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Automatic Control (AC) | Automatic Control (AC) | ||
− | Question 3: Optimization | + | Question 3: Optimization |
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− | :Student answers and discussions for [[ | + | :Student answers and discussions for [[QE2013_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]] |
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'''1.(20 pts) In some of the optimization methods, when minimizing a given function f(x), we select an intial guess <math>x^{(0)}</math> and a real symmetric positive definite matrix <math>H_{0}</math>. Then we computed <math>d^{(k)} = -H_{k}g^{(k)}</math>, where <math>g^{(k)} = \nabla f( x^{(k)} )</math>, and <math>x^{(k+1)} = x^{(k)} + \alpha_{k}d^{(k)}</math>, where''' | '''1.(20 pts) In some of the optimization methods, when minimizing a given function f(x), we select an intial guess <math>x^{(0)}</math> and a real symmetric positive definite matrix <math>H_{0}</math>. Then we computed <math>d^{(k)} = -H_{k}g^{(k)}</math>, where <math>g^{(k)} = \nabla f( x^{(k)} )</math>, and <math>x^{(k+1)} = x^{(k)} + \alpha_{k}d^{(k)}</math>, where''' | ||
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<math> \alpha_{k} = arg\min_{\alpha \ge 0}f(x^{(k)} + \alpha d^{(k)}) .</math> | <math> \alpha_{k} = arg\min_{\alpha \ge 0}f(x^{(k)} + \alpha d^{(k)}) .</math> | ||
<br> | <br> | ||
− | Suppose that the function we wish to minimize is a standard quadratic of the form, | + | '''Suppose that the function we wish to minimize is a standard quadratic of the form,''' |
<br> | <br> | ||
<math> f(x) = \frac{1}{2} x^{T} Qx - x^{T}b+c, Q = Q^{T} > 0. </math> | <math> f(x) = \frac{1}{2} x^{T} Qx - x^{T}b+c, Q = Q^{T} > 0. </math> | ||
− | <br> | + | <br><br> |
− | '''(i)(10 pts) Find | + | '''(i)(10 pts) Find a closed form expression for <math>\alpha_k</math> in terms of <math>Q, H_k, g^{(k)}</math>, and <math>d^{(k)}; </math>''' |
− | <math> | + | |
− | \ | + | |
− | + | ||
− | + | ||
− | </math> | + | |
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− | + | ||
<br> | <br> | ||
+ | '''(ii)(10 pts) Give a sufficient condition on <math>H_k</math> for <math>\alpha_k</math> to be positive.''' | ||
+ | :'''Click [[QE2013_AC-3_ECE580-1|here]] to view [[QE2013_AC-3_ECE580-1|student answers and discussions]]''' | ||
+ | ---- | ||
− | + | '''Problem 2. (20 pts) [(i) (10 pts)] Consider the one-point crossover of a chromosome in the schema''' | |
+ | <br> | ||
+ | H = * 1 * 0 1 0 * | ||
+ | <br> | ||
+ | '''where the probability that a chromosome is chosen for crossover is <math>p_c = 0.5.</math> Find an upper bound on the probability that a chromosome from H will be destroyed by the one-point crossover.''' | ||
+ | <br><br> | ||
+ | '''[(ii) (10 pts)] Consider a chromosome in the schema''' | ||
+ | <br> | ||
+ | H = * 1 * 0 * * * | ||
+ | <br> | ||
+ | '''Find the probability that the mutation operation destroys the schema, where the probability of random change of each symbol of the chromosome is <math>p_m = 0.1</math> independently.''' | ||
− | + | :'''Click [[QE2013_AC-3_ECE580-2|here]] to view [[QE2013_AC-3_ECE580-2|student answers and discussions]]''' | |
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− | '''Problem | + | '''Problem 3. (20 pts) [(i) (10 pts)] Convert the following optimization problem into a linear programming problem and solve it; ''' |
− | + | <br> | |
− | + | maximize <math> -|x_1| -|x_2| -|x_3| </math> | |
− | + | <br> | |
− | \begin{bmatrix} | + | subject to |
− | 1 & 1 \\ | + | <br> |
− | 1 & | + | <math>\begin{bmatrix} |
− | \end{bmatrix} | + | 1 & 1 &-1 \\ |
+ | 0 & -1 & 0 | ||
+ | \end{bmatrix} \begin{bmatrix} | ||
+ | x_1 \\ | ||
+ | x_2 \\ | ||
+ | x_3 | ||
+ | \end{bmatrix} = \begin{bmatrix} | ||
2 \\ | 2 \\ | ||
1 | 1 | ||
− | \end{bmatrix} | + | \end{bmatrix} .</math> |
+ | <br><br> | ||
+ | '''[(ii) (10 pts)] Construct the dual program of the linear program above and solve it. ''' | ||
− | + | :'''Click [[QE2013_AC-3_ECE580-3|here]] to view [[QE2013_AC-3_ECE580-3|student answers and discussions]]''' | |
− | + | ---- | |
− | :'''Click [[ | + | '''Problem 4. (20pts) Consider the following model of a discrete-time system, ''' |
+ | <br> | ||
+ | <math> x(k+1) = x(k) + 2u(k), x(0) = 3, 0 \le k \le 2</math> | ||
+ | <br> | ||
