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== '''AC - 3 August 2012 QE''' ==
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[[Category:ECE]]
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[[Category:QE]]
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[[Category:CNSIP]]
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[[Category:problem solving]]
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[[Category:automatic control]]
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[[Category:optimization]]
  
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<center>
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<font size= 4>
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[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
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</font size>
  
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<font size= 4>
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Automatic Control (AC)
  
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Question 3: Optimization
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</font size>
  
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'''1. (20 pts)
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August 2012
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</center>
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(i) (10 pts) Find the factor by which the uncertainty range is reduced when using the Fibonacci method. Assume that the last step has the form
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----
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----
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:Student answers and discussions for [[QE2012_AC-3_ECE580-1|Part 1]],[[QE2012_AC-3_ECE580-2|2]],[[QE2012_AC-3_ECE580-2|3]],[[QE2012_AC-3_ECE580-4|4]],[[QE2012_AC-3_ECE580-5|5]]
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----
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'''1.(20 pts)'''
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<br>
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'''(i)(10 pts) Find the factor by which the uncertainty range is reduced when using the Fibonacci method. Assume that the last step has the form'''
 
<math>  
 
<math>  
 
\begin{align}
 
\begin{align}
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</math>  
 
</math>  
  
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where <math>N - 1 </math> is the number of steps performed in the uncertainty range reduction process.'''
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'''where <span class="texhtml">''N'' − 1</span> is the number of steps performed in the uncertainty range reduction process. '''
  
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<br>
  
  
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<br> '''(ii)(10 pts)''' It is known that the minimizer of a certain function f(x) is located in the interval[-5, 15]. What is the minimal number of iterations of the Fibonacci method required to box in the minimizer within the range 1.0? Assume that the last useful value of the factor reducing the uncertainty range is 2/3, that is
  
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Solution: 
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<math> 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, </math>
  
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    The reduction factor is <math> (1- \rho_{1})(1- \rho_{2})(1- \rho_{3})...(1- \rho_{N-1}) </math>
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:'''Click [[QE2012_AC-3_ECE580-1|here]] to view [[QE2012_AC-3_ECE580-1|student answers and discussions]]'''
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    Since
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----
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    <math> 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, </math>
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    we have
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    <math> 1- \rho_{N-2} = \frac{F_{3}}{F_{4}}</math>    and so on.
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    Then, we have
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'''Problem 2. (20pts) Employ the DFP method to construct a set of Q-conjugate directions using the function'''
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    <math> (1- \rho_{1})(1- \rho_{2})(1- \rho_{3})...(1- \rho_{N-1}) = \frac{F_{N}}{F_{N+1}}  \frac{F_{N-1}}{F_{N}}  ...  \frac{F_{2}}{F_{3}} = \frac{F_{2}}{F_{N+1}}</math>
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    Therefore, the reduction factor is
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    <math>\frac{2}{F_{N+1}}</math>
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'''(ii)(10 pts)
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It is known that the minimizer of a certain function f(x) is located in the interval[-5, 15]. What is the minimal number of iterations of the Fibonacci method required to box in the minimizer within the range 1.0? Assume that the last useful value of the factor reducing the uncertainty range is 2/3, that is
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<math> 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, </math>'''
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Solution:
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      Final Range: 1.0; Initial Range: 20.
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      <math> \frac{2}{F_{N+1}} \le \frac{1.0}{20}</math>, or <math> F_{N+1} \ge 40</math>
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      So, <math> N+1 = 9</math>
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      Therefore, the minimal iterations is N-1 or 7.
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'''2. (20pts) Employ the DFP method to construct a set of Q-conjugate directions using the function
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       <math>f = \frac{1}{2}x^TQx - x^Tb+c </math>
 
       <math>f = \frac{1}{2}x^TQx - x^Tb+c </math>
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        <math>  =\frac{1}{2}x^T  
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      <math>  =\frac{1}{2}x^T  
 
\begin{bmatrix}
 
\begin{bmatrix}
 
   1 & 1 \\
 
   1 & 1 \\
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  \end{bmatrix} + 3.</math>
 
  \end{bmatrix} + 3.</math>
  
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Where <math>x^{(0)} </math> is arbitrary.'''
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Where <span class="texhtml">''x''<sup>(0)</sup></span> is arbitrary.  
  
