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| − | = [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]] in "Automatic Control" (AC)= | + | = [[ECE PhD Qualifying Exams|ECE Ph.D. Qualifying Exam]] in "Automatic Control" (AC) = |
| − | =Question 3, Part 4, August 2011 = | + | |
| − | :[[ECE- | + | = Question 3, Part 4, August 2011 = |
| + | |||
| + | :[[ECE-QE AC3-2011 solusion-1|Part 1]],[[ECE-QE AC3-2011 solusion-2|2]],[[ECE-QE AC3-2011 solusion-3|3]],[[ECE-QE_AC3-2011_solusion-4|4]],[[ECE-QE AC3-2011 solusion-5|5]] | ||
| + | |||
---- | ---- | ||
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---- | ---- | ||
| − | + | ||
| + | '''Discussion:''' | ||
| + | |||
| + | This problem need a bit interpretation before getting the standard form to apply KKT condition. | ||
| + | |||
| + | Theorem about KKT, SONC, SOSC please see "Part 5". | ||
| + | |||
---- | ---- | ||
| + | |||
<math>\color{blue}\text{Solution 1:}</math> | <math>\color{blue}\text{Solution 1:}</math> | ||
<math>\left.\begin{matrix} | <math>\left.\begin{matrix} | ||
| − | x\left ( 1 \right )=2x\left ( 0 \right )+ | + | x\left ( 1 \right )=2x\left ( 0 \right )+u\left ( 0\right )\\ |
| − | x\left ( 2 \right )=2x\left ( 1 \right )+ | + | x\left ( 2 \right )=2x\left ( 1 \right )+u\left ( 1\right )\\ |
| − | x\left ( 3 \right )=2x\left ( 2 \right )+ | + | x\left ( 3 \right )=2x\left ( 2 \right )+u\left ( 2\right )\\ |
x\left ( 0 \right )=0 | x\left ( 0 \right )=0 | ||
\end{matrix}\right\} \Rightarrow | \end{matrix}\right\} \Rightarrow | ||
\left\{\begin{matrix} | \left\{\begin{matrix} | ||
| − | x\left ( 1 \right )= | + | x\left ( 1 \right )=u\left ( 0 \right )\\ |
| − | x\left ( 2 \right )= | + | x\left ( 2 \right )=2u\left ( 0 \right )+u\left ( 1\right )\\ |
| − | x\left ( 3 \right )= | + | x\left ( 3 \right )=4u\left ( 0 \right )+2u\left ( 1\right )+u\left ( 2 \right )=7 |
\end{matrix}\right.</math><br> | \end{matrix}\right.</math><br> | ||
<font face="serif"><math>\text{The problem is equivalent to minimize } J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2}</math><br></font> | <font face="serif"><math>\text{The problem is equivalent to minimize } J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2}</math><br></font> | ||
| − | <font face="serif"></font> <math>\text{subject to } | + | <font face="serif"></font> <math>\text{subject to } 4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)=7</math> |
| − | <math>\text{Let } h( | + | <math>\text{Let } h(u )=4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)-7</math><br> |
<math>\text{FONC: } \left\{\begin{matrix} | <math>\text{FONC: } \left\{\begin{matrix} | ||
| − | l\left( | + | l\left(u,\lambda \right)=\nabla J\left ( u \right )+\lambda\nabla h\left( u \right)=\begin{pmatrix} |
| − | + | u\left ( 0 \right )\\ | |
| − | + | u\left ( 1 \right )\\ | |
