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Traffic Flow Modelling & Control Origin-Destination Estimation and Traffic Assignment Dr. Apostolos Kotsialos School of Engineering & Computing Sciences South Road, DH1 3LE, Durham Tel: 0191 334 2399 Fax: 0191 334 2408 E-mail: [email protected] March 7, 2014 Contents Objectives: • to introduce the notion of the Origin-Destination 1 Introduction (OD) matrix; 2 The OD Estimation Problem • to describe the problem of OD estimation; 2.1 The OD Structure and Information . . 2.2 Static OD Estimation . . . . . . . . . . • to introduce the notion of traffic assignment; 2.3 Dynamic OD Estimation in a Roundabout . . . . . . . . . . . . . . . . . . . • to introduce the notions of user and system optimum. 3 The Traffic Assignement Problem 3.1 The Basic Setting . . . . . . . . . . . . Learning outcomes: 3.2 User Optimum for the Single-Origin– Single-Destination Network . . . . . . • to understand the idea of the OD matrix in the 3.3 System Optimum for the Singletraffic management framework; Origin–Single-Destination Network . • be able to estimate a static OD matrix based on the ideas of minimising its distance from a prior matrix; 1 Introduction 1 . . 2 2 3 . 5 . 6 6 . 7 . 7 • be able to estimate a static OD matrix based on In terms of Figure 1, this lecture is concerned with the ideas of maximising entropy; the problem of estimating two of the possible distur• understand the use of Kalman Filter for dynamic bances to the traffic flow process, hence the methods that are discussed here can be placed into the OD estimation; “Surveillance” module. These two disturbances are • to be able to calculate the system and user op- the Origin-Destination (OD) matrix and the traffic timums for the single-origin–single-destination assignment. network with known travel times as a function Since both are considered as disturbances, they of volume. cannot be controlled, but they can be estimated and 1 Figure 2: A general road network with origins and destinations. Table 1: OD matrix structure Dest. 1 Dest. 2 ... Dest. M Origin 1 d1,1 d1,2 . . . d1,M Origin 2 d2,1 d2,2 . . . d2,M ... Origin N dN,1 dN,2 . . . dN,M Figure 1: The traffic control loop. predicted by use of traffic sensors. The OD estimation/prediction and the traffic assignment problems are coupled, in the sense that that they are complementary. The OD provides information that describes the interaction between the road traffic system and its environment, i.e. it refers to the boundary between the network and the environment that generates demand and supply. The traffic assignment problem is concerned with modelling the drivers’ route choices inside the network. This behaviour is formed from knowledge and past experience, but can also be formed in such a way so as to serve a certain purpose. In this case, the traffic assignment problem becomes a traffic control problem, rather than a disturbance estimation and prediction problem. Hence, the traffic assignment can emerge from drivers’ own experience and understanding (disturbance) or can be guided to an optimum one (control). What is necessary here is to understand the area of relevance of the OD and traffic assignment problems. ODs describe what happens at the network boundaries. Traffic assignment describes what happens inside the network. The OD estimation problem is described in section 2 and the traffic assignment problem in section 3. 2 2.1 origin to the destination. This means that an origin can be associated with many destinations, as there may be many different routes towards a number of destinations. In order to organise this information the notion of the OD matrix is used. Let us assume that the general network of Figure 2 has N origins and M destinations. Traffic volumes di,j originate at an origin i having j as destination. Since there may be many destinations reachable from i, grouped together into the set Ai the total flow measured by and inductive loop detection station placed at the origin’s entrance, qi is given by X di,j . (1) qi = j∈Ai The OD matrix has the structure of Table 1. Some of the entries in the OD matrix may be zero due to two reasons. First if there is no feasible route from a specific origin to a specific destination and second, if the location of the origin is very close to the location of the destination so it makes no sense to do the trip (except in some special cases). Furthermore, there are two equivalent ways of filling out the matrix’s entries. In the first one, the entries di,j represent traffic subflows of the total flow qi originating at i and have flow units, e.g. veh/h or veh/min. In the second case, the entries are pure numbers and they represent the percentages ϑi,j of the total traffic flow qi that is destined to j . Obviously the two representations are equivalent because The OD Estimation Problem The OD Structure and Information Let us consider the general road network shown in Figure 2. This network consists of links, junctions and also points, shown as nodes, where the traffic demand originates (origins) and points where the demand is heading. An origin is associated with a destination if there is at least one feasible route from the Dr. A. Kotsialos di,j = ϑi,j qi . 