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:'''Click [[QE2012_AC-3_ECE580-2|here]] to view [[QE2012_AC-3_ECE580-2|student answers and discussions]]'''
  
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----
  
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Solution:
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'''Problem 3. (20pts) For the system of linear equations,<math> Ax=b </math> where '''
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      <math>f = \frac{1}{2}x^TQx - x^Tb+c </math>
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      <math>Use</math> <math>initial</math>  <math> point</math> <math>x^{(0)} = [0, 0]^T and</math> <math> H_0 = I_2</math>
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      <math>In</math> <math>this</math> <math>case,</math>
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      <math>g^{(k)} = \begin{bmatrix}
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  1 & 1 \\
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  1 & 2
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\end{bmatrix} x^{(k)} - \begin{bmatrix}
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  2  \\
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  1
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\end{bmatrix}</math>
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      <math> Hence,</math>
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<math>A = \begin{bmatrix}
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      <math>g^{(0)} = \begin{bmatrix}
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   1 & 0 &-1 \\
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   -2 \\
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   0 & 1 & 0 \\
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   -1
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   0 & -1& 0
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\end{bmatrix},</math>  <math>d^{(0)} = -H_0g^{(0)} =- \begin{bmatrix}
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  \end{bmatrix}, b = \begin{bmatrix}
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  1 & 0 \\
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   0 \\
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   0 & 1
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   2 \\
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  \end{bmatrix}\begin{bmatrix}
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   -2 \\
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  -1
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\end{bmatrix} = \begin{bmatrix}
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   2 \\
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   1
 
   1
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  \end{bmatrix}</math>
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  \end{bmatrix}</math>  
  
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      <math>Because</math> <math>f</math> <math>is</math> <math>a</math> <math>quadratic</math> <math>function,</math>
 
  
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      <math>\alpha_0 = argminf(x^{(0)} + \alpha d^{(0)}) = - \frac{g^{(0)^T}d^{(0)}}{d^{(0)^T}Qd^{(0)}} = - \frac{\begin{bmatrix}
 
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  -2 & -1
 
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\end{bmatrix}\begin{bmatrix}
 
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  2  \\
 
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  1
 
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\end{bmatrix}}{\begin{bmatrix}
 
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  2 & 1\end{bmatrix}\begin{bmatrix}
 
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  1 & 1 \\
 
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  1 & 2
 
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\end{bmatrix}\begin{bmatrix}
 
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  2  \\
 
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  1
 
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\end{bmatrix}} = \frac{1}{2}</math>
 
  
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      <math>x^{(1)} = x^{(0)} + \alpha d^{(0)} = \frac{1}{2} \begin{bmatrix}
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'''Find the minimum length vector <math>x^{(\ast)}</math> that minimizes <math>\| Ax -b\|^{2}_2</math> '''
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  2 \\
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  1
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\end{bmatrix} = \begin{bmatrix}
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  1  \\
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  \frac{1}{2}
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\end{bmatrix}</math>
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      <math>\Delta x^{(0)} = x^{(1)}- x^{(0)} = \begin{bmatrix}
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:'''Click [[QE2012_AC-3_ECE580-3|here]] to view [[QE2012_AC-3_ECE580-3|student answers and discussions]]'''
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  1  \\
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----
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  \frac{1}{2}
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'''Problem 4. (20pts) Use any simplex method to solve the following linear program. '''
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\end{bmatrix}</math>
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      <math>g^{(1)} =\begin{bmatrix}
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  1 & 1 \\
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  1 & 2
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\end{bmatrix} x^{(1)} - \begin{bmatrix}
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  2  \\
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  1
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\end{bmatrix}= \begin{bmatrix}
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  -\frac{1}{2}  \\
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  1
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\end{bmatrix}</math>
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      <math>\Delta g^{(0)} = g^{(1)} - g^{(0)} = \begin{bmatrix}
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            <span class="texhtml">''Maximize''</span>'''    <span class="texhtml">''x''<sub>1</sub> + 2''x''<sub>2</sub></span>'''
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  -\frac{3}{2}  \\
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        <span class="texhtml">''S'ubject to''</span>'''    <math>-2x_1+x_2 \le 2</math>'''
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  2
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                        <math>x_1-x_2 \ge -3</math>
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\end{bmatrix} </math>
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                        <math>x_1 \le -3</math>
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                        <math>x_1 \ge 0, x_2 \ge 0.</math>
  
