4.7 Article

Solving the combined modal split and traffic assignment problem with two types of transit impedance function

期刊

EUROPEAN JOURNAL OF OPERATIONAL RESEARCH
卷 257, 期 3, 页码 870-880

出版社

ELSEVIER SCIENCE BV
DOI: 10.1016/j.ejor.2016.08.019

关键词

Combined modal split and traffic assignment problem; Elastic demand; User equilibrium; Gradient projection; Bi-modal networks

资金

  1. Basic Science Research Program through the National Research Foundation (NRF) of Korea by the Ministry of Science, ICT & Future Planning [NRF-2016R1C1B2016254, NRF-2010-0029443]
  2. Research Committee of the Hong Kong Polytechnic University [1-ZE5T]
  3. Research Grants Council of the Hong Kong Special Administrative Region [15267116]
  4. Transportation Research Center for Livable Communities (TRCLC)
  5. Mountain-Plains Consortium (MPC) - U.S. Department of Transportation

向作者/读者索取更多资源

The gradient projection (GP) algorithm has been shown as a successful path-based algorithm for solving various traffic assignment problems. In this paper, the GP algorithm is adapted for solving the combined modal split and traffic assignment (CMSTA) problem, which can be viewed as an elastic demand traffic equilibrium problem (EDTEP) with two modes. Using the excess-demand formulation of EDTEP, the CMSTA problem is reformulated and solved by a modified GP algorithm. Numerical results based on a real bi-modal network in the city of Winnipeg, Canada are provided to demonstrate the efficiency and robustness of the modified path-based GP algorithm for solving the CMSTA problem. In addition, the CMSTA problem is investigated with two types of impedance function for the transit mode and with different degrees of dispersion for the modal split function. The computational results show the modified GP algorithm outperforms the classical Evan's algorithm for both types of transit impedance function, and it can be as efficient as the original GP algorithm for solving the traffic assignment problem with fixed demand. (C) 2016 Elsevier B.V. All rights reserved.

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