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Traffic issues are an important topic in science, not only because there is a lot of research going on, but mostly because it affects our daily life when we drive to the office and when we come back ...

Traffic issues are an important topic in science, not only because there is a lot of research going on, but mostly because it affects our daily life when we drive to the office and when we come back home. In this bachelor thesis , two kind of traffic situations are discussed: a two one-way crossroads and a roadblock. Both situations are modeled and optimized by defining and then minimizing the total waiting time. This way of traffic optimization is not always the best, especially if one thinks from the driver\'s perspective. Therefore, in the second part, both traffic scenarios (two one-way crossroads and a roadblock) are discussed from a very different angle: the driver\'s perspective. Obviously, now quantities like maximum capacity of a road, maximum traffic flux, minimum waiting time are less relevant. Instead, the focus is on the driver\'s irritation produced by the surrounding traffic.

A discrete traffic flow model is used to capture the effect of the traffic lights policy on both queue lengths and on the driver\'s irritation. The simlation results recover real-life data describing the traffic flow on the Leidsestraat in Hillegom, The Netherlands.

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    Presentation bachelor thesis Presentation bachelor thesis Presentation Transcript

    • Maximum traffic flow vs.minimal driver’s irritationChristian Vleugels Where innovation starts
    • Introduction
    • Outline1 Introduction2 Model Two one-way crossroads Roadblock model3 Irritation4 Numerical Results5 Real traffic situation: Leidsestraat in Hillegom6 Conclusion
    • Two one-way crossroads
    • Mathematical NotationCycle: - Phase I: light 1 green, light 2 red: T1 - Phase II: light 1 red, light 2 green: T − T1Notation: - Cycle number: n - Arrival rates: α1 , α2 - Passing rates: β1 , β2 - Queue lengths: N1 , N2 (w,1) (w,2) - Waiting times: Tn→n+1 , Tn→n+1 (w,1) (w,2) - Total waiting time: T w = Tn→n+1 + Tn→n+1
    • ModelGoal:Find the value for T1 for which the total waiting time is minimizedAssumptions: - The arrival rates α1 , α2 are constant - The passing rates β1 , β2 are constant - The number of passengers in each car are the sameQueue lengthsAfter phase I: - N1 (nT + T1 ) = max{N1 (nT ) + α1 T1 − β1 T1 , 0} - N2 (nT + T1 ) = N2 (nT ) + α2 T1After phase II: - N1 ((n + 1)T ) = N1 (nT + T1 ) + α1 (T − T1 ) - N2 ((n + 1)T ) = max{N2 (nT + T1 ) + α2 − β2 (T − T1 ), 0}
    • ModelTotal waiting timeThe number of cars that have to wait during a red phase times the length ofthat red phaseWaiting time for direction 1: (w,1) 1 Tn→n+1 = N1 (nT + T1 )(T − T1 ) + α1 (T − T1 )2 2Waiting time for direction 2: (w,2) 1 2 Tn→n+1 = N2 (nT )T1 + α2 T1 2
    • Traffic statesLight trafficAll the cars in the queue are able to pass in one green timeHeavy trafficCars have to wait multiple cyclesCombination of light and heavy trafficOne direction light traffic, the other direction heavy traffic
    • Traffic states: Light trafficConditionsAll the cars in the queue are able to pass in one green time: α1 T − β1 T1 ≤ 0, α2 T − β2 (T − T1 ) ≤ 0Consequence 1:After each green phase, the queue is emptyConsequence 2:The total waiting time only consists of the contribution of the cars that arriveduring a red phase
    • Traffic states: Light trafficConsequence 2:The total waiting time only consists of the contribution of the cars that arriveduring a red phase: (w,1) 1 Tn→n+1 = α1 (T − T1 )2 2 (w,2) 1 2 Tn→n+1 = α2 T1 2Value for T1 which minimizes T w : α1 T1 = T α1 + α2
    • Traffic states: Light trafficConditions α1 T − β1 T1 ≤ 0, α2 T − β2 (T − T1 ) ≤ 0Value for T1 which minimizes T w : α1 T1 = T α1 + α2Relations on α1 , α2 , β1 , β2 : β1 ≥ α1 + α2 , β2 ≥ α1 + α2
