CH-13-14 Evaluating Transportation Alternatives – Traffic and Highway Engineering : Book Guide

This Document Contains Chapters 13 to 14 Chapter 13 Evaluating Transportation Alternatives 13-1 What is the main objective of conducting a transportation project evaluation? The main objective of conducting a transportation project evaluation is to furnish the appropriate information about the outcome of each alternative so that a selection can be made. This evaluation process should be viewed as an activity in which information that is relevant to the selection is made available to the person or group responsible for making a decision. 13-2 Describe four basic issues that should be considered prior to selection of an evaluation procedure. The following four basic issues should be considered prior to the selection of an evaluation procedure: Who will use the information and what is their viewpoint? A clear definition of whose viewpoint is being considered in the evaluation is necessary if proper consideration is to be given as to how the stakeholders will be impacted, either positively or negatively, by each proposed alternative. What are the relevant criteria and how should these be measured? A transportation project is intended to accomplish one or more goals and objectives. These are measured as criteria, and the numerical or relative results for each criterion are called measures of effectiveness. It is important that the criteria be related as closely as possible to the stated objective in order to properly evaluate each alternative. What measures of effectiveness are to be used in the evaluation process itself? One approach is to convert each measure of effectiveness to a common unit, and then for each alternative compute the summation for all measures. A common unit is money, and it might be possible to make a transformation of the relevant criteria to equivalent dollars and then compare each alternative from an economic viewpoint. Another common unit approach is to convert each measure of effectiveness to a numeric score. Otherwise, the measures of effectiveness can be simply reported for each alternative in a matrix form with no attempt made to combine them. The measures of effectiveness must not only be relevant to the problems but should be easy to measure and be sensitive to changes made in each alternative. How well will the evaluation process assist in making a decision? The decision maker will want to know what the costs of the project will be. This alone will usually determine the outcome. Also of importance is if the benefits justify the expenditure of funds for transportation or would the money be best spent elsewhere. The decision maker will also want to know if the proposed project is likely to produce the stated results. The evaluation process requires that engineers have all the appropriate facts about a proposed project and be able to convey these in a clear and logical manner so that decision making is facilitated. 13-3 List the basic criteria used for evaluating transportation alternatives. What units are used for measurement? The basic criteria used in evaluating transportation projects, along with their corresponding units of used for measurement, are listed below: Capital costs (dollars) Preliminary engineering Construction Right-of-way Vehicles Maintenance costs (dollars) Facility operating costs (dollars) (d) Travel time costs (dollars) Total hours or cost (dollars) Average door-to-door speed (miles per hour) Distribution of door-to-door speed, variance (miles per hour) Vehicle operating costs (dollars) Crash costs: lives, injuries, property damage (dollars) 13-4 Average demand on a rural roadway ranges from zero to 500 veh/day when the cost per trip goes from $1.50 to zero. Calculate the net user benefits per year if the cost decreases from $1.00 to $0.75/trip (assume a linear demand function). Compare the value calculated in (a) with the benefits as calculated in typical highway studies. (a) Construct the demand curve as shown in the figure below. From this curve, one can find V1 and V2 from the prices provided. V1 = 167 vehicles per day V2 = 250 vehicles per day Determine the net user benefits using Equation 13.1: Net user benefits = (P1 – P2) (V1 + V2) / 2 Net user benefits = [(1.00 – 0.75) × (162 + 250)] / 2 Net user benefits = $52.13 per day Next, determine the net user benefits per year. Yearly net user benefits = 365 × net user benefits Yearly net user benefits = 365 × 52.13 Yearly net user benefits = $19,027.45 Therefore, the net user benefits per year will be $19,027. (b) The benefits calculated in typical highway studies are not based on the concept of consumer surplus. They are usually based on travel time reduction as a result of an improvement. In this manner, the annual user benefit would be: Annual User Benefit = (P1 – P2) × (V2) × 365 Annual User Benefit = ($1.00 – $0.75) × (250) × 365 Annual User Benefit = $22,812.50 This method used in highway studies will usually overestimate the user benefits as shown above. 13-5 A ferry is currently transporting 300 veh/day at a cost of $1.50/vehicle. The ferry can attract 600 more veh/day when the cost /veh is $1.00. Calculate the net user benefits/year if the cost /veh decreases from $1.10 to $0.95. When 300 veh/day, the cost is $1.50/vehicle When 900 veh/day, the cost is $1.00/vehicle Construct the demand curve as shown in the figure below. From this curve, one can find V1 and V2 from the prices provided (from $1.10 to $0.95). V2 = 960 vehicles per day Determine the net user benefits using Equation 13.1: Net user benefits = (P1 – P2) (V1 + V2) / 2 Net user benefits = [(1.10 – 0.95) × (780 + 960)] / 2 Net user benefits = $130.50 per day Next, determine the net user benefits per year. Yearly net user benefits = 365 × net user benefits Yearly net user benefits = 365 × 130.50 Yearly net user benefits = $47,632.50 Therefore, the net user benefits per year will be $47,632.50. 13-6 What are the two components of the cost of a transportation facility improvement? Describe each. The cost of a transportation facility improvement includes two components: first cost (also called initial cost) and continuing costs for annual maintenance, operation, and administration. The first cost for a highway or transit project may include engineering design, right-of-way, and construction. Design parameters specific to each project will dictate which items will be required and their costs. Maintenance and operating costs of the facility must also be determined. These are recurring costs, or continuing costs, that will be incurred over the life of the facility and are usually based on historical data for similar projects. 13-7 Estimate the average unit costs for (a) operating a standard vehicle on a level roadway, (b) travel time for a truck company, (c) single-vehicle property damage, (d) personal injury, and (e) fatality. The following are estimated average unit costs for: Operating a standard vehicle on a level roadway. 