# Choosing robust constraints for MPC by using Lyapunov equation?

For a linear discrete discrete state space model:

$$x(k+1) = Ax(k) + Bu(k)$$ $$y = Cx(k) + Du(k)$$

I can choose the best inputs $U = \begin{bmatrix} u(k) & u(k+1) & u(k+2) & ... & u(k+ N_c) \end{bmatrix}^T$

By minimizing a cost function

$$J = X^TQX + U^TRU$$

Where:

$$X = F x(k) + \Phi U = \begin{bmatrix} x(k+1) & x(k+2) & x(k+3) & ... & x(k+1+N_p) \end{bmatrix}^T$$

Where:

$$F = \begin{bmatrix} A\\ A^2\\ A^3\\ \vdots \\ A^{N_p} \end{bmatrix} , \Phi = \begin{bmatrix} B &0 &0 &\cdots & 0\\ AB & B & 0 & \cdots & 0\\ A^2B& AB & 0 &\cdots &0 \\ \vdots & \vdots & \vdots & \vdots &\vdots \\ A^{N_p-1}B & A^{N_p-2}B & A^{N_p-3}B & \cdots & A^{N_p-N_c}B \end{bmatrix}$$

The variables $N_p$ and $N_c$ are prediction horizon and control horizon. The matricides $Q > 0$ and $R > 0$ are tuning matricides.

My goal is to minimize this cost function $$J = X^TQX + U^TRU$$

Where I say that the state vector $x(k)$ cannot be below or higher some values, e.i constraints. In reality, determining the maximum and minimum position for the system is not difficult, but the velocity is far more difficult.

And here stability margins comes in. If the state vector(trajectory) groes from one position to a another position with a certain velocity, the system can be unstable. Just because it's a predictive controller doesn't mean that the system would be 100% robust against overshoot.

To recive stability margins, e.i robustness. I would use lyapunov equation:

$$AXA^T -P + Q = 0$$

Where we want to solve $P$. Where $Q = Q^T > 0$, e.g $Q = I$ (not the same as the cost function) and $A$ which is system matrix from the discrete state space model.

Then we pick up the lyapunov function candidate

$$V(x(k)) = \frac{1}{2}x(k)^TPx(k)$$

Then we take the derivative of $V(x(k))$. Just take the difference between the past computation of lyapunov function and the current computation of lyapunov function. Only works in the discrete case.

$$\dot{V}(x(k)) = V(x(k)) - V(x(k-1))$$

If $\dot{V}(x(k))$ goes up -> bad, if $\dot{V}(x(k))$ goes down -> good

Question:

As I said...the goal is to minimize the cost function

$$J = X^TQX + U^TRU$$

To receive the best input signals for the system. Setting limits(constraints) for position is easy, but limits for velocity can be difficult. Then my question is:

Can I use the derivative of lyapunov function candidate

$$\dot{V}(x(k)) = V(x(k)) - V(x(k-1))$$

To compute the constraints for velocity before my quadratic solver minimize the cost function?

If I see $\dot{V}(x(k))$ is increasing, then I will decreasing the velocity limits(constraints) for the quadratic solver. If I see $\dot{V}(x(k))$ is decreasing, then I will increasing the velocity limits(constraints) for the quadratic solver.

A normal cost function for QP looks like this:

$$J = \frac{1}{2}x^TQx + c^Tx$$

With subject to:

$$Ax \leq b$$

Where $b$ is the constratins. Assume that we have the constratins written instead for the velocity:

$$Ax \leq b - \dot{V}(x(k))$$

• Very unclear what you are trying to do. Why do you think it is hard to constrain derivatives in the systems? If you mean that you easily run into infeasibility due to too hard constraints on velocities (and don't want to complicate matters with guaranteed stability approaches), you simply relax those constraints and use soft constraints instead – Johan Löfberg Apr 18 '18 at 18:37
• @JohanLöfberg It's very clear because I have remove all the theory behind. I assume that if the system gets to much speed, it can be unstable. What is soft constraints? – Daniel Mårtensson Apr 18 '18 at 18:42
• @JohanLöfberg I have added more information to the question. – Daniel Mårtensson Apr 18 '18 at 18:51
• To me it sounds like you simply don't have any hard constraints on velocity. Hence, you should not have any. Thus, you either penalize large velocities through the objective function, or you use soft constraints if you have some desired but not hard constraint. The hack you present here seems odd, and I would advice against such nonstandard solutions. – Johan Löfberg Apr 19 '18 at 5:44
• @JohanLöfberg Is there any standard solutions in control engineering? Not what I have seen :) All I see is papers on how to create different controllers with different solutions, but the results are the same. All theory and no practical solutions who works in real life. For me, it's very important to know pros and coins for the controllers I want to use. – Daniel Mårtensson Apr 19 '18 at 18:10

As Johan Löfberg mentioned, you can define soft constraints.

The conventional constraints are hard constraints which means that they should not be violated under any circumstance:

$$x_{\min}\le x_i \le x_{\max}$$

However, soft constraint means defining a constraint which should not be violated unless with a heavy penalty:

$$x_{\min}-s_v\le x_i \le x_{\max}+s_v$$

where $s_v$ is called a slack variable. Sometimes this variable is shown by $\epsilon$ too.

As we like $s_v=0$, we need to heavily penalize it in the cost function.

$$J_{\text{new}}=J_{\text{previous}}+\rho ||s_v||^n$$

where $\rho$ is a heavy penalty. Here $n$ can be $1$ or $2$.

Hence in your case, let the constraints on $V$ as hard and allow velocity exceeds boundaries with a heavy penalty.

• have a look at this and this. – Arash Apr 19 '18 at 2:50
• Does this requrie that I need to do analysis for compute ngram the best soft constraints or can I tunnelbanan in the soft constraints after I have implement the controller to the system ? – Daniel Mårtensson Apr 19 '18 at 9:32
• @DanielMårtensson, if I have understood what you mean, calculation of $s_v$ as a variable is performed together with the other variables. Not after control input calculation. – Arash Apr 19 '18 at 9:52
• Opps! I mean if I need to do analysis in my PC to compute the soft constraints? I seeking the practical usage for MPC. – Daniel Mårtensson Apr 19 '18 at 10:14
• The MPC problem with soft constraints introduced by penalties on slacks (using linear or quadratic penalties) all boil down to quadratic programming. There is essentially nothing new compared to standard MPC in terms of implementation. However, your talk about robustness and better stability margins is a bit difficult to answer. The issue addressed with slacks is lack of feasibility (which is a requirement for stability of course), but adding slacks to MPC complicates closed-loop stability theory significantly (almost unsolved I would say). – Johan Löfberg Apr 19 '18 at 20:02