CH-11-12 The Transportation Planning Process – Traffic and Highway Engineering : Book Guide

This Document Contains Chapters 11 to 12 Chapter 11 The Transportation Planning Process 11-1 Explain why the transportation planning process is not intended to furnish a decision or give a single result. The transportation planning process is a rational one that intends to furnish unbiased information about the effects that proposed transportation projects will have on the community, its environmental impacts, and its expected users. This process is not intended to furnish or to give a single result that must be followed but to provide information to the general public and to the political bodies responsible for deciding whether proposed transportation projects should be developed. 11-2 Describe the steps that an engineer must follow if asked to determine the need for a grade-separated railroad grade crossing that would replace an at-grade crossing of a two-lane highway with a rail line. The first step the engineer will need to follow is to define the situation. The engineer must fully understand the reason why the grade-separated structure is said to be needed. The next step is to define the problem in terms of the objectives to be accomplished by the project. The objectives for this project might be to improve safety by reducing or eliminating vehicle-train interaction. Another quantifiable objective might include the reduction in travel time to the motorists by eliminating delay due to crossing trains. The third step for the engineer is to identify solutions to the perceived problem. Possible solutions to the problem might include the installation of active warning devices at the grade crossing, the diversion of motorists to other close-by grade crossings, or to construct a grade separated structure. A “no-build” or “donothing” alternative is typically considered as a possible solution. The fourth step in the process is to analyze the expected performance of each of the alternatives under present and future conditions. This analysis should employ quantified measures of effectiveness such as travel time delay reductions and crash rate reductions. The analysis should also take into consideration costs such as the construction, maintenance, and operation costs of each alternative. The fifth step in the process is to compare the alternatives against the objectives developed in step 2 of this process. The advantages and disadvantages of each must be studied. The sixth step is to recommend one of the alternatives after considering all of the factors involved. The final step in the process is to design the facility after one of the alternatives has been selected. This step also moves the project out of the planning process and into project development. 11-3 Describe the basic steps in the transportation planning process. The basic steps in the transportation planning process are as follows: Step 1 – Situation definition is where the present situation is analyzed and described. Step 2 – Problem definition is where the problem is described in terms of objectives to be accomplished and establish criteria by which effectiveness of the project can be measured. Step 3 – Search for solutions is the brainstorming stage where many options may be proposed for testing and evaluating. Preliminary feasibility studies are included to narrow the field to the most promising. Step 4 – Analysis of performance is used to estimate how the proposed alternatives, under present and future conditions, operate using the criteria established earlier. Step 5 – Evaluation of alternatives is where each alternative is judged by the criteria as to how the objectives are accomplished. This involves the calculation of benefits and costs, economic evaluation, and comparison of cost effectiveness. Step 6 – Choice of project is where the selection is made considering all factors. It may be based on cost alone in simple cases or it may involve hearings, political considerations, or other issues. Step 7 – Specification and construction is the final stage in the process. Once the project has been selected it moves into the design phase of project development and ultimately into construction. 11-4 Select a current transportation problem in your community or state. Briefly describe the situation and the problem. Indicate options available and the major impacts of each option on the community. Problem Situation: Traffic congestion and safety on Interstate 81 is widely identified as a problem needing to be addressed. This aging highway, most of which was built in the 1960s, carries a relatively high percentage of trucks through mountainous terrain, resulting in less than desirable service provided by the facility. Options available: There are three general categories of options available to mitigate the problem Add lanes to existing highway Upgrade parallel rail line to improve its capacity, minimizing or eliminating the need to add lanes Do nothing alternative Impacts: Increasing the number of travel lanes will decrease the level of congestion and improve safety. However, this may require additional right-of-way, result in increased demand and therefore increased air and noise pollution. It would cause some inconvenience to motorists during the construction phase. Upgrading the parallel rail line will minimize and possibly eliminate in some rural sections, the need to modify the existing highway. This would result in increased rail traffic and impacts to communities along the rail line. If no investment is made, congestion would increase and safety deteriorate; the roadway will operate at a reduced level of service. This would result in increased vehicle delay and driver dissatisfaction 11-5 Evaluate a proposal to increase tolls on existing roads and bridges. Describe the general planning analysis used. In the seven-step process for transportation planning analysis, the following issues would be addressed: Situation definition: What is the need for increased tolls? Where are the toll roads and bridges and what are the traffic volumes on these facilities? Who are the primary users of these facilities and what are the alternative routes, if any? Problem definition: Why should tolls be raised and what would be the benefits of increased revenues? How will the increased tolls help the situation? Search for solutions: Can tolls be increased only on the roads with existing tolls? Should more toll roads be added? Can tolls remain constant and other solutions be sought? There might be viable combinations of increasing certain tolls and leaving other tolls constant. Determine feasibility of the alternatives developed. Analysis of performance: Analyze alternatives in terms of cost, travel demand, time delays, and other measures of effectiveness. Evaluate alternatives: Compare and the alternatives. Choose the best alternative after consulting with groups of policy makers. Implementation: Implement the recommendations. 11-6 Prepare a study to consider improvements to transportation between an airport and the city it serves.
Inventory of existing facilities
Collect data on existing transportation systems. Determine destinations of travelers arriving by air. Determine the purpose and need for potential improvements, including available resources and time period. Determine existing and projected travel patterns. Determine existing and future land use and socioeconomic conditions. Determine trips made. Identify available modes of travel between airport and city and distribute trips among them. Distribute trips across modal networks. Determine and evaluate alternatives. Select a project or set of projects for development. 11-7 What caused transportation planning to become institutionalized in urban areas, and what does the process need to be based on? The 1962 Federal Aid Highway Act institutionalized transportation planning in urban areas. This act required that all highway projects in urbanized areas with populations of 50,000 or more be based on a transportation planning process that is continuous, comprehensive, and cooperative (“3C”). The core of the process is the travel demand forecasting process, consisting of trip generation, trip distribution, modal split, and traffic assignment. 11-8 Explain the three “C”s concept in the transportation planning process, as mandated in the Federal-Aid Highway Act of 1962. The three “C”s are continuing, comprehensive, and cooperative term Continuing implies that the process be revisited frequently and viewed as an ongoing concern. Comprehensive in this context ensures that all transportation modes are addressed. A cooperative process indicates that the state (or states) and all municipalities in an urbanized area work together. 11-9 Explain the difference between transportation planning and transportation programming, including differences between the key documents in each process and their respective time horizons. Transportation planning entails the generation of plans for various types of transportation facilities and programs. Such studies may include long-term (20- or 25- year) comprehensive plans and short-term transportation improvement plans created by a regional planning body or specific jurisdiction, project-specific studies of a particular corridor or location, and statewide plans. The creation of these plans typically includes some degree of public involvement, such as public hearings, citizen surveys, or meetings with citizen committees. This process is also guided by federal requirements for state and MPO planning processes. For example, the metropolitan area’s long-range plan must be financially constrained such that the projects recommended by the plan do not exceed the forecast for revenue that will be available. Transportation programming is the process of reconciling recommended projects from the planning process with the amount of funds available over an expected period of time, usually between three and six years. The resulting transportation improvement program is thus a list of projects and costs that can be supported by the expected revenues within the specified time frame, which is substantially shorter than that for the long-range transportation plan. 11-10 Urban transportation is concerned with two separate time horizons. Briefly describe each and provide examples of the types of projects that can be categorized in each horizon. The two time horizons that urban transportation planners are concerned with are short-term projects and long-term projects. The short-term projects are those that can be implemented within a one- to three-year period. These projects are designed to provide better management of existing facilities by making them as efficient as possible. Examples of short-term projects might include re-timing of traffic signals to improve traffic flow, transit improvements, and car and van pooling initiatives to reduce traffic congestion. The second time horizon deals with long-range transportation needs of an area and identifies the projects to be constructed over a 20- or 25-year period. Typical long-term projects include the addition of new highway segments, additional lanes on existing highways, rapid transit systems and extensions, and access roads to airports. 