Difference between revisions of "QE2012 AC-3 ECE580" - Rhea

Revision as of 05:31, 26 January 2013

AC - 3 August 2012 QE

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

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

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

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.

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 $\| Ax -b\|^{2}_2$

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

            $Maximize$      $x 1 + 2 x 2$ 
         $S'ubject to$      $-2x_1+x_2 \le 2$ 
                        $x_1-x_2 \ge -3$ 
                        $x_1 \le -3$ 
                        $x_1 \ge 0, x_2 \ge 0.$

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.

@@ Line 17: / Line 17: @@
 <br>
-<br> Solution:
-    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
- <math> 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, </math>
- we have
- <math> 1- \rho_{N-2} = \frac{F_{3}}{F_{4}}</math>     and so on.
-    Then, we have
- <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>
- Therefore, the reduction factor is
- <math>\frac{2}{F_{N+1}}</math>
-<br>
-Solution 2:
-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>
-<font color="#ff0000"><span style="font-size: 19px;"><math>\color{blue}
-\text{In this specific case, we use only N-1 iterations. }
-</math></span></font>
-<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
@@ Line 46: / Line 22: @@
 <math> 1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3}, </math>
-<br> Solution:
-       Final Range: 1.0; Initial Range: 20.
-       <math> \frac{2}{F_{N+1}} \le \frac{1.0}{20}</math>, or <math> F_{N+1} \ge 40</math>
-       So, <span class="texhtml">''N'' + 1 = 9</span>
-       Therefore, the minimal iterations is N-1 or 7.
-<br>
-Solution 2:
-Since the final range is 1 and the initial range is 20, we can say
-<math>
-\frac{2}{F_{N+1}} \le \frac{1.0}{20} \text{or equivalently } F_{N+1} \ge 40
-</math>
-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
-<font color="#0000FF "><span style="font-size: 19px;"><math>\color{blue}
-F_9=55 \text{ and } F_8=34
-</math></span></font>
-<br>
 Problem 2. (20pts) Employ the DFP method to construct a set of Q-conjugate directions using the function'''
@@ Line 88: / Line 40: @@
 <br>
-Solution:
-       <math>f = \frac{1}{2}x^TQx - x^Tb+c </math>
-    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 \\
-& 2
- \end{bmatrix} x^{(k)} - \begin{bmatrix}
-  \\
- \end{bmatrix}</math>
-       Hence
-    <math>g^{(0)} = \begin{bmatrix}
-  -2  \\
-  -1
- \end{bmatrix},</math>  <math>d^{(0)} = -H_0g^{(0)} =- \begin{bmatrix}
-& 0 \\
-& 1
- \end{bmatrix}\begin{bmatrix}
-  -2  \\
-  -1
- \end{bmatrix} = \begin{bmatrix}
-  \\
- \end{bmatrix}</math>
-<br>
-       Because f is a quadratic function
-       <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}
-  \\
- \end{bmatrix}}{\begin{bmatrix}
-& 1\end{bmatrix}\begin{bmatrix}
-& 1 \\
-& 2
- \end{bmatrix}\begin{bmatrix}
-  \\
- \end{bmatrix}} = \frac{1}{2}</math>
-       <math>x^{(1)} = x^{(0)} + \alpha d^{(0)} = \frac{1}{2} \begin{bmatrix}
-  \\
- \end{bmatrix} = \begin{bmatrix}
-  \\
-  \frac{1}{2}
- \end{bmatrix}</math>
-       <math>\Delta x^{(0)} = x^{(1)}- x^{(0)} = \begin{bmatrix}
-  \\
-  \frac{1}{2}
- \end{bmatrix}</math>
-    <math>g^{(1)} =\begin{bmatrix}
-& 1 \\
-& 2
- \end{bmatrix} x^{(1)} - \begin{bmatrix}
-  \\
- \end{bmatrix}= \begin{bmatrix}
-  -\frac{1}{2}  \\
- \end{bmatrix}</math>
-       <math>\Delta g^{(0)} = g^{(1)} - g^{(0)} = \begin{bmatrix}
