CIV3703 Transport Engineering. Module 2 Transport Modelling
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1 CIV3703 Transport Engineering Module Transport Modelling
2 Objectives Upon successful completion of this module you should be able to: carry out trip generation calculations using linear regression and category analysis methods perform trip distributions using the Average Factor, Gravity and Fratar Methods evaluate simple transport studies outline the trends occurring in transport planning.
3 . Traditional Four-step Transport model Trip generation (Trip production and Trip attraction) How many trips will the population, employment, etc. of the study area generates? Trip distribution Where will these trips go? Modal split Which mode of travel will be used? Traffic assignment Which route will these trips take?
4 . Trip Generation Models In this modelling, measures of urban activity are converted into number of trips. TRIP: one way person movement by one or more modes of travel Surveys usually separate trips into Home-base or non-home based Two methods for trip generation analysis. Regression methods. Category analysis
5 .. Multiple linear regression method In the past, regression method has been used extensively. Studies shown residential land-use is an important parameter for trip generation The regression equation could be developed What are the parameters which will determine the number of trips per household generated from a zone? Household income? Swimming pool?
6 Example Parameters: Car ownership; family income; family size. Y = Trips per household X = car ownership X = family income X 3 = family size Y = A + B X + B X + B 3 X 3 A, B, B & B 3 = constants derived by calibration Variables X, X and X3 vary from study to study and are selected by using current trip information
7 Assumptions for regression equations All the variables are independent of each other. (Are car ownership, family income and family size independent of each other?) All the variables are continuous and normally distributed. (Is car ownership normally distributed?)
8 Example. A study area consisting of eight (8) zones has been surveyed and the following data has been found for zonal trip production in a day (Y) and zonal car ownership (X). Zone Number Trip production Car ownership Fit a linear regression model and hence predict the number of trips in a day which will be produced when a zone grows to a car ownership level of 650. (i.e, car ownership level given here is the number of people with car per 000 population)
9 Linear Regression Equations Y = A + B X S Y S X Where A = B ---- n n n S XY - S X S Y and B = n S X - ( S X) where n = number of observations (zones)
10 Continue n S XY - S X S Y B = n S X - ( S X) 8x X5800 B = x (050) Zone Number Trip production (Y) Car ownership (X) X^ XY B =.48 In this case, computation gives B =.48 and A = 89.5
11 Example - Regression model Therefore the regression model is Y = X If car ownership (X) grows to 650: Y = x 650 = 70 vehicle trips per day
12 .. Category analysis method Model uses the household as the fundamental (generating) unit considers household characteristics (must be easily measured). Also known as cross-classification analysis
13 Category analysis method Originally, 3 parameters used in UK in 960: Car ownership 3 classes (0,, >) Income 6 classes (say) Family structure 6 classes (see table below) Class Adults employed Adults not employed None None More than 3 4 More than 5 More than 6 More than More than Total 3 x 6 x 6 = 08 categories
14 Car ownership-income-trips Source: Papacostas, C.S & Prevedouros, P.D. 005 Transportation engineering and Planning, Prentice Hall
15 Category analysis method Later studies: 3 car ownership groups; 6 household structure groups = 8 groups Advantages: Simpler computational processes Disaggregate data a better representation of human behaviour earlier study results can be used for some extend
16 Example. (Study Book) Twenty households in a city were sampled for household income, cars per household and trips produced.. Develop matrices connecting income to car ownership, and draw a graph connecting trips per household to income.. Estimate how many trips per day will occur for a household with an income of $40,000 owning one car. Household Trips per day Income ($) Cars
17 Example. - SOLUTION (a) Set up a matrix which shows the household numbers ( to 0), according to income and car ownership Income Car ownership 0 or more Household Trips per day Income ($) Cars , ,8 7,9,0, 3,4,5 6,7 3,6,
18 Example. - SOLUTION (b) Then calculate the average number of trips per household for each category, based on the household data in that category e.g. households in income $45,000 to $65,000 and car are households 7, 9 and 0, which have trip numbers of 7, 8 and 9 respectively. The average is therefore 8 (= [7+8+9]/3). Income Income Car ownership 0 or more 4 5 Car ownership 0 or more, ,8 7,9,0, ,4,5 6,7 3,6, Household Trips per day Income ($) Cars
19 Example. - SOLUTION (c) Plot the number of trips versus average income level for each car ownership category Income Car ownership 0 or more (d) From the plot it can be estimated that a household with an income of $40,000 owning one car will make 7 trips per day.