11-11 In gathering data to support urban transportation planning, name the unit of geographic area at which data supporting an inventory of existing conditions are summarized. Note the factors that are considered in delineating these geographic units. The unit of a defined geographic area is called a transportation analysis zone (TAZ). Often, census tracts or census enumeration districts are used as TAZs. Within each TAZ, inventories and surveys are made to determine traffic volumes, land uses, origins and destinations of travelers, population, employment, and economic activity. Inventories are made of existing transportation facilities, both highway and transit. Capacity, speed, travel time, and traffic volume are determined. Criteria used in the delineation of TAZs include: Socioeconomic characteristics should be homogeneous. Intra-zonal trips should be minimized. Physical, political, and historical boundaries should be utilized where possible. Zones should not be created within other zones. The zone system should generate and attract approximately equal trips, households, population, or area. For example, labor force and employment should be similar. Zones should use census tract boundaries where possible. 11-12 An existing highway-rail at-grade crossing is to be upgraded. Plans were developed in 2001; the cost estimate for that improvement was $570,000 at that time. Due to funding constraints, construction of the improvement was delayed until 2005. Using the data given in Table 11.3, estimate the construction cost in 2005 dollars. Because this is a highway-rail crossing improvement, select the highway cost index from Table 11.3. Using Equation 11.1: Estimatein 2005 dollars = (Estimatein 2001 dollars) 2005 index 2001index Estimate in 2005 dollars= ($570,000) = $722,734 Thus, the improvements will cost approximately $722,734 in 2005 dollars. 11-13 Given the information in Problem 11-12, assume that construction was delayed until 2014. Using the compound growth rate (see Chapter 13) that can be derived from the data given in Table 11.3, estimate the construction cost in 2014 dollars. Use the following compound interest equation (see Chapter 13): F = P(1+i)n Rearrange to solve for i (compound growth rate) 1 i = F n −1  P  Treat the 2001 index value as P, and the 2005 index value as F. 1 i = F n − =1 183.6  − =1 0.06115  P  144.8 Now use this compound growth rate to carry forward the 2001 cost estimate to 2014. F = P(1+ =i)n 570,000(1.06115)13 = $1,233,028 Thus, the improvements will cost approximately $1,233,028 in 2014 dollars. 11-14 Describe the three forms of environmental impact analysis documentation. There are three forms of environmental impact analysis documentation: a full environmental impact statement (EIS), a simpler environmental assessment (EA), or a cursory checklist of requirements known as a categorical exclusion (CE). A categorical exclusion is typically applicable for small-scale projects that do not involve substantial new construction or environmental permits. In such cases no additional analysis or documentation is required. At the other extreme, an EIS is developed for large-scale projects for which it is apparent that environmental impacts will occur and therefore need to be documented. On some projects it may be readily apparent whether significant environmental impacts will occur, in which case an EA, simpler than an EIS is developed. The findings of the EA will either indicate that a full EIS is warranted, or that there is no significant impact, in which case, a FONSI is the final documentation. 11-15 Identify the types of impacts and effects that might be addressed in an analysis of environmental impacts. Federal and/or state regulations may require that the environmental impacts of proposed projects be assessed. These impacts may include effects on air quality, noise levels, water quality, wetlands, and the preservation of historic sites of interest. The analytical process through which these effects are identified can take one of three forms, depending on the scope of the proposed project: a full environmental impact statement (EIS), a simpler environmental assessment (EA), or a cursory checklist of requirements known as a categorical exclusion (CE). 11-16 What is the purpose of performing inventories and surveys for each defined geographic unit or transportation analysis zone within a study area? Inventories and surveys are made to determine traffic volumes, land-uses, the origins and destinations of travelers, population, employment, and economic activity. These inventories are made of existing transportation facilities, both highway and transit, their capacity, average speeds, travel time, and traffic volumes. 11-17 What are the four basic elements that make up the urban transportation forecasting process? The four basic elements that make up the urban transportation planning process are: (1) data collection, (2) analysis of existing conditions and calibration of forecasting techniques, (3) forecast of future travel demand, and (4) analysis of results. 11-18 In the data collection phase of the urban transportation forecasting process, what type of information should the data reveal for a transportation analysis zone? The data collection phase should provide the transportation planner with information about the city and its people. The data should reveal, specifically, the area's economic activity (such as employment, sales volume, and household income), types of land-uses and densities, travel characteristics (trip and traveler profiles), and physical transportation facility characteristics (such as capacity, travel speeds, and travel times). 11-19 Define the following terms: (a) link, (b) node, (c) centroid, and (d) network. Link – A link is a portion of the highway system that can be described by its capacity, lane width, and speed. Node – A node is the end point of a link and represents an intersection or location where a link changes direction, capacity, width, or speed. Centroid – A centroid is the location within a zone where trips are considered to begin and end. Network – A network is a combination of the above elements to represent a highway system. 11-20 Draw a link-node diagram of the streets and highway within your neighborhood or campus. For each link, show travel times and distances (to the nearest 0.1 mile). The following link-node diagram is of the roadways surrounding the University of Virginia. The travel times provided are for automobiles operating at the posted speed. 11-21 Explain the role that geographic information systems have come to play in transportation planning and key reasons as to why this has happened. Almost all network data are organized within some type of Geographic Information System (GIS). A GIS is a spatially-oriented database management system containing location and attribute information for manmade and natural features and supporting related queries with these features. There are three reasons that explain the popularity of GIS for transportation planning: First, a GIS is scale-able, meaning it may support analysis for a wide range of geographic scales, ranging from the macroscopic level of a state or region to the microscopic level of a single neighborhood. Second, a GIS contains information used by other professions, enabling planners to access data that already have been collected for other purposes. Third, a GIS offers strong spatial analytical capabilities that make use of the point, line, and area features contained within the GIS. 11-22 The initial zone structure for regional travel demand model has been proposed as summarized below. Name three potential problems with this structure. Zones: 50 Employment: 100,000 Labor Force: 60,000 Population: 200,000 Trips produced by the study area: 100,000 Trips attracted by the study area: 150,000 The three problems are: The ratio of labor force to employment is 60,000/100,000 = 0.60, meaning that a substantial number of workers are imported into the study area but not considered within the regional model. With 200,000 people and 50 zones, we have an average of about 4,000 people per zone, whereas an average of 1,000 people per zone is desired. The number of trips produced and attracted by the area are not equal. (If students have already discussed productions and attractions, then it may be noted that the ratio of productions to attractions is 100,000/150,000 or about 0.67, which is outside of the desired heuristic rule of 0.9 to 1.1.) 11-23 Explain the flaw in each of these traffic analysis zones (TAZs) TAZ 1 contains dormitories, a research park, and 300 single family detached dwelling units. TAZ 2 straddles an interstate highway, with half of the zone east of the highway and half to the west of the highway. TAZ 3 contains 5,000 people The first zone contains multiple land uses—ideally a zone should predominantly contain a single land use. The second zone includes a major topographical barrier—the interstate highway. Ideally zones should not include such barriers. The third zone is probably too large for analysis (containing more than 1,000 people). That in itself is not a firm rule, but in practice such a zone could easily contain two major parallel routes which may adversely affect the traffic assignment process. 11-24 Define these four acronyms and explain how they affect the transportation planning process: MPO, CLRP, TIP, and STIP. Describe the composition of one of them. A Metropolitan Planning Organization (MPO) is a transportation policymaking organization made up of representatives from local government and transportation authorities. For example, the Thomas Jefferson Planning District Commission generally staffs the Charlottesville-Albemarle MPO. Serving on the MPO Policy Board are five voting members: two from the Charlottesville city council, two from the Albemarle County Board of Supervisors, and one from the Virginia Department of Transportation. Nonvoting members on the Policy Board include local transit providers (Jefferson Area United Transportation and Charlottesville Transit Service), local and regional entities (the Thomas Jefferson Planning District Commission and the University of Virginia), state agencies (the Virginia Department of Rail and Public Transportation), and federal agencies (the Federal Highway Administration, the Federal Aviation Administration, and the Federal Transit Administration). MPOs affect the planning process because they are given specific responsibilities under federal reauthorizations. One of these is to develop and update the Constrained Long Range Plan (CLRP). This plan has a 20 year planning horizon and must meet several requirements such as (1) it must identify the projected demand of persons and goods in the MPO over the period of the plan and (2) it must identify walkway and bicycle transportation facilities. A Transportation Improvement Program (TIP) is a series of projects (four years out at least) that the MPO approves. The TIP must be financially constrained, that is, funds must reflect those available. Ideally, the TIP is a way of allocating limited transportation resources among the various capital and operating needs of the area. (Under TEA-21 TIPs had a minimum three year horizon, now under SAFETEA-LU it is four years). Projects must be in the TIP to receive federal funds. Without the MPOs approval, the project cannot receive federal funds. Thus if an MPO did not approve a particular bypass, for example and did not include this project in its TIP, then the project could not receive federal funds. The State Transportation Improvement Program (STIP) is a programming document that includes all projects planned for implementation with the funds expected from FHWA and FTA for the upcoming three years. The STIP also includes each MPO’s TIP and all of the projects included in the first (now four) years of that TIP. Without FHWA approval of the STIP, the project cannot be funded. 