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Flight Itinerary Problem ICS 311 Fall 2006 Matt Freeburg
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Problem Formulation: Graph Airports as vertices Flights as directed edges Flight time as edge weight ORD LAX HNL 8 hrs. 54 min.
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Advantages Easy to understand Existing mathematics Existing algorithms Existing implementations
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Special Considerations Edge weight is more than flight time Later flights cannot connect to earlier flights Two flights without enough layover time cannot connect either
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Input A graph G = (V,E) A set of vertices V, one for each airport A set of edges E, one for each flight A starting vertex v s, an ending vertex v f, and a starting time t s
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Input (cont.) A function A(e) that returns the airline a i for edge e A function T d (e) that returns the departure time for edge e A function T a (e) that returns the arrival time for edge e A function C(v, a i, a j ) that returns the minimum layover time, C ≥ 0
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Input (cont.) The weight of an edge f is defined as the sum of the flight time and the layover time, given by w(f) = ((T a (f) – T d (f)) + (T d (f) – T a (e)) where e is the previous edge in a path w(f) reduces to w(f) = T a (f) – T a (e)
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Input Constraint Two edges e and f are considered adjacent if and only if e = (x, y) and f = (y, z) T d (f) – T a (e) ≥ C(y, A(e), A(f))
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Output A path P = {e 1, e 2, e 3, …,e m } consisting of m edges, beginning at v s and ending at v f, where the total weight of the path is minimized among all such possible paths
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Output (cont.) The total weight of P is given by m W(P) = ∑ ( T a (e i ) – T a (e i-1 ) ) i = 1 i = 1 where T a (e 0 ) = t s
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