| − | + | u\left ( 2 \right ) | |
\end{pmatrix}+\lambda\begin{pmatrix} | \end{pmatrix}+\lambda\begin{pmatrix} | ||
4\\ | 4\\ | ||
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1 | 1 | ||
\end{pmatrix} =0\\ | \end{pmatrix} =0\\ | ||
| − | h( | + | h(u )=4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)-7=0 |
\end{matrix}\right. | \end{matrix}\right. | ||
\Rightarrow | \Rightarrow | ||
\left\{\begin{matrix} | \left\{\begin{matrix} | ||
| − | + | u\left(0 \right)=\frac{4}{3}\\ | |
| − | + | u\left(1 \right)=\frac{2}{3}\\ | |
| − | + | u\left(2 \right)=\frac{1}{3}\\ | |
\lambda=-\frac{1}{3} | \lambda=-\frac{1}{3} | ||
\end{matrix}\right.</math> | \end{matrix}\right.</math> | ||
| − | <math>\text{SOSC: } L\left( | + | <math>\text{SOSC: } L\left( u,\lambda \right)=\nabla l\left( u,\lambda \right)=\begin{bmatrix} |
1 & 0 & 0\\ | 1 & 0 & 0\\ | ||
0 & 1 & 0\\ | 0 & 1 & 0\\ | ||
0 & 0 & 1 | 0 & 0 & 1 | ||
| − | \end{bmatrix}>0</math> | + | \end{bmatrix}>0</math> |
| − | The sequence <math> | + | The sequence <math>u\left( 0 \right)=\frac{4}{3},u\left( 1 \right)=\frac{2}{3},u\left( 2 \right)=\frac{1}{3}</math> satisfies SOSC. It is the optimal sequence. |
| − | <br> | + | <math>\color{red} \text{This solution did not specifically state the complete SOSC.}</math><br> |
---- | ---- | ||
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<math>\color{blue}\text{Solution 2:}</math> | <math>\color{blue}\text{Solution 2:}</math> | ||
| − | <math>x\left ( 1 \right )= | + | <math>x\left ( 1 \right )=u\left ( 0 \right )</math><br> |
| − | <math>x\left ( 2 \right )= | + | <math>x\left ( 2 \right )=2u\left ( 0 \right )+u\left ( 1 \right )</math> |
| − | <math>x\left ( 3 \right )= | + | <math>x\left ( 3 \right )=4u\left ( 0 \right )+2u\left ( 1\right )+u\left ( 2 \right )=7</math> |
| − | <math>\text{The problem transfer to min } J\left ( | + | <math>\text{The problem transfer to min } J\left ( u \right )=\frac{1}{2} u \left ( 0 \right )^{2}+\frac{1}{2} u \left ( 1 \right )^{2}+\frac{1}{2} u \left ( 2 \right )^{2}</math> |
| − | <math>\text{subject to } h( | + | <math>\text{subject to } h(u )=4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)-7=0</math> |
| − | <math>\text{Apply KKT condition: } Dl\left( | + | <math>\text{Apply KKT condition: } Dl\left( u ,\lambda \right)=DJ\left(u \right)+\lambda Dh\left(u \right)=\left[ u\left(0 \right)+4\lambda,u\left(1 \right)+2\lambda,u\left(2 \right)+\lambda \right]=0</math> |
<math>\left\{\begin{matrix} | <math>\left\{\begin{matrix} | ||
| − | + | u\left(0 \right)+4\lambda=0\\ | |
| − | + | u\left(1 \right)+2\lambda=0\\ | |
| − | + | u\left(2 \right)+\lambda=0\\ | |
| − | + | 4u\left(0 \right)+2u\left(1 \right)+u\left(2 \right)-7=0 | |
\end{matrix}\right. | \end{matrix}\right. | ||
\Rightarrow | \Rightarrow | ||
\left\{\begin{matrix} | \left\{\begin{matrix} | ||