11-25 Name the steps in the four step travel demand forecasting process where feedback can occur. Which of these feedback loops directly affects land use? Hint: See Figure 11.7. There are three loops where this feedback commonly occurs and three loops where this feedback may occur: Congested highway times from traffic assignment affect the trip distribution and mode choice steps Transit times and costs from traffic assignment affect the mode choice step Transit times and costs from traffic assignment affect trip distribution where transit and travel time form a composite impedance for travel cost Highway and transit costs may influence employment and residential locations (possibly in the form of a land use model) Auto occupancy may be a function of the time of day Highway and transit times and costs may influence the auto ownership model. The one feedback loop that directly affects land use is the fourth one named above: travel costs (whether in the form of monetary costs or travel time) may ultimately influence employment and residential locations. 11-26 Name and briefly explain the tools that may be available to local governments that may affect how land use decisions influence travel demand. Tools that are available to local governments may take the form of regulation, incentives, or both. Concurrency requirements are a regulatory technique, where a state or locale requires that sufficient transportation infrastructure be present to accommodate anticipated growth. One mechanism through which a state may implement concurrency is to require that localities adopt set of performance standards for various transportation facilities in the county. For roadways, these standards may be based on level of service, functional classification, and location. Grants for public and private entities are examples of incentives. In contrast with residential development, commercial sites tend to generate more tax revenue than what they require in the form of services, which may encourage localities to have zoning that favors commercial over residential development. It is possible that such a situation may result in longer commuting distances if workers are unable to find affordable housing in proximity to their employment. To encourage an increase in housing stock in areas that have relatively high employment, grants may be provided by a state to localities which (1) have a relatively high ratio of jobs to housing and (2) issue a certain number of residential building permits. Priority funding areas are a mix of regulation and incentives. For example, in an effort to limit or manage the geographical growth of an urban area, a state may require that certain types of transportation investments, such as statefunded infrastructure capacity expansions, only be undertaken in regions that meet criteria in terms of residential density, water and sewer availability, or other indicators of land development. 11-27 Briefly explain the differences in the factors considered in freight transportation planning versus passenger transportation planning. The analysis underpinning freight transportation planning considers three factors that differ from that of passenger transportation planning. First, unlike passenger freight, some commodities are shipments that are not time sensitive, thereby allowing the shipper’s choice of mode for those commodities to be made solely on the basis of cost and convenience. Second, freight-movement data are generally studied at a larger geographic scale than passenger movements, with county-to-county or state-to-state flows commonly analyzed. Third, whereas each passenger chooses his or her own mode and travel path, a single decision maker makes this choice for a large number of freight parcels. While an equilibrium network assignment for passenger travel may be defined as the assignment where no passenger can reduce his/her travel time (which is not necessarily the lowest system cost), a freight logistics provider may be able to consider the total system cost, where some parcels may have a longer delivery time than others. Chapter 12 Forecasting Travel Demand 12-1 Identify and briefly describe the two basic demand forecasting situations in transportation planning. There are two basic demand forecasting situations in transportation planning. The first involves travel demand studies for urban areas, and the second deals with intercity travel demand. The urban travel demand forecasts, when first developed in the 1950s and 1960s, required that extensive data sets be prepared using home interview and/or roadside interview surveys. The information gathered provided useful insight concerning the characteristics of the trip maker, such as age, sex, income, auto ownership, the land-use at each end of the trip, and the mode of travel. Today much of this information can be obtained through U.S. Census Bureau transportation planning products and from privately assembles data sets, in addition to the approaches used since the advent of urban travel forecasting. In the intercity case, travel patterns are forecast between two cities or metropolitan areas. Such data generally are aggregated to a greater extent than for urban travel forecasting, such as city population, average city income, and travel time or travel cost between city pairs. 12-2 Identify the three factors that affect demand for urban travel. The three factors that affect the demand for urban travel are: 1) the location and intensity of land-use, 2) the socio-economic characteristics of the people living in the area, and 3) the extent, cost, and quality of available transportation services. Land-use characteristics are a primary determinant in travel demand. The amount of traffic generated by a parcel of land depends on how the land is used. Socio-economic characteristics of the people living within the city also influence the demand for transportation. Lifestyles and values affect how people decide to use their resources for transportation. The availability of transportation facilities and services also affects the demand for travel. Travelers are sensitive to the levels of service provided by alternative transportation modes. 12-3 Define the following terms: home-based work (HBW) trips home-based other (HBO) trips nonhome-based (NHB) trips production attraction origin destination Home-based Work (HBW) trip – a trip for which the purpose is to go from home to work or from work to home. Home-based Other (HBO) trip – a trip for which the purpose is to go from home to another location other than work (e.g., shopping, school, theater) or from non-work locations to home. Non-Home Based (NHB) trip – a trip for which neither trip end is at home. A trip falling into this category might be one that goes from work to shopping, or school to work. Production – the ability of a zone to generate trip ends. For all non-home based trips, productions are synonymous with origins. Attraction – the ability of a zone to generate trip ends. For non-home based trips, attractions in a zone can be considered synonymous with trip destinations in that zone. Origin – point at which a trip begins. Destination – point at which a trip ends. 12-4 Given: Cross-classification data for the Jeffersonville Transportation Study Area Develop the family of cross-classification curves. Determine the number of trips produced (by purpose) for a traffic zone containing 500 houses with an average household income of $35,000. (Use high = $55,000; medium = $25,000; low = $15,000.)
Income ($) Percent of Households Autos per Household Trip Rate per Auto Trips (%)
High Med Low 0 1 2 3 0 1 2 3+ HBW HBO NHB
10000 0 30 70 48 48 4 0 2.0 6.0 11.5 17.0 38 34 28
20000 0 50 50 4 72 24 0 2.5 7.5 12.5 17.5 38 34 28
30000 10 70 20 2 53 40 5 4.0 9.0 14.0 19.0 35 34 31
40000 20 75 5 1 32 52 15 5.5 10.5 15.5 20.5 27 35 38
50000 50 50 0 0 19 56 25 7.5 12.0 17.0 22.0 20 37 43
60000 70 30 0 0 10 60 30 8.0 13.0 18.0 23.0 16 40 44
From the graphs above, the following tables can be produced. Number of HH per Level of Income
Income Percent HH/Zone Total HH
Low 11 500 55
Medium 75 500 375
High 14 500 70
Total 100 500
Percentage of HH Owning x Vehicles
Auto Ownership Income
Low Medium High
0 19 3 0
1 68 64 14
2 13 32 58
3+ 0 1 28
Total 100 100 100
Trips per HH per Income Level and Auto Ownership
Auto Ownership Income
Low Medium High
0 2 3 8
1 7 8 13
2 12 13 18
3+ 17 18 23
Next, develop a table to show the number of households owning x vehicles. This can be accomplished by multiplying the values found in the table of the number of HH per level of income and the percentage of HH owning x vehicles together. For example: For a low income level and 0 auto ownership = (55) × 19% = 11 For medium income level and 0 auto ownership = (375) × 3% = 11 For high income level and 0 auto ownership = (70) × 0% = 0 This procedure is used to generate the following table. Number of HH Owning x Vehicles
Auto Ownership Income
Low Medium High
0 11 11 0
1 37 240 10
2 7 120 41
3 0 4 19
Total 55 375 70
The total numbers of trips made by each income level are then determined as follows: Low Income = (2 × 11) + (7 × 37) + (12 × 7) + (17 × 0) = 365 Medium Income = (3 × 11) + (8 × 240) + (13 × 120) + (18 × 4) = 3,585 High Income = (8 × 0) + (13 × 10) + (18 × 41) + (23 × 19) = 1,305 Total Number of Trips = 365 + 3,585 + 1,305 = 5,255 Trips by Purpose (%)
Trip Purpose Income
Low Medium High
HBW 38 37 18
HBO 34 34 38
NHB 28 29 44
TOTAL 100 100 100
The number of trips by purpose can then be calculated as follows: HBW = (0.38 × 365) + (0.37 × 3585) + (0.18 × 1305) = 1,700 HBO = (0.34 × 365) + (0.34 × 3585) + (0.38 × 1305) = 1,839 NHB = (0.28 × 365) + (0.29 × 3585) + (0.44 × 1305) = 1,716 12-5 Given: A person travels to work in the morning and returns home in the evening. Determine: Productions and attractions generated in the work and residence zones. Home - 2 Productions, 0 Attractions. Work - 0 Productions, 2 Attractions 12-6 Describe and illustrate cross-classification procedures for (a) trip production and (b) trip attraction. The FHWA method for estimating trip productions and attractions is based on the use of cross classification. Cross classification is a method used to determine the number of trips that begin or end at the home. Relationships are developed based on socioeconomic data (typically income data) and origindestination surveys. Determining the number of productions and attractions involves a process of 5 steps which determine the percentage of trips by purpose, i.e. the number of trips that are home-based work (HBW), home-based other (HBO), and non-home based (NHB). From these percentages the number of productions and attractions can be determined. Step 1: Determine the percentage of households in each economic category (low, medium, high). Develop a plot of average zonal income versus income distribution. Step 2: Determine the distribution of auto ownership per household for each category. Draw a curve showing percent of households, at each income level, that owns x autos. Step 3: Determine the number of trips per household for each income-auto ownership category. Draw a curve showing the relationship between trips per household, household income and ownership. Step 4: Calculate the total number of trips generated in the zone. This is done by computing the number of households in each income-auto ownership category and multiplying this result by the number of trips per household, as determined in step 3, and summing the results. Thus, Ti = (HH) × (Ii) × [Σ(Aij) × (T / HH)ij] Where: Ti = trips generated by income group i HH = number of households in the zone Ii = percentage of households in the zone with income level i Aij = percentage of households in income level i with j autos per household (T / HH)ij = number of trips produced in a household at income level i and auto ownership j. Total trips generated: TT = ΣTi Step 5: Determine the percentage of trips by purpose.
