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Maryam Pourebadi Kent State University April 2016.

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Presentation on theme: "Maryam Pourebadi Kent State University April 2016."— Presentation transcript:

1 Maryam Pourebadi Kent State University April 2016

2 Definition of the real-world problem construct a graph from this R-W problem Graph special properties Solution to the Problem Increase efficiency of the solution Conclusion References

3 Application of circular-arc graphs to the traffic light phasing problem

4 The problem is to install traffic lights at a road junction in such a way that traffic at the junction flows safely, Smoothly efficiently. With increasing concern about energy use, the latter goal is becoming of increased importance.

5 Approaching the traffic intersection are various traffic stream, patterns or routes through the intersection which traffic takes. Specific Example: A traffic intersection has a two-way street meeting a two-way street. Labeled with the letters “a” through “f”.

6 Application of circular-arc graphs to the traffic light phasing problem

7 Certain traffic streams may be termed compatible if their simultaneous flow would not result in any accidents. The decision about compatibility is made ahead of time, by a traffic engineer, and may be based on estimated volume of traffic in a stream as well as the traffic pattern.

8 The compatibility information can be summarized in a graph G, the Compatibility Graph. The vertices of G are the traffic streams, and two streams are joined by an edge if and only if they are judged compatible.

9 In traffic light phasing, we wish to assign a period of time to each stream during which it receives a green light, and to do it in such a way that only compatible traffics streams can get green lights at the same time. There is a cycle of green and red lights, and then after the cycle is finished, it begins again, over and over. Solving the problem Cycle is hard on general graphs, so, it leads us to circular arc graph.

10 Circle arc graphs are intersection graphs of arcs on a given circle. Assume that a circle perimeter corresponds to the total cycle period, and duration when a given traffic stream gets a green light corresponds to an arc of this circle. Every vertex is represented by an arc, such that two vertices are adjacent if and only if the corresponding arcs overlap. (we assume that a given stream receives only one continues green light during each cycle.)

11 Application of circular-arc graphs to the traffic light phasing problem

12 We call G a Helly circular-arc graph if there exists a circular-arc representation for G which satisfies the Helly property. An undirected graph G is a Helly circular arc graph if and only if its clique matrix has the circular I's property for columns. An undirected graph G is a circular-arc graph if its augmented adjacency matrix has the circular I's property for columns.

13 We call G a proper circular-arc graph if there exists a circular-arc representation for G in which no arc properly contains another. (i.e., no two arcs share a common endpoint and no two arcs together cover the entire circle) An undirected graph G is a proper circular-arc graph if and only if its augmented adjacency matrix has the circular I's property for columns and, for every permutation of the rows and columns that is a cyclic shift or inversion of their circular I's order, the last 1 in the first column does not occur after the last 1 of the second column, excluding columns which are either all zeros or all ones.

14 Circular-arc graphs are not always perfect, as the odd chordless cycles C5, C7, etc., are circular-arc graphs.

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16 A circular-arc representation of an undirected graph G which fails to cover some point p on the circle will be topologically the same as an interval representation of G. Specifically, we can cut the circle at p and straighten it out to a line, arcs becoming intervals. It is easy to see, therefore, that every interval graph is a circular-arc graph. The converse, is false. P

17 NO! Assume arcs of our circular-arc cover all the circle, then, at least during the night, yellow light start to flash. So, all streams can stop for a second. That point of the time can be considered as mentioned p, and we can cut the circle at p. It turns out that always there is a point p on the circle that has no arc.

18 Interval Graph is the intersection graph of a family of intervals on the real line. G is an interval graph if its vertices can be put into one-to-one correspondence with a set of intervals of a linearly ordered set. The four graph problems (Clique number, Stability number, Chromatic number, Clique cover) can be solved efficiently in linear time on Interval graphs.

19 Application of circular-arc graphs to the traffic light phasing problem

20 A feasible green light assignment consist of an assignment of an arc of the circle to each traffic stream so that only compatible streams are allowed to receive overlapping arcs. The resulting circular arc graph may not be the compatibility graph because There may be two compatible streams but they need not get a green light at the same time.