10 cents per mile for depreciation 4 cents per mile for fuel 8 cents per mile for insurance and registration 3 cents per mile for maintenance 25 cents per mile on level ground The estimated average unit cost to operate a vehicle on level ground is 25 cents per mile. Travel time for a truck company. Using estimated costs of 35 cents per mile to operate a truck and 25 cents per mile for its driver, the unit cost is approximately 60 cents per mile. At an average speed of 50 mi/h, the travel time cost is: ($0.60/mi)(50 mi/h) = $30 per hour of travel. Single vehicle property damage. According to a sample of crash records in Virginia, the average property damage per crash is approximately $1,800, and the average number of vehicles involved in a crash is 1.7. Therefore the average unit cost for single vehicle property damage is ($1,800 / 1.7) = $1,059. Personal injury. If a single X-ray costs $225 and a visit to the emergency room costs $175, it can be estimated that the average cost for a personal injury (not including costs associated with treatment) is approximately $400. (e) Fatality. The average unit cost of a fatality is difficult to determine; however, values between $100,000 and $4.5 million have been used. 13-8 An incident that occurs on a particular highway results in traffic incurring an average delay of one hour per vehicle. An estimated 2,000 vehicles are impacted by the incident. If the distribution of traffic is comprised of 5% five-axle trucks, 2% three-axle trucks, 4% six-tire trucks, 2% four-tire trucks, 4% medium auto for business purposes, 3% small auto for business purposes, and the remainder are automobiles for personal purposes, what is the value of the total delay incurred by the traffic. Determine the traffic volume for each vehicle type: Five axle trucks: 5% × 2000 = 100 Three axle trucks: 2% × 2000 = 40 Six tire trucks: 4% × 2000 = 80 Four tire trucks: 2% × 2000 = 40 Medium auto for business purposes: 4% × 2000 = 80 Small auto for business purposes: 3% × 2000 = 60 Small auto for personal purposes: 80% × 2000 = 1600 Select the value of one hour of travel time from Table 13.2 and multiply these values by the traffic volumes in each vehicle type category. Five axle trucks: $46.83 × 100 = $4,683.00 Three axle trucks: $40.08 × 40 = $1,603.20 Six tire trucks: $37.02 × 80 = $2,961.60 Four tire trucks: $22.37 × 40 = $894.80 Medium auto for business purposes: $41.96 × 80 = $3,356.80 Small auto for business purposes: $41.52 × 60 = $2,491.20 Small auto for personal purposes: $18.95 × 1600 = $30,320.00 Therefore, the value of total delay is $46,310.60 (the sum of the values for each vehicle type category. 13-9 Derive the equation to compute the equivalent annual cost given the capital cost of a highway, such that A = (A/P) x P, where A/P is the capital recovery factor. Compute the equivalent annual cost if the capital cost of a transportation project is $100,000, annual interest = 10%, and n = 15 years. Knowing that: F = A(1 + i)n–1 + A(1 + i)n–2 + … + A(1 + i) + A Multiplying by (1 + i), (1 + i)F = A(1 + i)n + A(1 + i)n–1 + … + A(1 + i) Subtracting the first equation from the second, i F = A(1 + i)n – A F = A [(1 + i)n – 1] / i Knowing that: F = P (1 + i)n A [(1 + i)n – 1 ] / i = P (1 + i)n A/P = i (1 + i)n / [(1 + i)n – 1 ] From Table 13.3, or using the equation derived above, determine the appropriate (A/P) term. EUAC = P(A/P, 10%, 15) EUAC = 100,000 × (0.1315) EUAC = $13,150.00 Therefore, the EUAC for this project will be $13,150.00. 13-10 A highway project is expected to cost $1,700,000 initially. The annual operating and maintenance cost after the first year is $5,000 and will increase by $500 each year for a project lifespan of 20 years. At the end of the tenth year, the project must be resurfaced at a cost of $300,000. Calculate the present worth of costs for this project over a 20-year period if the annual interest rate is 5%. Convert the value obtained in (a) to equivalent uniform annual costs. (a) Calculate the present worth using Equation 13.4: PW =nN=0 (1C+ni)n PW = 1,700,000 + + + + + + + + + + + + + + + + + + + + PW = 1,700,000 + 4,762 + 4,989 + 5,183 + 5,348 + 5,485 + 5,597 + 5,685 + 5,753 + 5,801 + 5,832 + 5,847 + 5,847 + 5,834 + 5,808 + 5,772 + 5,726 + 5,672 + 5,610 + 5,540 + 5,465 + 5,384 + 184,173 = 1,995,729 Therefore, the present worth (cost) for this project is $1,995,729. (b) Convert the above present worth cost to EUAC using Equation 13.6. Since the project is analyzed for 20 years, n = 20. EUAC = PW(A/P, 5%, 20) EUAC = (1995729) [0.05 (1 + 0.05)20] / [(1 + 0.05)20 – 1] EUAC = (1995729) (0.080243) EUAC = 160142 Therefore, the EUAC for this project will be $160,142. 13-11 The concepts applied in economic evaluation to address the time-dependent value of money can also be applied to other phenomena, such as growth in traffic volume. The traffic volume on a local highway was 3,510 vehicles per day on an average day in 2012. Based on historical trends, the traffic volume is expected to increase by 3% per year for the foreseeable future. What is the expected traffic volume in 2022? Calculate the future value in 10 years F = 3,510 (F/P, 3%, 10) Select value from Table 13.3. (F/P, 3%, 10) = 1/0.7441 = 1.3439 F = 3,510 (F/P, 3%, 10) = 3,510 × 1.3439 = 4,717 Therefore, the expected traffic volume in 2022 is 4,718 vehicles per day on average. 13-12 Three transportation projects have been proposed to increase the safety in and around a residential neighborhood. Each project consists of upgrading existing street signing to highly retroreflective sheeting to increase visibility. The following table shows the initial construction costs, annual operating costs, useful life of the sheeting, and the salvage values for each alternative. Assume that the discount rate is 10%. Calculate the present worth for each alternative and determine the preferred project based on the economic criteria.
Alternative Initial Construction Cost ($) Annual Operations and Maintenance Costs ($) Useful Life (years) Salvage Value ($)
1 19,000 2,500 10 3,000
2 8,000 4,000 5 900
3 20,000 2,500 10 4,600
Use Equation 13.4. Calculate the present worth for Alternative I. PWI = 19,000 + 2,500(P/A, 10%, 10) – 3,000(P/F, 10%, 10) PWI = 19,000 + 2,500(6.145) – 3000(0.3855) PWI = 33,206 Therefore, the present worth of Alternative I is $33,206. Calculate the present worth of Alternative II. In order to compare this alternative to the others, Alternative II must be analyzed using the prevailing useful life values in the other alternatives (10 years). PWII = 8000 + 4000(P/A, 10%, 10) + (8000 – 900)(P/F, 10%, 5) – 900(P/F, 10%, 10) PWII = 8000 + 4000(6.145) + 7100(0.6209) – 900(0.3855) PWII = 36641 Therefore, the present worth of Alternative II is $36,641. Calculate the present worth for Alternative III. PWIII = 20,000 + 2,500(P/A, 10%, 10) – 4,600(P/F, 10%, 10) PWIII = 20,000 + 2,500(6.145) – 4,600(0.3855) PWIII = 33,589.2 Therefore, the present worth of Alternative III is $33,589. Since Alternative I has the lowest net present worth of costs, it is the best alternative to choose based on the economic criteria provided. 13-13 Two designs have been proposed for a short span bridge in a rural area, as shown in the following table. The first proposal is to construct the bridge in two phases (Phase I now and Phase II in 25 years). The second alternative is to construct it in one phase. Assuming that the annual interest rate is 4%, determine which alternative is preferred using present worth analysis.
