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Akshay Agrawal
commited on
Commit
·
89f83d9
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Parent(s):
5a78583
minimum fuel optimal control
Browse files
optimization/03_minimum_fuel_optimal_control.py
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import marimo
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__generated_with = "0.11.0"
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app = marimo.App()
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@app.cell
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def _():
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import marimo as mo
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return (mo,)
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@app.cell(hide_code=True)
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def _(mo):
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mo.md(
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r"""
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# Minimal fuel optimal control
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This notebook includes an application of linear programming to controlling a
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physical system, adapted from [Convex
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Optimization](https://web.stanford.edu/~boyd/cvxbook/) by Boyd and Vandenberghe.
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We consider a linear dynamical system with state $x(t) \in \mathbf{R}^n$, for $t = 0, \ldots, T$. At each time step $t = 0, \ldots, T - 1$, an actuator or input signal $u(t)$ is applied, affecting the state. The dynamics
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of the system is given by the linear recurrence
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\[
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x(t + 1) = Ax(t) + bu(t), \quad t = 0, \ldots, T - 1,
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\]
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where $A \in \mathbf{R}^{n \times n}$ and $b \in \mathbf{R}^n$ are given and encode how the system evolves. The initial state $x(0)$ is also given.
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The _minimum fuel optimal control problem_ is to choose the inputs $u(0), \ldots, u(T - 1)$ so as to achieve
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a given desired state $x_\text{des} = x(T)$ while minimizing the total fuel consumed
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\[
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F = \sum_{t=0}^{T - 1} f(u(t)).
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\]
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The function $f : \mathbf{R} \to \mathbf{R}$ tells us how much fuel is consumed as a function of the input, and is given by
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\[
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f(a) = \begin{cases}
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|a| & |a| \leq 1 \\
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2|a| - 1 & |a| > 1.
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\end{cases}
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\]
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This means the fuel use is proportional to the magnitude of the signal between $-1$ and $1$, but for larger signals the marginal fuel efficiency is half.
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**This notebook.** In this notebook we use CVXPY to formulate the minimum fuel optimal control problem as a linear program. The notebook lets you play with the initial and target states, letting you see how they affect the planned trajectory of inputs $u$.
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First, we create the **problem data**.
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"""
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)
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return
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@app.cell
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def _():
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import numpy as np
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return (np,)
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@app.cell
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def _():
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n, T = 3, 30
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return T, n
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@app.cell
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def _(np):
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A = np.array([[-1, 0.4, 0.8], [1, 0, 0], [0, 1, 0]])
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b = np.array([[1, 0, 0.3]]).T
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return A, b
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@app.cell(hide_code=True)
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def _(mo, n, np):
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import wigglystuff
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x0_widget = mo.ui.anywidget(wigglystuff.Matrix(np.zeros((1, n))))
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xdes_widget = mo.ui.anywidget(wigglystuff.Matrix(np.array([[7, 2, -6]])))
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_a = mo.md(
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rf"""
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Choose a value for $x_0$ ...
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{x0_widget}
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"""
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)
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_b = mo.md(
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rf"""
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... and for $x_\text{{des}}$
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{xdes_widget}
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"""
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)
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mo.hstack([_a, _b], justify="space-around")
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return wigglystuff, x0_widget, xdes_widget
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@app.cell
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def _(x0_widget, xdes_widget):
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x0 = x0_widget.matrix
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xdes = xdes_widget.matrix
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return x0, xdes
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@app.cell(hide_code=True)
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def _(mo):
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mo.md(r"""**Next, we specify the problem as a linear program using CVXPY.** This problem is linear because the objective and constraints are affine. (In fact, the objective is piecewise affine, but CVXPY rewrites it to be affine for you.)""")
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return
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@app.cell
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def _():
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import cvxpy as cp
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return (cp,)
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@app.cell
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def _(A, T, b, cp, mo, n, x0, xdes):
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X, u = cp.Variable(shape=(n, T + 1)), cp.Variable(shape=(1, T))
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objective = cp.sum(cp.maximum(cp.abs(u), 2 * cp.abs(u) - 1))
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constraints = [
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X[:, 1:] == A @ X[:, :-1] + b @ u,
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X[:, 0] == x0,
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X[:, -1] == xdes,
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]
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fuel_used = cp.Problem(cp.Minimize(objective), constraints).solve()
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mo.md(f"Achieved a fuel usage of {fuel_used:.02f}. 🚀")
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return X, constraints, fuel_used, objective, u
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@app.cell(hide_code=True)
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def _(mo):
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mo.md(
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"""
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Finally, we plot the chosen inputs over time.
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**🌊 Try it!** Change the initial and desired states; how do fuel usage and controls change? Can you explain what you see? You can also try experimenting with the value of $T$.
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"""
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)
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return
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@app.cell
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def _(plot_solution, u):
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plot_solution(u)
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return
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@app.cell
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def _(T, cp, np):
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def plot_solution(u: cp.Variable):
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import matplotlib.pyplot as plt
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plt.step(np.arange(T), u.T.value)
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plt.axis("tight")
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plt.xlabel("$t$")
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plt.ylabel("$u$")
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return plt.gca()
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return (plot_solution,)
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if __name__ == "__main__":
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app.run()
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