Difference between revisions of "CDS 110b: Optimal Control"

This lecture provides an overview of optimal control theory. Beginning with a review of optimization, we introduce the notion of Lagrange multipliers and provide a summary of the Pontryagin's maximum principle.

Q: Could you please explain what the psi function is or what it means if psi(x(T))=0 versus what it means if psi(x(T))=x(T)?

The psi function represents a general form of terminal constraint for the state variables. It gives a way of indicating which states have a terminal cost attached to them. For example, by defining psi_i(x(T))=x_i(T)-x_i,f for i=1,2,...n, we can impose terminal costs on all states (a fully constrained case) by letting p=n (n being the # of states). When we optimize over time and want x(T)=x_f, then x(T)-x_f=0, and so psi(x(T))=0. If we take x_f=0, then psi(x(T))=x(T).

Luis Soto, 22 Jan 08

Q: In Problem 2.4d, are the boundary conditions for the differentially-flat trajectory correct?

Please ignore the boundary conditions given in part 2.4d for the differentially-flat trajectory and instead use x(0)=1 for the initial condition and x(1)=0 for the condition at final time t=1. Moreover, use c=100 instead of c=1. Note: the x(t_f) of the optimal solution won't be exactly 0, but will be close enough for the intent of this problem.

Luis Soto, 21 Jan 08

Q: In problem 2.4(d) of the homework, to what positive value should the parameter b be set?

Use b = 1 for part d when solving for and comparing the two trajectories found symbolically in previous parts.

Julia Braman, 18 Jan 08

Q: In the example on Bang-Bang control discussed in the lecture, how is the control law for $$u$$ obtained?

Pontryagin's Maximum Principle says that $$u$$ has to be chosen to minimise the Hamiltonian $$H(x,u,\lambda)$$ for given values of $$x$$ and $$\lambda$$. In the example, $$H = 1 + ({\lambda}^TA)x + ({\lambda}^TB)u$$. At first glance, it seems that the more negative $$u$$ is the more $$H$$ will be minimised. And since the most negative value of $$u$$ allowed is $$-1$$, $$u=-1$$. However, the co-efficient of $$u$$ may be of either sign. Therefore, the sign of $$u$$ has to be chosen such that the sign of the term $$({\lambda}^TB)u$$ is negative. That's how we come up with $$u = -sign({\lambda}^TB)$$.

Shaunak Sen, 12 Jan 06

Q: Notation question for you: In the Lecture notes from Wednesday, I'm assuming that $$T$$ is the final time and $$T$$ (superscript T) is a transpose operation. Am I correct in my assumption?

Yes, you are correct.

Jeremy Gillula, 07 Jan 05