Alternative Construction Costs ($) Annual Maintenance Costs ($) Service Period (yr)
A (Phase 1) 14,200,000 75,000 1-50
A (Phase 2) 12,600,000 25,000 26-50
B 22,400,000 100,000 1-50
Calculate the present worth of Alternative I. Note that the Phase II has an additional annual maintenance cost to that of Phase I. This will require that this annualized costs be "brought" back to the present condition at year 25, and then this value needs to be brought back to year 0. Using Equation 13.4: PWI = 14200000 + 75000(P/A, 4%, 25) + 12600000(P/F, 4%, 25) + ((25000(P/A, 4%, 25))(P/F, 4%, 25)) PWI = 14200000 + 75000(21.482) + 12600000(0.3751) + 25000(15.622)(0.3751) PWI = 14200000 + 1611150 + 4726260 + 146495.31 PWI = 20683905.31 Therefore, the present worth of Alternative I is $20,683,905. Next, calculate the present worth of Alternative II. PWII = 22400000 + 100000(P/A, 4%, 50) PWII = 22400000 + 100000(21.482) PWII = 24548200 Therefore, the present worth of Alternative II is $24,548,200. Based on the above economic analysis, Alternative I, consisting of a two-phase construction process should be chosen. Alternative I has a lower net present worth of costs than does Alternative II. 13-14 Three designs have been proposed to improve traffic flow at a major intersection in a heavily traveled suburban area. The first alternative involves improved traffic signaling. The second alternative includes traffic-signal improvements and intersection widening for exclusive left turns. The third alternative includes extensive reconstruction, including a grade separation structure. The construction costs, as well as annual maintenance and user costs, are listed in the following table for each alternative. Determine which alternative is preferred based on economic criteria if the analysis period is 20 years and the annual interest rate is 15%. Show that the result is the same using the present worth, equivalent annual cost, benefit– cost ratio, and rate-of-return methods.
Alternative Capital Costs ($) Annual Maintenance Costs ($) Annual User Costs ($) Salvage Value ($)
Present condition 0 15,000 600,000 0
Traffic signals 340,000 10,000 450,000 25,000
Intersection widening 850,000 5,000 300,000 12,000
Grade separation 2,120,000 5,000 225,000 0
Calculate the present worth for each of the alternatives. PWpresent conditions = 0 + 15000(P/A, 15%, 20) + 600000(P/A, 15%,20) PWpresent conditions = 0 + 15000(6.259) + 600000(6.259) PWpresent conditions = $3,849,285 PWtraffic signals = 340000 + 10000(P/A, 15%, 20) + 450000(P/A, 15%, 20) – 25,000(P/F, 15%, 20) PWtraffic signals = 340000 + 10000(6.259) + 450000(6.259) – 25000(0.0611) PWtraffic signals = $3,217,613 PWint. widening = 850000 + 5000(P/A, 15%, 20) + 300000(P/A, 15%, 20) – 12000(P/F, 15%, 20) PWint. widening = 850000 + 5000(6.259) + 300000(6.259) – 12,000(0.0611) PWint. widening = $2,758,262 PWgrade separation = 2120000 + 5000(P/A, 15%, 20) + 225000(P/A, 15%, 20) PWgrade separation = 2120000 + 5000(6.259) + 225000(6.259) PWgrade separation = $3,559,570 Based on the present worth analysis, Alternative III (Intersection widening) is preferred due to it having the lowest net present worth of costs among all alternatives. Calculate the equivalent uniform annual cost (EUAC) for each alternative. EUACpresent conditions = 0 + 15000 + 600000 EUACpresent conditions = $615,000 EUACtraffic signals = 340000(A/P, 15%, 20) + 10000 + 450000 – 25000(A/F, 15%, 20) EUACtraffic signals = 340000(0.1598) + 10000 + 450000 - 25000(0.0098) EUACtraffic signals = $514,087 EUACint. widening = 850000(A/P, 15%, 20) + 5000 + 300000 – 12000(A/F, 15%, 20) EUACint. widening = 850000(0.1598) + 5000 + 300000 – 12,000(0.0098) EUACint. widening = $440,712 EUACgrade separation = 2120000(P/A, 15%, 20) + 5000 + 225000 EUACgrade separation = 2120000(0.1598) + 5000 + 225000 EUACgrade separation = $568,776 Based on the EUAC analysis, Alternative III (Intersection widening) is preferred. This is because it has the lowest equivalent uniform annual cost among the alternatives. Calculate the benefit-cost ratio for each alternative. First, compare the traffic signal alternative to the present condition alternative. BCRII/I = [3,849,285 – (10000 + 450000)(6.259)] / [340000 – (25000)(0.0611) – 0] BCRII/I = 2.866 Since BCR is greater than 1.00, the traffic signal alternative is preferred to present conditions. Next, compare the intersection widening to the traffic signal alternative. BCRIII/II = [(10000 + 450000)(6.259) – (5000 + 300000)(6.259)] / [850000 – (12000)(0.0611) – (340000 – 25000 × 0.0611)] BCRIII/II = 1.899 Since BCR is greater than 1, the intersection widening is preferred to the traffic signal alternative. Next, compare the grade separation alternative to the traffic signal alternative. BCRIV/III = [(5000 + 300000)(6.259) – (5000 + 225000)(6.259)] / [2120000 – (850000 – 12000 × 0.0611)] BCRIV/II = 0.369 Since BCR is less than 1, the grade separation alternative is not preferred to the intersection widening alternative. From the benefit cost ratio analysis, the Intersection widening alternative is preferred. Calculate the rate of return for each alternative. First, compare the traffic signal alternative to the present condition alternative. NPWII/I = 0 = –[(340000 – 25000 × 0.0611) – 0] + [(15000 + 600000) – (450000 + 10000)](P/A, i, 20) NPWII/I = 0 = –338473 + 155000(P/A, i, 20) (P/A, i, 20) = 338473 / 155000 (P/A, i, 20) = 2.184 i = 45.8% (by interpolation or by trial and error) Since the rate of return is greater than 15%, the traffic signal alternative is preferred to the present conditions alternative. Next, compare the intersection widening to the traffic signal alternative. NPWIII/II = 0 = –[(850000 – 12000 × 0.0611) – (338473)] + ((450000 + 10000) – (5000 + 300000))(P/A, i, 20) NPWII/I = 0 = –510794 + 155000(P/A, i, 20) (P/A, i, 20) = 510794 / 155000 (P/A, i, 20) = 3.295 i = 30.2% (by interpolation or by trial and error) Since the rate of return is greater than 15%, the intersection widening alternative is preferred. Next, compare the grade separation to the traffic signal alternative. NPWIV/III = 0 = –[2120000 – (849267)] + [(300000+5000) – (5000 + 225000)](P/A, i, 20) NPWII/I = 0 = –1270733 + 75000(P/A, i, 20) (P/A, i, 20) = 1270733 / 75000 (P/A, i, 20) = 16.94 i = 1.6% (by interpolation or by trial and error) Since the rate of return is less than 15%, the grade separation alternative is not preferred. Therefore, based on the four different economic analyses performed above Intersection Widening Alternative is the best alternative. 13-15 A road is being proposed to facilitate a housing development on a scenic lake. Two alternatives have been suggested. One of the roadway alignments is to go around the lake and slightly impact a wetland. The second alternative will also go around the lake and will significantly impact two wetlands. The following table shows the anticipated costs for each alternative. Assuming that the annual interest rate is 7%, determine which alternative is preferred using equivalent annual cost analysis.