20 .3 Trip Distribution
21 Methods Growth Factor Methods Constant Factor Method Average Factor Method Fratar Method Synthetic Methods Gravity type models Opportunity models Growth factor methods are simpler than synthetic methods.
22 .3. Growth Factor Methods The growth rate with in each zone will increase in a uniform manner. Assumes profile of trip making remains substantially the same as at present. Volume of trips increases according to growth in generating or attracting zones.
23 Constant Factor Method Future trips from a zone i to a zone j = Present trips between the zones x Growth Factor (E) Total future trips in study area E = Total present trips in study area Example: Traffic in Toowoomba expected to grow 0% in next 0 years. (E=0%) Trips between zones currently 800 trips per day Future trips = 880 trips per day
24 Example.3 (modified) Zone of destination 3 4 Zone of Origin Travel is expected to increase by 5% over the next decade Zone of destination 3 4 Zone of Origin
25 Problems of Constant Factor Method Overestimates future trips between currently densely developed zones. Underestimates future trips between currently underdeveloped zones. Fails to make provision for zones which currently have no development.
26 Average Factor Method Attempts to account for varying rates of trip making (generation )expected in different zones. Future trips between zones = Present trips between the zones x Average of growth factors for the zones ( (Ei + Ej) / )
27 Example.4 Given: Trip generation after 0 years Zone 3 4 Production Attraction Current travel pattern (from example.3) Zone of destination 3 4 Zone of Origin Predict the travel pattern in 0 years time using the Average Factor Method
28 Example.4 Zone of destination 3 4 p i Production P i E i Zone of Origin a J Attraction, A J E J T ij = T ij E i +E j T = T E +E = 700 x ( )/ = 844
29 Next iteration Zone of destination 3 4 p i Production P i Zone of Origin a J Attraction, A J E J E i Iteration procedure is used to achieve a convergence of the T ij values so that sum of trips from the trip distribution model equal those predicted by trip generation.
30 Fratar Method Method developed by T. J. Fratar to overcome disadvantages inherent in constant factor and average factor methods. t ' ij t ij P p i i A a j j n n A a t k k ik t ik Assumes that the existing trips (t ij ) will increase in proportion to E i and in proportion to E j.
31 Example.5 Zone Of Destination 3 4 Current p i Future P i Zone Of Origin Current a j Future A j Similarly, T = = T = = 83
32 After st iteration Zone Of Destination 3 4 Current p i Future P i Zone Of Origin Current a j Future A j The sums of the rows equal the figures obtained from trip generation, but the sums of the columns do not equal the figures predicted by trip generation. Start: nd iteration T = =
33 After nd iteration Zone Of Destination 3 4 Current p i Future P i Zone Of Origin a j Future A j After the second iteration there is reasonable agreement between the sums of the columns and the attractions predicted by trip generation.
34 Disadvantages of Growth factor methods Present trip distribution matrix has to be obtained first from survey (O D studies) The errors in base data collected gets magnified None of the methods provide a measure of the resistance to travel. For example, it is not possible to consider the effect of travel impedance factors such as new road facilities or the negative effects of congestion.
