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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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Part 1 =
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[[Category:CNSIP]]
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[[Category:problem solving]]
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:[[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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[[Category:automatic control]]
 +
[[Category:optimization]]
  
 +
<center>
 +
<font size= 4>
 +
[[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]]
 +
</font size>
  
 +
<font size= 4>
 +
Automatic Control (AC)
  
 +
Question 3: Optimization
 +
</font size>
  
 +
August 2012
 +
</center>
 +
----
 +
----
 +
: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]]
 +
----
 +
'''1.(20 pts)'''
 +
<br>
 +
'''(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>  
  
−
where <span class="texhtml">''N'' − 1</span> 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. '''
  
 
<br>  
 
<br>  
  
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<br> Solution:
 
−
 
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    The reduction factor is <span class="texhtml">(1 − ρ<sub>1</sub>)(1 − ρ<sub>2</sub>)(1 − ρ<sub>3</sub>)...(1 − ρ<sub>''N'' − 1</sub>)</span>
 
−
Since
 
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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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<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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<br>
 
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Solution 2:
 
−
 
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The uncertainty interval is reduced by <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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<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}
 
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\text{In this specific case, we use only N-1 iterations. }
 
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</math></span></font>
 
−
 
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<br>
 
  
 
<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  
 
<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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<math> 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, </math>  
 
<math> 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, </math>  
  
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<br> Solution:  
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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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      Final Range: 1.0; Initial Range: 20.
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−
 
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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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−
 
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      So, <span class="texhtml">''N'' + 1 = 9</span>
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−
 
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      Therefore, the minimal iterations is N-1 or 7.
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−
 
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<br>
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Solution 2:
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Since the final range is 1 and the initial range is 20, we can say
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<math>
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\frac{2}{F_{N+1}} \le \frac{1.0}{20} \text{or equivalently } F_{N+1} \ge 40
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−
</math>
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−
 
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From the inequality, we get <math> F_{N+1} \ge 40 , \text{ so } N+1=9 </math>. Therefore the minimum number of iteration is N-1=7
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−
 
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<font color="#0000FF "><span style="font-size: 19px;"><math>\color{blue}
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F_9=55 \text{ and } F_8=34
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</math></span></font>
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−
 
+
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<br>
+
  
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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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'''Problem 2. (20pts) Employ the DFP method to construct a set of Q-conjugate directions using the function'''  
  
 
       <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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Where <span class="texhtml">''x''<sup>(0)</sup></span> is arbitrary.  
 
Where <span class="texhtml">''x''<sup>(0)</sup></span> is arbitrary.  
  
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<br>
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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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Solution:
+
----
−
 
+
−
      <math>f = \frac{1}{2}x^TQx - x^Tb+c </math>
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    Use initial point x<sup>(0)</sup> = [0,0]<sup>T</sub> and H<sub>0</sub> = I<sub>2</sub>
+
−
   
+
−
In this case
+
−
    <math>g^{(k)} = \begin{bmatrix}
+
−
  1 & 1 \\
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−
  1 & 2
+
−
\end{bmatrix} x^{(k)} - \begin{bmatrix}
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−
  2  \\
+
−
  1
+
−
\end{bmatrix}</math>
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−
 
+
−
      Hence
+
−
    <math>g^{(0)} = \begin{bmatrix}
+
−
  -2  \\
+
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  -1
+
−
\end{bmatrix},</math>  <math>d^{(0)} = -H_0g^{(0)} =- \begin{bmatrix}
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−
  1 & 0 \\
+
−
  0 & 1
+
−
\end{bmatrix}\begin{bmatrix}
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−
  -2  \\
+
−
  -1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix}</math>
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−
 
+
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<br>
+
−
 
+
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      Because f is a quadratic function
+
−
 
+
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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}
+
−
  -2 & -1
+
−
\end{bmatrix}\begin{bmatrix}
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−
  2  \\
+
−
  1
+
−
\end{bmatrix}}{\begin{bmatrix}
+
−
  2 & 1\end{bmatrix}\begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix}\begin{bmatrix}
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−
  2  \\
+
−
  1
+
−
\end{bmatrix}} = \frac{1}{2}</math>
+
−
 