    • Traffic states: Heavy trafficConditionsCars in the queue have to wait more than one cycle: α1 T − β1 T1 > 0, α2 T − β2 (T − T1 ) > 0Consequence 1:Every cycle, the queue lengths growConsequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase
    • Traffic states: Heavy trafficConsequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase: (w,1) Tn→n+1 = n(T − T1 )(α1 T − β1 T1 ) (w,2) Tn→n+1 = nT1 (α2 T − β2 (T − T1 ))Value for T1 which minimizes T w : α1 − α2 + β1 + β2 T1 = T 2β1 + 2β2
    • Traffic states: Heavy trafficConditions α1 T − β1 T1 > 0, α2 T − β2 (T − T1 ) > 0Value for T1 which minimizes T w : α1 − α2 + β1 + β2 T1 = T 2β1 + 2β2Relations on α1 , α2 , β1 , β2 : 2 2 α1 β1 + α2 β1 + 2α1 β2 > β1 + β1 β2 , α1 β2 + α2 β2 + 2α2 β1 > β1 β2 + β2
    • Traffic states: Combination trafficDirection 1 light, direction 2 heavy: α1 T − β1 T1 ≤ 0, α2 T − β2 (T − T1 ) > 0Waiting times: (w,1) 1 Tn→n+1 = α1 (T − T1 )2 2 (w,2) Tn→n+1 = nT1 (α2 T − β2 (T − T1 ))
    • Traffic states: Combination trafficValue for T1 which minimizes T w : β2 − α2 T1 = T 2β2T1 has to be positive: β2 > α2Substituting the value for T1 into the conditions for combination traffic:β2 < α2Consequence:The total waiting time is then minimized when the crossroads uses its fullcapacity: α1 β2 − α2 T1 = max , T β1 β2
    • Roadblock
    • Mathematical NotationCycle: - Phase I: light 1 green, light 2 red: T1 - Phase II: light 1 red, light 2 red: τ - Phase III: light 1 red, light 2 green: T − T1 − 2τ - Phase IV: light 1 red, light 2 red: τGoal:Find the value for T1 for which the total waiting time is minimized
    • ModelWaiting timeFor direction 1: (w,1) 1 Tn→n+1 = N1 (nT + T1 )(T − T1 ) + α1 (T − T1 )2 2Waiting timeFor direction 2: (w,2) 1 Tn→n+1 = N2 (nT )(T1 + τ ) + N2 ((n + 1)T − τ )τ + α2 (T1 + 2τ )2 2
    • Traffic states: Light trafficLight trafficAll the cars in the queue are able to pass in one green time.The total waiting time only consists of the contribution of the cars that arriveduring a red phase (w,1) 1 Tn→n+1 = α1 (T − T1 )2 2 (w,2) 1 Tn→n+1 = α2 (T1 + 2τ )2 2Value for T1 which minimizes T w : α1 T − 2α2 τ T1 = α1 + α2
    • Traffic states: Light trafficValue for T1 in the two one-way crossroads model α1 T1 = T α1 + α2Value for T1 in the roadblock model α1 T − 2α2 τ T1 = α1 + α2τ = 0 leads to the same value for T1 as in the two one-way crossroadsmodel
    • Traffic states: Heavy trafficConditionsCars in the queue have to wait more than one cycle: α1 T − β1 T1 > 0, α2 T − β2 (T − T1 − 2τ ) > 0Consequence 1:Every cycle, the queue lengths growConsequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase
    • Traffic states: Heavy trafficConsequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase: (w,1) Tn→n+1 = n(T − T1 )(α1 T − β1 T1 ) (w,2) Tn→n+1 = n(T1 + 2τ )(α2 T − β2 (T − T1 − 2τ ))Value for T1 which minimizes T w : (α1 − α2 + β1 + β2 )T − 4β2 τ T1 = 2β1 + 2β2
    • Traffic states: Heavy trafficValue for T1 in the two one-way crossroads model α1 − α2 + β1 + β2 T1 = T 2β1 + 2β2Value for T1 in the roadblock model (α1 − α2 + β1 + β2 )T − 4β2 τ T1 = 2β1 + 2β2Again τ = 0 leads to the same value for T1 as in the two one-waycrossroads model
    • Irritation (two one-way crossroads)Goal:Modeling the driver’s irritation and minimizing it with the use of smart trafficlight settingsDifferent kind of irritation: Irritation per cycle; Irritation per direction; Irritation per car;
    • Irritation (two one-way crossroads)Irritation I per cycle n:Is defined as the sum of the irritation per cycle of direction 1 and 2: (1) (2) In→n+1 = In→n+1 + In→n+1
    • Irritation (two one-way crossroads)There are two moments during a cycle where the irritation isdefined: - At the end of phase I: nT + T1 - At the end of phase II: (n + 1)TWe call the moment that the light switches from green to red the vitalmoment.