35 .3. Synthetic Methods Allow effect of differing planning strategies to be incorporated (e.g., travel cost, distance of travel). Most widely used: Gravity model
36 Newton s Law of Gravity The force between any two masses having masses m and m separated by a distance of r is an attraction acting along the line joining the masses and has a magnitude directly proportional to the mass of each body, and inversely proportional to the square of the distance between the bodies. m F F m r F G m m r
37 Gravity Model Applied to Transport Basis: Trip interchange between zones is proportional to attractiveness of zones, and inversely proportional to a function of the physical separation of the zones. P i A j T ij = k z n T ij = trips from zone I to zone j P i = trips produced from zone I A j = trips attracted to zone j z = distance between zones I and j (more normally travel time or distance) n commonly taken as k a constant found by calibration
38 Gravity Model Form T ij i n P A j j A F j ij F ij K ij K ij T ij = trips from zone I to zone j P i = trips produced from zone I A j = trips attracted to zone j F ij = travel time or friction factor K ij = zone-to-zone socio-economic adjustment factor n = number of zones
39 F Values F = Travel impedance factor Example: F = / Distance to a power F ij Z (or travel time to a power) n ij Typically: power, n =
40 K Values K = socio-economic adjustment factor Example: Low socio-economic area on fringe of CBD. Gravity model expectation lots of trips. Reality CBD jobs: white collar Need: downgrade trip quantity Procedure of determining suitable K values for each zone-to-zone combination is part of the model calibration
41 Calculation Procedure As with Average Factor method and Fratar Method, the Gravity model requires successive iterations in the calculation procedure, in order to arrive at a reasonable solution.
42 Example.6 (Study Book) (Includes data from examples.4 and.5) A town has the following current total trips, and predicted total trips for 0 years hence: Zone Of Destination 3 4 Current p i Future P i Zone Of Origin Current a j Future A j
43 Example.6 A travel time matrix has been established as follows (times in minutes): Zone of Destination 3 4 Zone Of Origin Investigation of the study area reveals all K values may be accepted as equal to.0 except K =. and K 43 =.0
44 Example.6 - SOLUTION. Determine F ij matrix. Assume F ij = / (travel time) Construct a matrix of F ij values (and multiply values by 000 for convenience) Zone of Destination 3 4 Zone Of Origin
45 Example.6 - SOLUTION. Establish K ij matrix Zone of origin 3 4 Zone of destination Calculate F ij K ij matrix (by multiplying cell values) Zone of origin 3 4 Zone of destination
46 Example.6 - SOLUTION 4. Apply Gravity model for first iteration t ij P A i n j j A F j ij F ij K ij K ij Zone Of Destination 3 4 Curre nt p i Futur e P i Zone Of Origi n a. Start with origin, i.e. P = 500 a j Future A j Destination Aj FjKj AjFjKj tj S = S = x x where t = = 804; t = =
47 Example.6 - SOLUTION 4b. Repeat with origin s, 3 and 4, and obtain a complete matrix: Zone of origin 3 4 S Aj Zone of destination S Pi It should be noted that the sums of the rows (i.e. trip productions from each zone of origin) agree with the target figures P i but the sums of the columns (i.e trip attractions to each zone) do not agree with the target figures A j. Further iterations are required. These are performed by adjusting the A j values to be used in the next iteration.
48 Example.6 - SOLUTION Attraction used in st iteration Adjusted A j = A j = x Desired Attraction Actual attraction resulted from st iteration 300 Therefore A = x 300 = 37. Similarly A = 784; A 3 = 490; A 4 = Destination Aj FijKij AjFijKij tij S = S = Apply Gravity model for nd iteration Zone of destination S Pi 3 4 Zone of origin 3 4 S Aj
49 Example.6 - SOLUTION 7. The sum of the columns still do not agree with target values but they are closer. Repeat the procedure using new adjusted A j values. Attraction used in nd iteration Adjusted A j = A j = x Desired Attraction Actual attraction from nd iteration 37 Thus A = x 300 = 348; and A = 83, A 3 = 497, A 4 = Apply Gravity model for the 3 rd iteration Zone of destination S Pi 3 4 Zone of origin 3 4 S Aj
50 Example.6 - SOLUTION 9. Sums of columns are now within % of targets so sufficient accuracy has been achieved and no further iterations are required. Summary of calculation progression (sum of columns): Zone of Destination S 3 4 st Value Error +.3% -0.% +0.4% +6.3% nd Value Error -0.6% -.0% +0.3% -.0% 3 rd Value Error -0.3% -0.% 0% +0.7% Target
51 Opportunities Model Criticism of Gravity Model: takes no account of individual behaviour. Opportunity models use probability theory to model the selection process used by an individual in making decisions about a trip.
