solveStepBruteForce

Method of discreteTimeMdss.

Solve one time step of a discrete-time mdss model.

Source: src/simulator/@discreteTimeMdss/solveStepBruteForce.m

What is solved

At fixed current state $x_0=x_k$ and input $u$, find the successor state $\dot x \equiv x_{k+1}$, the algebraic signals $y$ and the binaries $z$ that satisfy

$$F = \Phi_{\mathrm{eq}}\, m(x_{k+1}, x_0, u, y, z) = 0, \qquad G = \Phi_{\mathrm{ineq}}\, m(x_{k+1}, x_0, u, y, z) \le 0 ,$$

with $\Phi=\Phi_c+\Phi_{+1}-\Phi_{-1}$ and $m\in\mathbb R^{R}$ the norm-1 monomial vector. Substituting a value $v$ for the signal in structure-matrix row $n$ (row vectors $t_n,f_n,c_n$) updates $m$ multiplicatively:

$$m \;\leftarrow\; m \odot \big(1 + t_n f_n + t_n(v-1) - f_n v\big)\odot\big(1 - |c_n| + c_n v\big).$$

Algorithm

The known $x_0$ and $u$ are folded into $m$, then the solving order is walked. Each row is one of three cases.

Explicit continuous unknown ($x_{k+1,n}$ or $y_n$): its equation $e$ is affine in that unknown, $F_e = A\,v + B$, so

$$v = -\frac{B}{A},\quad A=\Phi_e\!\left[(t_n-f_n+c_n)\odot m\right],\quad B=\Phi_e\!\left[\big(1-|c_n+t_nf_n-t_n|\big)\odot m\right].$$

Explicit binary $z_n$: evaluate its equation value at the previous value $z_n^0$ and flip on a positive value,

$$F_0 = \Phi_e\, m\big|_{z_n=z_n^0},\qquad z_n = \begin{cases} z_n^0, & \operatorname{sign}(F_0)\le 0,\\[2pt] \lnot\, z_n^0, & \text{otherwise.}\end{cases}$$

Coupled subproblem: delegated to solveSubProblemVariableZ.

Finally $F=\Phi_{\mathrm{eq}}m$ and $G=\Phi_{\mathrm{ineq}}m$ are returned. Called once per sample by discreteSimulation.

Inputs / outputs

[xp, y, z, F, G] = solveStepBruteForce(sys, x0, u, solvingOrder, yguess, xpguess, zguess) — returns the successor state xp ($x_{k+1}$), algebraic row y, binary row z, and the equality/inequality function-value rows F, G.

See also

discreteTimeMdss · solveSubProblemVariableZ · discreteSimulation · msim

MyToolbox Documentation | Generated automatically by CI/CD pipeline