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.

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Answer 1

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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

Answer to Problem 14P

Alternative 3- Intersection widening

Explanation of Solution

Given:

Analysis period = 20 years

Annual interest rate = 15 percent

MindTap Engineering for Garber/Hoel's Traffic and Highway Engineering, 5th Edition, [Instant Access], 1 term (6 months), Chapter 13, Problem 14P

Calculation:

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

PW1=−340,000−(10,000+450,000)(P/A−15−20)+25,000(P/F−15−20)(P/A−15−20)= ( 1+i )n−1i ( 1+i )n= ( 1+0.15 ) 20−10.15 ( 1+0.15 ) 20=6.26(P/F−15−20)=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/A−15−20)+12,000(P/F−15−20)(P/A−15−20)= ( 1+i )n−1i ( 1+i )n= ( 1+0.15 ) 20−10.15 ( 1+0.15 ) 20=6.26(P/F−15−20)=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/A−15−20)(P/A−15−20)= ( 1+i )n−1i ( 1+i )n= ( 1+0.15 ) 20−10.15 ( 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 ) n−1]EUAW1=−3218072.50 [0.15 ( 1+0.15 ) 20 ( 1+0.15 ) 20−1]EUAW1=$−514891.60

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

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

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

EUAW=NPW[i ( 1+i ) n ( 1+i ) n−1]EUAW3=−3559800 [0.15 ( 1+0.15 ) 20 ( 1+0.15 ) 20−1]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/A−15−20)+25,000( P/F−15−20)−340,000(P/A−15−20)= ( 1+i )n−1i ( 1+i )n= ( 1+0.15 ) 20−10.15 ( 1+0.15 ) 20=6.26(P/F−15−20)=1 ( 1+i )n=1 ( 1+0.15 ) 20=0.0611BCR1=−2878072.5−340,000=8.465

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

BCR2=−( 5,000+300,000)( P/A−15−20)+12,000( P/F−15−20)−850,000(P/A−15−20)= ( 1+i )n−1i ( 1+i )n= ( 1+0.15 ) 20−10.15 ( 1+0.15 ) 20=6.26(P/F−15−20)=1 ( 1+i )n=1 ( 1+0.15 ) 20=0.0611BCR2=−1908566.8−850,000=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/A−15−20)−2,120,000(P/A−15−20)= ( 1+i )n−1i ( 1+i )n= ( 1+0.15 ) 20−10.15 ( 1+0.15 ) 20=6.26BCR3=−1439800−2,120,000=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:

BCR2−1=1,908,566.8−2,878,072.5850,000−340,000 =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=(850,000−340,000)−(5,000+150,000)(P/A−i−20)+(25,000−12,000)(P/F−i−20)

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/A−i−20)+13,000(P/F−i−20)

Try i=30 %,

−155,000(P/A−30−20)+13,000(P/F−30−20)=513911.62

For i=31 %

−155,000(P/A−31−20)+13,000(P/F−31−20)=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=(2,120,000−340,000)−(5,000+225,000)(P/A−i−20)+(25,000−0)(P/F−i−20)

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/A−i−20)+25,000(P/F−i−20)

Try i=11 %,

−230,000(P/A−11−20)+25,000(P/F−11−20)=1828464.62

For i=12 %

−230,000(P/A−12−20)+25,000(P/F−12−20)=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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