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:'''Click [[QE2012_AC-3_ECE580-4|here]] to view [[QE2012_AC-3_ECE580-4|student answers and discussions]]'''
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----
  
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      <math>Observe</math> <math>that:</math>
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<br> '''Problem 5.(20pts) Solve the following problem:'''
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      <math>\Delta x^{(0)} \Delta x^{(0)^T} = \begin{bmatrix}
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  1  \\
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  \frac{1}{2}
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\end{bmatrix} \begin{bmatrix}
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  1  & \frac{1}{2}
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\end{bmatrix} = \begin{bmatrix}
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  1 & \frac{1}{2}  \\
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  \frac{1}{2}  & \frac{1}{4}
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\end{bmatrix} </math>
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      <math> \Delta x^{(0)^T} \Delta g^{(0)} = \begin{bmatrix}
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  1  & \frac{1}{2}
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\end{bmatrix}\begin{bmatrix}
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  \frac{3}{2}  \\
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  2
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\end{bmatrix}  = \frac{5}{2}</math>
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      <math>H_0 \Delta g^{(0)} = \begin{bmatrix}
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  1 & 0 \\
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  0 & 1
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\end{bmatrix} \begin{bmatrix}
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  \frac{3}{2}  \\
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  2
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\end{bmatrix} = \begin{bmatrix}
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  \frac{3}{2}  \\
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  2
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\end{bmatrix},</math> <math>(H_0 \Delta g^{(0)})(H_0 \Delta g^{(0)})^T = \begin{bmatrix}
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  \frac{9}{4}  & 3 \\
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  3 & 4
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\end{bmatrix}</math>
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      <math>\Delta g^{(0)^T}H_0 \Delta g^{(0)} = \begin{bmatrix}
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  \frac{3}{2}  & 2
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\end{bmatrix} \begin{bmatrix}
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  1 & 0 \\
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  0 & 1
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\end{bmatrix} \begin{bmatrix}
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  \frac{3}{2}  \\ 2
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\end{bmatrix} = \frac{25}{4}</math>
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      <math>Using</math> <math>the</math> <math>above, </math> <math>now</math> <math>we</math> <math>compute</math>
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      <math>H_1 = H_0 + \frac{\Delta x^{(0)} \Delta x^{(0)^T}}{\Delta x^{(0)^T} \Delta g^{(0)}} - \frac{(H_0 \Delta g^{(0)})(H_0 \Delta g^{(0)})^T}{\Delta g^{(0)^T}H_0 \Delta g^{(0)} }  = \begin{bmatrix}
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  1 & 0 \\
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  0 & 1
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\end{bmatrix} + \begin{bmatrix}
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  \frac{2}{5} & \frac{1}{5} \\
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  \frac{1}{5} & \frac{1}{10}
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\end{bmatrix} - \frac{25}{4}\begin{bmatrix}
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  \frac{9}{4} & 3 \\
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  3 & 4
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\end{bmatrix} = \begin{bmatrix}
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  \frac{26}{25} & -\frac{7}{25} \\
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  -\frac{7}{25} & \frac{23}{50}
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\end{bmatrix}</math>
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      <math>Then</math> <math>we</math> <math>have,</math>
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            <span class="texhtml">''Minimize''</span>'''    <math>-x_1^2 + 2x_2</math>'''
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      <math>d^{(1)} = -H_1 g^{(0)} = - \begin{bmatrix}
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        <span class="texhtml">''Subject to''</span>'''    <math>x_1^2+x_2^2 \le 1</math>'''
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  \frac{26}{25} & -\frac{7}{25} \\
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                        <math> x_1 \ge 0</math>
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  -\frac{7}{25} & \frac{23}{50}
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                        <math>x_2 \ge 0</math>
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\end{bmatrix} \begin{bmatrix}
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  -\frac{1}{2}  \\
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  1
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\end{bmatrix} = \begin{bmatrix}
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  \frac{4}{5}  \\
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  -\frac{3}{5}
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\end{bmatrix}</math>
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      <math>\alpha_0 = argminf(x^{(0)} + \alpha d^{(0)}) = - \frac{g^{(0)^T}d^{(0)}}{d^{(0)^T}Qd^{(0)}} = - \frac{\begin{bmatrix}
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'''(i)(10pts) Find the points that satisfy the KKT condition.'''
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  -2 & -1
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\end{bmatrix}\begin{bmatrix}
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  2  \\
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  1
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<br> '''(ii)(10pts)Apply the SONC and SOSC to determine the nature of the critical points from the previous part.'''
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\end{bmatrix}}{\begin{bmatrix}
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  2 & 1\end{bmatrix}\begin{bmatrix}
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:'''Click [[QE2012_AC-3_ECE580-5|here]] to view [[QE2012_AC-3_ECE580-5|student answers and discussions]]'''
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  1 & 1 \\
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----
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  1 & 2
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[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]]
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\end{bmatrix}\begin{bmatrix}
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  2  \\
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  1
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\end{bmatrix}} = \frac{1}{2}</math>
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Latest revision as of 09:17, 13 September 2013