12-7 Given: Socioeconomic data for the Jeffersonville Transportation Study Area as follows: Population = 72,173 Area = 70 sq mi Registered vehicles = 26,685 Single-family housing units = 15,675 Apartment units = 7,567 Retail employment = 5,502 Nonretail employment = 27,324 Student attendance = 28,551 Average household income = $17,500 Transportation analysis zones = 129 The results of the cross-classification analysis are as follows: Total trips produced for study area = 282,150 per day Home-to-work trips: 13% (36,680) Home-to-nonwork trips: 62% (174,933) Non-home trips: 25% (70,537) The attraction rates for the area have been developed using the following assumptions: 100% of home-to-work trips go to employment locations. Home-to-nonwork trips are divided into the following types: Visit friends: 10% Shopping: 60% School: 10% Nonretail employment: 20% Non-home trips are divided into the following types: Other employment area (nonretail): 60% Shopping: 40% Determine: The number of home-to-work, home-to-nonwork, and nonhome-based trips attracted to a zone with the following characteristics: population = 1,440; dwelling units = 630; retail employment = 40; nonretail employment = 650; school attendance = 0. Home-to-work trips attracted to zone (Home-to-work trips in study area) × (Total employ. in zone/Total employ. in study area) Home-to-work trips attracted to zone = 36,680 × [(650 + 40) / (27,324 + 5,502)] Home-to-work trips attracted to zone = 771 trips Home-to-nonwork trips attracted to zone Visiting Friends: (0.10) × [(Total home-to-nonwork trips) × (Total dwelling units in zone/Total dwelling units in study area)] Visiting Friends = (0.10) × [(174,933 × (630 / (15,675 + 7,567)))] Visiting Friends = 474 trips Shopping: (0.60) × [(Total home-to-nonwork trips) × (Total retail employ. in zone/Total retail employ. in study area)] Shopping = (0.60) × [(174,933 × (40 / 5,502))] Shopping = 763 trips School: (0.10) × [(Total home-to-nonwork trips) × (School attendance in zone/School attendance in study area)] School = (0.10) × [(174,933 × (0 / 28,551))] School = 0 trips Non-retail employment: (0.20) × [(Total home-to-nonwork trips) × (Non-retail employ. in zone/Non-retail employ. in study area)] Non-retail employment = (0.20) × [(174,933 × (650 / 27,324))] Non-retail employment = 832 trips Total home-to-nonwork trips in zone = 474 + 763 + 0 + 832 = 2,069 trips Non-home based trips attracted to zone Non-retail employment: (0.60) × [(Total non-home based trips) × (Non-retail employ. zone/Non-retail employ. in study area)] Non-retail employment = (0.60) × [(70,537 × (650 / 27,324))] Non-retail employment = 1,007 trips Shopping: (0.40) × [(Total home-to-nonwork trips) × (Total retail employ. in zone/Total retail employ. in study area)] Shopping = (0.40) × [(70,537 × (40 / 5,502))] Shopping = 205 trips Total non-home based trips in zone = 1,007 + 205 = 1,212 12-8 Given: Small town with three transportation analysis zones, and origin-destination survey results. Provide a trip distribution calculation using the gravity model for two iterations; assume Kij = 1. The following table shows the number of productions and attractions in each zone:
Zone 1 2 3 Total
Productions 250 450 300 1000
Attractions 395 180 425 1000
The survey’s results for the zones’ travel time in minutes were as follows:
Zone 1 2 3
1 6 4 2
2 2 8 3
3 1 3 5
The following table shows travel time versus friction factor.
Time (min) 1 2 3 4 5 6 7 8
Friction factor 82 52 50 41 39 26 20 13
The mathematical formulation for the gravity model as provided as Equation 12.3: (AF Kj ij ij ) T Pij = i (A F Kj ij ij ) Since Kij = 1, this factor does not affect calculations. The iterative application of Equation 12.3 is as follows: Iteration 1 T11 = 250 × ((395 × 26) / ((395 × 26) + (180 × 41) + (425 × 52))) T11 = 250 × (10,270 / 39,750) T11 = 65 T12 = 250 × ((180 × 41) / ((395 × 26) + (180 × 41) + (425 × 52))) T12 = 250 × (7,380 / 39,750) T12 = 46 T13 = 250 × ((425 × 52) / ((395 × 26) + (180 × 41) + (425 × 52))) T13 = 250 × (22,100 / 39,750) T13 = 139 T21 = 450 × ((395 × 52) / ((395 × 52) + (180 × 13) + (425 × 50))) T21 = 450 × (20,540 / 44,130) T21 = 209 T22 = 450 × ((180 × 13) / ((395 × 52) + (180 × 13) + (425 × 50))) T22 = 450 × (2,340 / 44,130) T22 = 24 T23 = 450 × ((425 × 50) / ((395 × 52) + (180 × 13) + (425 × 50))) T23 = 450 × (21,250 / 44,130) T23 = 217 T31 = 300 × ((395 × 82) / ((395 × 82) + (180 × 50) + (425 × 39))) T31 = 300 × (32,390 / 57,965) T31 = 168 T32 = 300 × ((180 × 50) / ((395 × 82) + (180 × 50) + (425 × 39))) T32 = 300 × (9,000 / 57,965) T32 = 46 T33 = 300 × ((425 × 39) / ((395 × 82) + (180 × 50) + (425 × 39))) T33 = 300 × (16,575 / 57,965) T33 = 86 Trip Matrix for Iteration 1
Zone 1 2 3 Productions
1 65 46 139 250
2 209 24 217 450
3 168 46 86 300
Computed Attractions 442 116 442
Given Attractions 395 180 425
Next, calculate the adjusted attraction factors using Equation 12.4. Aj Ajk = Aj k( −1) C j k( −1) Zone 1 Ajk = (395 / 442) × 395 Ajk = 353 Zone 2 Ajk = (180 / 116) × 180 Ajk = 279 Zone 3 Ajk = (425 / 442) × 425 Ajk = 409 Now apply the gravity model formula for Iteration 2 using the above adjusted attraction factors. Iteration 2 T11 = 250 × ((353 × 26) / ((353 × 26) + (279 × 41) + (409 × 52))) T11 = 250 × (9,178 / 41,885) T11 = 55 T12 = 250 × ((279 × 41) / ((353 × 26) + (279 × 41) + (409 × 52))) T12 = 250 × (11,439 / 41,885) T12 = 68 T13 = 250 × ((409 × 52) / ((353 × 26) + (279 × 41) + (409 × 52))) T13 = 250 × (21,268 / 41,885) T13 = 127 T21 = 450 × ((353 × 52) / ((353 × 52) + (279 × 13) + (409 × 50))) T21 = 450 × (18,356 / 42,433) T21 = 195 T22 = 450 × ((279 × 13) / ((353 × 52) + (279 × 13) + (409 × 50))) T22 = 450 × (3,627 / 42,433) T22 = 38 T23 = 450 × ((409 × 50) / ((353 × 52) + (279 × 13) + (409 × 50))) T23 = 450 × (20,450 / 42,433) T23 = 217 T31 = 300 × ((353 × 82) / ((353 × 82) + (279 × 50) + (409 × 39))) T31 = 300 × (28,946 / 58,847) T31 = 148 T32 = 300 × ((279 × 50) / ((353 × 82) + (279 × 50) + (409 × 39))) T32 = 300 × (13,950 / 58,847) T32 = 71 T33 = 300 × ((409 × 39) / ((353 × 82) + (279 × 50) + (409 × 39))) T33 = 300 × (15,951 / 58,847) T33 = 81 Trip Matrix for Iteration 2
Zone 1 2 3 Productions
1 55 68 127 250
2 195 38 217 450
3 148 71 81 300
Computed Attractions 398 177 425
Given Attractions 395 180 425
Observe that the computed attractions approximately equal the given attractions. A total convergence would be expected in another iteration. 12-9 Given: Study area with four transportation analysis zones, and origin-destination survey results. Provide a trip distribution calculation using the gravity model for two iterations; assume Kij = 1. Travel time (min)
District Productions Attractions 1 2 3 4
1 3400 2800 4 11 15 10
2 6150 6500 11 6 6 9
3 3900 2550 15 6 6 11
4 2800 Friction factors are 4400 as follows: 10 9 11 4
Travel time (minutes) 4 5 6 7 8 9 10 11 12 13 14 15
Fij 1.51 1.39 1.30 1.22 1.13 1.04 0.97 0.92 0.85 0.80 0.75 0.71
