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An efficient algorithm for solving linear equality-constrained LQR problems

João Sousa-Pinto, Dominique Orban

TL;DR

This work tackles linear-quadratic regulator problems with linear equality constraints (CLQR) by introducing a parallelizable reduction framework that eliminates linear equalities while preserving positive definiteness. The method sequentially reduces D_i and E_i to full row rank and simple forms (e.g., $D_i = I$ or $E_i = I$), eliminates $C_i x_i + D_i u_i + d_i = 0$ and $E_i x_i + e_i = 0$, and, where possible, removes entire stages with $B_i = 0$, ultimately yielding an unconstrained LQR solved with known parallel algorithms. All eliminations are accompanied by systematic updates to problem data and a tractable way to recover the eliminated variables via affine transformations and bookkeeping. The resulting approach achieves linear sequential runtime in the number of stages and logarithmic parallel runtime, offering substantial speedups for constrained discrete-time optimal control and enabling efficient subroutines for nonlinear problems that rely on constrained LQR subproblems.

Abstract

We present a new algorithm for solving linear-quadratic regulator (LQR) problems with linear equality constraints, also known as constrained LQR (CLQR) problems. Our method's sequential runtime is linear in the number of stages and constraints, and its parallel runtime is logarithmic in the number of stages. The main technical contribution of this paper is the derivation of parallelizable techniques for eliminating the linear equality constraints while preserving the standard positive (semi-)definiteness requirements of LQR problems.

An efficient algorithm for solving linear equality-constrained LQR problems

TL;DR

This work tackles linear-quadratic regulator problems with linear equality constraints (CLQR) by introducing a parallelizable reduction framework that eliminates linear equalities while preserving positive definiteness. The method sequentially reduces D_i and E_i to full row rank and simple forms (e.g., or ), eliminates and , and, where possible, removes entire stages with , ultimately yielding an unconstrained LQR solved with known parallel algorithms. All eliminations are accompanied by systematic updates to problem data and a tractable way to recover the eliminated variables via affine transformations and bookkeeping. The resulting approach achieves linear sequential runtime in the number of stages and logarithmic parallel runtime, offering substantial speedups for constrained discrete-time optimal control and enabling efficient subroutines for nonlinear problems that rely on constrained LQR subproblems.

Abstract

We present a new algorithm for solving linear-quadratic regulator (LQR) problems with linear equality constraints, also known as constrained LQR (CLQR) problems. Our method's sequential runtime is linear in the number of stages and constraints, and its parallel runtime is logarithmic in the number of stages. The main technical contribution of this paper is the derivation of parallelizable techniques for eliminating the linear equality constraints while preserving the standard positive (semi-)definiteness requirements of LQR problems.
Paper Structure (16 sections, 2 theorems, 43 equations)

This paper contains 16 sections, 2 theorems, 43 equations.

Table of Contents

  1. Introduction
  2. Constrained LQR
  3. Background
  4. Related Work
  5. Results
  6. Reduction to full row rank $D_i$
  7. Reduction $D_i = G_iI$ or $D_i = I$
  8. Eliminating $C_i x_i + D_i u_i + d_i = 0$
  9. Eliminate stage $i+1$ whenever $B_i = 0$
  10. Reduction to full row rank $E_i$
  11. Reduction to $E_i = F_iI$ or $E_i = I$
  12. Eliminating $E_i x_i + e_i = 0$
  13. Solving the LQR problem
  14. Retrieve the eliminated state and control variables
  15. Computational Complexity
  16. ...and 1 more sections

Key Result

Lemma 1

Let $f(x, u) = \frac{1}{2} x^T Q x + x^T M u + \frac{1}{2} u^T R u + q^T x + r^T u$ and $g(x) = \min \limits_{u} f(x, u)$. Then $g(x) = \frac{1}{2} x^T (Q - M R^{-1} M^T) x + (q - M R^{-1} r)^T x - \frac{1}{2} r^T R^{-1} r$, corresponding to $u = -R^{-1}(r + M^T x)$.

Theorems & Definitions (3)

  • Lemma 1
  • Theorem 2
  • proof