Alternative First Cost ($) Annual Maintenance Cost ($) Service Life (yr) Salvage Value ($) Annual Wetland Costs ($) Annual Lighting Costs ($)
I 75,000 3,000 15 45,000 7,500 1,500
II 125,000 2,000 15 25,000 2,500 2,500
First, calculate the EUAC of Alternative I. EUACI = 75000(A/P, 7%, 15) + 3000 + 7500 + 1500 – 45000 (A/F, 7%, 15) EUACI = 75000(0.1098) + 12000 – 45000(0.0398) EUACI = $18,444 Therefore, the EUAC of Alternative I is $18,444. Next, calculate the EUAC of Alternative II. EUACII = 125000(A/P, 7%, 15) + 2000 + 2500 + 2500 – 25000 (A/F, 7%, 15) EUACII = 125000(0.1098) + 7000 – 25000(0.0398) EUACII = $19,730 Therefore, the EUAC of Alternative II is $19,730. Alternative I is preferred since it has a lower EUAC. 13-16 Two alternatives are under consideration for maintenance of a bridge. Select the most cost-effective alternative using present worth analysis. Assume an interest rate of 10% per year and a design life of 50 years for each alternative. Alternative A consists of annual maintenance costs of $5,000 per year for the design life except for: Year 20, in which bridge deck repairs will cost $20,000 Year 30, in which a deck overlay and structural repairs will cost $105,000 Alternative B consists of annual maintenance costs of $3,000 per year for the design life except for: Year 20, in which bridge deck repairs will cost $35,000 Year 30, in which a deck overlay and structural repairs will cost $85,000 NPWA = – {5000(P/A,10%,50) + 15000(P/F,10%,20) + 100000(P/F,10%,30)} NPWA = – {5000(1/0.1009) + 15000(0.1486) + 100000(0.0573)} = – $57,513 NPWB = – {3000(P/A,10%,50) + 32000(P/F,10%,20) + 82000(P/F,10%,30)} NPWB = – {3000(1/0.1009) + 32000(0. 1486) + 82000(0.0573)} = – $39,186 Alternative B is the most cost-effective (higher net present worth). 13-17 Two pavement maintenance plans have been proposed for a county road. Select the most cost-effective alternative based on equivalent annual cost. Assume an interest rate of 5% per year and a design life of 25 years. Show cash flow diagrams for each alternative. Alternative A entails expenses of $500 per year for the entire lifespan plus an additional $1000 in year 5, $1500 in year 10, $2000 in year 15, and $2500 in year 20. Alternative B entails expenses of $600 per year for the entire lifespan plus an additional $100 per year starting in year 16 and continuing through year 25 ($700 per year during this period). The cash flow diagram for Alternative A is shown as follows: EUACA = AA + 1000(P/F, 5%, 5)(A/P, 5%, 25) + 1500(P/F, 5%, 10)(A/P, 5%, 25) + 2000(P/F, 5%, 15)(A/P, 5%, 25) + 2500(P/F, 5%, 20)(A/P, 5%, 25) = 500 + 1000 × 0.7835 × 0.0710 + 1500 × 0.6139 × 0.0710 + 2000 × 0.4810 × 0.0710 + 2500 × 0.3769 × 0.0710 = 756.21 The cash flow diagram for Alternative B is shown as follows: EUACB = AB + 100(P/F, 5%, 16)(A/P, 5%, 25) + 100(P/F, 5%, 17)(A/P, 5%, 25) + … + 100(P/F, 5%, 25)(A/P, 5%, 25) = 600 + 3.25 + 3.10 + 2.95 + 2.81 + 2.68 + 2.55 + 2.43 + 2.31 + 2.20 + 2.09 = 626.37 Since Alternative B cost less, this maintenance plan is preferred. 13-18 Two highway capacity improvement plans have been proposed for a congested suburban arterial. Select the most cost-effective alternative using present worth analysis. Assume an interest rate of 5% per year and a design life of 20 years for each alternative. Alternative A entails improvements to traffic signals at an initial cost of $82,000 and a salvage value of $5,000. Annual maintenance costs will be $700 per year. Alternative B entails traffic signal improvements and addition of a left-turn lane at an initial cost of $72,000 and no salvage value. Annual maintenance costs will be $1,200 per year, except in year 10 in which a rehabilitation will cost $14,000. NPWA = – {82000 – 5000(P/F,5%,20) + 700(P/A,5%,20)} NPWA = – {82000 – 5000(0.3769) + 700(1/0.0802)} = – $92,613 NPWB = – {72000 + 1200(P/A,5%,20) + 12800(P/F,5%,10)} NPWB = – {72000 + 1200(1/0.0802) + 12800(0.6139)} = – $94,820 Alternative A is the most cost-effective (higher net present worth). 13-19 The light-rail transit line described in this chapter is being evaluated by another group of stakeholders. Using the revised information, determine the weighted score for each alternative and comment on your result. Objective Ranking 5 3 1 2 4 First, determine the relative weight. For a ranking of 1, the highest, assign n where n is the number of objectives. In this case n = 5. Therefore, for a ranking of 2, the next highest, assign n – 1 = 5 – 1 = 4. This process is then continued for the remainder of the objectives. Next, determine the weighting factors using Equation 13.9. The weighting factor for objective I is: K1 = relative / sum of relative weights K1 = [1 / (5 + 4 + 3 + 2 + 1)] × 100 K1 = (1 / 15) × 100 K1 = 6.667 = 7 Objective II K2 = (3 / 15) × 100 K2 = 20 Objective III K3 = (5 / 15) × 100 K3 = 33.333 = 33 Objective IV K4 = (4 / 15) × 100 K4 = 26.667 = 27 Objective V K5 = (2 / 15) × 100 K5 = 13.333 = 13
Objective Ranking Relative Weight Weighting Factor
1 5 1 7
2 3 3 20
3 1 5 33
4 2 4 27
5 4 2 13
Total 15 100
Next, use the estimated values for measures of effectiveness found in Table 13.5 to produce the point score for the candidate transit lines.