2 (2) TFMC, 2013–2014 The OD matrix is not necessarily constant and in fact it changes over time. This is manifested more vividly when comparing the OD patterns in the morning peak hours when drivers are commuting towards their workplace and the evening peak hours when drivers are returning back home. The OD information is used for different applications, including: • Transportation planning (e.g. for designing public transport routes). • Forecasting/Modelling (e.g. for predicting the flows the development of a shopping mall will Figure 3: A network link used by flows from many origins create (gravitational models)). to different destinations. • Control (e.g. for real time route guidance). It may be measured directly by using number of subflows di,j (unknown) which have originated at origin i and are heading to destination j and are using link m. Let pm ij denote the percentage of the volume dij that uses link m to reach destination j (alternative routes may be feasible to be used), hence 0 ≤ pm ij ≤ 1. Then X qm = pm (3) ij dij . • roadside interviews; • toll networks; • route guidance systems; • license plate surveys (automated or not). (i,j) If they are not measured directly, then they need to be estimated based on traffic volume measurements. The problem of the OD matrix estimation consists of estimating and tracking over time, the quantities qi,j or ϑi,j based on measurements from the origins and from internal network points. Static or dynamic estimation problem formulations can be used depending on the user requirements. pm ij The fractions are considered for every network link m, they describe the traffic assignment in the network, i.e. how each route is utilised. If the fractions they are known, then there are linear constraints in the problem and the OD estimation problem takes the form q=P·d 2.2 where q is a vector where the measured link flows are organised, P is a matrix of routing fractions, which depends on the network topology structure and knowledge of the pm i.j s and d is the vector of unknown OD subflows. If pm i,j are unknown, they they have to be estimated along with the OD demands, which results to a nonlinear problem formulation. If they are known, then (4) is a system of linear equations, which most often than not has more unknowns than equations. Let us consider the example network shown in Figure 4. Nodes 1 and 2 are origins where demand originates towards reachable destinations at nodes 3 and 4. Link junctions are also indicated as nodes. In some (not all) links, there are detection stations which provide traffic flow measurements. The link defined by nodes 1 and 5 has volume q1,5 = 8; the units can represent veh/h or 1000×veh/h or veh/30min etc. Two more measurements are known, q5,6 = 15 and q6,3 = 10. The OD matrix for this system has the following form. d1,3 d1,4 OD = . (5) d2,3 d2,4 Static OD Estimation The static OD problem is concerned with estimating an OD matrix that does not change over the considered time horizon. The solutio to this problem is based on available historical measurements. The basic problem encountered is that in most cases there are less measurements available than required OD information. This results to a system of equations where the number of equations is lower than number of unknowns The literature offers a wide range of approaches and methodologies. A basic approach consists of comparing a revised OD matrix to a prior known matrix. In this case, the objective is the minimisation of the distance between the updated and the prior matrices, The distance is defined as the sum of the squared differences between corresponding entries in the two matrices. The solution possibly can be combined with traffic assignment considerations (route choice). Let us consider the network shown in Figure 3, with N origins and M destinations. Assume that the traffic volume qm of link m is known, i.e. is measured (e.g. by loop detectors). This total flow consists of a Dr. A. Kotsialos (4) 3 TFMC, 2013–2014 the outcome of an optimisation problem which aims at minimising the distance between the new OD estimate and the prior one. The