CDS 110b: Optimal Control: Difference between revisions

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== Frequently Asked Questions ==
== Frequently Asked Questions ==
'''Q: In Problem 2.4d, are the boundary conditions for the differentially-flat trajectory correct?'''
<blockquote>
<p>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(tf) of the optimal solution won't be exactly 0,
but will be close enough for the intent of this problem.</p>
<p>Luis Soto, 21 Jan 08</p>
</blockquote>


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

Revision as of 23:22, 21 January 2008

CDS 110b Schedule Project Course Text

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.

References and Further Reading

Frequently Asked Questions

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(tf) 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 obtained?

Pontryagin's Maximum Principle says that has to be chosen to minimise the Hamiltonian for given values of and . In the example, . At first glance, it seems that the more negative is the more will be minimised. And since the most negative value of allowed is , . However, the co-efficient of may be of either sign. Therefore, the sign of has to be chosen such that the sign of the term is negative. That's how we come up with .

Shaunak Sen, 12 Jan 06

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

Yes, you are correct.

Jeremy Gillula, 07 Jan 05