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STAQ based Matrix estimation - initial concept (presented at hEART conference 2014)
- 1. 1 Challenge the future MatrixEstimation using STAQ (Static Traffic Assignment with Queuing) Luuk Brederode Consultant DAT.mobility / PhD student Delft University
- 2. 2 Challenge the future Contents • Motivation • From static and dynamic assignment models to STAQ • Differences in response functions • The matrix estimation problem • Example showing differences in response functions • Proposed method: lower level • Intermezzo: the node model as a function • Proposed method: upper level and whole bilevel problem • Application example • Conclusions
- 3. 3 Challenge the future Motivation (1): why not use unconstrained? 0 10 20 30 40 50 60 70 80 90 0 500 1000 1500 2000 2500 3000 3500 speed (km/u) flow (veh/u) Fundamentaldiagram Traveltimefunction Observed flow and speed Modelled flow and speed (prior matrix) Modelled flow and speed (posterior matrix estimation using static assignment model) 0 10 20 30 40 50 60 70 80 90 0 500 1000 1500 2000 2500 3000 3500 speed (km/u) flow (veh/u) Fundamentaldiagram Traveltimefunction When observed flow in free flow branch: When observed flow in congested branch
- 4. 4 Challenge the future Motivation (2): why not use unconstrained? • In STAQ there is no time dimension, reducing dimensionality of the problem • Assignment matrix can be derived from link reduction factors which are readily available from STAQ • No need to (iteratively) run full simulation model to approximate response function, instead only marginal runs of node model are used • Approximation errors due to marginal runs can be made explicit and constrained
- 5. 5 Challenge the future From static (STA) and dynamic (DTA) assignment models to STAQ STA models • Speed-flow curve • Stationary travel demand • Single time period • No hard capacity constraints due to lack of node model First order DTA models • Fundamental diagram • Variable travel demand • Multiple time periods • Hard capacity constraints due to explicit node model STAQ (Brederode et al. 2010, Bliemer et al. 2012) • Fundamental diagram • Stationary travel demand • Single time period • Hard capacity constraints due to explicit node model
- 6. 6 Challenge the future From STA and DTA models to STAQ STA models • Monotonically increasing • Separable over space • Separable over time (no time dimension) First order DTA models • Not monotonically increasing • Inseparable over space • Inseparable over time STAQ (Brederode et al. 2010, Bliemer et al. 2012) • Not monotonically increasing • Inseparable over space • Separable over time (no time dimension) Differences in response function (link flows as a function of ODdemand)
- 7. 7 Challenge the future The Matrix estimation problem (STA) Cournot-Nash Game Lower level: STA( ) Upper level: : current ODMatrix : prior ODMatrix : vector of modelled link flows : vector of observed link flows f1 and f2: distance functions A(D): assignment matrix D D D 0 D y y * 1 0 2 argmin argmin ( , ) ( ( ), ) F f f D D D D D y D y ( ) ( ) y D A D D
- 8. 8 Challenge the future The Matrix estimation problem ((s)DTA) Stackelberg Game Lower level: (s)DTA( ) Upper level: : current ODMatrix : prior ODMatrix : vector of modelled link flows : vector of observed link flows f1 and f2: distance functions A(D): assignment matrix D D D 0 D y y * 1 0 2 argmin argmin ( , ) ( ( ), ) F f f D D D D D y D y d ( ) ( ) ( ) d D A D y D A D D D
- 9. 9 Challenge the future The Matrix estimation problem O1 D1 D2 O2 Demand O2-D2 (D2) Demand O1-D1 (D1) Link b Assume: D2 = 2*D1 D2 = 2*Ca D1 = Cb Cb’ = Cb’’ Example showing differences in response function Link a’
- 10. 10 Challenge the future Link b The Matrix estimation problem: STA O1 D1 D2 O2 1 2 3 a b rs b rs RS y D D D C 2 2 a a rs a rs RS y D D C Example showing differences in response function Demand O2-D2 (D2) Demand O1-D1 (D1) Assume: D2 = 2*D1 D2 = 2*Ca D1 = Cb Cb’ = Cb’’
