Exponentially Convergent Receding Horizon Strategy for Constrained Optimal Control

Abstract

Receding horizon control has been a widespread method in industrial control engineering as well as an extensively studied subject in control theory. In this work, we consider a lag L receding horizon strategy that applies the initial L optimal controls from each quadratic program to each receding horizon. We investigate a discrete-time and time-varying linear-quadratic optimal control problem that includes a nonzero reference trajectory and constraints on both state and control. We prove that, under boundedness and controllability conditions, the solution obtained by the receding horizon strategy converges to the solution of the full problem interval exponentially fast in the length of the receding horizon for some lag L. The exponential rate of convergence provides a systematic way of choosing the receding horizon length given a desired accuracy level. We illustrate our theoretical findings using a small, synthetic production cost model with real demand data.

Appendix A: Proofs of Results in Sections 2 and 3

1.1 A.1 Proof of Proposition 2

For any $x_{q}\in \mathbb {R}^{n}$, consider the standard linear-quadratic problem:

$$ \begin{array}{@{}rcl@{}} \min && \sum\limits_{k=q}^{n_{2}-1} {u_{k}^{T}}R_{k}u_{k}+{x_{k}^{T}}\hat{Q}_{k}x_{k}+x_{n_{2}}^{T}Q_{n_{2}}x_{n_{2}} \end{array} $$

(44a)

$$ \begin{array}{@{}rcl@{}} \text{s.t.} && x_{k+1}=\hat{A}_{k}x_{k}+\hat{B}_{k}u_{k}, \qquad q\le k\le n_{2}-1. \end{array} $$

(44b)

For k ≥ q, successively applying (44b) gives, for j ≥ 0,

$$ x_{q+j}-\left( \prod\limits_{l=0}^{j-1}\hat{A}_{q+l}\right)x_{q}=\left[\begin{array}{lll} \hat{B}_{q+j-1}&\hat{A}_{q+j-1}\hat{B}_{q+j-2}& {\ldots} \left( \prod\limits_{l=1}^{j-1}\hat{A}_{q+l}\right)\hat{B}_{q} \end{array}\right] \left[\begin{array}{cc} u_{q+j-1}\\ \vdots\\ u_{q} \end{array}\right], $$

(45)

and for j = t, (45) reduces to

$$ x_{q+t}-\left( \prod\limits_{l=0}^{t-1}\hat{A}_{q+l}\right)x_{q}=C_{q,t}\left[\begin{array}{cc} u_{q+t-1}\\ \vdots\\ u_{q} \end{array}\right]. $$

The index set being UCC(λ_C) implies that C_q,t is uniformly completely controllable and in particular that C_q,t has full row rank. Therefore, there exists $\hat {u}=(\hat {u}_{q}^{T},\dots ,\hat {u}_{q+t-1}^{T})^{T}$ so that

$$ -\left( \prod\limits_{l=0}^{t-1}\hat{A}_{l}\right)x_{q}=C_{q,t}\left[\begin{array}{cc} \hat{u}_{q+t-1}\\ \vdots\\ \hat{u}_{q} \end{array}\right]. $$

(46)

Several $\hat {u}$ satisfy this relationship; we consider the one defined by

$$ \hat{u}=-C_{q,t}^{T}\left( C_{q,t}C_{q,t}^{T}\right)^{-1}\left( \prod\limits_{l=0}^{t-1}\hat{A}_{q+l}\right)x_{q}. $$

Denote the corresponding states generated with $\hat {u}_{q:q+t-1}$ as $\hat {x}_{q:q+t}$. Then $\hat {x}_{q+t}=\mathbf {0}$ by (46).