52 .4 Modal Choice Choice of mode of travel - known as modal choice or mode split. Eg - travel Brisbane suburb to CBD: private car bus train Significance in urban areas
53 Factors affecting modal split Trip maker characteristics car ownership, income, net residential density Trip characteristics home-work, home-school System characteristics evaluated by considering cost of travel, including time
54 Modal Choice Model Basis: probability of a user selecting a particular mode Probability based on utility (i.e. ability of a service to satisfy human needs) Model form: Where U mk U mk = S b mn Z kmn = total utility provided by mode m to a traveller k b mn = coefficient estimated from traveller data for mode m, corresponding to a characteristic n Z kmn = traveller or mode characteristic n (e.g. income, travel time of mode, etc) for mode m of traveller k
55 Opposite of utility is disutility the cost the user experiences in using the service Thus probability = u k / S u k or e Uk / S e Uk Where u k = utility of an alternative And e is the exponential Exponential form known as logit model
56 Example.7 (Study Book) A mode choice model is estimated from journey-to-work data for an urban area. The mode choices available are: The utility functions derived are: car (c) carpool (p) bus (b) Uc =.0 0. (cost c) 0.03 (travel time c) Up = (cost p) 0.03 (travel time p) Ub = (cost b) 0.0 (travel time b) {costs in $ ; time in minutes} Five thousand (5000) workers travel from a residential area to an industrial area during the peak hour. For all workers the cost of driving a car to work is $5.00 with a travel time of 0 minutes, while the bus fare is $0.50 with a travel time of 5 minutes. The car pooling option results in an average of travellers sharing cost equally. How many workers travel by each mode?
57 Example.7 - SOLUTION Substituting cost and travel time data into the utility expressions: Uc =.0 0.x5 0.03x0 = 0.4 Up = x x0 = - 0. Ub = -0.3x x5 = Substituting these values in the Logit model equation: e P c = = = 0.5 P p = 0.75 P b = 0.5 e e e Multiplying these probabilities by 5000 (number of workers) gives: car travellers,500 car poolers,375 bus travellers,5
58 .5 Traffic Assignment All-or-Nothing Diversion Curves Capacity Restrained Multipath Proportional
59 All-or-nothing Assignment Route of least cost between zone centroids calculated. All future trips allocated to the route of least cost. Problem: route may become easily congested.
60 Example.8- All or nothing Demand, Q = 5 vph Two options: Route X - travel time 0 minutes Route Y travel time 5 minutes X Q Y Qx = 5 vph Qy = 0
61 Diversion Curve Assignment Diversion curve developed on basis of cost, to consider proportion of traffic using a new / improved route compared to existing route.
62 Example Passenger cars can use either freeway or arterial roads between two sub-urban areas. Freeway: Travel distance - 5km, speed limit 00 kmh. Alternative arterial road: Travel distance - 0 km, an average speed limit of 60kmph. Allocate the traffic using the diversion curve developed by Bureau of Public Roads, USA. Travel time ratio = Freeway(d/V)/Arterial (d/v) = (5km/00kph)/(0km/60kph) = 0.75 Answer: About 80% will use the Freeway
63 Capacity Restrained Assignment Problem with all-or-nothing: link may become overloaded. Practice: this doesn t occur because drivers shift trips elsewhere. Total trips distributed over several routes using speed flow relationships. Start with all-or-nothing and in successive iterations shift traffic to other routes so all routes have equal travel time, cost, etc.
64 Example.0a
65 Example.0b
66 Solution
67 Multipath Proportional Assignment Used in urban areas where many alternate travel paths are available. Drivers perceive best route in different ways e.g. minimum travel time; minimum congestion; minimum number of traffic signals. Future traffic allocated over all feasible routes.
68 Summary Four step model Trip generation Production Attraction Trip distribution Modal Split Route assignment
69 A simple explanation- four step model output
70 End Module
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