ECE Ph.D. Qualifying Exam

Automatic Control (AC)

Question 3: Optimization

August 2012

Student answers and discussions for Part 1,2,3,4,5

1.(20 pts)
(i)(10 pts) Find the factor by which the uncertainty range is reduced when using the Fibonacci method. Assume that the last step has the form $ \begin{align} 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, \end{align} $

where N − 1 is the number of steps performed in the uncertainty range reduction process.




(ii)(10 pts) It is known that the minimizer of a certain function f(x) is located in the interval[-5, 15]. What is the minimal number of iterations of the Fibonacci method required to box in the minimizer within the range 1.0? Assume that the last useful value of the factor reducing the uncertainty range is 2/3, that is

$ 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, $

Click here to view student answers and discussions

Problem 2. (20pts) Employ the DFP method to construct a set of Q-conjugate directions using the function 

      $ f = \frac{1}{2}x^TQx - x^Tb+c  $
     $   =\frac{1}{2}x^T  \begin{bmatrix}   1 & 1 \\   1 & 2  \end{bmatrix}x-x^T\begin{bmatrix}   2  \\   1  \end{bmatrix} + 3. $

Where x(0) is arbitrary.

Click here to view student answers and discussions

Problem 3. (20pts) For the system of linear equations,$ Ax=b $ where

$ A = \begin{bmatrix} 1 & 0 &-1 \\ 0 & 1 & 0 \\ 0 & -1& 0 \end{bmatrix}, b = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix} $


Find the minimum length vector $ x^{(\ast)} $ that minimizes $ \| Ax -b\|^{2}_2 $

Click here to view student answers and discussions

Problem 4. (20pts) Use any simplex method to solve the following linear program. 

           Maximize    x1 + 2x2
        S'ubject to    $ -2x_1+x_2 \le 2 $
                       $ x_1-x_2 \ge -3 $
                       $ x_1 \le -3 $
                       $ x_1 \ge 0, x_2 \ge 0. $
Click here to view student answers and discussions


Problem 5.(20pts) Solve the following problem: 

           Minimize    $ -x_1^2 + 2x_2 $
        Subject to    $ x_1^2+x_2^2 \le 1 $
                       $  x_1 \ge 0 $
                       $ x_2 \ge 0 $

(i)(10pts) Find the points that satisfy the KKT condition.



(ii)(10pts)Apply the SONC and SOSC to determine the nature of the critical points from the previous part.

Click here to view student answers and discussions

Back to ECE QE page

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