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.

1. Maximum traffic flow vs.
minimal driver’s irritation
Christian Vleugels
Where innovation starts

2. Introduction

3. Outline
1 Introduction
2 Model
Two one-way crossroads
Roadblock model
3 Irritation
4 Numerical Results
5 Real traffic situation: Leidsestraat in Hillegom
6 Conclusion

4. Two one-way crossroads

5. 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
(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

6. 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 lengths
After phase I: - N1 (nT + T1 ) = max{N1 (nT ) + α1 T1 − β1 T1 , 0}
- N2 (nT + T1 ) = N2 (nT ) + α2 T1
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}

7. Model
Total waiting time
The number of cars that have to wait during a red phase times the length of
that red phase
Waiting time for direction 1:
(w,1) 1
Tn→n+1 = N1 (nT + T1 )(T − T1 ) + α1 (T − T1 )2
2
Waiting time for direction 2:
(w,2) 1 2
Tn→n+1 = N2 (nT )T1 + α2 T1
2

8. Traffic states
Light traffic
All the cars in the queue are able to pass in one green time
Heavy traffic
Cars have to wait multiple cycles
Combination of light and heavy traffic
One direction light traffic, the other direction heavy traffic

9. Traffic states: Light traffic
Conditions
All the cars in the queue are able to pass in one green time:
α1 T − β1 T1 ≤ 0, α2 T − 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 arrive
during a red phase

10. Traffic states: Light traffic
Consequence 2:
The total waiting time only consists of the contribution of the cars that arrive
during a red phase:
(w,1) 1
Tn→n+1 = α1 (T − T1 )2
2
(w,2) 1 2
Tn→n+1 = α2 T1
2
Value for T1 which minimizes T w :
α1
T1 = T
α1 + α2

11. Traffic states: Light traffic
Conditions
α1 T − β1 T1 ≤ 0, α2 T − β2 (T − T1 ) ≤ 0
Value for T1 which minimizes T w :
α1
T1 = T
α1 + α2
Relations on α1 , α2 , β1 , β2 :
β1 ≥ α1 + α2 , β2 ≥ α1 + α2

12. Traffic states: Heavy traffic
Conditions
Cars in the queue have to wait more than one cycle:
α1 T − β1 T1 > 0, α2 T − 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 number
of cars waiting at the start of a red phase

13. Traffic states: Heavy traffic
Consequence 2:
For large cycle numbers the total waiting time only depends on the number
of 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

14. Traffic states: Heavy traffic
Conditions
α1 T − β1 T1 > 0, α2 T − β2 (T − T1 ) > 0
Value for T1 which minimizes T w :
α1 − α2 + β1 + β2
T1 = T
2β1 + 2β2
Relations 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

15. Traffic states: Combination traffic
Direction 1 light, direction 2 heavy:
α1 T − β1 T1 ≤ 0, α2 T − β2 (T − T1 ) > 0
Waiting times:
(w,1) 1
Tn→n+1 = α1 (T − T1 )2
2
(w,2)
Tn→n+1 = nT1 (α2 T − β2 (T − T1 ))

16. Traffic states: Combination traffic
Value for T1 which minimizes T w :
β2 − α2
T1 = T
2β2
T1 has to be positive: β2 > α2
Substituting 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 full
capacity:
α1 β2 − α2
T1 = max , T
β1 β2

17. Roadblock

18. 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

19. Model
Waiting time
For direction 1:
(w,1) 1
Tn→n+1 = N1 (nT + T1 )(T − T1 ) + α1 (T − T1 )2
2
Waiting time
For direction 2:
(w,2) 1
Tn→n+1 = N2 (nT )(T1 + τ ) + N2 ((n + 1)T − τ )τ + α2 (T1 + 2τ )2
2

20. Traffic states: Light traffic
Light traffic
All 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 arrive
during a red phase
(w,1) 1
Tn→n+1 = α1 (T − T1 )2
2
(w,2) 1
Tn→n+1 = α2 (T1 + 2τ )2
2
Value for T1 which minimizes T w :
α1 T − 2α2 τ
T1 =
α1 + α2

21. Traffic states: Light traffic
Value for T1 in the two one-way crossroads model
α1
T1 = T
α1 + α2
Value for T1 in the roadblock model
α1 T − 2α2 τ
T1 =
α1 + α2
τ = 0 leads to the same value for T1 as in the two one-way crossroads
model

22. Traffic states: Heavy traffic
Conditions
Cars in the queue have to wait more than one cycle:
α1 T − β1 T1 > 0, α2 T − 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 number
of cars waiting at the start of a red phase

23. Traffic states: Heavy traffic
Consequence 2:
For large cycle numbers the total waiting time only depends on the number
of 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

24. Traffic states: Heavy traffic
Value for T1 in the two one-way crossroads model
α1 − α2 + β1 + β2
T1 = T
2β1 + 2β2
Value for T1 in the roadblock model
(α1 − α2 + β1 + β2 )T − 4β2 τ
T1 =
2β1 + 2β2
Again τ = 0 leads to the same value for T1 as in the two one-way
crossroads model

25. Irritation (two one-way crossroads)
Goal:
Modeling the driver’s irritation and minimizing it with the use of smart traffic
light settings
Different kind of irritation:
Irritation per cycle;
Irritation per direction;
Irritation per car;

26. 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

27. Irritation (two one-way crossroads)
There are two moments during a cycle where the irritation is
defined:
- 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 vital
moment.

28. 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

29. 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

30. 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

31. Irritation per car
Waiting time
The 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 direction
The smaller the number of cars that are waiting in the other direction, the
higher the irritation
1
i (1) (k ) ∼ and i (2) (k ) ∼ 1
N1 +1
N2 + 1

32. 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 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 ) ∼ k
Function:
1
- Case I: f (k ) = k
- Case II: f (k ) = k

33. 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

34. 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 ) + 1
Where:
N1 (nT +T1 ) N2 ((n+1)T )
1 1
HN1 := and HN2 := )
k k
k =1 k =1

35. 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 ) + 1
Where:
N1 (nT +T1 ) N2 ((n+1)T )
SN1 := k and SN2 := k
k =1 k =1

36. Remarks on the irritation
Dimensions
Temperature (Degrees Celsius)
Pressure (Pascal)

37. 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 no
irritation at the moment that a direction gets red light, because all the cars
have been able to pass.
Consequence:
The only irritation arises from the waiting time of cars that arrive during a
red phase.
Result:
For the light traffic state, the total waiting time and the irritation are minimal
for the same value for T1 .

38. 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

39. Numerical results: heavy traffic
Minimizing T w :
(α1 − α2 + β1 + β2 )T − 4β2 τ
T1 = = 16.0
2β1 + 2β2
This also means that the other direction gets a green time of only 4.0
seconds

40. 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 30
Minimizing T w numerically:
T1 = 15.9846

41. 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 30
Minimizing I numerically for both cases:
Case I: T1 = 12.2157;
Case II: T1 = 11.1761;

42. 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

43. Numerical results: heavy traffic
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 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;

44. 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

45. Real traffic situation: Hillegom

46. Real traffic situation: Hillegom
Data:
Consists of traffic intensities during morning and evening rush hour
These 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;

47. 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 20
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;

48. 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 20
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;

49. 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.

50. 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?

51. Questions?