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A parallel quadratic programming method for dynamic optimization problems

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Abstract

Quadratic programming problems (QPs) that arise from dynamic optimization problems typically exhibit a very particular structure. We address the ubiquitous case where these QPs are strictly convex and propose a dual Newton strategy that exploits the block-bandedness similarly to an interior-point method. Still, the proposed method features warmstarting capabilities of active-set methods. We give details for an efficient implementation, including tailored numerical linear algebra, step size computation, parallelization, and infeasibility handling. We prove convergence of the algorithm for the considered problem class. A numerical study based on the open-source implementation qpDUNES shows that the algorithm outperforms both well-established general purpose QP solvers as well as state-of-the-art tailored control QP solvers significantly on the considered benchmark problems.

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Notes

  1. We make use of the symmetry and positive semidefiniteness of \(\mathcal {M}\, {\tilde{\mathcal {M}}}^{-1} = \mathcal {I}- \delta \, {\tilde{\mathcal {M}}}^{-1}\).

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Acknowledgments

This research was supported by Research Council KUL: PFV/10/002 Optimization in Engineering Center OPTEC, GOA/10/09 MaNet and GOA/10/11 Global real-time optimal control of autonomous robots and mechatronic systems. Flemish Government: IOF/KP/SCORES4CHEM, FWO: PhD/postdoc grants and projects: G.0320.08 (convex MPC), G.0377.09 (Mechatronics MPC); IWT: PhD Grants, projects: SBO LeCoPro; Belgian Federal Science Policy Office: IUAP P7 (DYSCO, Dynamical systems, control and optimization, 2012–2017); EU: FP7- EMBOCON (ICT-248940), FP7-SADCO ( MC ITN-264735), ERC ST HIGHWIND (259 166), Eurostars SMART, ACCM.

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Authors and Affiliations

  1. Institute of Mathematical Optimization, Otto von Guericke University Magdeburg, Magdeburg, Germany

    Janick V. Frasch & Sebastian Sager

  2. Department of Microsystems Engineering, University of Freiburg, Freiburg, Germany

    Moritz Diehl

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  1. Janick V. Frasch
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  2. Sebastian Sager
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Correspondence to Janick V. Frasch.

Appendix

Appendix

Proof

(Theorem 6) The proof is done by calculation. We start from the the Cholesky recursion property, apply the assumed relation between Cholesky and Riccati iterates, and transform the expression into the form of the Riccati recursion. We have

$$\begin{aligned} X_{k-1}&= W_{k-1} - U_{k-1} X_k^{-1} U_{k-1}^\top \\ \Leftrightarrow \quad P^{-1} + {\textit{CH}}^{-1} C^\top&= {\textit{CH}}^{-1} C^\top + {\textit{EH}}^{-1} E^\top \\&\quad - {\textit{EH}}^{-1} C^\top \left( P^{-1} + {\textit{CH}}^{-1} C^\top \right) ^{-1} {\textit{CH}}^{-1} E^\top \, , \end{aligned}$$

and therefore

$$\begin{aligned} P^{-1}&= {\textit{EH}}^{-1} E^\top - \textit{EH}^{-1} C^\top \left( P^{-1} + {\textit{CH}}^{-1} C^\top \right) ^{-1} {\textit{CH}}^{-1} E^\top . \end{aligned}$$
(38)

Using the Schur complement \(\bar{Q} = Q - S^\top R^{-1} S\) it is well known from elementary linear algebra that the inverse \(H^{-1}\) can be expressed by

$$\begin{aligned} H^{-1} = \begin{bmatrix} Q&S^\top \\ S&R \end{bmatrix}^{-1} = \begin{bmatrix} \bar{Q}^{-1}&- \bar{Q}^{-1} S^\top R^{-1}\\ - R^{-1} S \bar{Q}^{-1} ~&~ R^{-1} + R^{-1} S \bar{Q}^{-1} S^\top R^{-1} \end{bmatrix}. \end{aligned}$$

Using this, \(C = \begin{bmatrix}\, A&B \,\end{bmatrix}\), and the special structure of \(E = \begin{bmatrix}\, I&0 \,\end{bmatrix}\), we first see that the identities

$$\begin{aligned} {\textit{EH}}^{-1} E^\top&= \bar{Q}^{-1}, \\ {\textit{CH}}^{-1} E^\top&= \left( A - {\textit{BR}}^{-1} S \right) \bar{Q}^{-1} =: \bar{C} \bar{Q}^{-1} \end{aligned}$$

and

$$\begin{aligned} {\textit{CH}}^{-1} C^\top&= \left( A - {\textit{BR}}^{-1} S \right) \bar{Q}^{-1} \left( A^\top - S^\top R^{-1} B^\top \right) + {\textit{BR}}^{-1} B^\top \\&= \bar{C} \bar{Q}^{-1} \bar{C}^\top + {\textit{BR}}^{-1} B^\top \end{aligned}$$

hold. Thus, (38) can be written as

$$\begin{aligned} P^{-1}&= \bar{Q}^{-1} - \bar{Q}^{-1} \bar{C}^\top \left( P^{-1} + B R^{-1} B^\top + \bar{C} \bar{Q}^{-1} \bar{C}^\top \right) ^{-1} \bar{C} \bar{Q}^{-1} \, . \end{aligned}$$

