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December  2018, 11(6): 1259-1282. doi: 10.3934/dcdss.2018071

## A study of structure-exploiting SQP algorithms for an optimal control problem with coupled hyperbolic and ordinary differential equation constraints

 1 Institut für Mathematik und Rechneranwendung (LRT-1), Universität der Bundeswehr München, Werner-Heisenberg-Weg 39, 85577 Neubiberg/München, Germany 2 Center for Computational Mathematics, University of California, San Diego, 9500 Gilman Drive, La Jolla, CA 92093-0112, USA

* Corresponding author: Sven-Joachim Kimmerle

Received  April 2017 Revised  September 2017 Published  June 2018

In this article, structure-exploiting optimisation algorithms of the sequential quadratic programming (SQP) type are considered for optimal control problems with control and state constraints. Our approach is demonstrated for a 1D mathematical model of a vehicle transporting a fluid container. The model involves a fully coupled system of ordinary differential equations (ODE) and nonlinear hyperbolic first-order partial differential equations (PDE), although the ideas for exploiting the particular structure may be applied to more general optimal control problems as well. The time-optimal control problem is solved numerically by a full discretisation approach. The corresponding nonlinear optimisation problem is solved by an SQP method that uses exact first and second derivative information. The quadratic subproblems are solved using an active-set strategy. In addition, two approaches are examined that exploit the specific structure of the problem: (A) a direct method for the KKT system, and (B) an iterative method based on combining the limited-memory BFGS method with the preconditioned conjugate gradient method. Method (A) is faster for our model problem, but can be limited by the problem size. Method (B) opens the door for a potential extension of the truck-container model to three space dimensions.