Lemma 1 implies that

$$ \begin{array}{@{}rcl@{}} &&\max\limits_{1\le j\le t}\left\|\left[\begin{array}{lll} \hat{B}_{q+j-1}&\hat{A}_{q+j-1}\hat{B}_{q+j-2}& {\ldots} \left( \prod\limits_{l=1}^{j-1}\hat{A}_{q+l}\right)\hat{B}_{q} \end{array}\right]\right\|_{2}\\ &&\le \max\limits_{1\le j\le t}\left( C_{B}+C_{A}C_{B}+\dots+C_{A}^{j-1}C_{B}\right)\\ &&\le \frac{C_{B}\left( 1-{C_{A}^{t}}\right)}{1-C_{A}}\stackrel{\Delta}{=}M. \end{array} $$

Then from Definition 6 and Lemma 1, we have

$$ \|\hat{u}\|\le \frac{M}{\lambda_{C}}{C_{A}^{t}}\|x_{q}\|. $$

(47)

From (45), we have, for 1 ≤ j ≤ t − 1,

$$ \|\hat{x}_{q+j}\|\le {C_{A}^{j}}\|x_{q}\|+M\|\hat{u}\|\le \left( {C_{A}^{j}}+\frac{M^{2}}{\lambda_{C}}{C_{A}^{t}}\right)\|x_{q}\|. $$

(48)

Now we let $\hat {u}_{k}=\mathbf {0}$ for k ≥ q + t. Then it follows that $\hat {x}_{k}=\mathbf {0}$ for k ≥ q + t. Also note that since (A. 1) is a standard linear-quadratic regulator problem, the optimal value is given by ${x_{q}^{T}}K_{q}x_{q}$ [2]. As a result, we have the following.

$$ \begin{array}{@{}rcl@{}} {x_{q}^{T}}K_{q}x_{q}&=&\min_{u_{k}}\sum\limits_{k=q}^{n_{2}-1}{x_{k}^{T}}\hat{Q}_{k}x_{k}+{u_{k}^{T}}R_{k}u_{k}+x_{n_{2}}^{T}Q_{n_{2}}x_{n_{2}}\\ &\le &\sum\limits_{k=q}^{n_{2}-1}\hat{x}_{k}^{T}\hat{Q}_{k}\hat{x}_{k}+\hat{u}_{k}^{T}R_{k}\hat{u}_{k}+\hat{x}_{n_{2}}^{T}Q_{n_{2}}\hat{x}_{n_{2}}\\ &\le &\sum\limits_{k=q}^{q+t-1}\hat{x}_{k}^{T}\hat{Q}_{k}\hat{x}_{k}+\hat{u}_{k}^{T}R_{k}\hat{u}_{k}\\ &\le &C_{Q}\sum\limits_{k=q}^{q+t-1}\|\hat{x}_{k}\|^{2}+C_{R}\sum\limits_{k=q}^{q+t-1}\|\hat{u}_{k}\|^{2}\\ &\overset{(47),~(48)}{\le} &C_{Q}\left( 1+\sum\limits_{i=1}^{t-1}\left( {C_{A}^{i}}+\frac{M^{2}}{\lambda_{C}}{C_{A}^{t}}\right)^{2}\right)\|x_{q}\|^{2}+C_{R}\frac{M^{2}C_{A}^{2t}}{{\lambda^{2}_{C}}}\|x_{q}\|^{2}. \end{array} $$

Letting

$$ \beta=C_{Q}\left( 1+\sum\limits_{i=1}^{t-1}\left( {C_{A}^{i}}+\frac{M^{2}}{\lambda_{C}}{C_{A}^{t}}\right)^{2}\right)+C_{R}\frac{M^{2}C_{A}^{2t}}{{\lambda^{2}_{C}}} $$

completes the proof. Note that β depends only on the quantities in Assumption 1, Definitions 1 and 6, and Lemma 1, which are independent of n₁, n₂, and the particular choice of $\mathcal {I}$ given it is UDB(λ_H) and UCC(λ_C).

1.2 A.2 Proof of Proposition 3

Define $L_{k}=-W_{k}^{-1}\hat {B}_{k}^{T}K_{k+1}\hat {A}_{k}$. Then from Lemma 5 and (5d) we have $D_{k}=\hat {A}_{k}+\hat {B}_{k}L_{k}$. In [2] the recursion (15b) is shown to be equivalent to

$$ K_{k}={D_{k}}^{T}K_{k+1}D_{k}+\hat{Q}_{k}+{L_{k}}^{T}R_{k}L_{k}. $$

(49)