21 The intersection graph corresponding to any feasible green light assignment (corresponding to the arcs of the circle), is a sub-graph of the compatibility graph(G). It means perhaps some of the edges deleted, so called a spanning sub-graph (H). Note that the compatibility graph (G) is not necessary interval, However, the spanning sub-graph (H) is an interval graph. So we have to take all spanning sub- graph of G in to account and choose from them the spanning sub-graph that has the most maximal clique.

22 Maximal Clique is a clique that cannot be extended by including one more adjacent vertex. the cliques of the intersection graph is a solution which is done by dividing cliques as signal groups. The cliques of mentioned intersection graph are: C1= { a, b, e}C3= { c, g} C2= { b, d, e}C4= { f, g} There are 4 phases and if we consider the cycle time to be 120 seconds each can be allowed to move once in the cycle for 30 seconds which is the trivial solution.

23 Since the intersection graph of the phasing traffic light problem is an interval graph, finding maximal cliques can be done efficiently in linear time. C1C1C2C2 C4C4C3C3

24 Application of circular-arc graphs to the traffic light phasing problem

25 We might wish to minimize a weighted sum of red light times by weighting more heavily the red light time for heavily traveled traffic streams. As Stoffers points out, we might have some information about expected arrival times of different traffic streams, and we might wish to penalize starting times for being far from the traffic stream’s expected arrival time and minimize the penalties. We might wish to minimize the total amount of waiting time, total amount of red light time in cycle.

26 The cliques of mentioned intersection graph were: C1= { a, b, e}C3= { c, g} C2= { b, d, e}C4= { f, g} Let us consider the cliques of the intersection graph H as the vertices of the graph F and an edge joining the vertices if there exist a non empty intersection. Thus the cliques set correspond to the family of sets. F = {C1,C2,C3,C4 }; where Ci; i = 1,2,3,4.

27 Now if we want to phase these sets of cliques, we can put C1 & C2 in a set and C3 & C4 in another set: S1= { C1,C2 } S2= { C3, C4 } Again considering the cycle time to be 120 seconds; each of the signal groups can be allowed to move for 60 seconds each which is a better solution as compared to the earlier one as most of the traffic streams are moving which reduces the waiting time.

28 Display new traffic streams:

29 assume that at the time of starting, all lights are red Ci corresponds to a phase during which all streams in that clique receive green lights. In phase1, traffic streams a, b, c, in phase2, traffic streams b, d receive a green light. Suppose we assign to each phase Ci a duration Di. Our aim is to determine the Di s (≥ 0 ) so that the total waiting time is minimum. Further, we may assume that the minimum green light time for any stream is 20 seconds.

30 C1= { a, b, c} C2= { b, d} a: d2 b: - c: d2d: d1 the total red light time of all the streams in one cycle is: Z= d2 + - + d2 + d1 = d1 + 2d2 Minimize Z, subject to: D1 >= 20, D2>=20, D1 + d2 = 120 The optimal solution to this problem is d1=100, d2=20, min Z=140. The phasing that corresponds to this least value would then be the best phasing of the traffic lights.

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33 Application of circular-arc graphs to the traffic light phasing problem

34 Many of the traffic real world problems can be solved by graph theory. Phasing of traffic lights are done by considering the cliques of the intersection graph as signal groups, and are made more efficient by using intersection graph of a family of sets. Recent days technologies are used to get efficient solutions to the traffic problems to provide safer, cheaper and better life.

35 Application of circular-arc graphs to the traffic light phasing problem

36 F. S. Roberts. (1978).Graph Theory and Its Application to Social Science, Regional Conference Series in Applied Mathematics. Martin. Charles. Golumbic. (2004). Algorithmic Graph Theory and Perfect Graphs, Elsevier. Karl E. Stoffers (1968), Scheduling of traffic lights- A new approach. Arun Kumar Baruah (2013), Intersection graph in traffic control problems, JMCAR. S. Mohsen Hosseini (2009), Phasing of Traffic Lights at a Road Junction, Applied Mathematical Sciences. Abdulhakeem Mohammed (2014), Traffic Flow and Circular-arc Graph, PPT. Ahmed Al-Baghdadi (2015), Traffic flow control, PPT.

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38 Maryam Pourebadi Kent State University April 2016


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