Alternatives
MOE I II III IV V
1 6.1 6.5 5.1 6.3 7.0
2 20.0 18.4 16.0 14.4 13.6
3 16.5 23.1 26.4 33.0 33.0
4 27.0 23.6 20.3 16.9 16.9
5 13.0 11.1 7.4 5.6 5.6
Total 82.6 82.7 75.2 76.2 76.1
The ranking of alternatives in order of preference is II, I, IV, V, and III. Alternatives I and II are clearly superior to the others. 13-20 Three alternatives to replace an existing two-lane highway with a four-lane highway that will meet current design standards are proposed. The selected alternative will provide a more direct route between two towns that are 12.0 miles apart along the existing highway. With each alternative operating speeds are expected to be at or near the design speed of 60 miles per hour. Develop the scores for each alternative and recommend a preferred alternative for development. The following scoring method, developed by the transportation oversight board, is to be used:
Evaluation Criterion Performance Measure Weight (%)
Mobility Travel time of shortest travel time alternative divided by travel time of alternative i 25
Safety Annual reduction in number of crashes of alternative i divided by highest annual reduction in number of crashes among all alternatives 25
Costeffectiveness Project development cost of least expensive alternative (in $ per mile) divided by project development cost of alternative i (in $ per mile) 20
Environmental impacts Area of wetlands impacted of least-impacting alternative divided by area of wetlands impacted by alternative i 15
Community impacts Number of business and residences displaced by least-impacting alternative divided by number of businesses and residences displaced by alternative i 15
The following information has been estimated for each alternative by the planning staff:
Property Cost of development Length Annual crash reduction Alt. 1 $10,900,000 11.2 miles 10 Alt. 2 $18,400,000 9.8 miles 17 Alt. 3 $16,900,000 10.1 miles 19
Business displacements 3 4 5
Residential displacements 4 3 3
Wetlands impacted 1.5 acres 3.9 acres 3.9 acres
Use the approach given in Equation 13.9, in which the score for an alternative is the sum of the products of the weight for each evaluation criterion and the relative score or value for each criterion. First, the input values are to be used to calculate the relative scores, Vij . For example, to evaluate the performance measure for the criterion mobility, the travel time of each alternative must be calculated. For example, for alternative 1, the travel time is:  hr 11.2mi = 0.1867hr  60mi  Performance Measure Alt. 1 Alt. 2 Alt. 3
Travel time 0.1867 0.1633 0.1683
Annual crash reduction 10.0000 17.0000 19.0000
Cost per mile 973214 1877551 1673267
Wetlands impacted 1.5000 3.9000 3.9000
Total displacements 7.0000 7.0000 8.0000
The relative scores, Vij, can then be calculated. For example, for the evaluation criterion mobility, the travel time of the shortest travel time alternative divided by travel time of alternative i must be determined for each alternative. The shortest travel time is for alternative 2, at 0.1633 hours. For alternative 1, the travel time is 0.1867 hr, and the relative score is 0.1633/0.1867 = 0.8750.
Alt. 1 Alt. 2 Alt. 3
Evaluation Criterion Kj Relative Scores (Vij)
Mobility 25 0.8750 1.0000 0.9703
Safety 25 0.5263 0.8947 1.0000
Cost-effectiveness 20 1.0000 0.5183 0.5816
Environmental impacts 15 1.0000 0.3846 0.3846
Community impacts 15 1.0000 1.0000 0.8750
The scores for each alternative i can then be found by finding the products of the weights (Kj) and relative scores (Vij) and then summing these products. Alt. 1 Alt. 2 Alt. 3 21.8750 25.0000 24.2574 13.1579 22.3684 25.0000 20.0000 10.3668 11.6325 15.0000 5.7692 5.7692 15.0000 15.0000 13.1250 Scores 85.033 78.504 79.784 The recommended alternative is Alternative 1, with a score of 85.03, the highest score among the alternatives. 13-21 Four alternatives to replace an existing two-lane highway with a four-lane highway that will meet current design standards are proposed. The selected alternative will provide a more direct route between two cities that are 17.0 miles apart along the existing highway. With each alternative operating speeds are expected to be at or near the design speed of 70 miles per hour. Develop the scores for each alternative and recommend a preferred alternative for development. The following scoring method, developed by the transportation oversight board, is to be used:
Evaluation Criterion Mobility Safety Cost- effectiveness Environmental impacts Community impacts Performance Measure Travel time of shortest travel time alternative divided by travel time of alternative i Annual reduction in number of crashes of alternative i divided by highest annual reduction in number of crashes among all alternatives Project development cost of least expensive alternative (in $ per mile) divided by project development cost of alternative i (in $ per mile) Area of wetlands impacted of leastimpacting alternative divided by area of wetlands impacted by alternative i Number of business and residences displaced by least-impacting alternative divided by number of businesses and residences displaced by alternative i Weight (%) 28 30 18 14 10
The following information has been estimated for each alternative by the planning staff:
Property Alt. A Alt. B Alt. C Alt. D
Cost of development $12,800,000 $12,300,000 $11,900,000 $12,900,000
Length 16.2 miles 15.6 miles 16.1 miles 16.5 miles
Annual crash reduction 19 24 23 27
Business displacements 2 1 4 4
Residential displacements 10 12 8 6
Wetlands impacted 1.9 acres 4.2 acres 2.9 acres 2.9 acres
Use the approach given in Equation 13.9, in which the score for an alternative is the sum of the products of the weight for each evaluation criterion and the relative score or value for each criterion. First, the input values are to be used to calculate the relative scores, Vij . For example, to evaluate the performance measure for the criterion mobility, the travel time of each alternative must be calculated. For example, for alternative 1, the travel time is:  hr 16.2mi = 0.2314hr  70mi  Performance Alt. A Alt. B Alt. C Alt. D Measure
Travel time 0.2314 0.2229 0.2300 0.2357
Annual crash reduction 19 24 23 27
Cost per mile 790123 788462 739130 781818
Wetlands impacted 1.9 4.2 2.9 2.9
Total displacements 2 1 4 4
The relative scores, Vij, can then be calculated. For example, for the evaluation criterion mobility, the travel time of the shortest travel time alternative divided by travel time of alternative i must be determined for each alternative. The shortest travel time is for alternative 2, at 0.1633 hours. For alternative 1, the travel time is 0.1867 hr, and the relative score is 0.1633/0.1867 = 0.8750.