optimisation problem in this case is as follows. min1,5 J = di,j ,p1,3 Figure 4: Example network for static OD estimation, [3]. (6) q5,6 = 15 = p1,5 1,3 d1,3 + d1,4 + d2,3 + d2,4 (7) q6,3 = 10 = p1,5 1,3 d1,3 + d2,3 . dpi,j − di,j 2 wi,j (10) (i,j) subject to eqns. (6)–(8) di,j ≥ 0 (11) p1,5 1,3 (12) 0≤ ≤1 where dpi,j is the prior OD matrix entry for the flow from i to j and wi,j are weights that assign different degrees of importance to certain OD pairs. This is a nonlinear optimisation problem which can be solved analytically for simple cases and by use of appropriate solvers in more complex situations. Another possibility for selecting a feasible OD matrix is to select the one that adds as less information as possible to the existing information given by the measurements. This means to select an OD matrix that maximises the system’s entropy W , given by the following expression. X W =− [di,j ln(di,j ) − di,j ] . (13) The problem at hand is to estimate the entris of the OD matrix di,j of (5) based on the available three measurements. In order to formulate the problem, the fraction p1,5 1,3 of the flow that originates in node 1 and is destined to destination 3 and uses link (1, 5) to reach its destination, needs to be introduced. Then, the principle of vehicle conservation can be utilised to formulate the problem’s equations. q1,5 = 8 = p1,5 1,3 d1,3 + d1,4 X (i,j) (8) Let us consider the example of Figure 4, but with the following assumptions: Equation 6 states that the measured flow in link (1, 5) equals the flow that originates at 1 destined to 3 and uses link (1, 5) instead of link (1, 3), plus the whole flow that originates at 1 destined to destination 2. The flow in link (5, 6) (eqn. (7)) consists of the flows in (1, 5) plus the flow that originates in 2 irrespective of its destination, since both have to pass through (5, 6). Finally, (8) states the fact that the flow in (6, 3) consists of the flow originating in 1 destined to 2 and using (6, 3) plus the flow originating in 2 destined to 3. The system of equations (6)–(8) has three equations and five unknowns (p1,5 1,3 , d1,3 , d1,4 , d2,3 and d2,4 ). Furthermore, it is nonlinear as two of the unknowns are multiplied in the equations. The system can be brought into the matrix form 1,5 " # d1,3 p1,3 1 0 0 8 d1,4 15 = p1,5 (9) 1,3 1 1 1 d2,3 1,5 10 p1,3 0 1 0 d2,4 Because of the integer values assumption, there are only six feasible matrices that satisfy eqns. (6)–(8), namely 8 0 7 1 6 2 1. 2. 3. 2 5 3 4 4 3 5 3 4 4 3 5 4. 5. 6. 5 2 6 1 1 0 Since there are more unknowns than equations, there is an infinite number of feasible solutions that satisfy (9). The question is how to choose one of these possible solutions. A first idea is to utilise past knowledge about the OD matrix, in the form of a previously known or estimated OD matrix. Then the new OD could be as Hence, it is possible to solve the optimisation problem (10)–(12) by exhaustive search over the space of feasible matrices. For each of the possible solutions, the objective function value of (10) is calculated using the data provided by the prior OD matrix. For each of the six possible matrices the distance is calculated (wi,j = 1). Dr. A. Kotsialos • p1,5 1,3 = 1, i.e. link (1, 3) is not used at all; • only integer values of volume are considered; • the prior OD matrix is " # (Dest. 3) (Dest. 4) (Orig. 1) 3 2 ; (Orig. 2) 1 3 • the weights wi,j = 1 for every OD pair. 4 TFMC, 2013–2014 • Matrix 1: (8 − 3)2 + (0 − 2)2 + (2 − 1)2 + (3 − 5)2 = 24 + 4 + 1 + 4 = 34. • Matrix 2: (7 − 3)2 + (2 − 1)2 + (3 − 1)2 + (4 − 3)2 = 16 + 1 + 4 + 1 = 22. • Matrix 3: 18. This is the optimum matrix. • Matrix 4: 25. • Matrix 5: 34. • Matrix 6: 54. Hence, if the chosen criterion is the minimisation of the distance to the prior matrix, feasible OD matrix number 3 should be chosen. If the chosen criterion is the maximisation of the system’s entropy, then expression (13) needs to be evaluated for each of the feasible OD matrices and the one with the largest value be selected. Part of the relevant calculations are shown below. Figure 5: A roundabout with four approaches, [2]. • Matrix 1: −11.07. In a general setting, assume that there are m in• Matrix 2: −[7 ln(7) − 7] − [ln(1) − 1] − [3 ln(3) − 3] − flows and n outflows. Then instead of the estimation [4 ln(4) − 4] = −7.462. of the subflows, the splitting portions bi,j (k) ∈ [0, 1], • Matrix 3: −[6 ln(6) − 6] − [2 ln(2) − 2] − [4 ln(4) − are estimated. A Kalman filter approach can be used to track slowly varying real-time O-D changes. The 4] − [3 ln(3) − 3] = −5.977. output equations in this case are • Matrix 4: −5.78. This is the optimum matrix. • Matrix 5: −6.84. yj = • Matrix 6: −9.96. m X dij (k) = i=1 T m X qj (k)bij (k) + ζj (k) i=1 Notice that the entropy value for each candidate so= q(k) bj (k) + ζj (k) (14) lution is negative, hence the selected maximum will be the minimum in absolute value. When the en- where ζj is the measurement error with known cotropy criterion is used, then the solution selected is variance Rj and q and b are vectors where the flow OD matrix 4. measurements and splitting portions (or rates) are organised. The modelling equation is given by 2.3 Dynamic OD Estimation in a Roundabout bj (k + 1) = bj (k) + g(k) The OD estimation problem comes in different flavours depending on the setting. The example of Figure 4 referred to a relative large network, an OD estimation problem rises at a roundabout. Consider the four approaches roundabout shown in Figure 5. There are four entrances and four exits from this system. All entrances and all exits are monitored by detection stations which provide entry flows qj (k) and exit flows yj (k), i = 1, . . . , 4 and k the discrete time index. The flow that enter the roundabout from entry point 1, is split into subflows d1,2 , d1,3 and d1,4 assuming that there is no flow returning and exiting the roundabout from approach 1. Flow at entry point 2 consists of subflows d2,1 , d2,3 and d2,4 and so forth for the rest approaches. The problem at hand is to calculate the subflows di,j (k) during time step k based on the known measurements qj (k) and yj (k). Dr. A. Kotsialos (15) where g is the modelling error with known covariance matrix Q. The Kalman Filter iterative scheme is then as follows [2] h i b̂j (k + 1) = b̂j (k) + Kj (k) yj (k) − q(k)T b̂j (k) (16) −1 Kj (k) = Pj (k)q(k) q(k)T Pj (k)q(k) + Rj Pj (k + 1) = [I − Kj (k)q(k)] Pj (k) + Q (17) (18) where b̂(k) is the estimate of b at k . Truncation and normalisation is applied so that the portions are less than one and the sum up to 1 for each origin, i.e. bij (k) ≥ 0 5 (19) TFMC, 2013–2014 flows q1 using route 1 and q2 using route 2. The traffic assignment problem consists of determining the values of q1 and q2 . These may take some default values based on the average driver behaviour and past experience. They can also be determined as a desired traffic flow pattern that a traffic control centre will try to effectuate. This means that the discussion on the traffic assignment problem stands on the margin between traffic control strategy design and disturbance estimation/prediction. Indeed, there is a very strong relationship between traffic assignment and route guidance. Two basic notions of traffic assignment exist. • User Optimum Principle (Wardrop): The travel times on all routes actually used are equal between them and equal or less than those which would be experienced by a single vehicle on any unused route. In other words, a driver does not have a motive to diverge from the indicated route, as this would result to increased delays. Figure 6: Kalman Filter tracking of OD matrix individual entries, [2]. • System Optimum Principle (normative control): The total travel time is minimum. In this case, the total travel time spent in the network, i.e. the time for queuing and travel time inside the network, is a measure of “social” or system efficiency, as opposed to the driver’s travel time, which is a measure of individual efficiency. Figure 7: Simple network topology for the traffic assignment problem, from . System optimum and user optimum are partially (20) competitive conditions. The former are favoured by traffic operators, while the latter by drivers. For j=1 more detailed treatment of the traffic assignment The results for approach 4 and portions destined to and its extensions, see [1]. 1 and 3 are shown in Figure 6 This method faces difficulties for larger networks.Large networks have long sample times and 3.2 User Optimum for the Single-Origin– Single-Destination Network thus less information is available. Furthemore, congestion results to even longer sample times and Let us consider the network of Figure 7 and let us queues at traffic lights result to reduced correlations assume that the travel times τ along route i = 1, 2 i between inflows and outflows. At any rate, for slowly are given by known functions of the flows qi , fi (qi ). changing ODs this KF approach can be used to track Then the problem at hand is to determine the flows it. q and q that will equalise the travel times of these n X bij (k) = 1∀i = 1, . . . , m. 1 3 3.1 The Traffic Assignement Problem The Basic Setting As mentioned before, the OD matrix provides information about what happens at the network boundaries. The drivers’ routing behaviour is describe in terms of the traffic assignment. Let