Citation: Jan-Hendrik Webert, Philip E. Gill, Sven-Joachim Kimmerle, Matthias Gerdts. A study of structure-exploiting SQP algorithms for an optimal control problem with coupled hyperbolic and ordinary differential equation constraints. Discrete & Continuous Dynamical Systems - S, 2018, 11 (6) : 1259-1282. doi: 10.3934/dcdss.2018071
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##### References:
Model of the truck and fluid tank, cf. [7].
Optimal solution: control force $u(t)$ (left) and fluid column height $h(t, x) + b(x)$ (right) for $N = 300$, $M = 10$ (final time minimised, $T = 10.8s$) and a straight fluid tank bottom.
Optimal solution: control force $u(t)$ (left) and fluid column height $h(t, x) + b(x)$ (right) for $N = 300$, $M = 10$ (final time and fluid-level deviation minimised, $T = 19.1s$) and a straight fluid tank bottom.
Optimal solution: control force $u(t)$ (left) and fluid column height $h(t, x) + b(x)$ (right) for $N = 150$, $M = 5$ (final time minimised, $T = 11.5s$). The bottom of the fluid tank is given by the linear function $b(x) = - x/8 + 1/2, \, 0 \leq x \leq L = 4$.
Solution output of Problem (1) with $M = 5$ and $N = 150$. $\texttt{CV}$ and $\texttt{KKT}$ denote the norm of the constraint violation and the norm of the gradient of the Lagrangian, respectively. $\texttt{QPIT}$ is the number of QP iterations and $\texttt{OBJ}$ the value of the objective function.
 $\texttt{ITER}$ $\texttt{QPIT}$ $\texttt{ALPHA}$ $\texttt{OBJ}$ $\texttt{CV}$ $\texttt{KKT}$ $\texttt{0}$ $\texttt{0}$ $\texttt{0.0000E+00}$ $\texttt{0.100000050000E+01}$ $\texttt{0.9893E+02}$ $\texttt{0.1000E+00}$ $\texttt{1}$ $\texttt{224}$ $\texttt{0.1000E+01}$ $\texttt{0.102710574197E+01}$ $\texttt{0.1811E+02}$ $\texttt{0.2648E-01}$ $\texttt{2}$ $\texttt{6}$ $\texttt{0.1000E+01}$ $\texttt{0.107819921922E+01}$ $\texttt{0.2086E+00}$ $\texttt{0.5244E-02}$ $\texttt{3}$ $\texttt{24}$ $\texttt{0.1000E+01}$ $\texttt{0.109015429623E+01}$ $\texttt{0.4723E-02}$ $\texttt{0.2969E-03}$ $\texttt{4}$ $\texttt{5}$ $\texttt{0.1000E+01}$ $\texttt{0.109049950972E+01}$ $\texttt{0.3198E-05}$ $\texttt{0.4477E-06}$ $\texttt{5}$ $\texttt{3}$ $\texttt{0.1000E+01}$ $\texttt{0.109049961114E+01}$ $\texttt{0.4253E-11}$ $\texttt{0.4570E-12}$ $\texttt{END OF SQP METHOD}$
 $\texttt{ITER}$ $\texttt{QPIT}$ $\texttt{ALPHA}$ $\texttt{OBJ}$ $\texttt{CV}$ $\texttt{KKT}$ $\texttt{0}$ $\texttt{0}$ $\texttt{0.0000E+00}$ $\texttt{0.100000050000E+01}$ $\texttt{0.9893E+02}$ $\texttt{0.1000E+00}$ $\texttt{1}$ $\texttt{224}$ $\texttt{0.1000E+01}$ $\texttt{0.102710574197E+01}$ $\texttt{0.1811E+02}$ $\texttt{0.2648E-01}$ $\texttt{2}$ $\texttt{6}$ $\texttt{0.1000E+01}$ $\texttt{0.107819921922E+01}$ $\texttt{0.2086E+00}$ $\texttt{0.5244E-02}$ $\texttt{3}$ $\texttt{24}$ $\texttt{0.1000E+01}$ $\texttt{0.109015429623E+01}$ $\texttt{0.4723E-02}$ $\texttt{0.2969E-03}$ $\texttt{4}$ $\texttt{5}$ $\texttt{0.1000E+01}$ $\texttt{0.109049950972E+01}$ $\texttt{0.3198E-05}$ $\texttt{0.4477E-06}$ $\texttt{5}$ $\texttt{3}$ $\texttt{0.1000E+01}$ $\texttt{0.109049961114E+01}$ $\texttt{0.4253E-11}$ $\texttt{0.4570E-12}$ $\texttt{END OF SQP METHOD}$
Sequence of problems with gradually refined discretisations solved with $\texttt{sqpfiltertoolbox}$ Method (A) and $\texttt{SNOPT}$. $\texttt{SQPIT}$ states the number of SQP iterations.
 $\underline{\rm{\texttt{problem}}\ \rm{\texttt{data:}}}$ $\texttt{FEASTOL = 1.00E-008}$ $\alpha_0$ = $\texttt{1.00E-001}$ $\texttt{OPTTOL = 1.00E-008}$ $\alpha_1$ = $\texttt{0}$ $\mu$ = $\texttt{1.00E-012}$ $\alpha_2$ = $\texttt{1.00E-007}$ $\texttt{threshold = 1.00E-002}$ $\alpha_3$ = $\texttt{1.00E+000}$ $\texttt{initialisation by collocation}$ $\alpha_4$ = $\texttt{1.00E-007}$ $c_0$ = $\texttt{1.00E-003}$ $\texttt{N}$ $\texttt{M}$ $\tilde{J}$ $\texttt{SQPIT}$ $\texttt{QPIT}$ $t[s]$ $t[s]$ $\texttt{SNOPT}$ $\texttt{150}$ $\texttt{5}$ $\texttt{1.0905}$ $\texttt{5}$ $\texttt{263}$ $\texttt{10.4}$ $\texttt{1.6}$ $\texttt{300}$ $\texttt{10}$ $\texttt{1.0870}$ $\texttt{3}$ $\texttt{47}$ $\texttt{19.4}$ $\texttt{7.8}$ $\texttt{450}$ $\texttt{15}$ $\texttt{1.0840}$ $\texttt{3}$ $\texttt{49}$ $\texttt{91.6}$ $\texttt{25.4}$ $\texttt{600}$ $\texttt{20}$ $\texttt{1.0810}$ $\texttt{3}$ $\texttt{49}$ $\texttt{281.9}$ $\texttt{99.0}$ $\texttt{900}$ $\texttt{30}$ $\texttt{1.0751}$ $\texttt{4}$ $\texttt{110}$ $\texttt{2824.7}$ $\texttt{1541.0}$ $\texttt{1200}$ $\texttt{40}$ $\texttt{1.0705}$ $\texttt{4}$ $\texttt{289}$ $\texttt{22333.2}$ $\texttt{162545.6}$
 $\underline{\rm{\texttt{problem}}\ \rm{\texttt{data:}}}$ $\texttt{FEASTOL = 1.00E-008}$ $\alpha_0$ = $\texttt{1.00E-001}$ $\texttt{OPTTOL = 1.00E-008}$ $\alpha_1$ = $\texttt{0}$ $\mu$ = $\texttt{1.00E-012}$ $\alpha_2$ = $\texttt{1.00E-007}$ $\texttt{threshold = 1.00E-002}$ $\alpha_3$ = $\texttt{1.00E+000}$ $\texttt{initialisation by collocation}$ $\alpha_4$ = $\texttt{1.00E-007}$ $c_0$ = $\texttt{1.00E-003}$ $\texttt{N}$ $\texttt{M}$ $\tilde{J}$ $\texttt{SQPIT}$ $\texttt{QPIT}$ $t[s]$ $t[s]$ $\texttt{SNOPT}$ $\texttt{150}$ $\texttt{5}$ $\texttt{1.0905}$ $\texttt{5}$ $\texttt{263}$ $\texttt{10.4}$ $\texttt{1.6}$ $\texttt{300}$ $\texttt{10}$ $\texttt{1.0870}$ $\texttt{3}$ $\texttt{47}$ $\texttt{19.4}$ $\texttt{7.8}$ $\texttt{450}$ $\texttt{15}$ $\texttt{1.0840}$ $\texttt{3}$ $\texttt{49}$ $\texttt{91.6}$ $\texttt{25.4}$ $\texttt{600}$ $\texttt{20}$ $\texttt{1.0810}$ $\texttt{3}$ $\texttt{49}$ $\texttt{281.9}$ $\texttt{99.0}$ $\texttt{900}$ $\texttt{30}$ $\texttt{1.0751}$ $\texttt{4}$ $\texttt{110}$ $\texttt{2824.7}$ $\texttt{1541.0}$ $\texttt{1200}$ $\texttt{40}$ $\texttt{1.0705}$ $\texttt{4}$ $\texttt{289}$ $\texttt{22333.2}$ $\texttt{162545.6}$
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