Applying the Woodbury matrix identity with \(Y := P^{-1} + {\textit{BR}}^{-1} B^\top \) we can express the right-hand-side term by

$$\begin{aligned} P^{-1} = \left( \bar{Q} + \bar{C}^\top Y^{-1} \bar{C} \right) ^{-1} \,. \end{aligned}$$

Thus

$$\begin{aligned} P&= \bar{Q} + \bar{C}^\top Y^{-1} \bar{C} \\&= Q - S^\top R^{-1} S + \left( A^\top - S^\top R^{-1} B^\top \right) \left( P^{-1} + {\textit{BR}}^{-1} B^\top \right) ^{-1} \left( A - {\textit{BR}}^{-1} S \right) . \end{aligned}$$

Note that the inverse matrices of \(Q,R,P\) and \(Y\) exist (and are real-valued), since we assume that QP (1) is convex. Applying the Woodbury identity once again (however, in opposite direction) on \(\left( P^{-1} + {\textit{BR}}^{-1} B^\top \right) ^{-1}\), and introducing \(\bar{R} := \left( R + B^\top {\textit{PB}} \right) \), we get

$$\begin{aligned} P&= Q - S^\top R^{-1} S \nonumber \\&\quad + \left( A^\top - S^\top R^{-1} B^\top \right) \left( P - {\textit{PB}} \left( R + B^\top {\textit{PB}} \right) ^{-1} B^\top P \right) \left( A - {\textit{BR}}^{-1} S \right) \nonumber \\&= Q - S^\top R^{-1} S \nonumber \\&\quad + A^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) A \nonumber \\&\quad - S^\top R^{-1} B^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) A \nonumber \\&\quad - A^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) {\textit{BR}}^{-1} S \nonumber \\&\quad + S^\top R^{-1} B^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) {\textit{BR}}^{-1} S. \end{aligned}$$
(39)

Using the identity \(I = \bar{R} \bar{R}^{-1} = \bar{R}^{-1} \bar{R}\), we further have

$$\begin{aligned}&S^\top R^{-1} B^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) {\textit{BR}}^{-1} S - S^\top R^{-1} S \\&\quad = S^\top R^{-1} B^\top P {\textit{BR}}^{-1} S - S^\top R^{-1} B^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S - S^\top R^{-1} S \\&\quad = S^\top R^{-1} \bar{R} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\qquad - S^\top R^{-1} B^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S - S^\top \bar{R}^{-1} \bar{R} R^{-1} S \\&\quad = S^\top R^{-1} \left( R + B^\top {\textit{PB}} \right) \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\qquad - S^\top R^{-1} B^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S - S^\top \bar{R}^{-1} \left( R + B^\top {\textit{PB}} \right) R^{-1} S \\&\quad = S^\top R^{-1} R \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S + S^\top R^{-1} B^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\qquad - S^\top R^{-1} B^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\qquad - S^\top \bar{R}^{-1} R R^{-1} S - S^\top \bar{R}^{-1} B^\top {\textit{PB}} R^{-1} S \\&\quad = - S^\top \bar{R}^{-1} S \end{aligned}$$

and

$$\begin{aligned}&- A^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) {\textit{BR}}^{-1} S \\&\quad = - A^\top P {\textit{BR}}^{-1} S + A^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\quad = - A^\top {\textit{PB}} \bar{R}^{-1} \bar{R} R^{-1} S + A^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\quad = - A^\top {\textit{PB}} \bar{R}^{-1} \left( R + B^\top {\textit{PB}} \right) R^{-1} S + A^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\quad = - A^\top {\textit{PB}} \bar{R}^{-1} {\textit{RR}}^{-1} S - A^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S + A^\top {\textit{PB}} \bar{R}^{-1} B^\top P {\textit{BR}}^{-1} S \\&\quad = - A^\top {\textit{PB}} \bar{R}^{-1} S . \end{aligned}$$

Analogously,

$$\begin{aligned} - S^\top R^{-1} B^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) A = - S^\top \bar{R}^{-1} B^\top P A . \end{aligned}$$

Therefore (39) is equivalent to

$$\begin{aligned} P&= Q - S^\top \bar{R}^{-1} S \\&\quad - S^\top \bar{R}^{-1} B^\top P A - A^\top {\textit{PB}} \bar{R}^{-1} S + A^\top \left( P - {\textit{PB}} \bar{R}^{-1} B^\top P \right) A \\&= Q+ A^\top P A - \left( S^\top + A^\top {\textit{PB}} \right) \bar{R}^{-1} \left( S + B^\top P A \right) , \end{aligned}$$

which concludes the proof. \(\square \)

Proof

(Corollary 7 ) The proof follows exactly the lines of the proof to Theorem 6, but keeps the matrix block indices. In particular, one has

$$\begin{aligned} P_{k-1}^{-1}&= E_{k-1} H_{k-1}^{-1} E_{k-1}^\top - E_{k-1} H_{k-1}^{-1} C_{k-1}^\top \left( P_{k}^{-1} + C_{k-1} H_{k-1}^{-1} C_{k-1}^\top \right) ^{-1}\\&\quad \times C_{k-1} H_{k-1}^{-1} E_{k-1}^\top \end{aligned}$$

in place of (38) and transforms it into (19b) using the same matrix identities. \(\square \)

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Frasch, J.V., Sager, S. & Diehl, M. A parallel quadratic programming method for dynamic optimization problems. Math. Prog. Comp. 7, 289–329 (2015). https://doi.org/10.1007/s12532-015-0081-7

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