+
−
      <math>x^{(1)} = x^{(0)} + \alpha d^{(0)} = \frac{1}{2} \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  1  \\
+
−
  \frac{1}{2}
+
−
\end{bmatrix}</math>
+
−
 
+
−
      <math>\Delta x^{(0)} = x^{(1)}- x^{(0)} = \begin{bmatrix}
+
−
  1  \\
+
−
  \frac{1}{2}
+
−
\end{bmatrix}</math>
+
−
    <math>g^{(1)} =\begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix} x^{(1)} - \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix}= \begin{bmatrix}
+
−
  -\frac{1}{2}  \\
+
−
  1
+
−
\end{bmatrix}</math>
+
−
 
+
−
      <math>\Delta g^{(0)} = g^{(1)} - g^{(0)} = \begin{bmatrix}
+
−
  -\frac{3}{2}  \\
+
−
  2
+
−
\end{bmatrix} </math>
+
−
 
+
−
<br>
+
−
 
+
−
      Observe that
+
−
    <math>\Delta x^{(0)} \Delta x^{(0)^T} = \begin{bmatrix}
+
−
  1  \\
+
−
  \frac{1}{2}
+
−
\end{bmatrix} \begin{bmatrix}
+
−
  1  & \frac{1}{2}
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  1 & \frac{1}{2}  \\
+
−
  \frac{1}{2}  & \frac{1}{4}
+
−
\end{bmatrix} </math>
+
−
    <math> \Delta x^{(0)^T} \Delta g^{(0)} = \begin{bmatrix}
+
−
  1  & \frac{1}{2}
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  \frac{3}{2}  \\
+
−
  2
+
−
\end{bmatrix}  = \frac{5}{2}</math>
+
−
    <math>H_0 \Delta g^{(0)} = \begin{bmatrix}
+
−
  1 & 0 \\
+
−
  0 & 1
+
−
\end{bmatrix} \begin{bmatrix}
+
−
  \frac{3}{2}  \\
+
−
  2
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  \frac{3}{2}  \\
+
−
  2
+
−
\end{bmatrix},</math> <math>(H_0 \Delta g^{(0)})(H_0 \Delta g^{(0)})^T = \begin{bmatrix}
+
−
  \frac{9}{4}  & 3 \\
+
−
  3 & 4
+
−
\end{bmatrix}</math>
+
−
    <math>\Delta g^{(0)^T}H_0 \Delta g^{(0)} = \begin{bmatrix}
+
−
  \frac{3}{2}  & 2
+
−
\end{bmatrix} \begin{bmatrix}
+
−
  1 & 0 \\
+
−
  0 & 1
+
−
\end{bmatrix} \begin{bmatrix}
+
−
  \frac{3}{2}  \\ 2
+
−
\end{bmatrix} = \frac{25}{4}</math>
+
−
    <font face="serif" size="3">''Using the above, now we have''</font>
+
−
    <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}
+
−
  1 & 0 \\
+
−
  0 & 1
+
−
\end{bmatrix} + \begin{bmatrix}
+
−
  \frac{2}{5} & \frac{1}{5} \\
+
−
  \frac{1}{5} & \frac{1}{10}
+
−
\end{bmatrix} - \frac{25}{4}\begin{bmatrix}
+
−
  \frac{9}{4} & 3 \\
+
−
  3 & 4
+
−
\end{bmatrix} = \begin{bmatrix}
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−
  \frac{26}{25} & -\frac{7}{25} \\
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−
  -\frac{7}{25} & \frac{23}{50}
+
−
\end{bmatrix}</math>
+
−
 
+
−
      <span class="texhtml">''T'hen we have''</span><span class="texhtml">''','''</span>
+
−
    <math>d^{(1)} = -H_1 g^{(0)} = - \begin{bmatrix}
+
−
  \frac{26}{25} & -\frac{7}{25} \\
+
−
  -\frac{7}{25} & \frac{23}{50}
+
−
\end{bmatrix} \begin{bmatrix}
+
−
  -\frac{1}{2}  \\
+
−
  1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix}</math>
+
−
 
