Difference between Model Predictive Control and Rolling Horizon Optimization

model predictive controloptimization

Lately I've been reading numerous papers regarding Energy Hub optimization, and often the authors talk about rolling horizon optimization for taking into account uncertainty.

For instance:

"A rolling horizon optimization framework for the simultaneous energy supply and demand planning in microgrids"
"Peer-to-peer energy sharing through a two-stage aggregated battery control in a community Microgrid"

To me, this seems similar to the Model Predictive Control approach.

Is there any difference between Rolling Horizon optimization approach and Model Predictive Control approach? Are these two terms interchangeable?

Best Answer

Yes, effectively the same thing. Also called receding horizon control.

Related Solutions

[Math] Model Predictive Control

In addition to the answer of @Johan at point 5), note I use a different cost function, with which you are probably more familiar with

Define the cost function $\mathbf{V}(\mathbf{x},\mathbf{U})$ as

$$ \mathbf{V}(\mathbf{x},\mathbf{U}) := \mathbf{x}_N^\mathrm{T} \mathbf{P} \mathbf{x}_N + \sum_{i=0}^{N-1}\left( \mathbf{x}_i^\mathrm{T} \mathbf{Q} \mathbf{x}_i + \mathbf{u}_i^\mathrm{T} \mathbf{R} \mathbf{u}_i\right) $$

Herein $N$ is the control horizon, $\mathbf{P}$ is the terminal state weight, $\mathbf{Q}$ is the state weight and $\mathbf{R}$ is the input weight. Remark: $\mathbf{P}$, $\mathbf{Q}$ and $\mathbf{R}$ are positive semidefinite. Furthermore, $\mathbf{U}$ is the stacked input vector, that is,

$$ \mathbf{U} := \begin{bmatrix} \mathbf{u}_0^T & \mathbf{u}_1^T & \ldots & \mathbf{u}_{N-1}^T \end{bmatrix}^T $$

The optimal control $\mathbf{U}^*$ input is given by

$$ \mathbf{U}^* = \arg \mathbf{V}^*(\mathbf{x},\mathbf{U}) = \arg \min_{\mathbf{U}} \mathbf{V}(\mathbf{x},\mathbf{U}) $$

The sum in the cost function can be rewritten in a matrix equation

$$ \begin{gathered} \mathbf{V}(\mathbf{x},\mathbf{U}) = \mathbf{x}_0^\mathrm{T} \mathbf{Q} \mathbf{x}_0 + \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \vdots \\ \mathbf{x}_N \end{bmatrix}^\mathrm{T} \begin{bmatrix} \mathbf{Q} & 0 & \cdots & 0 \\ 0 & \mathbf{Q} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{P} \end{bmatrix} \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \vdots \\ \mathbf{x}_N \end{bmatrix} + \begin{bmatrix} \mathbf{u}_0 \\ \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_{N-1} \end{bmatrix}^\mathrm{T} \begin{bmatrix} \mathbf{R} & 0 & \cdots & 0 \\ 0 & \mathbf{R} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \mathbf{R} \end{bmatrix} \begin{bmatrix} \mathbf{u}_0 \\ \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_{N-1} \end{bmatrix} \\ \Leftrightarrow \\ \mathbf{V}(\mathbf{x},\mathbf{U}) = \mathbf{x}^\mathrm{T}\mathbf{Q}\mathbf{x} + \mathbf{X}^\mathrm{T}\mathbf{\Omega}\mathbf{X} + \mathbf{U}^\mathrm{T}\mathbf{\Psi}\mathbf{U} \end{gathered} $$

Remark: $\mathbf{x} = \mathbf{x}_0$. Furthermore, $\mathbf{P} \succeq 0 \wedge \mathbf{Q} \succeq 0$ implies that $\mathbf{\Omega} \succeq 0$ and $\mathbf{R} \succ 0$ implies that $\mathbf{\Psi} \succ 0$. The cost function now still depends on the future state $\mathbf{X}$. To eliminate $\mathbf{X}$ the prediction matrices $\mathbf{\Phi}$ and $\mathbf{\Gamma}$ are computed such that the stacked state vector is a function of the initial state and the stacked input vector i.e. $\mathbf{X} = \mathbf{\Phi}\mathbf{x} + \mathbf{\Gamma}\mathbf{U}$.

$$ \mathbf{\Phi} = \begin{bmatrix} \mathbf{A} \\ \mathbf{A}^2 \\ \vdots \\ \mathbf{A}^N \end{bmatrix} \qquad \mathbf{\Gamma} = \begin{bmatrix} \mathbf{B} & 0 & \cdots & 0 \\ \mathbf{A}\mathbf{B} & \mathbf{B} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \mathbf{A}^{N-1}\mathbf{B} & \mathbf{A}^{N-2}\mathbf{B} & \cdots & \mathbf{B} \end{bmatrix} $$

The cost function $\mathbf{V}(\mathbf{x},\mathbf{U})$ can now be written as a function of the initial state $\mathbf{x}$ and the stacked input vector $\mathbf{U}$

$$ \mathbf{V}(\mathbf{x},\mathbf{U}) = \frac{1}{2} \mathbf{U}^\mathrm{T} \mathbf{G} \mathbf{U} + \mathbf{U}^\mathrm{T} \mathbf{F}\mathbf{x} + \mathbf{x}^\mathrm{T} \left( \mathbf{Q} + \mathbf{\Phi}^\mathrm{T} \mathbf{\Omega} \mathbf{\Phi} \right) \mathbf{x} $$