The mathematical formulation for the gravity model as provided as Equation 12.3: (AF Kj ij ij ) T Pij = i (A F Kj ij ij ) Since Kij = 1, this factor does not affect calculations. The iterative application of Equation 12.3 is as follows: Iteration 1 T11 = 3,400 × ((2,800 × 1.51) / ((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97))) T11 = 3,400 × (4,228 / 16287) T11 = 883 T12 = 3,400 × ((6,500 × 0.92) / ((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97))) T12 = 3,400 × (5,980 / 16,287) T12 = 1,248 T13 = 3,400 × ((2,550 × 0.71) / ((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97))) T13 = 3,400 × (1,811 / 16,287) T13 = 378 T14 = 3,400 × ((4,400 × 0.97)/ ((2,800 × 1.51) + (6,500 × 0.92) + (2,550 × 0.71) + (4,400 × 0.97))) T14 = 3,400 × (4268 / 16,287) T14 = 891 T21 = 6,150 × ((2,800 × 0.92) / ((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04))) T21 = 6,150 × (2,576 / 18,917) T21 = 837 T22 = 6,150 × ((6,500 × 1.30) / ((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04))) T22 = 6,150 × (8,450 / 18,917) T22 = 2747 T23 = 6,150 × ((2,550 × 1.30) / ((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04))) T23 = 6,150 × (3,315 / 18,917) T23 = 1078 T24 = 6,150 × ((4,400 × 1.04) / ((2,800 × 0.92) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 1.04))) T24 = 6,150 × (4576 / 18,917) T24 = 1488 T31 = 3,900 × ((2,800 × 0.71) / ((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92))) T31 = 3,900 × (1,988 / 17,801) T31 = 436 T32 = 3,900 × ((6,500 × 1.30) / ((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92))) T32 = 3,900 × (8,450 / 17,801) T32 = 1851 T33 = 3,900 × ((2,550 × 1.30) / ((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92))) T33 = 3,900 × (3,315 / 17,801) T33 = 726 T34 = 3,900 × ((4,400 × 0.92) / ((2,800 × 0.71) + (6,500 × 1.30) + (2,550 × 1.30) + (4,400 × 0.92))) T34 = 3,900 × (4,048 / 17,801) T34 = 887 T41 = 2,800 × ((2,800 × 0.97) / ((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51))) T41 = 2,800 × (2,716 / 18,466) T41 = 412 T42 = 2,800 × ((6,500 × 1.04) / ((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51))) T42 = 2,800 × (6,760 / 18,466) T42 = 1025 T43 = 2,800 × ((2,550 × 0.92) / ((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51))) T43 = 2,800 × (2,346 / 18,466) T43 = 356 T44 = 2,800 × ((4,400 × 1.51) / ((2,800 × 0.97) + (6,500 × 1.04) + (2,550 × 0.92) + (4,400 × 1.51))) T44 = 2,800 × (6,644 / 18,466) T44 = 1007 Trip Matrix for Iteration 1
Zone 1 2 3 4 Productions
1 883 1,248 378 891 3,400
2 837 2,747 1,078 1,488 6,150
3 436 1851 726 887 3,900
4 412 1,025 356 1,007 2,800
Computed Attractions 2,568 6,871 2,538 4,273
Given Attractions 2,800 6,500 2,550 4,400
Next, calculate the adjusted attraction factors using Equation 12.4. Aj Ajk = Aj k( −1) C j k( −1) Zone 1 Ajk = (2,800 / 2,568) × 2,800 Ajk = 3,053 Zone 2 Ajk = (6,500 / 6,871) × 6,500 Ajk = 6,149 Zone 3 Ajk = (2,550 / 2,538) × 2,550 Ajk = 2,562 Zone 4 Ajk = (4,400 / 4,273) × 4,400 Ajk = 4,531 Now apply the gravity model formula for Iteration 2 using the above adjusted attraction factors. Iteration 2 T11 = 3,400 × ((3,053 × 1.51) / ((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97))) T11 = 3,400 × (4,610 /16,481) T11 = 951 T12 = 3,400 × ((6,149 × 0.92) / ((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97))) T12 = 3,400 × (5,657 /16,481) T12 = 1167 T13 = 3,400 × ((2,562 × 0.71) / ((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97))) T13 = 3,400 × (1,819 /16,481) T13 = 375 T14 = 3,400 × ((4,531 × 0.97) / ((3,053 × 1.51) + (6,149 × 0.92) + (2,562 × 0.71) + (4,531 × 0.97))) T14 = 3,400 × (4,395 /16,481) T14 = 907 T21 = 6,150 × ((3,053 × 0.92) / ((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04))) T21 = 6,150 × (2,809 / 18,845) T21 = 917 T22 = 6,150 × ((6,149 × 1.30) / ((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04))) T22 = 6,150 × (7,994 / 18,845) T22 = 2609 T23 = 6,150 × ((2,562 × 1.30) / ((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04))) T23 = 6,150 × (3,331 / 18,845) T23 = 1087 T24 = 6,150 × ((4,531 × 1.04) / ((3,053 × 0.92) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 1.04))) T24 = 6,150 × (4,712 / 18,845) T24 = 1538 T31 = 3,900 × ((3,053 × 0.71) / ((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92))) T31 = 3,900 × (2,168 / 17,660) T31 = 479 T32 = 3,900 × ((6,149 × 1.30) / ((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92))) T32 = 3,900 × (7,994 / 17,660) T32 = 1765 T33 = 3,900 × ((2,562 × 1.30) / ((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92))) T33 = 3,900 × (3,331 / 17,660) T33 = 736 T34 = 3,900 × ((4,531 × 0.92) / ((3,053 × 0.71) + (6,149 × 1.30) + (2,562 × 1.30) + (4,531 × 0.92))) T34 = 3,900 × (4,169 / 17,660) T34 = 921 T41 = 2,800 × ((3,053 × 0.97) / ((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51))) T41 = 2,800 × (2,961 / 18,555) T41 = 447 T42 = 2,800 × ((6,149 × 1.04) / ((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51))) T42 = 2,800 × (6,395 / 18,555) T42 = 965 T43 = 2,800 × ((2,562 × 0.92) / ((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51))) T43 = 2,800 × (2,357 / 18,555) T43 = 356 T44 = 2,800 × ((4,531 × 1.51) / ((3,053 × 0.97) + (6,149 × 1.04) + (2,562 × 0.92) + (4,531 × 1.51))) T44 = 2,800 × (6,842 / 18,555) T44 = 1032 Trip Matrix for Iteration 2
Zone 1 2 3 4 Productions
1 951 1167 375 907 3,400
2 917 2609 1087 1538 6,150
3 479 1765 736 921 3,900
4 447 965 356 1,032 2,800
Computed Attractions 2,794 6,506 2,554 4,398
Given Attractions 2,800 6,500 2,550 4,400
Observe that the computed attractions approximately equal the given attractions. A total convergence would be expected in another iteration. 12-10 Given: Table with production and attraction data. Determine: The number of productions and attractions that should be used for each zone in the second iteration.
1 2 3 4
P 100 200 400 600
A 300 100 200 700
P' 100 200 400 600
A' 250 150 300 600
Calculate the adjusted attraction factors using Equation 12.4. Aj Ajk = Aj k( −1) C j k( −1) Zone 1 Ajk = (300 / 250) × 300 Ajk = 360 Zone 2 Ajk = (100 / 150) × 100 Ajk = 67 Zone 3 Ajk = (200 / 300) × 200 Ajk = 133 Zone 4 Ajk = (700 / 600) × 700 Ajk = 817 The attractions change with each iteration; there is no change in productions. The following table summarizes the results.
1 2 3 4
P” 100 200 400 600
A” 360 67 133 817
12-11 Given: Table with production and attraction data. Determine: The number of productions and attractions that should be used for each zone in the second iteration.
1 2 3 4 5
P 100 150 350 500 200
A 250 100 150 500 300
P' 100 150 350 500 200
A' 200 100 250 400 350
Calculate the adjusted attraction factors using Equation 12.4. Aj Ajk = Aj k( −1) C j k( −1) Zone 1 Ajk = (250 / 200) × 250 Ajk = 313 Zone 2 Ajk = (100 / 100) × 100 Ajk = 100 Zone 3 Ajk = (150 / 250) × 150 Ajk = 90 Zone 4 Ajk = (500 / 400) × 500 Ajk = 625 Zone 5 Ajk = (300 / 350) × 300 Ajk = 257 The attractions change with each iteration; there is no change in productions. The following table summarizes the results.
1 2 3 4 5
P” 100 150 350 500 200
A” 313 100 90 625 257
12-12 The Jeffersonville Transportation Study area has been divided into four traffic zones. The following data have been compiled. Complete the second iteration. Travel Time (min) District Productions Attractions 1 2 3 4 1000 1000 5 8 12 15 2000 700 8 5 10 8 3000 6000 12 10 5 7 2200 500 15 8 7 5
Travel Time 1 5 6 7 8 10 12 15
Fij 2.00 1.30 1.10 1.00 0.95 0.85 0.80 0.65
After the first iteration, the trip table was
District 1 2 3 4 Ps
1 183 94 677 46 1000
2 256 244 1372 128 2000
3 250 186 2404 160 3000
4 180 183 1657 180 2200
As 869 707 6110 514 8200