| − | + | u\left(0 \right)=\frac{4}{3}\\ | |
| − | + | u\left(1 \right)=\frac{2}{3}\\ | |
| − | + | u\left(2 \right)=\frac{1}{3}\\ | |
\lambda=-\frac{1}{3} | \lambda=-\frac{1}{3} | ||
\end{matrix}\right.</math> | \end{matrix}\right.</math> | ||
| − | <math>\text{Check SOSC: } L\left( | + | <math>\text{Check SOSC: } L\left( u,\lambda \right)=D^{2}l\left( u,\lambda \right)=\begin{bmatrix} |
1 & 0 & 0\\ | 1 & 0 & 0\\ | ||
0 & 1 & 0\\ | 0 & 1 & 0\\ | ||
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---- | ---- | ||
| − | Automatic Control (AC)- Question 3, August 2011 | + | <math>\color{blue} \text{Related Problem:}</math> |
| + | |||
| + | <math>\text{extremize } 9x_{1}+3x_{2}</math> | ||
| + | |||
| + | <math>\text{subject to } \frac{1}{2}x_{1}^{2}-x_{2}=0</math> | ||
| + | |||
| + | <math>\color{blue} \text{(i) Find point(s) that satisfy the FONC}</math> | ||
| + | |||
| + | <math>\color{blue} \text{(ii) Apply the SOSC to determine the nature of the critical point(s) from the previous part}</math> | ||
| + | |||
| + | <math>\color{blue} \text{Solution: }</math> | ||
| + | |||
| + | <math>l\left( u,\lambda \right)=9x_{1}+3x_{2}+\lambda\left( \frac{1}{2}x_{1}^{2}-x_{2} \right)</math> | ||
| + | |||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | Automatic Control (AC)- Question 3, August 2011 | ||
| + | |||
| + | Go to | ||
| − | + | *Problem 1: [[ECE-QE AC3-2011 solusion-1|solutions and discussions]] | |
| − | *Problem 1: [[ECE- | + | *Problem 2: [[ECE-QE AC3-2011 solusion-2|solutions and discussions]] |
| − | *Problem 2: [[ECE- | + | *Problem 3: [[ECE-QE AC3-2011 solusion-3|solutions and discussions]] |
| − | *Problem 3: [[ECE- | + | *Problem 4: [[ECE-QE_AC3-2011_solusion-4|solutions and discussions]] |
| − | *Problem 4: [[ECE-QE_AC3-2011_solusion-4|solutions and discussions]] | + | *Problem 5: [[ECE-QE AC3-2011 solusion-5|solutions and discussions]] |
| − | *Problem 5: [[ECE- | + | |
---- | ---- | ||
Revision as of 10:02, 29 June 2012
ECE Ph.D. Qualifying Exam in "Automatic Control" (AC)
Question 3, Part 4, August 2011
$ \color{blue}\text{4. } \left( \text{20 pts} \right) \text{ Consider the following model of a discrete-time system, } $
$ x\left ( k+1 \right )=2x\left ( k \right )+u\left ( k \right ), x\left ( 0 \right )=0, 0\leq k\leq 2 $
$ \color{blue}\text{Use the Lagrange multiplier approach to calculate the optimal control sequence} $
$ \left \{ u\left ( 0 \right ),u\left ( 1 \right ), u\left ( 2 \right ) \right \} $
$ \color{blue}\text{that transfers the initial state } x\left( 0 \right) \text{ to } x\left( 3 \right)=7 \text{ while minimizing the performance index} $
$ J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2} $
Discussion:
This problem need a bit interpretation before getting the standard form to apply KKT condition.
Theorem about KKT, SONC, SOSC please see "Part 5".