- 11. 11 Challenge the future Link b pagina 11 O1 D1 D2 O2 0.5 b b The Matrix estimation problem: STAQ 0.5 a 2 1 ˆ a rs rs b a a b b b rs RS y D D D C 2 ˆ a rs rs a a a a rs RS y D D C Example showing differences in response function Demand O2-D2 (D2) Demand O1-D1 (D1) Assume: D2 = 2*D1 D2 = 2*Ca D1 = Cb Cb’ = Cb’’
- 12. 12 Challenge the future Proposed method (lower level) Realising that: • elements in the assignment matrix: • can be reconstructed using the results of the node models of paths using the considered link by: • And is direct output of the STAQ model We can directly derive the assignment matrix from STAQ 1 1 1 1 ˆ ˆ ( ) ˆ ˆ RS RS y y A D ˆ rs rs a a a P a Calculate assignment matrix
- 13. 13 Challenge the future Proposed method (lower level) Realising that: • Elements in vectors within the derivative of the assignment matrix: • can be reconstructed using the product rule: • And can be approximated using the node model We approximate the derivative of the assignment matrix using marginal simulation of only the node model within STAQ 1 1 1 1 ˆ ˆ d d d d d ( ) d ˆ ˆ d d d d RS D RS y y D D A D D D D ˆ / rs rs rs rs a a a rs a P a P a d d dD dD / rs a d dD Calculate derivative of the assignment matrix
- 14. 14 Challenge the future Intermezzo: the node model as a function • Assume directional capacity constrained node model described in Tampère et al. 2011 and Flötteröd and Rohde 2011 • Using the example as used in Tampère et al. 2011: O1 D1 300 150 50 300 150 50 68 100 D4 0 1096 1600 O2 0 205 300 100 81 0 0 0 O4 800 645 0 0 0 D2 800 645 0 0 600 100 100 600 100 100 D3 O3 1000 1996 795 249 1000 2000 1000 2000 1000 2000 1000 2000
- 15. 15 Challenge the future Intermezzo: the node model as a function • Approximation of 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 200 400 600 800 alpha demand on turn O1-->D4 O1 O2 O3 O4 14 ( ) a D 14 304 D 14 305 390 D 14 391 456 D 14 457 800 D 2 1 3 4 1 2 3 4
- 16. 16 Challenge the future O1 D1 300 150 50 300 150 50 68 100 D4 0 1096 1600 O2 0 205 300 100 81 0 0 0 O4 800 645 0 0 0 D2 800 645 0 0 600 100 100 600 100 100 D3 O3 1000 1996 795 249 1000 2000 1000 2000 1000 2000 1000 2000 Intermezzo: the node model as a function (interval 1) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 200 400 600 800 alpha demand on turn O1-->D4 O1 O2 O3 O4 1 D D S S
- 17. 17 Challenge the future Intermezzo: the node model as a function (interval 2) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 200 400 600 800 alpha demand on turn O1-->D4 O1 O2 O3 O4 2 O1 D1 350 150 50 350 150 50 68 100 D4 0 1096 1600 O2 0 205 300 100 81 0 0 0 O4 800 645 0 0 0 D2 800 645 0 0 554 92 92 600 100 100 D3 O3 1000 2000 787 241 1000 2000 1000 2000 1000 2000 1000 2000 D S S S
- 18. 18 Challenge the future Intermezzo: the node model as a function (interval 3) O1 D1 400 150 50 400 150 50 68 100 D4 0 1089 1600 O2 0 204 300 100 81 0 0 0 O4 800 646 0 0 0 D2 800 646 0 0 511 85 85 600 100 100 D3 O3 1000 2000 781 234 1000 2000 1000 2000 1000 2000 1000 2000 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 200 400 600 800 alpha demand on turn O1-->D4 O1 O2 O3 O4 3 S S S D
- 19. 19 Challenge the future Intermezzo: the node model as a function (interval 4) 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 200 400 600 800 alpha demand on turn O1-->D4 O1 O2 O3 O4 4 O1 D1 600 150 50 484 121 40 65 100 D4 0 1032 1600 O2 0 194 300 100 86 0 0 0 O4 800 685 0 0 0 D2 800 685 0 0 484 81 81 600 100 100 D3 O3 1000 2000 806 231 1000 2000 1000 2000 1000 2000 1000 2000 S S S S
- 20. 20 Challenge the future Intermezzo: the node model as a function : • is continuous on its positive domain • can be constructed piece wise • is differentiable almost everywhere • For each piece wise interval the reduction factor of an inlink is determined by the same constraint • At each non-differentiable point a switch of one active constraint occurs • is either monotonically increasing or decreasing on its positive domain, depending on the effect that the considered turndemand has on the (directed capacity) share of the turn for the constraining outlink. • can only be increasing when the considered inlink a is not the same as the inlink of the considered turn rs. ( ) rs a D