Calculate the adjusted attraction factors using Equation 12.4. Aj Ajk = Aj k( −1) C j k( −1) Zone 1 Ajk = (1,000 / 869) × 1,000 Ajk = 1,151 Zone 2 Ajk = (700 / 707) × 700 Ajk = 693 Zone 3 Ajk = (6,000 / 6,110) × 6,000 Ajk = 5,892 Zone 4 Ajk = (500 / 514) × 500 Ajk = 486 Now apply the gravity model formula for Iteration 2 using the above adjusted attraction factors. Iteration 2 T11 = 1,000 × ((1,151 × 1.3) / ((1,151 × 1.3) + (693 × 0.95) + (5,892 × 0.8) + (496 × 0.65))) T11 = 1,000 × (1,496 / 7,190) T11 = 208 T12 = 1,000 × ((693 × 0.95) / ((1,151 × 1.3) + (693 × 0.95) + (5,892 × 0.8) + (496 × 0.65))) T12 = 1,000 × (658 / 7,190) T12 = 92 T13 = 1,000 × ((1,151 × 1.3) / ((1,151 × 1.3) + (693 × 0.95) + (5,892 × 0.8) + (496 × 0.65))) T13 = 1,000 × (4,714 / 7,190) T13 = 655 T14 = 1,000 × ((1,151 × 1.3) / ((1,151 × 1.3) + (693 × 0.95) + (5,892 × 0.8) + (496 × 0.65))) T14 = 1,000 × (322 / 7,190) T14 = 45 T21 = 2,000 × ((1,151 × 0.95) / ((1,151 × 0.95) + (693 × 1.3) + (5,892 × 0.85) + (496 × 0.95))) T21 = 2,000 × (1,093 / 7,473) T21 = 293 T22 = 2,000 × ((1,151 × 0.95) / ((1,151 × 0.95) + (693 × 1.3) + (5,892 × 0.85) + (496 × 0.95))) T22 = 2,000 × (901 / 7,473) T22 = 241 T23 = 2,000 × ((1,151 × 0.95) / ((1,151 × 0.95) + (693 × 1.3) + (5,892 × 0.85) + (496 × 0.95))) T23 = 2,000 × (5,008 / 7,473) T23 = 1,340 T24 = 2,000 × ((1,151 × 0.95) / ((1,151 × 0.95) + (693 × 1.3) + (5,892 × 0.85) + (496 × 0.95))) T24 = 2,000 × (471 / 7,473) T24 = 126 T31 = 3,000 × ((1,151 × 0.80) / ((1,151 × 0.80) + (693 × 0.85) + (5,892 × 1.3) + (496 × 1))) T31 = 3,000 × (921 / 9,666) T31 = 286 T32 = 3,000 × ((693 × 0.85) / ((1,151 × 0.80) + (693 × 0.85) + (5,892 × 1.3) + (496 × 1))) T32 = 3,000 × (589 / 9,666) T32 = 183 T33 = 3,000 × ((5,892 × 1.3) / ((1,151 × 0.80) + (693 × 0.85) + (5,892 × 1.3) + (496 × 1))) T33 = 3,000 × (7,660 / 9,666) T33 = 2,377 T34 = 3,000 × ((496 × 1) / ((1,151 × 0.80) + (693 × 0.85) + (5,892 × 1.3) + (496 × 1))) T34 = 3,000 × (496 / 9,666) T34 = 154 T41 = 2,200 × ((1,151 × 0.65) / ((1,151 × 0.65) + (693 × 0.95) + (5,892 × 1) + (496 × 1.3))) T41 = 2,200 × (748 / 7,943) T41 = 207 T42 = 2,200 × ((693 × 0.95) / ((1,151 × 0.65) + (693 × 0.95) + (5,892 × 1) + (496 × 1.3))) T42 = 2,200 × (658 / 7,943) T42 = 182 T43 = 2,200 × ((5,892 × 1) / ((1,151 × 0.65) + (693 × 0.95) + (5,892 × 1) + (496 × 1.3))) T43 = 2,200 × (5,892 / 7,943) T43 = 1,632 T44 = 2,200 × ((496 × 1.3) / ((1,151 × 0.65) + (693 × 0.95) + (5,892 × 1) + (496 × 1.3))) T44 = 2,200 × (645 / 7,943) T44 = 179 Trip Matrix for Iteration 2
Zone 1 2 3 4 Productions
1 208 92 655 45 1,000
2 293 241 1,340 126 2,000
3 286 183 2,377 154 3,000
4 207 182 1,632 179 2,200
Computed Attractions 994 698 6,004 504
Given Attractions 1,000 700 6,000 500
Observe that the computed attractions approximately equal the given attractions. A total convergence would be expected in another iteration. 12-13 For the travel pattern illustrated in Figure 12.18, develop the Fratar method of trip distribution for two iterations. First Iteration
Zones Present Totals Growth Factor Estimated Future Totals
A 100 3 300
B 250 4 1,000
C 400 2 800
D 300 1 300
Next, use Equation 12.5 to solve the problem. TAB = 300 × ((25 × 4) / [(25 × 4) + (50 × 2) + (25 × 1)]) TAB = 300 × (100 / 225) TAB = 133.33 TAC = 300 × ((50 × 2) / [(25 × 4) + (50 × 2) + (25 × 1)]) TAC = 300 × (100 / 225) TAC = 133.33 TAD = 300 × ((25 × 1) / [(25 × 4) + (50 × 2) + (25 × 1)]) TAD = 300 × (25 / 225) TAD = 33.33 TBA = 1,000 × ((25 × 3) / [(25 × 3) + (150 × 2) + (75 × 1)]) TBA = 1,000 × (75 / 450) TBA = 166.67 TBC = 1,000 × ((150 × 2) / [(25 × 3) + (150 × 2) + (75 × 1)]) TBC = 1,000 × (300 / 450) TBC = 666.67 TBD = 1,000 × ((75 × 1) / [(25 × 3) + (150 × 2) + (75 × 1)]) TBD = 1,000 × (75 / 450) TBD = 166.67 TCA = 800 × ((50 × 3) / [(50 × 3) + (150 × 4) + (200 × 1)]) TCA = 800 × (150 / 950) TCA = 126.32 TCB = 800 × ((150 × 4) / [(50 × 3) + (150 × 4) + (200 × 1)]) TCB = 800 × (600 / 950) TCB = 505.26 TCD = 800 × ((200 × 1) / [(50 × 3) + (150 × 4) + (200 × 1)]) TCD = 800 × (200 / 950) TCD = 168.42 TDA = 300 × ((25 × 3) / [(25 × 3) + (75 × 4) + (200 × 2)]) TDA = 300 × (75 / 775) TDA = 29.03 TDB = 300 × ((75 × 4) / [(25 × 3) + (75 × 4) + (200 × 2)]) TDB = 300 × (300 / 775) TDB = 116.13 TDC = 300 × ((200 × 2) / [(25 × 3) + (75 × 4) + (200 × 2)]) TDC = 300 × (400 / 775) TDC = 154.84 Next, calculate the movement between zones as well as the new growth factors. A - B = (133.33 + 166.67) / 2 = 150 A - C = (133.33 + 126.32) / 2 = 129.83 - D = (33.33 + 29.03) / 2 = 31.18 TOTAL = 311.01 New Growth factor for A = 300 / 311.01 = 0.965 - A = (133.33 + 166.67) / 2 = 150 B - C = (666.67 + 505.26) / 2 = 585.97 - D = (166.67 + 116.13) / 2 = 141.40 TOTAL = 877.37 New Growth factor for B = 1,000 / 877.37 = 1.140 - A = (133.33 + 126.32) / 2 = 129.83 C - B = (666.67 + 505.26) / 2 = 585.97 - D = (154.84 + 168.42) / 2 = 161.63 TOTAL = 877.43 New Growth factor for C = 800 / 877.43 = 0.912 - A = (33.33 + 29.03) / 2 = 31.18 D - B = (166.67 + 116.13) / 2 = 141.40 D - C = (154.84 + 168.42) / 2 = 161.63 TOTAL = 334.21 New Growth factor for D = 300 / 334.21 = 0.898 Second Iteration
Zones Present Totals Growth Factor Estimated Future Totals
A 311.01 0.965 300
B 877.37 1.140 1,000
C 877.43 0.912 800
D 334.21 0.898 300
Next, use Equation 12.5 to solve the problem. TAB = 300 × ((150 × 1.14) / [(150 × 1.1 4) + (129.83 × 0.912) + (31.18 × 0.898)]) TAB = 300 × (171 / 317.40) TAB = 161.63 TAC = 300 × ((129.83 × 0.912) / [(150 × 1.1 4) + (129.83 × 0.912) + (31.18 × 0.898)]) TAC = 300 × (118.40 / 317.40) TAC = 111.91 TAD = 300 × ((31.18 × 0.898) / [(150 × 1.1 4) + (129.83 × 0.912) + (31.18 × 0.898)]) TAD = 300 × (28 / 317.40) TAD = 26.46 TBA = 1,000 × ((150 × 0.965) / [(150 × 0.965) + (585.97 × 0.912) + (141.4 × 0.898)]) TBA = 1,000 × (144.75 / 806.13) TBA = 179.56 TBC = 1,000 × ((585.97 × 0.912) / [(150 × 0.965) + (585.97 × 0.912) + (141.4 × 0.898)]) TBC = 1,000 × (534.40 / 806.13) TBC = 662.92 TBD = 1,000 × ((141.40 × 0.898) / [(150 × 0.965) + (585.97 × 0.912) + (141.4 × 0.898)]) TBD = 1,000 × (126.98 / 806.13) TBD = 157.51 TCA = 800 × ((129.83 × 0.965) / [(129.83 × 0.965) + (585.97 × 1.14) + (161.63 × 0.898)]) TCA = 800 × (125.29 / 938.44) TCA = 106.80 TCB = 800 × ((585.97 × 1.14)/ [(129.83 × 0.965) + (585.97 × 1.14) + (161.63 × 0.898)]) TCB = 800 × (668.01 / 938.44) TCB = 569.46 TCD = 800 × ((161.63 × 0.898)/ [(129.83 × 0.965) + (585.97 × 1.14) + (161.63 × 0.898)]) TCD = 800 × (145.14 / 938.44) TCD = 123.73 TDA = 300 × ((31.18 × 0.965) / [(31.18 × 0.965) + (141.4 × 1.14) + (161.63 × 0.912)]) TDA = 300 × (30.09 / 338.69) TDA = 26.65 TDB = 300 × ((141.4 × 1.14) / [(31.18 × 0.965) + (141.4 × 1.14) + (161.63 × 0.912)]) TDB = 300 × (161.2 / 338.69) TDB = 142.78 TDC = 300 × ((161.63 × 0.912) / [(31.18 × 0.965) + (141.4 × 1.14) + (161.63 × 0.912)]) TDC = 300 × (147.41 / 338.69) TDC = 130.57 Next, calculate the movement between zones. A - B = (161.63 + 179.56) / 2 = 170.60 A - C = (111.91 + 106.80) / 2 = 109.36 - D = (26.46 + 26.65) / 2 = 26.56 TOTAL = 306.52 - A = (161.63 + 179.56) / 2 = 170.60 B - C = (662.92 + 569.46) / 2 = 616.19 - D = (157.51 + 142.78) / 2 = 150.15 TOTAL = 936.94 - A = (111.91 + 106.80) / 2 = 109.36 C - B = (662.92 + 569.46) / 2 = 616.19 - D = (123.73 + 130.57) / 2 = 127.15 TOTAL = 852.70 - A = (26.46 + 26.66) / 2 = 26.56 D - B = (157.51 + 142.78) / 2 = 150.15 D - C = (123.73 + 130.57) / 2 = 127.15 TOTAL = 303.86 Results
Zone New Total Desired Total
A 306 300
B 937 1,000
C 853 800
D 304 300
These results could serve as the starting point for a third iteration. 12-14 Redo Problem 12-13 using the average growth factor method. First Iteration
Zones Present Totals Growth Factor Estimated Future Totals
A 100 3 300
B 250 4 1,000
C 400 2 800
D 300 1 300
Next, use Equation 12.5a to solve the problem. In this step, the movement between zones is calculated. TAB = 25 × ((3 + 4) / 2) TAB = 87.5 TAC = 50 × ((3 + 2) / 2) TAC = 125 TAD = 25 × ((3 + 1) / 2) TAD = 50 TBC = 150 × ((4 + 2) / 2) TBC = 450 TBD = 75 × ((4 + 1) / 2) TBD = 187.5 TCD = 200 × ((2 + 1) / 2) TCD = 300 Next, sum the trip ends in each zone and develop the new growth factors. TA = TAB + TAC + TAD = 87.5 + 125 + 50 = 262.5 TB = TBA + TBC + TBD = 87.5 + 450 + 187.5 = 725 TC = TCA + TCB + TCD = 125 + 450 + 300 = 875 TD = TDA + TDB + TDC = 50 + 187.5 + 300 = 537.5 New growth factor for A = 300 / 262.5 = 1.143 New growth factor for B = 1,000 / 725 = 1.379 New growth factor for C = 800 / 875 = 0.914 New growth factor for D = 300 / 537.5 = 0.558 Second Iteration
Zones Present Totals Growth Factor Estimated Future Totals
A 262.5 1.143 300
B 725 1.379 1,000
C 875 0.914 800
D 537.5 0.558 300
