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Sampled–data model predictive control: Adaptive time–mesh refinement algorithms and guarantees of stability

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  • This article addresses the problem of controlling a constrained, continuous–time, nonlinear system through Model Predictive Control (MPC). In particular, we focus on methods to efficiently and accurately solve the underlying optimal control problem (OCP). In the numerical solution of a nonlinear OCP, some form of discretization must be used at some stage. There are, however, benefits in postponing the discretization process and maintain a continuous-time model until a later stage. This is because that way we can exploit additional freedom to select the number and the location of the discretization node points.We propose an adaptive time–mesh refinement (AMR) algorithm that iteratively finds an adequate time–mesh satisfying a pre–defined bound on the local error estimate of the obtained trajectories. The algorithm provides a time–dependent stopping criterion, enabling us to impose higher accuracy in the initial parts of the receding horizon, which are more relevant to MPC. Additionally, we analyze the conditions to guarantee closed–loop stability of the MPC framework using the AMR algorithm. The numerical results show that the proposed AMR strategy can obtain solutions as fast as methods using a coarse equidistant–spaced mesh and, on the other hand, as accurate as methods using a fine equidistant–spaced mesh. Therefore, the OCP can be solved, and the MPC law obtained, faster and/or more accurately than with discrete-time MPC schemes using equidistant–spaced meshes.

    Mathematics Subject Classification: Primary: 49J15; Secondary: 65L50, 93D20, 93C57, 93C10.

    Citation:

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  • Figure 1.  Illustration of the multi–level adaptive time–mesh refinement strategy

    Figure 2.  Illustration of the extended (time–dependent) time–mesh refinement strategy with different refinement thresholds

    Figure 3.  Illustration of the extended time–mesh refinement algorithm for MPC

    Figure 4.  Construction of the (extended) admissible control $ {\bf{\tilde u}} $ with $ \Pi = \{t_k\}_{k \in \mathbb{N}} $, $ t_k = k \delta $, and with $ \pi_r = \{s_i\}_{i \in 0, 1, \ldots N_r} $, $ s_i = i \delta/2 $

    Figure 5.  Car–like system geometry

    Figure 6.  Pathwise state constraints (13) for (PCP)

    Figure 7.  Optimal path computed in the initial coarse mesh

    Figure 8.  Discretization error estimate in the initial coarse mesh

    Figure 9.  Optimal path computed in the final mesh $ \pi_{\rm{AMR}} $

    Figure 10.  Optimal trajectory and control

    Figure 11.  Discretization error in the coarse mesh and the MPC refining levels

    Figure 12.  Path resulting from the AMR–MPC scheme

    Figure 13.  Trajectory and control resulting from the AMR–MPC scheme

    Table 1.  Results for problem (PCP) solved in each time-mesh

    $ \pi_j $ $ N_j $ $ \Delta t_j $ $ I_j $ $ \left|\left|\varepsilon_{\bf{x}}^{(j)}\right|\right|_\infty $ CPU time (s)
    Solver $ \varepsilon_{\bf{x}} $
    $ \pi_0 $ 21 $ 0.5 $ 42 $ 1.0016{\rm{E}}^{-4} $ $ 0.9816 $ $ 0.0563 $
    $ \pi_1 $ 82 $ 1/54 $ 42 $ 3.3801{\rm{E}}^{-7} $ $ 0.7061 $ $ 0.0642 $
    $ \pi_{\rm{AMR}} $ 82 $ 1/54 $ 84 $ 3.3801{\rm{E}}^{-7} $ $ 1.6877 $ $ 0.1205 $
    $ \pi_{\rm{F}} $ 541 $ 1/54 $ 403 $ 4.0358{\rm{E}}^{-7} $ $ 13.2473 $ $ 0.4675 $
     | Show Table
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    Table 2.  Results for each MPC and AMR iterations

    MPC Iter AMR Iter $ N_j $ $ \Delta t_j $ $ I_j $ $ \left|\left|\varepsilon_{\bf{x}}^{(j)}\right|\right|_\infty $ CPU time (s)
    Solver $ \varepsilon_{\bf{x}} $
    $ \pi_{0} $ 21 0.5 $ 42 $ $ 1.002{\rm{E}}^{-4} $ $ 0.982 $ $ 0.0563 $
    1 $ \pi_{1} $ 21 0.5 $ 8 $ $ 1.002{\rm{E}}^{-4} $ $ 0.105 $ $ 0.0156 $
    $ \pi_{2} $ 52 0.0625 $ 22 $ $ 3.525{\rm{E}}^{-6} $ $ 0.344 $ $ 0.0374 $
    $ \pi_{\rm{AMR}} $ 52 0.0625 $ 30 $ $ 3.525{\rm{E}}^{-6} $ $ 0.449 $ $ 0.0530 $
    2 $ \pi_{1}=\pi_{\rm{AMR}} $ 31 0.0625 $ 11 $ $ 3.525{\rm{E}}^{-6} $ $ 0.1564 $ $ 0.0230 $
    3 $ \pi_{1}=\pi_{\rm{AMR}} $ 21 0.5 $ 11 $ $ 2.042{\rm{E}}^{-7} $ $ 0.1639 $ $ 0.0139 $
    4 $ \pi_{1}=\pi_{\rm{AMR}} $ 21 0.5 $ 7 $ $ 4.321{\rm{E}}^{-7} $ $ 0.0936 $ $ 0.0126 $
    5 $ \pi_{1}=\pi_{\rm{AMR}} $ 21 0.5 $ 7 $ $ 4.515{\rm{E}}^{-7} $ $ 0.0912 $ $ 0.0123 $
     | Show Table
    DownLoad: CSV
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