• 제목/요약/키워드: Shortest aborescence

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공공차량 경로문제의 이중 최단나무 결합 해법 (Double Shortest Arborescence & Merging Algorithm for the Public Vehicle Routing Problem)

  • 장병만
    • 대한산업공학회지
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    • 제16권2호
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    • pp.71-79
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    • 1990
  • In this paper, the Double Shortest Arborescence & Merging method is presented as an efficient heuristic algorithm for the Public Vehicle Routing Problem which is to find the minimum total cost routes of M or less vehicles to traverse the required arcs(demand streets) at least once and return to their starting depot on a directed network. Double Shortest Arborescence which consists of forward shortest aborescence and backward one informs M or less shortest routes to traverse all required arcs. The number of these routes is reduced to M or less by merging routes. The computational experiment based on randomly generated networks reports that this algorithm is efficient.

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복수최단경로의 새로운 최적해법 (A New Algorithm for K Shortest Paths Problem)

  • 장병만
    • 한국경영과학회지
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    • 제26권3호
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    • pp.79-94
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    • 2001
  • This paper presents a new algorithm for the K shortest paths Problem which develops initial K shortest paths, and repeat to expose hidden shortest paths with dual approach and to replace the longest path in the present K paths. The initial solution comprises K shortest paths among shortest paths to traverse each arc in a Double Shortest Arborescence which is made from bidirectional Dijkstra algorithm. When a crossing node that have two or more inward arcs is found at least three time by turns in this K shortest paths, there may be some hidden paths which are shorter than present k-th path. To expose a hidden shortest path, one inward arc of this crossing node is chose by means of minimum detouring distance calculated with dual variables, and then the hidden shortest path is exposed with joining a detouring subpath from source to this inward arc and a spur of a feasible path from this crossing node to sink. If this exposed path is shorter than the k-th path, the exposed path replaces the k-th path. This algorithm requires worst case time complexity of O(Kn$^2$), and O(n$^2$) in the case k$\leq$3.

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