Alt. A Alt. B Alt. C Alt. D
Evaluation Criterion Kj Relative Scores (Vij)
Mobility 28 0.9633 1.0000 0.9691 0.9457
Safety 30 0.7037 0.8889 0.8519 1.0000
Cost-effectiveness 18 0.9355 0.9374 1.0000 0.9454
Environmental impacts 14 1.0000 0.4524 0.6552 0.6552
Community impacts 10 0.5000 1.0000 0.2500 0.2500
The scores for each alternative i can then be found by finding the products of the weights (Kj) and relative scores (Vij) and then summing these products. Alt. A Alt. B Alt. C Alt. D Mobility 26.97 28.00 27.13 26.48 Safety 21.11 26.67 25.56 30.00 Cost-effectiveness 16.84 16.87 18.00 17.02 Environmental impacts 14.00 6.33 9.17 9.17 Community impacts 5.00 10.00 2.50 2.50 Scores 83.92 87.87 82.36 85.17 The recommended alternative is Alternative B, with a score of 87.87, the highest score among the alternatives. 13-22 You have been hired as a consultant to a medium-sized city to develop and implement a procedure for evaluating whether or not to build a highway bypass around the CBD. Write a short report describing your proposal and recommendation as to how the city should proceed with this process. I. Collection of Required Data and Information In order for a consultant to develop and evaluate alternatives for a highway bypass, information is needed to decide upon an evaluation process. Establishing goals and objectives, identification of stakeholders (e.g., governing bodies, citizen groups), definition of purpose and need for a bypass, and its potential impacts (e.g., traffic flow, environmental effects) are all important steps toward selecting an evaluation process. The data required, such as socioeconomic data, and traffic data, should also be considered in selecting a suitable evaluation process. II. Use of Information Upon deciding on the data and information requirements, an evaluation process will be chosen that best fits the situation and any stated objectives and goals of the study. If cost is a critical factor, an economic evaluation that includes a life cycle cost analysis may be followed. The process will then be divided into components, the data gathered, and alternatives defined and evaluated. III. Results A recommendation would be included with the evaluation of the highway bypass. If several alternatives are relatively close, additional forms of evaluation should be considered to ensure a sound decision. 13-23 The following data have been developed for four alternative transportation plans for a high-speed transit line that will connect a major airport with the downtown area of a large city. Prepare an evaluation report for these proposals by considering the cost effectiveness of each attribute. Show your results in graphical form and comment on each proposal.
Measure of Effectiveness Rail Alternatives
Existing Service Plan A Plan B Plan C Plan D
Personal displaced 0 264 3200 3200 3200
Business displaced 0 23 275 275 275
Average door-to-door trip speed (mph) 10.2 38 45 46 48
Annual passengers (millions) 118.6 124.4 118.6 124.4 127.0
Annual cost (millions) – 16.4 20.2 23.8 22.7
Persons Displaced The existing service moves no one out of their home while plans B, C, and D displace 3,200 people each. Plan A, however, displaces only 264 people, 92% less than the other proposals. The graph below depicts the number of people displaced versus costs. This graph clearly shows that plan A is the preferred alternative when considering residential displacements. Businesses Displaced Plan A impacts far fewer businesses than does Plans B, C, or D, which will displace 275 businesses each. As shown in the graph below, Plan A is the preferred alternative when considering business impacts. Average Speed While the existing plan has an average speed of 10.2 mi/h, the four new proposals increase this value to almost 4 times the current travel speed (Plan A) and in one case almost 5 times (Plan D) the current rate. Out of all of the proposals, Plan D has the fastest rate and Plan A has the slowest. When one relates these statistics to annual passengers carried on each alternative, Plan D is preferred, but only slightly above the existing and other proposed alternatives. As the graph below indicates, all of the alternatives are relatively equal. Plan A provides the least improvement. Costs From observing the data gathered, Plan A has the lowest cost. Overall, Plan A fares best in three of the four criteria considered (excepting average speed). 13-24 A new carpool lane has replaced one lane of an existing six-lane highway. During peak hours, the lane is restricted to cars carrying three or more passengers. After five months of operation, the carpool lane handles 800 autos/h, whereas the existing lanes are operating at capacity levels of 1500 veh/h/ln at an occupancy rate of 1.2. How would you determine if the new carpool lane is successful or if the lane should be open to all traffic? This would be a good case to utilize a control versus experimental group analysis. By comparing the new level of traffic on the carpool highway with that of another highway without a carpool lane, it can be determined if the carpool lane is a success. In order to do this, one must first have traffic statistics for the number of automobiles and number of persons per hour on the roads before and after the modification to the facility. Next, one must establish a minimum acceptable rate of increase in persons traveling by carpool. If the capacity (persons carried per lane per hour) with the carpool modification is greater than that with natural growth plus the minimum acceptable rate, the carpool lane program is considered successful. Assume the High Occupancy Vehicles (HOV) in the carpool lane carry 3 passengers per vehicle, the average autos carry 1.2 persons, the acceptable rate of increase is 10% and the growth rate is 4% in the five months. 800(3) = 2,400 > 1,500(1.2)(1 + 0.04) = 1,872 Therefore, the carpool lane is a success. Chapter 14 Highway Surveys and Location 14-1 Describe the three categories of information gathered, in the office prior to any field survey activity, about the characteristics of the area of a proposed highway location. The information gathered in an office study of existing information can be divided into three categories: engineering, environmental, and economic. Engineering includes topography, geology, climate, and traffic volumes, and social and demographic, including land use and zoning patterns. Environmental includes types of wildlife; location of recreational, historic, and archeological sites; and the possible effects of air, noise, and water pollution. Economic includes unit costs for construction and the trend of agricultural, commercial, and industrial activities in the proposed location area. 14-2 Briefly discuss factors that are of specific importance in the location of scenic routes. The following factors are of specific importance in the location of scenic routes: Special provisions should be provided to discourage fast driving as design speeds are usually low (e.g. provide narrow road bed). Conflict between driver's attention on the road and the need to enjoy the scenic view should be minimized. This is achieved by providing turn-outs with wide shoulders and adequate turning space at regular intervals, or by providing only straight alignment when the view is exceptional. Only minimum disruption to the area should be caused as a result of the scenic route construction. 