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:} $
extremize 9x1 + 3x2
$ \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) $
Applying the FONC
$ D_{x} l\left( x,\lambda \right) = \begin{bmatrix} 9+\lambda x_{1} & 3-\lambda \end{bmatrix} =\begin{bmatrix} 0 & 0 \end{bmatrix} $
$ \Rightarrow \lambda^{*}=3 \text{ and } x^{*}_{1}=-3 $
$ x_{2}^{*}=0.5x_{1}^{*2}=9/2 $
$ \therefore \text{the point that satisfies the FONC is } x^{*}= \begin{bmatrix} -3\\ 9/2 \end{bmatrix} $
$ D_{x} L\left( x^{*},\lambda^{*} \right) = F\left( x^{*} \right)+ \lambda^{*} H\left( x^{*} \right) = \begin{bmatrix} 3 & 0\\ 0 & 0 \end{bmatrix} $
$ T\left( x^{* }\right)= \left \{ y: \begin{bmatrix} -3 & -1 \end{bmatrix}y=0 \right \} = \left \{ y =\begin{bmatrix} a & -3a \end{bmatrix}^{T}, a\in\Re \right \} $
$ \text{Hence } y^{T}L\left ( x^{\ast },\mu ^{\ast } \right )y=\begin{bmatrix} a & -3a \end{bmatrix}\begin{bmatrix} 3 & 0\\ 0 & 0 \end{bmatrix}\begin{bmatrix} a \\ -3a \end{bmatrix}=3a^{2}>0 $
Therefore, x* is a strict local minimizer.
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
