CIV3703 Transport Engineering Module Transport Modelling
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.
. 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?
. 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
.. 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?
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
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?)
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 3 4 5 6 7 8 Trip production 500 300 300 400 00 900 000 00 Car ownership 00 50 500 00 400 300 400 00 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)
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)
Continue n S XY - S X S Y B = ----------------------- n S X - ( S X) 8x975000-050 X5800 B = --------------------------------- 8x 7500 - (050) Zone Number Trip production (Y) Car ownership (X) X^ 500 00 40000 00000 300 50 500 5000 3 300 500 50000 650000 4 400 00 0000 40000 5 00 400 60000 480000 6 900 300 90000 70000 7 000 400 60000 400000 8 00 00 0000 0000 5800 050 7500 975000 XY B =.48 In this case, computation gives B =.48 and A = 89.5
Example - Regression model Therefore the regression model is Y = 89.5 +.48 X If car ownership (X) grows to 650: Y = 89.5 +.48 x 650 = 70 vehicle trips per day
.. 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
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
Car ownership-income-trips Source: Papacostas, C.S & Prevedouros, P.D. 005 Transportation engineering and Planning, Prentice Hall
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
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 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 0 4 0 5 5 5 7 4 5 8 9 0 0 8 8 9 0 000 30 000 85 000 55 000 000 75 000 47 000 36 000 8 000 76 000 7 000 84 000 8 000 40 000 4 000 5 000 60 000 44 000 5 000 60 000 0 0 0 3 0 3
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 5 000 5 000 5 000 45 000 45 000 65 000 65 000 85 000,8 4 5 9,8 7,9,0, 3,4,5 6,7 3,6,0 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 0 4 0 5 5 5 7 4 5 8 9 0 0 8 8 9 0 000 30 000 85 000 55 000 000 75 000 47 000 36 000 8 000 76 000 7 000 84 000 8 000 40 000 4 000 5 000 60 000 44 000 5 000 60 000 0 0 0 3 0 3
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 $65,000 and car are households 7, 9 and 0, which have trip numbers of 7, 8 and 9 respectively. The average is therefore 8 (= [7+8+9]/3). Income 5 000 5 000 5 000 45 000 45 000 65 000 65 000 85 000 Income 5 000 5 000 5 000 45 000 45 000 65 000 65 000 85 000 Car ownership 0 or more 4 5 Car ownership 0 or more,8 4 5 9,8 7,9,0, 5 6.5 8 8.5 3,4,5 6,7 3,6,0 0.3.5. Household Trips per day Income ($) Cars 3 4 5 6 7 8 9 0 3 4 5 6 7 8 9 0 4 0 5 5 5 7 4 5 8 9 0 0 8 8 9 0 000 30 000 85 000 55 000 000 75 000 47 000 36 000 8 000 76 000 7 000 84 000 8 000 40 000 4 000 5 000 60 000 44 000 5 000 60 000 0 0 0 3 0 3
Example. - SOLUTION (c) Plot the number of trips versus average income level for each car ownership category Income Car ownership 0 or more 5 000 5 000 5 000 45 000 45 000 65 000 65 000 85 000 4 5 5 6.5 8 8.5 0.3.5. (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.
.3 Trip Distribution
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.
.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.
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
Example.3 (modified) Zone of destination 3 4 Zone of Origin 600 700 00 400 000 500 500 300 3 400 800 900 0 4 900 300 600 000 Travel is expected to increase by 5% over the next decade Zone of destination 3 4 Zone of Origin 630 735 0 40 050 55 55 35 3 40 840 945 0 4 945 35 630 050
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.
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) / )
Example.4 Given: Trip generation after 0 years Zone 3 4 Production 500 500 700 300 Attraction 300 500 500 700 Current travel pattern (from example.3) Zone of destination 3 4 Zone of Origin 600 700 00 400 900 000 500 500 300 00 3 400 800 900 00 00 4 900 300 600 000 800 900 300 00 700 Predict the travel pattern in 0 years time using the Average Factor Method
Example.4 Zone of destination 3 4 p i Production P i E i Zone of Origin 600 700 00 400 900 500.3 000 500 500 300 00 500.4 3 400 800 900 00 00 700.3 4 900 300 600 000 800 300.4 a J 900 300 00 700 Attraction, A J 300 500 500 700 E J.0.09.4.59 T ij = T ij E i +E j T = T E +E = 700 x (.3 +.09)/ = 844
Next iteration Zone of destination 3 4 p i Production P i Zone of Origin 76 844 46 58 398 500.04 0 558 570 73 5 500 0.99 3 466 98 066 4 60 700.04 4 008 335 684 365 339 300 0.94 a J 330 665 566 36 09 Attraction, A J 300 500 500 700 E J 0.96 0.94 0.97.4 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.
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.