+
−
      <math>\alpha_1 = argminf(x^{(1)} + \alpha d^{(1)}) = - \frac{g^{(1)^T}d^{(1)}}{d^{(1)^T}Qd^{(1)}} = - \frac{\begin{bmatrix}
+
−
  -2 & 1
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix}}{\begin{bmatrix}
+
−
  \frac{4}{5} & -\frac{3}{5}\end{bmatrix}\begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix}} = \frac{5}{2}</math>
+
−
 
+
−
      <math>x^{(2)} = x^{(1)} + \alpha_1 d^{(1)} = \begin{bmatrix}
+
−
  1  \\
+
−
  \frac{1}{2}
+
−
\end{bmatrix} + \frac{5}{2}\begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  3  \\
+
−
  -1
+
−
\end{bmatrix} </math>
+
−
 
+
−
      <math>\Delta x^{(1)} = x^{(2)} - x^{(1)} = \begin{bmatrix}
+
−
  2  \\
+
−
  -\frac{3}{2}
+
−
\end{bmatrix}</math>
+
−
    <math>g^{(2)} = \begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix} x^{(0)} - \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  0  \\
+
−
  0
+
−
\end{bmatrix}</math>
+
−
 
+
−
      <span class="texhtml">''Note that we have''</span> <math>d^{(0)^T}Qd^{(0)} = 0;</math>
+
−
    <span class="texhtml">''that is''</span>, <math>d^{(0)} = \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix}</math> and  <math>d^{(1)}  = \begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix}</math> <span class="texhtml">''are Q-conjugate directions''</span><span class="texhtml">'''.'''</span>
+
−
 
+
−
 
+
−
Solution 2:
+
−
 
+
−
<math>
+
−
\text{Let the initial point be } x^{(0)}= \begin{bmatrix} 0\\ 0 \end{bmatrix}
+
−
\text{and initial Hessian be } H_0=\begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}
+
−
 
+
−
</math>
+
−
 
+
−
<math>g^{(k)} = \begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix} x^{(k)} - \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix} , \text{so}</math>
+
−
 
+
−
<math>g^{(0)} = \begin{bmatrix}
+
−
  -2  \\
+
−
  -1
+
−
\end{bmatrix},</math>  <math>d^{(0)} = -H_0 g^{(0)} =- \begin{bmatrix}
+
−
  1 & 0 \\
+
−
  0 & 1
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  -2  \\
+
−
  -1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>\alpha_0 = - \frac{g^{(0)^T}d^{(0)}}{d^{(0)^T}Qd^{(0)}} = - \frac{\begin{bmatrix}
+
−
  -2 & -1
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix}}{\begin{bmatrix}
+
−
  2 & 1\end{bmatrix}\begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix}} = \frac{1}{2}</math>
+
−
 
+
−
<math>x^{(1)} = x^{(0)} + \alpha d^{(0)} = \frac{1}{2} \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  1  \\
+
−
  \frac{1}{2}
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>\Delta x^{(0)} = x^{(1)}- x^{(0)} = \begin{bmatrix}
+
−
  1  \\
+
−
  \frac{1}{2}
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>g^{(1)} =\begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix} x^{(1)} - \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix}= \begin{bmatrix}
+
−
  -\frac{1}{2}  \\
+
−
  1
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>\Delta g^{(0)} = g^{(1)} - g^{(0)} = \begin{bmatrix}
+
−
  -\frac{3}{2}  \\
+
−
  2
+
−
\end{bmatrix} </math>
+
−
 
+
−
<math>
+
−
\text{If we plug in the above numbers in the formula, we can get}
+
−
</math>
+
−
 