For q ≤ j ≤ n₂ − 1, define x_j+ 1 = D_jx_j. Then (49) and Proposition 2 imply that

$$ \begin{array}{@{}rcl@{}} {x_{j}^{T}}K_{j}x_{j}&\ge& x_{j+1}^{T}K_{j+1}x_{j+1}+{x_{j}^{T}}\hat{Q}_{j}x_{j}\\ &\stackrel{\text{Prop}.~2}{\ge}& x_{j+1}^{T}K_{j+1}x_{j+1}+\frac{\lambda_{Q}}{\beta}{x_{j}^{T}}K_{j}x_{j}\\ &\stackrel{(49)}{\ge}& \left( 1+\frac{\lambda_{Q}}{\beta}\right)x_{j+1}^{T}K_{j+1}x_{j+1}. \end{array} $$

(50)

Here we used the bounds from Lemma 1 and the fact that ${x_{j}^{T}}K_{j}x_{j} \geq x_{j+1}^{T}K_{j+1}x_{j+1}$, as implied by (49) and the positive definiteness of $\hat {Q}_{k}$ and R_k. Also we have

$$ {x_{j}^{T}}K_{j}x_{j}\stackrel{(49), \text{Lemma}~3}{\ge} {x_{j}^{T}}\hat{Q}_{j}x_{j}\ge \lambda_{Q}\|x_{j}\|^{2}. $$

(51)

As a result, for n₂ − 1 ≥ j ≥ q, we have the following:

$$ \begin{array}{@{}rcl@{}} \left\|\prod\limits_{l=q}^{j}D_{l}x_{q}\right\|^{2}=\|x_{j+1}\|^{2}&\stackrel{(51)}{\le} & \frac{1}{\lambda_{Q}}x_{j+1}^{T}K_{j+1}x_{j+1}\\ &\stackrel{(50)}{\le} &\frac{1}{\lambda_{Q}(1+\lambda_{Q}/\beta)}{x_{j}^{T}}K_{j}x_{j}\\ &\stackrel{(50)}{\le} &\frac{1}{\lambda_{Q}}\left( \frac{1}{1+\lambda_{Q}/\beta}\right)^{j-q+1}{x_{q}^{T}}K_{q}x_{q}\\ &\stackrel{\text{Prop}.~2}{\le} &\frac{\beta}{\lambda_{Q}}\left( \frac{1}{1+\lambda_{Q}/\beta}\right)^{j-q+1}\|x_{q}\|^{2}, \end{array} $$

where the third inequality is obtained by repeatedly applying (50).

1.3 A.3 Proof of Lemma 9

Let

$$ L(y,\theta)=y^{T}Gy/2+y^{T}c(\theta)+\lambda^{T}(Ay-r)+\phi^{T}(By-d(\theta))+\theta^{T}F\theta +y^{T}c_{1}+\theta^{T}c_{2}+C $$

be the Lagrangian of problem (31). Then we have

$$ \nabla_{(y,\theta)}^{2}L=\left[\begin{array}{ll} G & \nabla_{\theta} c\\ \nabla_{\theta}^{T} c & \ast \end{array}\right]. $$

Since G and F are positive definite and LICQ holds at y₀, then from [6, Theorem 5.53] and [6, Remark 5.55] we have

$$ \begin{array}{@{}rcl@{}} D_{p}y(\theta_{0})&=&\text{argmin}_{h\in S}\left[\begin{array}{ll} h^{T} & p^{T} \end{array}\right] \left( \nabla_{(y,\theta)}^{2}L(y_{0},\theta_{0})\right)\left[\begin{array}{ll} h\\ p \end{array}\right]\\ &=&\text{argmin}_{h\in S} h^{T}Gh/2+p^{T}\left( \nabla_{\theta}^{T}c(\theta_{0})\right)h, \end{array} $$

where S is the solution of the following linearized problem,

$$ \begin{array}{@{}rcl@{}} \min\limits_{s} &&\left( Gy_{0}+c(\theta_{0})+c_{1}\right)^{T} s+\left( \nabla_{\theta}^{T} c(\theta_{0}){y_{0}}+2F\theta_{0}+c_{2}\right)^{T}p\\ \text{s.t.} && Bs-\left( \nabla_{\theta} d(\theta_{0})\right)p=0, \\ && A_{I(y_{0},\theta_{0})}s \le 0, \end{array} $$

and S is given by

$$ S=\left\{s:~\left[\begin{array}{ll} B & -\nabla_{\theta} d(\theta_{0}) \end{array}\right]\left[\begin{array}{ll} s\\ p \end{array}\right] =0,~~ \left[\begin{array}{ll} A_{I_{+}(y_{0},\theta_{0},\bar{\lambda})} & 0 \end{array}\right]\left[\begin{array}{ll} s\\ p \end{array}\right]=0,~~ \left[\begin{array}{ll} A_{I_{0}(y_{0},\theta_{0},\bar{\lambda})} & 0 \end{array}\right]\left[\begin{array}{ll} s\\ p \end{array}\right]\le 0\right\}. $$

