calculation:

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

  PW1=340,000(10,000+450,000)(P/A1520)+25,000(P/F1520)(P/A1520)= ( 1+i )n1i ( 1+i )n= ( 1+0.15 ) 2010.15 ( 1+0.15 ) 20=6.26(P/F1520)=1 ( 1+i )n=1 ( 1+0.15 ) 20=0.0611PW1=340,000(10,000+450,000)(6.26)+25,000(0.0611)=$3218072.50

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

  PW2=850,000(5,000+300,000)(P/A1520)+12,000(P/F1520)(P/A1520)= ( 1+i )n1i ( 1+i )n= ( 1+0.15 ) 2010.15 ( 1+0.15 ) 20=6.26(P/F1520)=1 ( 1+i )n=1 ( 1+0.15 ) 20=0.0611PW2=850,000(5,000+300,000)(6.26)+12,000(0.0611)=$2758566.80

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

  PW3=2,120,000(5,000+225,000)(P/A1520)(P/A1520)= ( 1+i )n1i ( 1+i )n= ( 1+0.15 ) 2010.15 ( 1+0.15 ) 20=6.26PW3=2,120,000(5,000+225,000)(6.26)=$3559800

The project with the highest present worth is alternative 2.

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

  EUAW=NPW[i ( 1+i ) n ( 1+i ) n1]EUAW1=3218072.50[0.15 ( 1+0.15 ) 20 ( 1+0.15 ) 201]EUAW1=$514891.60

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

  EUAW=NPW[i ( 1+i ) n ( 1+i ) n1]EUAW2=2758566.80[0.15 ( 1+0.15 ) 20 ( 1+0.15 ) 201]EUAW2=$441370.688

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

  EUAW=NPW[i ( 1+i ) n ( 1+i ) n1]EUAW3=3559800[0.15 ( 1+0.15 ) 20 ( 1+0.15 ) 201]EUAW3=$569568

The project with the highest equivalent annual costis alternative 2.

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

  BCR1=( 10,000+450,000)( P/A1520)+25,000( P/F1520)340,000(P/A1520)= ( 1+i )n1i ( 1+i )n= ( 1+0.15 ) 2010.15 ( 1+0.15 ) 20=6.26(P/F1520)=1 ( 1+i )n=1 ( 1+0.15 ) 20=0.0611BCR1=2878072.5340,000=8.465

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

  BCR2=( 5,000+300,000)( P/A1520)+12,000( P/F1520)850,000(P/A1520)= ( 1+i )n1i ( 1+i )n= ( 1+0.15 ) 2010.15 ( 1+0.15 ) 20=6.26(P/F1520)=1 ( 1+i )n=1 ( 1+0.15 ) 20=0.0611BCR2=1908566.8850,000=2.245>1

Since BCR >1, we would select Alternative 2.

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

  BCR3=( 5,000+225,000)( P/A1520)2,120,000(P/A1520)= ( 1+i )n1i ( 1+i )n= ( 1+0.15 ) 2010.15 ( 1+0.15 ) 20=6.26BCR3=14398002,120,000=0.68<1

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

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

  BCR21=1,908,566.82,878,072.5850,000340,000=1.9

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

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

  NPW=0=(850,000340,000)(5,000+150,000)(P/Ai20)+(25,00012,000)(P/Fi20)

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(P/Ai20)+13,000(P/Fi20)

Try i=30%,

  155,000(P/A3020)+13,000(P/F3020)=513911.62

For i=31%

  155,000(P/A3120)+13,000(P/F3120)=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=(2,120,000340,000)(5,000+225,000)(P/Ai20)+(25,0000)(P/Fi20)

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(P/Ai20)+25,000(P/Fi20)

Try i=11%,

  230,000(P/A1120)+25,000(P/F1120)=1828464.62

For i=12%

  230,000(P/A1220)+25,000(P/F1220)=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.

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