- 21. 21 Challenge the future Intermezzo: the node model as a function : • is 0 when • Inlink a is demand constrained • Inlink a is constrained by outlink other than the outlink of considered turn rs • is linear when inlink is supply constrained by an outlink to which at least one demand constrained turn exists • is of the form when inlink a is supply constrained by an outlink to which only supply constrained turns exist d /d rs a D 2 1/( 2 3) rs c c D c
- 22. 22 Challenge the future Proposed method (lower level) • Use current assignment matrix directly from one STAQ run • Use linear point derivative approximation of using node model • Constrain the allowed per iteration to • The piece wise interval within using a binary search • some error on alpha using a binary search (when nonlinear): -8.78E-04 0 0.2 0.4 0.6 0.8 1 1.2 0 200 400 600 800 1000 alpha demand on turn O4-->D3 O4 approx 43 D 43 D D d /d rs a D ( ) rs a D
- 23. 23 Challenge the future Proposed method (upper level) • Minimize: • Subject to • A non negativity constraint on demand • A constraint on link capacities • Constraints imposed by the lower level • w is a wheighting parameter chosen by the decision maker • θ is a normalisation parameter determined by ratio between objective space of first and second component: • (Nadir1-Utopia1)/(Nadir2’-Utopia2) 2 2 0, ( ) (1 ) ( ( ) ) rs rs a a a A rs RS w D D w y y D
- 24. 24 Challenge the future Application example • Same example • Added observed link flows on D1 and D2 O1 D1 300 150 50 498 300 150 50 68 100 D4 0 1096 1600 O2 0 205 300 100 81 0 0 0 O4 800 645 0 0 0 D2 800 645 0 0 600 100 100 1590 600 100 100 D3 O3 1000 1996 795 249 1000 2000 1000 2000 1000 2000 1000 2000
- 25. 25 Challenge the future Application example: only optimizing on link flows (w=0) • Varying with start solution and epsilon, different (valid) solutions are found; the problem is underspecified.
- 26. 26 Challenge the future Application example: w>0 • w=0.5 (unconstrained): • w=0.5 (constrained, ε = 0.01) • No convergence, but at least the weighting works… 0 0.02 0.04 0.06 0.08 0.1 0.12 1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132333435363738394041424344454647484950 objective function value Iteration # f1 f2 F 0 0.02 0.04 0.06 0.08 0.1 0.12 1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132333435363738394041424344454647484950 objective function value Iteration # f1 f2 F
- 27. 27 Challenge the future Application example: w>0 • w = 0.5 (constrained, ε = 0.01), development of alpha per inlink: • Epsilon is violated in every iteration; secondary ( ) and higher order interaction effects are the cause • From the definition of the node model, on a node with n turning movements, each inlink potentially needs n-1 order interaction effects 0 0.2 0.4 0.6 0.8 1 1.2 1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132333435363738394041424344454647484950 alpha iteration# O1 O2 O3 O4 2 ' ' d /d d rs r s a D D
- 28. 28 Challenge the future Conclusions • The proposed method makes the matrix estimation problem more tractable and potentially more scalable • Secondary (and possbily even higher order) interaction effects cannot be omitted, first order approximation is not enough on the node level • Derivatives of the directional capacity proportional node model can be analytically derived
- 29. 29 Challenge the future Further research • Use analytical derivatives for each piece wise segment instead of point derivatives (no more constraining for first order approximation errors needed) • How to efficiently determine, calculate and include relevant secondary (or even higher order) interaction effects • Test the generalisation to path level • Develop method to generalise to network level • Develop method to include propagation of spillback effects (or assume constant?) • Develop method to generalise to include route choice
- 30. 30 Challenge the future Thank you! • Full paper available on request • Questions? Now, or lbrederode@dat.nl
Editor's Notes
- Eigenlijk is het alpha_inlink_a en alpha_inlink_b ipv alpha_a en alpha_b