Thus the directional derivative D_py(𝜃₀) of y(𝜃) along direction p at 𝜃₀ is the solution of the problem

$$ \begin{array}{@{}rcl@{}} \min\limits_{h} && h^{T}Gh/2+p^{T}\left( \nabla_{\theta}^{T}c(\theta_{0})\right)h\\ \text{s.t.} && Bh-\left( \nabla_{\theta} d(\theta_{0})\right)p=0, \\ && A_{I_{+}(y_{0},\theta_{0},\bar{\lambda})}h= 0,\\ && A_{I_{0}(y_{0},\theta_{0},\bar{\lambda})}h\le 0. \end{array} $$

(52)

Let I₁ be the set of active inequality constraints of problem (52). Then $I_{1}\subset I_{0}(y_{0},\theta _{0},\bar {\lambda })$, and let $I^{\prime }(\theta _{0})=I_{1}\cup I_{+}(y_{0},\theta _{0},\bar {\lambda })$. The KKT condition of problem (52) is hence

$$ \widetilde{G} \stackrel{\Delta}{=}\left[\begin{array}{lll} G & A_{I^{\prime}(\theta_{0})}^{T} & B^{T}\\ A_{I^{\prime}(\theta_{0})} &0 &0 \\ B & 0& 0 \end{array}\right], \qquad \widetilde{G} \left[\begin{array}{ll} h^{\ast}\\ \phi_{1}^{\ast}\\ \phi_{2}^{\ast} \end{array}\right]=\left[\begin{array}{ll} -\nabla_{\theta} c(\theta_{0}) p\\ 0\\ \nabla_{\theta} d(\theta_{0}) p \end{array}\right] $$

for some Lagrange multipliers $\phi _{1}^{\ast }$ and $\phi _{2}^{\ast }$. Since LICQ holds at y₀, rows of $A_{I^{\prime }(\theta _{0})}$ and B are linearly independent. Together with the fact that G is positive definite, we have that $\widetilde {G}$ is invertible. Denote the first row of $\widetilde {G}^{-1}$ to be $\left [\begin {array}{lll} p_{11} & p_{12} & p_{13} \end {array}\right ]$. Then, we have

$$ D_{p}y(\theta_{0})=h^{\ast}=\left( -p_{11}\nabla_{\theta} c(\theta_{0})+p_{13}\nabla_{\theta} d(\theta_{0})\right)p. $$

On the other hand, for problem (32) with $I^{\prime }(\theta _{0})$ constructed above, the KKT condition is

$$ \widetilde{G}\left[\begin{array}{ll} y_{I^{\prime}(\theta_{0})}^{\ast}(\theta)\\ \psi_{1}^{\ast}\\ \psi_{2}^{\ast} \end{array}\right]=\left[\begin{array}{ll} -c(\theta)\\ r^{\prime}\\ d(\theta) \end{array}\right], $$

for some Lagrange multipliers $\psi _{1}^{\ast }$ and $\psi _{2}^{\ast }$. Since $\widetilde {G}$ is invertible, we have $y_{I^{\prime }(\theta _{0})}^{\ast }(\theta )=-p_{11}c(\theta )+p_{12}r^{\prime }+p_{13}d(\theta )$. It follows that

$$ \left.\frac{dy_{I^{\prime}(\theta_{0})}^{\ast}(\theta)}{d\theta}\right|_{\theta=\theta_{0}}=-p_{11}\nabla_{\theta} c(\theta_{0})+p_{13}\nabla_{\theta} d(\theta_{0}). $$

As a result, we have

$$ D_{p}y(\theta_{0})=\left( \left.\frac{dy_{I^{\prime}(\theta_{0})}^{\ast}(\theta)}{d\theta}\right|_{\theta=\theta_{0}}\right)p, $$

which proves the claim.