us consider the simplest road network topology with one origin, one destination and two alternative routes from the origin to the destination as shown in Figure 7. There is a traffic volume d that enters into the system from the origin, which subsequently splits into Dr. A. Kotsialos 2 two alternative routed based on the knowledge of the total demand d and the expressions of fi . The structure of the fi s is very important since they determine each route’s travel times. Figure 8 shows how they are interrelated. When the flow is zero or very light, route 1 is faster than route 2, reflecting the fact that route 1 is shorter (geometrically) from route 2, and hence assuming same free flow speed, the travel times will be shorter. As flow grows in route 1, there will be a point θ when its travel time will become equal to the travel time of an empty route 2. For a total demand that is lower than θ, it does not make sense for driver to divert to route 2. When d > θ, however, there is merit for some of the demand to be diverted to route 2. 6 TFMC, 2013–2014 The total travel time in the system is given by J = q1 τ1 + q2 τ2 ⇒ J = q1 (20q12 + 20) + q2 (15q22 + 100) The problem of determining those q1∗ and q2∗ that will minimise J is as follows. min J = q1 (20q12 + 20) + q2 (15q22 + 100) q1 ,q2 subject to q1 + q2 = d q1 ≥ 0 and q2 ≥ 0 Formulating the problem’s Lagrangian L using λ as multiplier gives Figure 8: Travel times relationship for the simple network topology of Figure 7. L = q1 (20q12 + 20) + q2 (15q22 + 100) − λ(q1 + q2 − d) or Hence, the user optimum is obtained by d if d < θ q1 = qu if d > θ L = 20q13 + 20q1 + 15q23 + 100q2 − λ(q1 + q2 − d) (21) First order optimality conditions yield ∂L ∂q1 ∂L ∂q2 ∂L ∂λ where qu is determined by solving the following equation for it. f1 (qu ) = f2 (d − qu ) (22) = 60q12 + 20 − λ = 0 = 45q22 + 100 − λ = 0 As an example, consider the case where f1 (q1 ) = = q1 + q2 − d = 0 20q12 + 20 and f2 (q2 ) = 15q22 + 100. Then f1 (θ) = f2 (0) ⇒ θ = 2; for d = 4 > θ = 2, qu = 2.422. Therefore for user optimum q1 = 2.422 and q2 = 4 − 4.422 = 1.578. From the first two equations The travel times are f1 (2.422) = 137.32 ≈ f2 (1.578) = 137.35. 60q12 + 20 = 45q22 + 100 ⇒ 3.3 ( by substituting q2 = q1 − d) 15q12 + 90dq1 − 45d2 − 80 = 0 System Optimum for the SingleOrigin–Single-Destination Network The discriminant of the quadratic equation is The system optimum problem for the network of Figure 7 requires the minimisation of the total time spent J = q1 τ1 + q2 τ2 . Thus, the following optimisation problem is formulated. ∆ = (90d)2 − 4 × 15(−45d2 − 80) ⇒ ∆ = 8, 100d2 + 300(9d2 + 16) Thus, the two roots of the quadratic expression are p min J = q1 f1 (q1 ) + q2 f2 (q2 ) (23) q1 ,q2 −90d ± 8, 100d2 + 300(9d2 + 16) q1,1,2 = subject to 2 × 15 q1 + q2 = d (24) and discarding the negative root q1 , q2 ≥ 0 (25) p −45d + 2, 025d2 + 75(9d2 + 16) For the simple network topology of Figure 7 and the q1∗ = 15 static demand assumption, this optimisation probp 2, 025d2 + 75(9d2 + 16) lem can be solved analytically. = −3d + . Assuming the same data as in the user optimum 15 example, each route i = 1, 2 has travel time τi = fi (qi ), given by f1 = 20q12 + 20 and f2 = 15q22 + 100. Obviously The total demand at the origin is d. q2∗ = d − q1∗ . Dr. A. Kotsialos 7 TFMC, 2013–2014 For d = 4 q1∗ p = −3 × 4 + ⇒ q1∗ = 2.048, 2, 025 × 16 + 75(9 × 16 + 16) 15 which means that q2∗ = 4 − 2.048 = 1.952. The travel times on the two alternative routes are f1∗ = f1 (q1∗ ) = 20 × 2.0482 + 20 = 103.89 f2∗ = f2 (q2∗ ) = 15 × 1.9522 + 100 = 157.15. However, is this solution indeed a minimum? Because there is only one linear equality constraint in the problem along with non-negativity constraints, it is sufficient to show that q∗ is a local minimum. To show this we have to verify that the Hessian matrix is positive-definite at q∗ or that " H= ∂2L ∂q12 ∂2L ∂q2 ∂q1 # ∂2L ∂q1 ∂q2 ∂2L ∂q22 120q1 0 0 90q2 >0⇔ q=q ∗ >0 q=q∗ which is true. The Hessian is positive-definite since the deteriminant det H = 10, 800q1 q2 > 0 for every positive q1 and q2 . References [1] G.E. Cantarella and A. Sforza. Traffic Assignment. In M. Papageorgiou, editor, Concise Encyclopedia of Traffic & Transportation Systems, pages 513–520. Pergamon Press, 1991. [2] M. Cremer. Origin-Destination Matrix: Dynamic Estimation. In M. Papageorgiou, editor, Concise Encyclopedia of Traffic & Transportation Systems, pages 310–315. Pergamon Press, 1991. [3] L.G. Willumsen. Origin-Destination Matrix: Dynamic Estimation. In M. Papageorgiou, editor, Concise Encyclopedia of Traffic & Transportation Systems, pages 315–322. Pergamon Press, 1991. Dr. A. Kotsialos 8 TFMC, 2013–2014