14-3 Describe the factors that significantly influence the location of highways in urban areas. Connection to local streets is primarily a factor with design of freeways and expressways such that traffic flow is made as efficient as possible. Right-of-way acquisition, particularly in commercial and industrial areas, can be a substantial expense; such costs often influence the available corridor width and the ultimate design. The interaction between the many travel modes available in urban areas requires coordination of the highway system with other transportation systems. Adequate provisions for pedestrians and bicycles is important as increased use of these modes can reduce energy use and traffic congestion. 14-4 What are three elements that highway surveys usually involve? Highway surveys usually involve measuring and computing horizontal and vertical angles, vertical heights (elevation), and horizontal distances. The surveys are then used to prepare base maps with contour lines and longitudinal cross sections. 14-5 Briefly describe the use of each of the following instruments in conventional ground surveys: total station level measuring tapes electronic distance measuring devices The total station is an electronic theodolite and distance measuring device. It is used for measuring angles in both the vertical and horizontal plane as well as distances. A level is used in conjunction with a graduated leveling rod to measure changes in elevation . Measuring tapes are used for the direct measurement of horizontal distances. An electronic distance measuring device (EDM) is used to measure distances. When an EDM is used in conjunction with a slope reduction calculator, it can also measure slope and height distances. These devices allow for distances and direction to be determined from a single instrument setup. 14-6 Briefly compare the factors that should be considered in locating an urban freeway with those for a rural highway. Factors that are similar when considering the location for both urban freeways and rural freeways include: Social and demographic characteristics of the area in which the freeway is to be located (including land use and zoning patterns). Environmental impacts (including recreation and historic sites, plant and animal life, wetlands, and air, noise, and water pollution). Serviceability of the route (to industrial and residential areas). Crossing of other transportation facilities. Terrain and soil conditions. Economic feasibility of the location. Directness of route. Factors that are considered for urban freeways and not usually considered for rural freeways include: Connection to local streets. Cost of right-of-way acquisition. Coordination of the urban freeway and other urban transportation systems. Adequate provisions for pedestrians. 14-7 Describe how each of the following could be used in highway survey location: aerial photogrammetry computer graphics conventional survey techniques Aerial Photogrammetry: Aerial photogrammetry is used in identifying suitable highway locations and in preparing contour maps suitable for cross sections. This involves obtaining the aerial photographs and determining distances and elevations from these photographs. Computer Graphics: Computer graphics are used in identifying suitable highway locations by combining photogrammetry and computer techniques. The procedure also involves the use of aerial photographs to determine distances and elevations. In addition, the horizontal and vertical alignment of a proposed centerline can be obtained and displayed on a monitor. This enables the designer to make alignment changes and to immediately see the effect of these changes. (c) Conventional Survey Techniques: Conventional surveys are used in identifying suitable locations by determining distances and elevations for all natural and man-made land features. Contour maps are then obtained which can be used to identify alternative locations. Profiles, showing the change in elevation along a proposed centerline, and cross sections at selected stations along the proposed centerline can also be developed. 14-8 A photograph is to be obtained at a scale of 1:12,000 by aerial photogrammetry. If the focal length of the camera to be used is 7.5", determine the height at which the aircraft should be flown if the average elevation of the terrain is 850 ft. Use Equation 14.1 and solve for H. S = f H −h (1 / 12,000) = ((7.5 in)/(ft/12in)) / (H – 850 ft) H – 850 = 7500 H = 8350 feet Therefore, the plane should fly at 8,350 feet. 14-9 The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. If the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken. Let the actual distance between the points MN be x feet M'N' = 3.6 in = 0.3 ft (M'N') / (MN) = (ON') / (ON) = Scale 1:24,000 1 / 24000 = 0.3 / x x = 7200 feet Now solve for the scale of the second photo. M'N = 4.5 in = 0.375 ft Scale = 0.375 / 7200 Scale = 1:19,200 Now determine the height from which the first photo was taken (H1). 1 / 19,200 = (6.0/12) / (H1 – 100) H1 = 9,700 feet Now determine the height from which the second photo was taken (H2). 1 / 24,000 = (6.0/12) / (H2 – 100) H2 = 12,100 feet Therefore, the scale of the first photograph is 1:19,200, the elevation it was taken from is 9,700 feet, and the second photograph was taken at an elevation of 12,100 feet. 14-10 The distance in the x direction between two control points on a vertical aerial photograph is 6.2". If the distance between these same two points is 4.9" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. If the focal length of the camera is 6"and the average elevation at these points is 300 ft, determine the flying height from which each photograph was taken. Let the actual distance between the points MN be x feet M'N' = 4.9 in = 0.40833 ft (M'N') / (MN) = (ON') / (ON) = Scale 1:24,000 1 / 24000 = 0.0.40833 / x x = 9800 feet Now solve for the scale of the second photo. M'N = 6.2 in = 0.5167 ft Scale = 0.5167 / 9800 Scale = 1:18,970 Now determine the height from which the first photo was taken (H1). 1 / 18,970 = (6.0/12) / (H1 – 300) H1 = 9,784 feet Now determine the height from which the second photo was taken (H2). 