    • Irritation (two one-way crossroads)At the end of phase I: N1 (nT +T1 ) (1) In→n+1 (nT + T1 ) = i (1) (k ) k =1 (2) 1 2 In→n+1 (nT + T1 ) = C2 α2 T1 2
    • Irritation (two one-way crossroads)At the end of phase II: N1 (nT +T1 ) (1) 1 In→n+1 ((n + 1)T ) = i (1) (k ) + C1 α1 (T − T1 )2 2 k =1 N2 ((n+1)T ) (2) 1 2 In→n+1 ((n + 1)T ) = C2 α2 T1 + i (2) (k ) 2 k =1
    • Irritation per carThe irritation per car i (1) (k ) and i (2) (k ) depend on: - the waiting time - the number of cars in the queue of the other direction - the position k in the queue
    • Irritation per carWaiting timeThe longer you have to wait, the higher the irritation i (1) (k ) ∼ (T − T1 ) and i (2) (k ) ∼ T1The number of cars that are waiting in the other directionThe smaller the number of cars that are waiting in the other direction, thehigher the irritation 1 i (1) (k ) ∼ and i (2) (k ) ∼ 1 N1 +1 N2 + 1
    • Irritation per carPosition in the queue - Case I: the closer you are to the traffic light when the light switches to red, the higher the irritation 1 i (1) (k ), i (2) (k ) ∼ k - Case II: the further away you are to the traffic light when the light switches to red, the higher the irritation i (1) (k ), i (2) (k ) ∼ kFunction: 1 - Case I: f (k ) = k - Case II: f (k ) = k
    • Irritation per car f (k ) i (1) (k ) = (T − T1 ), N2 (nT + T1 ) + 1 f (k ) i (2) (k ) = T1 . N1 ((n + 1)T ) + 1
    • Total irritation (two one-way crossroads)For case I HN1 (T − T1 ) In→n+1 = + C1 1 α1 (T − T1 )2 + 2 N2 (nT + T1 ) + 1 HN2 T1 + C2 1 α2 T1 2 2 N1 ((n + 1)T ) + 1Where: N1 (nT +T1 ) N2 ((n+1)T ) 1 1 HN1 := and HN2 := ) k k k =1 k =1
    • For case II SN1 (T − T1 ) In→n+1 = + C1 1 α1 (T − T1 )2 + 2 N2 (nT + T1 ) + 1 SN2 T1 + C2 1 α2 T1 2 2 N1 ((n + 1)T ) + 1Where: N1 (nT +T1 ) N2 ((n+1)T ) SN1 := k and SN2 := k k =1 k =1
    • Remarks on the irritationDimensions Temperature (Degrees Celsius) Pressure (Pascal)
    • Numerical results: light trafficFor light traffic holds that:All the cars in the queue are able to pass in one green time. There is noirritation at the moment that a direction gets red light, because all the carshave been able to pass.Consequence:The only irritation arises from the waiting time of cars that arrive during ared phase.Result:For the light traffic state, the total waiting time and the irritation are minimalfor the same value for T1 .