+
−
<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}
+
−
  1 & 0 \\
+
−
  0 & 1
+
−
\end{bmatrix} + \begin{bmatrix}
+
−
  \frac{2}{5} & \frac{1}{5} \\
+
−
  \frac{1}{5} & \frac{1}{10}
+
−
\end{bmatrix} - \frac{25}{4}\begin{bmatrix}
+
−
  \frac{9}{4} & 3 \\
+
−
  3 & 4
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  \frac{26}{25} & -\frac{7}{25} \\
+
−
  -\frac{7}{25} & \frac{23}{50}
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>d^{(1)} = -H_1 g^{(1)} = - \begin{bmatrix}
+
−
  \frac{26}{25} & -\frac{7}{25} \\
+
−
  -\frac{7}{25} & \frac{23}{50}
+
−
\end{bmatrix} \begin{bmatrix}
+
−
  -\frac{1}{2}  \\
+
−
  1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>\alpha_1 = - \frac{g^{(1)^T}d^{(1)}}{d^{(1)^T}Qd^{(1)}} = - \frac{\begin{bmatrix}
+
−
  -2 & 1
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix}}{\begin{bmatrix}
+
−
  \frac{4}{5} & -\frac{3}{5}\end{bmatrix}\begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix}\begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix}} = \frac{5}{2}</math>
+
−
 
+
−
<math>x^{(2)} = x^{(1)} + \alpha_1 d^{(1)} = \begin{bmatrix}
+
−
  1  \\
+
−
  \frac{1}{2}
+
−
\end{bmatrix} + \frac{5}{2}\begin{bmatrix}
+
−
  \frac{4}{5}  \\
+
−
  -\frac{3}{5}
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  3  \\
+
−
  -1
+
−
\end{bmatrix} </math>
+
−
 
+
−
<math>\Delta x^{(1)} = x^{(2)} - x^{(1)} = \begin{bmatrix}
+
−
  2  \\
+
−
  -\frac{3}{2}
+
−
\end{bmatrix}
+
−
</math>
+
−
 
+
−
<math>
+
−
g^{(2)} = \begin{bmatrix}
+
−
  1 & 1 \\
+
−
  1 & 2
+
−
\end{bmatrix} x^{(2)} - \begin{bmatrix}
+
−
  2  \\
+
−
  1
+
−
\end{bmatrix} = \begin{bmatrix}
+
−
  0  \\
+
−
  0
+
−
\end{bmatrix}</math>
+
−
 
+
−
<math>
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\text{When the gradient is 0, we reach the minimum point, which is } x^{(2)}=\begin{bmatrix}
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  3 \\
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  -1
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\end{bmatrix}
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</math>
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<br>
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'''Problem 3. (20pts) For the system of linear equations,<math> Ax=b </math> where  
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'''Problem 3. (20pts) For the system of linear equations,<math> Ax=b </math> where '''
  
 
<math>A = \begin{bmatrix}
 
<math>A = \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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'''Find the minimum length vector <math>x^{(\ast)}</math> that minimizes <math>\| Ax -b\|^{2}_2</math> '''
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+
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<br>
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 +
:'''Click [[QE2012_AC-3_ECE580-3|here]] to view [[QE2012_AC-3_ECE580-3|student answers and discussions]]'''
 +
----
 
'''Problem 4. (20pts) Use any simplex method to solve the following linear program. '''  
 
'''Problem 4. (20pts) Use any simplex method to solve the following linear program. '''  
  
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                         <math>x_1 \ge 0, x_2 \ge 0.</math>
 
                         <math>x_1 \ge 0, x_2 \ge 0.</math>
  
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<br>
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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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<br> '''Problem 5.(20pts) Solve the following problem:'''  
 
<br> '''Problem 5.(20pts) Solve the following problem:'''  
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<br> '''(ii)(10pts)Apply the SONC and SOSC to determine the nature of the critical points from the previous part.'''
 
<br> '''(ii)(10pts)Apply the SONC and SOSC to determine the nature of the critical points from the previous part.'''
 +
 +
:'''Click [[QE2012_AC-3_ECE580-5|here]] to view [[QE2012_AC-3_ECE580-5|student answers and discussions]]'''
 +
----
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[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]]

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

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