Example.5 Zone Of Destination 3 4 Current p i Future P i 600 700 00 400 900 500 Zone 000 500 500 00 00 500 Of 3 400 800 900 00 00 700 Origin 4 900 300 600 000 800 300 Current a j 900 300 00 700 900 Future A j 300 500 500 700 0900 Similarly, T = 600 500 900 300 900 600 + 700 + 00 + 400 = 74 300 500 500 700 600 + 700 + 00 + 900 300 00 700 400 T = 700 500 900 500 300 900 86 = 83
After st iteration Zone Of Destination 3 4 Current p i Future P i 74 83 48 695 499 500 Zone 089 538 56 33 50 500 Of 3 478 94 08 7 700 700 Origin 4 885 9 608 46 300 300 Current a j 376 603 55 596 0900 Future A j 300 500 500 700 0900 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 = 74 500 499 300 376 74 + 83 + 48 + 695 = 730 300 500 500 700 74 + 83 + 48 + 376 603 55 596 695
After nd iteration Zone Of Destination 3 4 Current p i Future P i 730 800 46 74 500 500 Zone 099 58 557 36 500 500 Of 3 488 98 3 8 700 700 Origin 4 879 79 593 45 300 300 a j 396 55 509 683 0900 Future A j 300 500 500 700 0900 After the second iteration there is reasonable agreement between the sums of the columns and the attractions predicted by trip generation.
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.
.3. Synthetic Methods Allow effect of differing planning strategies to be incorporated (e.g., travel cost, distance of travel). Most widely used: Gravity model
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
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
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
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 =
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
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.
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 900 500 Zone 00 500 Of 3 00 700 Origin 4 800 300 Current a j 900 300 00 700 900 Future A j 300 500 500 700 0900
Example.6 A travel time matrix has been established as follows (times in minutes): Zone of Destination 3 4 Zone Of Origin 3 6 0 0 5 4 7 8 3 9 8 4 0 4 0 7 9 3 Investigation of the study area reveals all K values may be accepted as equal to.0 except K =. and K 43 =.0
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 8 0 0 40 63 0 6 3 6 63 0 4 0 0
Example.6 - SOLUTION. Establish K ij matrix Zone of origin 3 4 Zone of destination 3 4..0 3. Calculate F ij K ij matrix (by multiplying cell values) Zone of origin 3 4 Zone of destination 3 4 40 0 34 63 6 0 0 0 63 4 0 6 0
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 804 43 7 37 900 500 Zone 00 500 Of 3 00 700 Origi n 4 800 300 4a. Start with origin, i.e. P = 500 a j 900 300 00 700 900 Future A j 300 500 500 700 0900 Destination Aj FjKj AjFjKj tj 3 4 3 00 500 500 700 34 0 0 355 00 85 000 5 000 7 000 804 43 7 37 S = 49 00 S = 500 500 x 35500 500 x 85000 where t = -------------------- = 804; t = ------------------ = 43 4900 49 00
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 3 4 804 845 394 3 3 75 3 00 43 040 4 36 45 500 7 330 68 435 50 500 37 85 77 7 870 700 S 500 500 700 3 00 Pi 500 500 700 3 00 5. 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.
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 = 540 375 Destination Aj FijKij AjFijKij tij 3 4 3 7 784 490 540 34 0 0 347 097 94 656 4 900 5 400 763 48 7 9 300 300 375 S = 49 053 S = 500 6. Apply Gravity model for nd iteration Zone of destination S Pi 3 4 Zone of origin 3 4 S Aj 763 800 383 33 379 300 48 455 46 474 500 7 38 60 446 493 500 9 60 60 05 754 700 500 500 700 3 00 500 500 700 3 00
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 = 490 379 8. Apply Gravity model for the 3 rd iteration Zone of destination S Pi 3 4 Zone of origin 3 4 S Aj 765 80 385 38 3 89 3 00 483 8 458 45 494 500 6 38 603 45 499 500 6 53 54 085 78 700 500 500 700 3 00
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 375 45 50 870 0900 Error +.3% -0.% +0.4% +6.3% nd Value 379 474 493 754 0900 Error -0.6% -.0% +0.3% -.0% 3 rd Value 389 494 499 78 0900 Error -0.3% -0.% 0% +0.7% Target 300 500 500 700 0900
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.
.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
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
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
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
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 = 0.9 0. (cost p) 0.03 (travel time p) Ub = - 0.3 (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?
Example.7 - SOLUTION Substituting cost and travel time data into the utility expressions: Uc =.0 0.x5 0.03x0 = 0.4 Up = 0.9 0.x.5 0.03x0 = - 0. Ub = -0.3x0.5 0.0x5 = - 0.4 Substituting these values in the Logit model equation: e 0.4.49 P c = -------------------------- = -------- = 0.5 P p = 0.75 P b = 0.5 e 0.4 + e -0. + e -0..4.98 Multiplying these probabilities by 5000 (number of workers) gives: car travellers,500 car poolers,375 bus travellers,5
.5 Traffic Assignment All-or-Nothing Diversion Curves Capacity Restrained Multipath Proportional
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.
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
Diversion Curve Assignment Diversion curve developed on basis of cost, to consider proportion of traffic using a new / improved route compared to existing route.
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
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.
Example.0a
Example.0b
Solution
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.
Summary Four step model Trip generation Production Attraction Trip distribution Modal Split Route assignment
A simple explanation- four step model output
End Module