$ \color{blue}\text{Solution 1:} $
$ \left.\begin{matrix} x\left ( 1 \right )=2x\left ( 0 \right )+u\left ( 0\right )\\ x\left ( 2 \right )=2x\left ( 1 \right )+u\left ( 1\right )\\ x\left ( 3 \right )=2x\left ( 2 \right )+u\left ( 2\right )\\ x\left ( 0 \right )=0 \end{matrix}\right\} \Rightarrow \left\{\begin{matrix} x\left ( 1 \right )=u\left ( 0 \right )\\ x\left ( 2 \right )=2u\left ( 0 \right )+u\left ( 1\right )\\ x\left ( 3 \right )=4u\left ( 0 \right )+2u\left ( 1\right )+u\left ( 2 \right )=7 \end{matrix}\right. $
$ \text{The problem is equivalent to minimize } J=\frac{1}{2}\sum\limits_{k=0}^2 u\left ( k \right )^{2} $
$ \text{subject to } 4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)=7 $
$ \text{Let } h(u )=4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)-7 $
$ \text{FONC: } \left\{\begin{matrix} l\left(u,\lambda \right)=\nabla J\left ( u \right )+\lambda\nabla h\left( u \right)=\begin{pmatrix} u\left ( 0 \right )\\ u\left ( 1 \right )\\ u\left ( 2 \right ) \end{pmatrix}+\lambda\begin{pmatrix} 4\\ 2\\ 1 \end{pmatrix} =0\\ h(u )=4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)-7=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} u\left(0 \right)=\frac{4}{3}\\ u\left(1 \right)=\frac{2}{3}\\ u\left(2 \right)=\frac{1}{3}\\ \lambda=-\frac{1}{3} \end{matrix}\right. $
$ \text{SOSC: } L\left( u,\lambda \right)=\nabla l\left( u,\lambda \right)=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}>0 $
The sequence $ u\left( 0 \right)=\frac{4}{3},u\left( 1 \right)=\frac{2}{3},u\left( 2 \right)=\frac{1}{3} $ satisfies SOSC. It is the optimal sequence.
$ \color{red} \text{This solution did not specifically state the complete SOSC.} $
$ \color{blue}\text{Solution 2:} $
$ x\left ( 1 \right )=u\left ( 0 \right ) $
$ x\left ( 2 \right )=2u\left ( 0 \right )+u\left ( 1 \right ) $
$ x\left ( 3 \right )=4u\left ( 0 \right )+2u\left ( 1\right )+u\left ( 2 \right )=7 $
$ \text{The problem transfer to min } J\left ( u \right )=\frac{1}{2} u \left ( 0 \right )^{2}+\frac{1}{2} u \left ( 1 \right )^{2}+\frac{1}{2} u \left ( 2 \right )^{2} $
$ \text{subject to } h(u )=4u \left(0 \right)+2u \left(1 \right)+u\left(2 \right)-7=0 $
$ \text{Apply KKT condition: } Dl\left( u ,\lambda \right)=DJ\left(u \right)+\lambda Dh\left(u \right)=\left[ u\left(0 \right)+4\lambda,u\left(1 \right)+2\lambda,u\left(2 \right)+\lambda \right]=0 $
$ \left\{\begin{matrix} u\left(0 \right)+4\lambda=0\\ u\left(1 \right)+2\lambda=0\\ u\left(2 \right)+\lambda=0\\ 4u\left(0 \right)+2u\left(1 \right)+u\left(2 \right)-7=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} u\left(0 \right)=\frac{4}{3}\\ u\left(1 \right)=\frac{2}{3}\\ u\left(2 \right)=\frac{1}{3}\\ \lambda=-\frac{1}{3} \end{matrix}\right. $
$ \text{Check SOSC: } L\left( u,\lambda \right)=D^{2}l\left( u,\lambda \right)=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}>0 $
$ \therefore \text{For all y, } y^{T}Ly\geq 0 $
$ \therefore \text{sequence } \left\{ \frac{4}{3},\frac{2}{3},\frac{1}{3} \right\} \text{ satisfy SOSC is a strict minimizer of the problem.} $
$ \color{blue} \text{Related Problem:} $
$ \text{extremize } 9x_{1}+3x_{2} $
$ \text{subject to } \frac{1}{2}x_{1}^{2}-x_{2}=0 $
$ \color{blue} \text{(i) Find point(s) that satisfy the FONC} $
$ \color{blue} \text{(ii) Apply the SOSC to determine the nature of the critical point(s) from the previous part} $
$ \color{blue} \text{Solution: } $
$ l\left( u,\lambda \right)=9x_{1}+3x_{2}+\lambda\left( \frac{1}{2}x_{1}^{2}-x_{2} \right) $
Automatic Control (AC)- Question 3, August 2011
Go to
- Problem 1: solutions and discussions
- Problem 2: solutions and discussions
- Problem 3: solutions and discussions
- Problem 4: solutions and discussions
- Problem 5: solutions and discussions
