Presentation bachelor thesis

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Maximum traffic flow vs.minimal driver’s irritation

Christian Vleugels

Introduction

Outline

1 Introduction

2 ModelTwo one-way crossroadsRoadblock model

3 Irritation

4 Numerical Results

5 Real traffic situation: Leidsestraat in Hillegom

6 Conclusion

Two one-way crossroads

Mathematical Notation

Cycle:- Phase I: light 1 green, light 2 red: T1- Phase II: light 1 red, light 2 green: T − T1

Notation:- Cycle number: n- Arrival rates: α1, α2- Passing rates: β1, β2- Queue lengths: N1,N2

- Waiting times: T (w,1)n→n+1,T

(w,2)n→n+1

- Total waiting time: T w = T (w,1)n→n+1 + T (w,2)

n→n+1

Model

Goal:Find the value for T1 for which the total waiting time is minimized

Assumptions:- The arrival rates α1, α2 are constant- The passing rates β1, β2 are constant- The number of passengers in each car are the same

Queue lengthsAfter phase I: - N1(nT + T1) = max{N1(nT ) + α1T1 − β1T1,0}

- N2(nT + T1) = N2(nT ) + α2T1

After 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}

Model

Total waiting timeThe number of cars that have to wait during a red phase times the length ofthat red phase

Waiting time for direction 1:

T (w,1)n→n+1 = N1(nT + T1)(T − T1) +

12α1(T − T1)

2

Waiting time for direction 2:

T (w,2)n→n+1 = N2(nT )T1 +

12α2T 2

1

Traffic states

Light trafficAll the cars in the queue are able to pass in one green time

Heavy trafficCars have to wait multiple cycles

Combination of light and heavy trafficOne direction light traffic, the other direction heavy traffic

Traffic states: Light traffic

ConditionsAll the cars in the queue are able to pass in one green time:

α1T − β1T1 ≤ 0, α2T − β2(T − T1) ≤ 0

Consequence 1:After each green phase, the queue is empty

Consequence 2:The total waiting time only consists of the contribution of the cars that arriveduring a red phase


Consequence 2:The total waiting time only consists of the contribution of the cars that arriveduring a red phase:

T (w,1)n→n+1 =

12α1(T − T1)

2

T (w,2)n→n+1 =

12α2T 2

1

Value for T1 which minimizes T w :

T1 =α1

α1 + α2T


Conditions

α1T − β1T1 ≤ 0, α2T − β2(T − T1) ≤ 0


T1 =α1

α1 + α2T

Relations on α1, α2, β1, β2:

β1 ≥ α1 + α2, β2 ≥ α1 + α2

Traffic states: Heavy traffic

ConditionsCars in the queue have to wait more than one cycle:

α1T − β1T1 > 0, α2T − β2(T − T1) > 0

Consequence 1:Every cycle, the queue lengths grow

Consequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase


Consequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase:

T (w,1)n→n+1 = n(T − T1)(α1T − β1T1)

T (w,2)n→n+1 = nT1(α2T − β2(T − T1))


T1 =α1 − α2 + β1 + β2

2β1 + 2β2T


Conditions

α1T − β1T1 > 0, α2T − β2(T − T1) > 0


T1 =α1 − α2 + β1 + β2

2β1 + 2β2T

Relations on α1, α2, β1, β2:

α1β1 + α2β1 + 2α1β2 > β21 + β1β2, α1β2 + α2β2 + 2α2β1 > β1β2 + β2

2

Traffic states: Combination traffic

Direction 1 light, direction 2 heavy:

α1T − β1T1 ≤ 0, α2T − β2(T − T1) > 0

Waiting times:

T (w,1)n→n+1 =

12α1(T − T1)

2

T (w,2)n→n+1 = nT1(α2T − β2(T − T1))

Traffic states: Combination traffic


T1 =β2 − α2

2β2T

T1 has to be positive: β2 > α2Substituting the value for T1 into the conditions for combination traffic:β2 < α2

Consequence:The total waiting time is then minimized when the crossroads uses its fullcapacity:

T1 = max(α1

β1,β2 − α2

β2

)T

Roadblock

Mathematical Notation

Cycle:- 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: τ

Goal:Find the value for T1 for which the total waiting time is minimized

Model

Waiting timeFor direction 1:

T (w,1)n→n+1 = N1(nT + T1)(T − T1) +

12α1(T − T1)

2

Waiting timeFor direction 2:

T (w,2)n→n+1 = N2(nT )(T1 + τ) + N2((n + 1)T − τ)τ + 1

2α2(T1 + 2τ)2


Light 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

T (w,1)n→n+1 =

12α1(T − T1)

2

T (w,2)n→n+1 =

12α2(T1 + 2τ)2


T1 =α1T − 2α2τ

α1 + α2


Value for T1 in the two one-way crossroads model

T1 =α1

α1 + α2T

Value for T1 in the roadblock model

T1 =α1T − 2α2τ

α1 + α2

τ = 0 leads to the same value for T1 as in the two one-way crossroadsmodel


ConditionsCars in the queue have to wait more than one cycle:

α1T − β1T1 > 0, α2T − β2(T − T1 − 2τ) > 0

Consequence 1:Every cycle, the queue lengths grow

Consequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase


Consequence 2:For large cycle numbers the total waiting time only depends on the numberof cars waiting at the start of a red phase:

T (w,1)n→n+1 = n(T − T1)(α1T − β1T1)

T (w,2)n→n+1 = n(T1 + 2τ)(α2T − β2(T − T1 − 2τ))


T1 =(α1 − α2 + β1 + β2)T − 4β2τ

2β1 + 2β2


Value for T1 in the two one-way crossroads model

T1 =α1 − α2 + β1 + β2

2β1 + 2β2T

Value for T1 in the roadblock model

T1 =(α1 − α2 + β1 + β2)T − 4β2τ

2β1 + 2β2

Again 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 settings

Different kind of irritation:Irritation per cycle;Irritation per direction;Irritation per car;


Irritation I per cycle n:Is defined as the sum of the irritation per cycle of direction 1 and 2:

In→n+1 = I(1)n→n+1 + I(2)n→n+1


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)T

We call the moment that the light switches from green to red the vitalmoment.


At the end of phase I:

I(1)n→n+1(nT + T1) =

N1(nT+T1)∑k=1

i(1)(k)

I(2)n→n+1(nT + T1) = C212α2T 2

1


At the end of phase II:

I(1)n→n+1((n + 1)T ) =

N1(nT+T1)∑k=1

i(1)(k) + C112α1(T − T1)

2

I(2)n→n+1((n + 1)T ) = C212α2T 2

1 +

N2((n+1)T )∑k=1

i(2)(k)

Irritation per car

The 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 car

Waiting timeThe longer you have to wait, the higher the irritation

i(1)(k) ∼ (T − T1) and i(2)(k) ∼ T1

The 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

i(1)(k) ∼ 1N2 + 1

and i(2)(k) ∼ 1N1+1

Irritation per car

Position in the queue- Case I: the closer you are to the traffic light when the light

switches to red, the higher the irritationi(1)(k), i(2)(k) ∼ 1

k- Case II: the further away you are to the traffic light when the light

switches to red, the higher the irritationi(1)(k), i(2)(k) ∼ k

Function:- Case I: f (k) = 1

k- Case II: f (k) = k

Irritation per car

i(1)(k) =f (k)

N2(nT + T1) + 1(T − T1),

i(2)(k) =f (k)

N1((n + 1)T ) + 1T1.

Total irritation (two one-way crossroads)

For case I

In→n+1 =HN1(T − T1)

N2(nT + T1) + 1+ C1

12α1(T − T1)

2+

HN2T1

N1((n + 1)T ) + 1+ C2

12α2T 2

1

Where:

HN1 :=

N1(nT+T1)∑k=1

1k

and HN2 :=

N2((n+1)T )∑k=1

1k)

For case II

In→n+1 =SN1(T − T1)

N2(nT + T1) + 1+ C1

12α1(T − T1)

2+

SN2T1

N1((n + 1)T ) + 1+ C2

12α2T 2

1

Where:

SN1 :=

N1(nT+T1)∑k=1

k and SN2 :=

N2((n+1)T )∑k=1

k

Remarks on the irritation

DimensionsTemperature (Degrees Celsius)Pressure (Pascal)

Numerical results: light traffic

For 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 traffic

Scenario: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


Minimizing T w :

T1 =(α1 − α2 + β1 + β2)T − 4β2τ

2β1 + 2β2= 16.0

This also means that the other direction gets a green time of only 4.0seconds


0 5 10 15 20 25 30T1 HsL

10 000

20 000

30 000

40 000

50 000

60 000

TWT HsL

Minimizing T w numerically:

T1 = 15.9846


0 5 10 15 20 25 30T1 HsL

100

200

300

400

Ir case I HsL

0 5 10 15 20 25 30T1 HsL

20 000

40 000

60 000

80 000

100 000

120 000

Ir case II HsL

Minimizing I numerically for both cases:Case I: T1 = 12.2157;Case II: T1 = 11.1761;


Table : Table of the total waiting time and the irritation (case I) for different T1.

T1 = 15.9846 T1 = 12.2157Total 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 = 11.1761Total Waiting Time 42656.2 43755.9

Irritation (case II) 25167.4 22084.4


Can 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 higherpassing rate;Note that the irritation depends on the queue length of the otherdirection;Therefore, a larger value of T1 would mean that more cars can leavequeue 1;This means that the irritation of queue 2 would be larger;Hence, a smaller value for T1 would result in a smaller irritation;


0 5 10 15 20 25 30T1 HsL

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Relative value

TWT�TWTH0L

Ir�IrH0L in Case II

Ir�IrH0L in Case I

Real traffic situation: Hillegom

Real traffic situation: Hillegom

Data: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

0 5 10 15 20T1 HsL

0.2

0.4

0.6

0.8

1.0

1.2

1.4

Relative value

TWT�TWTH0L



Minimizing: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

0 5 10 15 20T1 HsL

0.2

0.4

0.6

0.8

1.0

Relative value

TWT�TWTH0L



Minimizing: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-waycrossroads and the roadblock), we can conclude that both situationsare mathematically the same;Minimizing the total waiting time for the two one-way crossroads andthe roadblock leads to nice expressions for the green time(s);Minimizing the irritation for both models numerically leads to the samevalues for the green time(s) in the light traffic state and the combinationof 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 thebest 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 toinfinity?Is it possible to put the two different case into one single case?

Questions?

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