1 / 24,000 = (6.0/12) / (H2 – 300) H2 = 12,300 feet Therefore, the scale of the first photograph is 1:18,970, the elevation it was taken from is 9,784 feet, and the second photograph was taken at an elevation of 12,300 feet. 14-11 The scale at the image of a well-defined object on an aerial photograph is 1:24,000, and the elevation of the object is 1500 ft. The focal length of the camera lens is 6.5". If the air base (B) is 250 ft, determine the elevation of the two points A and C and the distance between them if the coordinates of A and C are as given below. A C x coordinate 5.5” 6.5” y coordinate 3.5” 5.0” Scale of photograph 1:13,000 1:17,400 First determine the flying height, H, using Equation 14.1. S = f H −h (1 / 24000) = (6.5/12) / (H – 1500) H – 1500 = 13000 H = 14,500 feet Now determine the elevation of point A, ha, using Equation 15.1. (1 / 13000) = (6.5/12) / (14500 – ha) 14500 – ha = 7042 ha = 7,458 feet Now determine the elevation of point B, hb, using Equation 15.1. (1 / 17,400) = (6.5/12) / (14500 – hb) 14500 – hb = 9425 hb = 5,075 feet Now determine the distance between points A and C. Use Equations 14.4 and 14.5 to solve for the X and Y coordinates, respectively. XA = (5.5 / 12) / (1 / 13000) XA = 5958 ft XB = (6.5 / 12) / (1 / 17400) XB = 9425 ft YA = (3.5 / 12) / (1 / 13000) YA = 3792 ft YB = (5.0 / 12) / (1/ 17400) YB = 7250 ft Now use Equation 14.6 to solve for the distance between these points. D= (XA −XB)2 +(Y YA − B)2 D= (5958 9425)− 2 +(3792 −7250)2 D = 4,897 ft. Therefore, the elevation at point A is 7,458 ft; point B is 5,075 ft and the distance between the two points is 4,897 ft. 14-12 A vertical photo has an air base of 2400 ft. Stereoscopic measurements of parallax at a point representing the top of a 200 ft tower is 0.278". The camera focal length is 6.5". Photos were taken at an elevation of 7500 ft. Determine the elevation of the base of the tower. Use Equation 14.2 to determine the elevation of the base of the tower. H − h = B f p 7500 −h = 200 (6.5/12) (0.278/12) 7500 – h = 4676 h = 2824 ft Since the elevation of the top of the tower is 2824 ft, the elevation at its base is 2624 ft (200 ft below the top of the tower). 14-13 The length of a runway at a national airport is 7500 ft long and at elevation 1500 ft above sea level. The airport was recently expanded to include another runway used primarily for corporate aircraft. It is desired to determine the length of this runway whose elevation is 1560 ft. An aerial photograph was taken of the airport. Measurements on the photograph for the national airport runway are 4.80" and for the corporate runway, are 3.4". The camera focal length is 6". Determine the length of the corporate runway. To calculate the length of the corporate runway, the scale of the photograph at the elevation of the corporate runway must be determined. To calculate the scale at any point on the photograph, the flying height from which the photograph was taken must be determined. The scale at the elevation of the national runway is: S(1500 ft) = (4.8 in)(1 ft / 12 in) / 7500 ft = 1/18750 Use Equation 14.1 to solve for the flying height, S = f = (6/12) = 1 H −h H −1500 18750 H – 1500 = 9375 H = 10,875 ft Use Equation 14.1 to solve for the scale at the elevation of the corporate runway, S = f = (6/12) = 1 H −h 10875−1560 18630 The length of the corporate runway can then be found, L = (3.4/12)(18630) L = 5,278.5 ft 14-14 Using an appropriate diagram, discuss the importance of side and forward overlaps in aerial photography. Side and forward overlaps are critical for stereoscopic viewing of aerial photographs. In order for the aerial photographs to be viewed in three dimensions, an object should be viewed by the observer’s left eye on the left photograph and the same object viewed by the observer’s right eye on the right photograph on a set of stereopairs. This requires that each object in the area be on at least two aerial photographs. This is achieved through the side overlap which provides for approximately 60 percent overlap in the direction of flight, and the overlap in the direction perpendicular to flight which provides for about 25 percent overlap as shown in Figure 14.6 in the text. 14-15 Under what conditions would the borrowing of new material from a borrow pit for a highway embankment be preferred over using material excavated from an adjacent section of the road? The conditions under which the borrowing of new material from a borrow pit is preferable to using excavated material from an adjacent section of highway might include the following: The engineering properties of the material from the adjacent section are not satisfactory. Excavation of this material may result in serious negative environmental impacts. Excavation of this material might result in serious drainage problems for the highway. Excavation of the material might be cost prohibitive. 14-16 Using the data given in Table 14.1, determine the total overhaul cost if the free haul is 700 ft and the overhaul cost is $10 per cubic yard station. Stations of the free haul lines are 1 + 80 and 8 + 80 and 10 + 20 and 17 + 20. The first step is to construct the mass diagram shown in Figure 14.17 from the data in Table 14.1. The data required to solve this problem using the method of moments are shown below. Note that the ordinate is zero at station 9+63, and other ordinates as given in Table 14.1 are shown. First, find the moments and overhaul distances about stations 1 + 80 and 8 + 80. About sta. 1 + 80: [(100/2 + 80)/100](130) + [(80/2)/100](374 – 130) = 266.6 yd3-sta Overhaul distance: (266.6 yd3-sta) / 374 yd3 = 0.713 sta About sta. 8 + 80: [(63/2 + 20)/100](299) + [(20/2)/100](374 – 299) = 161.5 yd3-sta Overhaul distance: (161.5 yd3-sta) / 374 yd3 = 0.432 sta The overhaul cost for the first section (between sta. 0 + 00 and sta. 9 + 63) can be calculated as: Overhaul cost = (0.713 sta + 0.432 sta)(374 yd3)($10.0/yd3-sta) = $4280 Then, find the moments and overhaul distances about stations 10+20 and 17+20. About sta. 10 + 20: [(37/2 + 20)/100](201) + [(20/2)/100](255 – 201) = 86.2 yd3-sta Overhaul distance: (86.2 yd3-sta) / 255 yd3 = 0.338 sta About sta. 17 + 20: [(49/2)/100](255) = 62.5 yd3-sta Overhaul distance: (62.5 yd3-sta) / 255 yd3 = 0.245 sta The overhaul cost for the second section (between sta. 10 + 20 and sta. 17 + 20) can be calculated as: Overhaul cost = (0.338 sta + 0.245 sta)(255 yd3)($10.0/yd3-sta) = $1487 Total overhaul cost = $4280 + $1487 = $5767. 14-17 The following table shows the stations and ordinates for a mass diagram. The free-haul distance is 600 ft. Overhaul cost is $12 per station yard. Station Ordinate (yd3)
0 + 00 0
1 + 00 50
2 + 00 75
2 + 20 95
4 + 00 135
6 + 00 150
7 + 00 125
8 + 20 95
9 + 00 85
10 + 00 55
10 + 30 0
Use the method of moments to compute the additional cost that must be paid to the contractor. Sketch the ground profile if the finished grade of this roadway section is level (0%). First, find the moments and overhaul distances about stations 2 + 20 and 8 + 20. About sta. 2+20: [(100/2 + 100 + 20)/100](50) + [(100/2 + 20)/100] (75 –50) + [(20/2)/100](95 – 75) = 104.5 yd3-sta Overhaul distance: (104.5 yd3-sta) / 95 yd3 = 1.10 sta About sta. 8 + 20: [(80/2)/100](90 – 85) + [(100/2+80)/100](85 – 55) + [(30/2 + 100 + 80)/100](55) = 152.25 yd3-sta Overhaul distance: (152.25 yd3-sta) / 90 yd3 = 1.691 sta The overhaul cost can be calculated as: Overhaul cost = (1.100 sta + 1.691 sta)(90 yd3)($12/yd3-sta) = $3014.28 Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157