    • Numerical results: heavy trafficScenario: Arrival rates (cars per second): α1 = 0.5 and α2 = 0.4; Passing rates (cars per second): β1 = 0.4 and β2 = 0.1; Cycle period (seconds): T = 30.0; Transit period (seconds): τ = 5.0; Cycle number: n = 100;The arrival and passing rates apply to a heavy traffic scenario
    • Numerical results: heavy trafficMinimizing T w : (α1 − α2 + β1 + β2 )T − 4β2 τ T1 = = 16.0 2β1 + 2β2This also means that the other direction gets a green time of only 4.0seconds
    • Numerical results: heavy traffic TWT s 60 000 50 000 40 000 30 000 20 000 10 000 T1 s 0 5 10 15 20 25 30Minimizing T w numerically: T1 = 15.9846
    • Numerical results: heavy traffic Ir case I s Ir case II s 400 120 000 100 000 300 80 000 200 60 000 40 000 100 20 000 T1 s T1 s 0 5 10 15 20 25 30 0 5 10 15 20 25 30Minimizing I numerically for both cases: Case I: T1 = 12.2157; Case II: T1 = 11.1761;
    • Numerical results: heavy traffic Table : Table of the total waiting time and the irritation (case I) for different T1 . T1 = 15.9846 T1 = 12.2157 Total Waiting Time 42656.2 43320.4 Irritation (case I) 184.517 178.062 Table : Table of the total waiting time and the irritation (case II) for different T1 . T1 = 15.9846 T1 = 11.1761 Total Waiting Time 42656.2 43755.9 Irritation (case II) 25167.4 22084.4
    • Numerical results: heavy trafficCan we explain these different values? They more or less have the same arrival rates; The passing rates are not the same, direction 1 has a much higher passing rate; Note that the irritation depends on the queue length of the other direction; Therefore, a larger value of T1 would mean that more cars can leave queue 1; This means that the irritation of queue 2 would be larger; Hence, a smaller value for T1 would result in a smaller irritation;
    • Numerical results: heavy traffic Relative value 1.4 1.2 Ir Ir 0 in Case I 1.0 0.8 Ir Ir 0 in Case II 0.6 0.4 TWT TWT 0 0.2 T1 s 0 5 10 15 20 25 30
    • Real traffic situation: Hillegom
    • Real traffic situation: HillegomData:Consists of traffic intensities during morning and evening rush hourThese traffic intensities are converted into arrival rates.Assumptions: We can categorize the arrival rates in the heavy traffic scenario; We assume both passing rates to be equal to 0.2; Cycle period: 30.0; Transit period: 5.0; The period of rush hour lasts for 2 hours, so n = 240;
    • Leidsestraat during morning rush hour Relative value 1.4 1.2 Ir Ir 0 in Case I 1.0 0.8 Ir Ir 0 in Case II 0.6 0.4 TWT TWT 0 0.2 T1 s 0 5 10 15 20Minimizing: T w leads to T1 = 5.8; I in case I leads to T1 = 5.4; I in case II leads to T1 = 3.7;
    • Leidsestraat during evening rush hour Relative value 1.0 Ir Ir 0 in Case I 0.8 0.6 Ir Ir 0 in Case II 0.4 TWT TWT 0 0.2 T1 s 0 5 10 15 20Minimizing: T w leads to T1 = 13.3; I in case I leads to T1 = 15.0; I in case II leads to T1 = 14.0;
    • Conclusion After analyzing the models of the two traffic situations (two one-way crossroads and the roadblock), we can conclude that both situations are mathematically the same; Minimizing the total waiting time for the two one-way crossroads and the roadblock leads to nice expressions for the green time(s); Minimizing the irritation for both models numerically leads to the same values for the green time(s) in the light traffic state and the combination of light and heavy traffic; In the heavy traffic state, we get different values for the green time(s) when we minimized the total waiting time and the irritation; Therefore, minimizing the total waiting time does not always lead to the best traffic light settings for the drivers.
    • Open questions Is the irritation an extensive or intensive property? What happens with the irritation if the cycle period goes to 0 or to infinity? Is it possible to put the two different case into one single case?
    • Questions?