Next, use Equation 12.5a to solve the problem. In this step, the movement between zones is calculated. TAB = 87.5 × ((1.143 + 1.379) / 2) TAB = 110.3 TAC = 125 × ((1.143 + 0.914) / 2) TAC = 128.6 TAD = 50 × ((1.143 + 0.558) / 2) TAD = 42.5 TBC = 450 × ((1.379 + 0.914) / 2) TBC = 515.9 TBD = 187.5 × ((1.379 + 0.558) / 2) TBD = 181.6 TCD = 300 × ((0.914 + 0.558) / 2) TCD = 220.9 Next, sum the trip ends in each zone and develop the new growth factors. TA = TAB + TAC + TAD = 110.3 + 128.6 + 42.5 = 281 TB = TBA + TBC + TBD = 110.3 + 515.9 + 181.6 = 808 TC = TCA + TCB + TCD = 128.6 + 515.9 + 220.9 = 865 TD = TDA + TDB + TDC = 42.5 + 181.6 + 220.9 = 445 New growth factor for A = (1.143)(262.5) / 281 = 1.066 New growth factor for B = (1.379)(725) / 808 = 1.238 New growth factor for C = (0.914)(875) / 865 = 0.924 New growth factor for D = (0.558)(537.5) / 445 = 0.674 Results
Zone New Total Desired Total
A 281 300
B 808 1,000
C 866 800
D 445 300
Additionally, these results could serve as the starting point for a third iteration. 12-15 What data are required in order to use (a) the gravity model and (b) the Fratar model? The data needed for the gravity model include: Total number of trips produced in each zone Number of trips attracted to each zone Travel time between each zone (to determine friction factor F) Socioeconomic adjustment factor for interchanges. The data required to use the Fratar model include: Present trip generation in each zone Growth factors of each zone • Present number of trips between zones. 12-16 The amount of lumber produced and consumed by three states is shown in the following table. Intrastate shipment distances are 200 miles and interstate distances are 800 miles (between states 1 and 2), 1,000 miles (between states 1 and 3), and 400 miles (between states 2 and 3). Assuming an impedance function of the form 1/d, estimate the tonnage of lumber that will travel between the three states: Tons of Lumber Produced and Consumed Per Year (Tons)
State Lumber Produced Lumber Consumed
1 5,880 980
2 3,300 10,000
3 9,800 8,000
Impedance factors are established as follows
State 1 2 3
1 1/200 = 0.005 1/800 = 0.00125 1/1000 = 0.001
2 1/800 = 0.00125 1/200 = 0.005 1/400 = 0.0025
3 1/1000 = 0.001 1/400 = 0.0025 1/200 = 0.005
The general form for the gravity model is P A Fi j ij Tij = n (A Fj ij ) j 1= For example, the number of trips from zone 1 to zone 2, T12, is T12 = = 2894 Thus the initial trip table is
1 2 3 Total
1 1134 2894 1852 5880
2 57 2317 927 3300
3 146 3713 5941 9800
Total 1337 8924 8720
Because computed attractions do not match given attractions, attractions are adjusted using Equation 12.4: Aj Ajk = Aj k( −1) C j k( −1) For example, attractions for the second iteration for zone 2 can be found as = (10,000)(10,000)/8,924 = 11,206. Similarly, the new attractions for zones 1 and 3 are found to be 719 and 7,340, respectively. After five iterations, the final flows are found as shown below.
1 2 3 Total
1 835 3320 1724 5880
2 39 2462 799 3300
3 106 4218 5476 9800
Total 980 10000 8000
12-17 Suppose that traffic congestion has rendered the distance-based impedance function unsuitable for Problem 12-16. A detailed survey yields present-day commodity flows shown below. In future, state #1 lumber production will increase to 18,000 and state #2 lumber consumption will increase to 21,800. Estimate the lumber flows between the three states.
State 1 2 3 Total Produced
1 200 1,800 4,000 6,000
2 100 200 3,000 3,300
3 800 7,000 1,200 9,000
Total Consumed 1,100 9,000 8,200
Because impedances are not available and an older OD matrix is available, the Fratar method is appropriate. In the application of Eq. 12.5, note that the growth factor Gi refers to lumber produced and the growth factor Gj refers to lumber consumed. Note also that tij is a true from/to matrix (e.g., lumber tons from zone i to zone j will not equal tons from zone j to zone i). The growth factors are as follows:
Zone Production Growth Factor (Gi) Consumption Growth Factor (Gj)
1 3.0 (18,000/6,000) 1.0 (No change)
2 1.0 (No change) 2.422 (21,800/9,000)
3 1.0 (No change) 1.0 (No change)
Equation 12.5 may be applied using the tonnage of lumber from zone 1 to zone 2 as an example: t G Tij = (t Gi i ) ij j t Gij j j T12 = (t G1 1) t G12 2 t G11 1 + t G12 2 + t G13 3 T12 =(6,000 3× ) = 9,168 Ten iterations of the Fratar method gives the following trip interchange matrix:
State 1 2 3 Total Produced
1 532 11,969 5,499 18,000
2 153 767 2,379 3,300
3 414 9,064 321 9,800
Total Consumed 1,100 21,800 8,200
12-18 Survey data suggest the trip interchange matrix shown below. Calibrate the friction factors for one iteration, assuming travel times are 2.0 minutes for interzonal trips and 1.0 minute for all intrazonal trips. Assume the friction factor takes the form t-b.
Given Trips
Zone 1 2 3
1 10 40 70
2 20 50 80
3 30 60 90
Based on the trip table, the productions and attractions are:
Zone Productions Attractions
1 120 60
2 150 150
3 180 240
The initial guess is to assume b = 1. Thus the intrazonal friction factors are 1/1 = 1 and the interzonal friction factors are 1/2 = 0.5 The gravity model is then applied with these initial friction factors. For example, the trips from zone 1 to zone 1 are found to be approximately 28. T11 = PA F1 1 11 = (120 60 1)( )( ) ) = 28 A F1 11 + A F2 12 + A F3 13 (60 1)( ) (+ 150 0.5)( ) (+ 240 0.5)( The remaining trips are found in a similar manner. 1 2 3
1 28 35 56
2 15 75 60
3 16 39 125
We now compare predicted and actual results for each trip impedance. In this particular example, we only have two trip impedances: impedances for intrazonal trips and impedances for interzonal trips. For intrazonal trips (T11, T22, T23) a perfect friction factor would have predicted 10 + 50 + 90 = 150 trips. The guess with b = 1 meant that the model showed 28 + 75 + 125 = 228 trips. For interzonal trips (T12,T13, T21, T23, T31, and T33) a perfect friction factor would have predicted 40 + 70 + 20 + 80 + 30 + 60 = 300 trips. The guess with b = 1 predicted, however, 35 + 56 + 15 + 60 + 16 + 39 = 222 trips The adjusted friction factors are: Intrazonal travel = (1)(150)/228 = 0.658 Interzonal travel = (0.5)(300)/222 = 0.676 The regression equation will be: F = t–b Ln(F) = – b ln (t) Ln (0.658) = –b ln(1) Ln (0.676) = –b ln(2) b = 0.565 Thus the friction factor based on one iteration is t–b. For intrazonal travel (with travel time = 1.0) the friction factor is thus 1.0. For interzonal travel time with travel time = 2.0, the friction factor is thus 2.0–0.565 = 0.676. 12-19 For Problem 12-18, a field study of travel times was conducted which yielded the travel times of 1.5 minutes for intrazonal trips and 4.0 minutes for interzonal factors. Perform the first iteration of friction factor calibration. Based on the trip table, the productions and attractions are:
Zone Productions Attractions
1 120 60
2 150 150
3 180 240