Where $\mathbf{G} = 2\left(\mathbf{\Psi} + \mathbf{\Gamma}^\mathrm{T} \mathbf{\Omega} \mathbf{\Gamma} \right)$, $\mathbf{F} = 2 \mathbf{\Gamma}^\mathrm{T} \mathbf{\Omega} \mathbf{\Phi}$ and $\mathbf{G} \succ 0$. The last term of the newly derived cost function is independent on $\mathbf{U}$. Therefore, it can be neglected when computing the optimal input sequence $\mathbf{U}^*(\mathbf{x})$

Now lets add some input and state constraints

$$ \mathbf{u}_\mathrm{min} \leq \mathbf{u}_k \leq \mathbf{u}_\mathrm{max} \wedge \mathbf{x}_\mathrm{min} \leq \mathbf{x}_k \leq \mathbf{x}_\mathrm{max} $$

This can be rewritten to

$$ \begin{bmatrix} 0 \\ 0 \\ -\mathbf{I}_n \\ \mathbf{I}_n \end{bmatrix} \mathbf{x}_k + \begin{bmatrix} -\mathbf{I}_m \\ \mathbf{I}_m \\ 0 \\ 0 \end{bmatrix} \mathbf{u}_k \leq \begin{bmatrix} -\mathbf{u}_\mathrm{min} \\ \mathbf{u}_\mathrm{max} \\ -\mathbf{x}_\mathrm{min} \\ \mathbf{x}_\mathrm{max} \end{bmatrix} \Leftrightarrow \mathbf{M}_k\mathbf{x}_k + \mathbf{E}_k\mathbf{u}_k \leq \mathbf{b}_k $$

Now note that this is only valid for $k = 0,1 \ldots, N-1$ so not up to the control horizon $N$. As such we add a terminal constraint for $N$ on the state.

$$ \begin{bmatrix} -\mathbf{I}_{n} \\ \mathbf{I}_{n} \\ \end{bmatrix} \mathbf{x}_N \leq \begin{bmatrix} -\mathbf{x}_\mathrm{min} \\ \mathbf{x}_\mathrm{max} \\ \end{bmatrix} \Rightarrow \mathbf{M}_N\mathbf{x}_N \leq \mathbf{b}_N $$

All constraints can now be combined to

$$ \begin{gathered} \begin{bmatrix} \mathbf{M}_0 \\ 0 \\ \vdots \\ 0 \end{bmatrix} \mathbf{x}_0 + \begin{bmatrix} 0 & \cdots & 0 \\ \mathbf{M}_1 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \mathbf{M}_N \end{bmatrix} \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \\ \vdots \\ \mathbf{x}_N \end{bmatrix} + \begin{bmatrix} \mathbf{E}_0 & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & \mathbf{E}_{N-1} \\ 0 & \cdots & 0 \end{bmatrix} \begin{bmatrix} \mathbf{u}_0 \\ \mathbf{u}_1 \\ \vdots \\ \mathbf{u}_N \end{bmatrix} \leq \begin{bmatrix} \mathbf{b}_0 \\ \mathbf{b}_1 \\ \vdots \\ \mathbf{b}_N \end{bmatrix} \\ \Leftrightarrow \\ \mathbf{\mathcal{D}}\mathbf{x} + \mathbf{\mathcal{M}}\mathbf{X} + \mathbf{\mathcal{E}}\mathbf{U} \leq \mathbf{c} \end{gathered} $$

Again this equation still depends on $\mathbf{X}$, which we eliminate using the same procedure as before. As a result, the above equation can be written as

$$\mathbf{J}\mathbf{U} \leq \mathbf{c} + \mathbf{W}\mathbf{x}$$

With $\mathbf{J} = \mathbf{\mathcal{M}}\mathbf{\Gamma} + \mathbf{\mathcal{E}}$ and $\mathbf{W} = -\mathbf{\mathcal{D}} - \mathbf{\mathcal{M}}\mathbf{\Phi}$. Now the constraints only depend on the stacked input vector and the initial state. The constrained MPC problem can then be formulated as

$$ \arg \underset{\mathbf{U}}{\min}\; \mathbf{V}(\mathbf{x},\mathbf{U}) \quad \text{s.t.} \quad \mathbf{J}\mathbf{U} \leq \mathbf{c} + \mathbf{W}\mathbf{x} $$

Now if you use a QP solver you will have the following QP problem

$$ \begin{aligned} \mathbf{U}^*(\mathbf{x}) = \arg \underset{\mathbf{U}}{\min}\;\;& \frac{1}{2}\mathbf{U}^\mathrm{T} \mathbf{G} \mathbf{U} + \mathbf{U}^\mathrm{T} \mathbf{F} \mathbf{x} \\[1ex] \text{subject to:}\;\;&\mathbf{x}_0 = \mathbf{x} \hspace{3cm} \text{Measurement} \nonumber \\ &\mathbf{J}\mathbf{U} \leq \mathbf{c} + \mathbf{W}\mathbf{x} \hspace{1.82cm} \text{Constraints} \nonumber \\ &\mathbf{Q} \succ 0, \mathbf{R} \succ 0 \hspace{1.99cm} \text{Cost weights} \end{aligned} $$

Now if you want to solve these kind of typical MPC problems, I would strongly advice to use the MPT toolbox, http://people.ee.ethz.ch/~mpt/3/, an example http://control.ee.ethz.ch/~mpt/3/UI/RegulationProblem. Furthermore, the slides from this workshop may help you as well https://www.kirp.chtf.stuba.sk/pc11/data/workshops/mpc/

[Math] What’s the difference between Generalized Predictive Control and Model Predictive Control

GPC is essentially finite-horizon unconstrained adaptive discrete-time LQ for a particular system model. As constraint-handling is the main reason MPC has become so successful, it is not that interesting today in an MPC context.

Yes, one could consider it obsolete.

Best Answer

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[Math] Model Predictive Control

[Math] What’s the difference between Generalized Predictive Control and Model Predictive Control

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