+ | '''Use the Lagrange multiplier approach to calculate the optimal control sequence''' | ||
+ | <br> | ||
+ | {u(0), u(1), u(2)} | ||
+ | <br> | ||
+ | ''' that transfers the initial state x(0) to x(3) = 9 while minimizing the performance index''' | ||
+ | <br> | ||
+ | <math> J = \frac{1}{2} \sum_{k=0}^2 u(k)^2 = \frac{1}{2}u^Tu. </math> | ||
+ | <br> | ||
+ | ''' Hint: Use the system model to obtain a constraint of the form, <math>Au = b, A \in R^{1 \times 3}. </math>''' | ||
+ | :'''Click [[QE2013_AC-3_ECE580-4|here]] to view [[QE2013_AC-3_ECE580-4|student answers and discussions]]''' | ||
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− | '''Problem | + | <br> '''Problem 5. (20pts) Find minimizers and maximizers of the function, ''' |
− | + | <br> | |
− | <math> | + | <math> f(x) = (a^Tx)(b^Tx), x \in R^3, </math> |
− | + | <br> | |
− | + | '''subject to''' | |
− | + | <br> | |
− | + | <math> x_1 + x_2 = 0 </math> | |
+ | <br> | ||
+ | <math> x_2 + x_3 = 0, </math> | ||
+ | <br> | ||
+ | '''where''' | ||
+ | <br> | ||
+ | <math>a = \begin{bmatrix} | ||
0 \\ | 0 \\ | ||
− | + | 1 \\ | |
+ | 0 | ||
+ | \end{bmatrix}</math> and <math>b = \begin{bmatrix} | ||
+ | 1 \\ | ||
+ | 0 \\ | ||
1 | 1 | ||
− | \end{bmatrix}</math> | + | \end{bmatrix}</math> |
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− | :'''Click [[ | + | :'''Click [[QE2013_AC-3_ECE580-5|here]] to view [[QE2013_AC-3_ECE580-5|student answers and discussions]]''' |
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[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]] | [[ECE_PhD_Qualifying_Exams|Back to ECE QE page]] |
Latest revision as of 11:23, 25 March 2015
Automatic Control (AC)
Question 3: Optimization
August 2013
1.(20 pts) In some of the optimization methods, when minimizing a given function f(x), we select an intial guess $ x^{(0)} $ and a real symmetric positive definite matrix $ H_{0} $. Then we computed $ d^{(k)} = -H_{k}g^{(k)} $, where $ g^{(k)} = \nabla f( x^{(k)} ) $, and $ x^{(k+1)} = x^{(k)} + \alpha_{k}d^{(k)} $, where
$ \alpha_{k} = arg\min_{\alpha \ge 0}f(x^{(k)} + \alpha d^{(k)}) . $
Suppose that the function we wish to minimize is a standard quadratic of the form,
$ f(x) = \frac{1}{2} x^{T} Qx - x^{T}b+c, Q = Q^{T} > 0. $
(i)(10 pts) Find a closed form expression for $ \alpha_k $ in terms of $ Q, H_k, g^{(k)} $, and $ d^{(k)}; $
(ii)(10 pts) Give a sufficient condition on $ H_k $ for $ \alpha_k $ to be positive.
- Click here to view student answers and discussions
Problem 2. (20 pts) [(i) (10 pts)] Consider the one-point crossover of a chromosome in the schema
H = * 1 * 0 1 0 *
where the probability that a chromosome is chosen for crossover is $ p_c = 0.5. $ Find an upper bound on the probability that a chromosome from H will be destroyed by the one-point crossover.
[(ii) (10 pts)] Consider a chromosome in the schema
H = * 1 * 0 * * *
Find the probability that the mutation operation destroys the schema, where the probability of random change of each symbol of the chromosome is $ p_m = 0.1 $ independently.
- Click here to view student answers and discussions
Problem 3. (20 pts) [(i) (10 pts)] Convert the following optimization problem into a linear programming problem and solve it;
maximize $ -|x_1| -|x_2| -|x_3| $
subject to
$ \begin{bmatrix} 1 & 1 &-1 \\ 0 & -1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} . $
[(ii) (10 pts)] Construct the dual program of the linear program above and solve it.
- Click here to view student answers and discussions
Problem 4. (20pts) Consider the following model of a discrete-time system,
$ x(k+1) = x(k) + 2u(k), x(0) = 3, 0 \le k \le 2 $
Use the Lagrange multiplier approach to calculate the optimal control sequence
{u(0), u(1), u(2)}
that transfers the initial state x(0) to x(3) = 9 while minimizing the performance index
$ J = \frac{1}{2} \sum_{k=0}^2 u(k)^2 = \frac{1}{2}u^Tu. $
Hint: Use the system model to obtain a constraint of the form, $ Au = b, A \in R^{1 \times 3}. $
- Click here to view student answers and discussions
Problem 5. (20pts) Find minimizers and maximizers of the function,
$ f(x) = (a^Tx)(b^Tx), x \in R^3, $
subject to
$ x_1 + x_2 = 0 $
$ x_2 + x_3 = 0, $
where
$ a = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} $ and $ b = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $
- Click here to view student answers and discussions