The initial guess is to assume b = 1. Thus the intrazonal friction factors are 1/1.5 = 0.67 and the interzonal friction factors are 1/4 = 0.25 The gravity model is then applied with these initial friction factors. For example, the trips from zone 1 to zone 1 are found to be approximately 35. P A F1 1 11 (120 60 0.67)( )( ) T11 = = )( ) = 35 A F1 11 + A F2 12 + A F3 13 (60 0.67)( ) (+ 150 0.25)( ) (+ 240 0.25 The remaining trips are found in a similar manner. 35.0 32.7 52.3 12.8 85.9 51.3 12.7 31.6 135.7 We now compare predicted and actual results for each trip impedance. In this particular example, we only have two trip impedances: impedances for intrazonal trips and impedances for interzonal trips. For intrazonal trips (T11, T22, T23) a perfect friction factor would have predicted 10 + 50 + 90 = 150 trips. The guess with b = 1 meant that the model showed 35.0 + 85.9 + 135.7 = 256.6 trips. For interzonal trips (T12,T13, T21, T23, T31, and T33) a perfect friction factor would have predicted 40 + 70 + 20 + 80 + 30 + 60 = 300 trips. The guess with b = 1 predicted, however, 32.7 + 52.3 + 12.8 + 51.3 + 12.7 + 31.6 = 193.4 trips The adjusted friction factors are: Intrazonal travel = (0.67)(150)/256.6 = 0.392 Interzonal travel = (0.25)(300)/193.4 = 0.388 The regression equation will be: F = t–b Ln(F) = –b ln (t) Ln (0.392) = –b ln(1.5) Ln (0.388) = –b ln(4) b = 2.31 for intrazonal trips b = 0.68 for interzonal trips Thus the friction factor based on one iteration is t-b. For intrazonal travel (with travel time = 1.5) the friction factor is thus (1.5)–2.31 = 0.392. For interzonal travel time with travel time = 4.0, the friction factor is thus (4.0)–0.68 = 0.218. 12-20 Determine the share (proportion) of person-trips by each of two modes (private auto and mass transit) using the multinomial logit model and given the following information:
Parameter Ta = access time (min.) Tw = waiting time (min.) Tr = riding time (min.) C = out-of-pocket cost (cents) Calibration constant, Ak Private auto 5 0 25 150 –0.01 Mass transit 10 15 40 100 -0.07
Utility function: Uk = Ak – 0.05 Ta – 0.04 Tw – 0.03 Tr – 0.014 C Uauto = – 0.01 – 0.05 (5) – 0.04 (0) – 0.03 (25) – 0.014 (150) = – 3.11 Utransit = – 0.07 – 0.05 (10) – 0.04 (15) – 0.03 (40) – 0.014 (100) = – 3.77 Pauto = U autoeU+autoeUtransit = e−3.11e−+3.11e−3.77 = 0.659 e Utransit −3.77 Ptransit = eU autoe+eUtransit = e−3.11e+e−3.77 = 0.341 12-21 For Problem 12-20, an increase in the price of fuel has changed the out-of-pocket costs for the auto mode to $1.80. Recalculate the mode shares accordingly. Utility function: Uk = Ak – 0.05 Ta – 0.04 Tw – 0.03 Tr – 0.014 C Uauto = – 0.01 – 0.05 (5) – 0.04 (0) – 0.03 (25) – 0.014 (180) = – 3.53 Utransit = – 0.07 – 0.05 (10) – 0.04 (15) – 0.03 (40) – 0.014 (100) = – 3.77 = eUautoeU+autoeUtransit e e−3.53 Pauto = −3.53 + e−3.77 = 0.560 Utransit −3.77 Ptransit = Uautoe + eUtransit = e−3.53e + e−3.77 = 0.440 e 12-22 A mode choice logit model is to be developed based on the following information. A survey of travelers in an area with bus service found the following data:
Model parameter Auto Bus
X1, waiting time (min.) 0 10
X2, travel time (min.) 20 35
X3, parking time (min.) 5 0
X4, out-of-pocket cost (cents) 225 100
Ak, calibration constant –0.33 –0.27
The following utility function was calibrated based on an observed mode split of 84.9% private auto use and 15.1% bus use. Utility function: Uk = Ak – 0.10 X1 – 0.13 X2 – 0.12 X3 – 0.0045 X4 After implementing service improvements to the buses, the mode split changed to 81.6% private auto use and 18.4% bus use. Determine a value for the calibration constant for the bus mode that reflects this shift in mode split. Using the given information prior to the bus service improvements, the utility functions are: Uauto = – 0.33 – 0.10 (0) – 0.13 (20) – 0.12 (5) – 0.0045 (225) = –4.54 Ubus = – 0.27 – 0.10 (10) – 0.13 (35) – 0.12 (0) – 0.0045 (100) = –6.27 Using the new mode split data, the calibration constant for the bus utility function will need to be recalculated. The logit function for bus service that reflects this shift can be written as: eUbus eUbus Ubus + Uauto = Ubus + −4.54 = 0.184 e e e e Ubus = –6.03 The bus utility function and new value for Abus can now be solved. Ubus = Abus – 0.10 (10) – 0.13 (35) – 0.12 (0) – 0.0045 (100) = 6.03 Abus = – 0.03 12-23 A previously developed elasticity model for improvements to a transit system found a midpoint arc elasticity value of 0.38 to describe the relationship between demand increase and route mileage. The system is now planning a 60% increase in route mileage. Estimate the expected demand increase. Using Equation 12.15, Midpoint Arc Elasticity, eM=ΔΔX XD//Davgavg 0.38 =
12-24 Determine the minimum path for nodes 1, 3, and 9 in Figure 12.19. Sketch the final trees. 12-25 Assign the vehicle trips shown in the O-D trip table to the network in Figure 12.20, using all-or-nothing assignment. Make a list of links in the network and indicate the volume assigned to each. Calculate the total vehicle-minutes of travel. Show the minimum path and assign traffic for each of the five nodes. Minimum Paths Nodes Link Nodes Link From - To Path From - To Path 1 2 1 - 2 4 1 4 - 5, 5 - 1 1 3 1 - 2, 2 - 3 4 2 4 - 2 1 4 1 - 5, 5 - 4 4 3 4 - 3 5 1 - 5 4 5 4 - 5 1 2 - 1 5 1 5 - 1 2 3 2 - 3 5 2 5 - 4, 4 - 2 2 4 2 - 4 5 3 5 - 4, 4 - 3 5 2 - 4, 4 - 5 5 4 5 - 4 1 3 - 2, 2 - 1 3 2 3 - 2 3 4 3 - 4 3 5 3 - 4, 4 - 5
Link Volume Travel Time Veh-Min of Travel
1 - 2 200 8 1,600
2 - 1 600 8 4,800
1 - 5 350 5 1,750
5 - 1 450 5 2,250
2 - 5 0 12 0
5 - 2 0 12 0
2 - 3 300 3 900
3 - 2 300 3 900
2 - 4 600 5 3,000
4 - 2 250 5 1,250
3 - 4 250 7 1,750
4 - 3 350 7 2,450
4 - 5 1,300 6 7,800
5 - 4 700 6 4,200
TOTAL 32,650
Traffic Assigned to Nodes
Node Volume Assigned
1 1,050
2 750
3 650
4 1,550
5 1,650
Minimum Path Trees
12-26 Figure 12.21 represents travel times on links connecting six zonal centroids. Determine the minimum path from zone to zone using all-or-nothing assignment based on the given trip table.
Link Volumes Total Volume for Link
1 - 2 500 500
2 - 1 500 500
1 - 5 0 0
5 - 1 0 0
1 - 6 650 + 550 + 200 + 500 1,900
6 - 1 650 + 200 + 550 + 500 1,900
2 - 3 0 0
3 - 2 0 0
2 - 6 600 + 525 + 550 + 350 2,025
6 - 2 600 + 525 + 550 + 350 2,025
3 - 5 575 + 550 + 350 + 500 1,975
5 - 3 575 + 550 + 350 + 500 1,975
3 - 6 200+200+550+525+800+350+500+350+550 4,025
6 - 3 200+200+550+525+800+350+500+350+550 4,025
3 - 4 600 + 350 + 200 + 200 1,350
4 - 3 600 + 350 + 200 + 200 1,350
4 - 5 400 400
5 - 4 400 400
5 - 6 0 0
6 - 5 0 0
Total Trips 24,350
Minimum Path Trees 12-27 Given the following information, and using the generalized capacity restraint link performance function, perform two iterations of multipath traffic assignment. A flow of 10,100 vehicles in the peak hour is to be distributed between three routes whose properties are given in the following table.
Link Performance Function Component Free flow travel time (min.) Capacity (103 veh/h) Α Β Route 1 17.0 3.8 0.2 2.8 Route 2 15.5 4.2 0.3 3.7 Route 3 12.5 6.6 0.4 4.3
For each path j, the proportion of traffic p(j) and volume v(j) must be found, and then using Equation 12.13a, the travel time t(j) can be determined. First Iteration p(1) == 0.2893  1 + 1 + 1  17 15.5 12.5 v(1) = (0.2893)(10,100) = 2922 t(1) =17 1 + 0.2 29222.8 =18.63min   3800  p(2) == 0.3173  1 + 1 + 1   17 15.5 12.5 v(2) = (0.3173)(10,100) = 3205 t(2) =15.5 1 + 0.3 32053.7 =17.21min   4200  p(3) == 0.3934  1 + 1 + 1   17 15.5 12.5 v(3) = (0.3934)(10,100) = 3973   39734.3 =13.06min t(3) =12.5 1 + 0.4     6600  Second Iteration p(1) == 0.2850    18.63 17.21 13.06 v(1) = (0.2850)(10,100) = 2878 t(1) =17 1 + 0.2 28782.8 =18.56min   3800  p(2) == 0.3085    18.63 17.21 13.06 v(2) = (0.3085)(10,100) = 3116 t(2) =15.5 1 + 0.3 3116 3.7 =17.04min   4200  p(3) == 0.4065    18.63 17.21 13.06 v(3) = (0.4065)(10,100) = 4106   41064.3 =13.15min t(3) =12.5 1 + 0.4     6600  12-28 Consider the user equilibrium assignment and the lowest system cost assignment discussed in section 12.5.1.4 (Figure 12.16). Zones 1, 2, and 3 represent a company’s warehouses all located on a privately-owned parcel of land. The vehicles shown are delivery trucks rather than passenger cars. Which method of traffic assignment is more appropriate? The system optimal assignment, where total travel costs are minimized, would be appropriate assuming the warehouse owner is able to schedule the movement of delivery trucks on the privately owned parcel of land. Solution Manual for Traffic and Highway Engineering Nicholas J. Garber, Lester A. Hoel 9781133605157