-  -\frac{3}{2}  \\
- \end{bmatrix} </math>
-<br>
-       Observe that
-    <math>\Delta x^{(0)} \Delta x^{(0)^T} = \begin{bmatrix}
-  \\
-  \frac{1}{2}
- \end{bmatrix} \begin{bmatrix}
-  & \frac{1}{2}
- \end{bmatrix} = \begin{bmatrix}
-& \frac{1}{2}  \\
-  \frac{1}{2}  & \frac{1}{4}
- \end{bmatrix} </math>
-    <math> \Delta x^{(0)^T} \Delta g^{(0)} = \begin{bmatrix}
-  & \frac{1}{2}
- \end{bmatrix}\begin{bmatrix}
-  \frac{3}{2}   \\
- \end{bmatrix}  = \frac{5}{2}</math>
-    <math>H_0 \Delta g^{(0)} = \begin{bmatrix}
-& 0 \\
-& 1
- \end{bmatrix} \begin{bmatrix}
-  \frac{3}{2}   \\
- \end{bmatrix} = \begin{bmatrix}
-  \frac{3}{2}   \\
- \end{bmatrix},</math> <math>(H_0 \Delta g^{(0)})(H_0 \Delta g^{(0)})^T = \begin{bmatrix}
-  \frac{9}{4}  & 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}
-& 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}
-& 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 \\
-& 4
- \end{bmatrix} = \begin{bmatrix}
-  \frac{26}{25} & -\frac{7}{25} \\
-  -\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}  \\
- \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 \\
-& 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}
-  \\
-  \frac{1}{2}
- \end{bmatrix} + \frac{5}{2}\begin{bmatrix}
-  \frac{4}{5}  \\
-  -\frac{3}{5}
- \end{bmatrix} = \begin{bmatrix}
-  \\
-  -1
- \end{bmatrix} </math>
-       <math>\Delta x^{(1)} = x^{(2)} - x^{(1)} = \begin{bmatrix}
-  \\
-  -\frac{3}{2}
- \end{bmatrix}</math>
-    <math>g^{(2)} = \begin{bmatrix}
-& 1 \\
-& 2
- \end{bmatrix} x^{(0)} - \begin{bmatrix}
-  \\
- \end{bmatrix} = \begin{bmatrix}
-  \\
- \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}
-  \\
- \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 \\
-& 2
- \end{bmatrix} x^{(k)} - \begin{bmatrix}
-  \\
- \end{bmatrix} , \text{so}</math>
-<math>g^{(0)} = \begin{bmatrix}
-  -2  \\
-  -1
- \end{bmatrix},</math>  <math>d^{(0)} = -H_0 g^{(0)} =- \begin{bmatrix}
-& 0 \\
-& 1
- \end{bmatrix}\begin{bmatrix}
-  -2  \\
-  -1
- \end{bmatrix} = \begin{bmatrix}
-  \\
- \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}
-  \\
- \end{bmatrix}}{\begin{bmatrix}
-& 1\end{bmatrix}\begin{bmatrix}
-& 1 \\
-& 2
- \end{bmatrix}\begin{bmatrix}
-  \\
- \end{bmatrix}} = \frac{1}{2}</math>
-<math>x^{(1)} = x^{(0)} + \alpha d^{(0)} = \frac{1}{2} \begin{bmatrix}
-  \\
- \end{bmatrix} = \begin{bmatrix}
-  \\
-  \frac{1}{2}
- \end{bmatrix}</math>
-<math>\Delta x^{(0)} = x^{(1)}- x^{(0)} = \begin{bmatrix}
-  \\
-  \frac{1}{2}
- \end{bmatrix}</math>
-<math>g^{(1)} =\begin{bmatrix}
-& 1 \\
-& 2
- \end{bmatrix} x^{(1)} - \begin{bmatrix}
-  \\
- \end{bmatrix}= \begin{bmatrix}
-  -\frac{1}{2}  \\
- \end{bmatrix}</math>
-<math>\Delta g^{(0)} = g^{(1)} - g^{(0)} = \begin{bmatrix}
-  -\frac{3}{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}
-& 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 \\
-& 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}  \\
- \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 \\
-& 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}
-  \\
-  \frac{1}{2}
- \end{bmatrix} + \frac{5}{2}\begin{bmatrix}
-  \frac{4}{5}  \\
-  -\frac{3}{5}
- \end{bmatrix} = \begin{bmatrix}
-  \\
-  -1
- \end{bmatrix} </math>
-<math>\Delta x^{(1)} = x^{(2)} - x^{(1)} = \begin{bmatrix}
-  \\
-  -\frac{3}{2}
- \end{bmatrix}
-</math>
-<math>
-g^{(2)} = \begin{bmatrix}
-& 1 \\
-& 2
- \end{bmatrix} x^{(2)} - \begin{bmatrix}
-  \\
- \end{bmatrix} = \begin{bmatrix}
-  \\
- \end{bmatrix}</math>
-<math>
-\text{When the gradient is 0, we reach the minimum point, which is } x^{(2)}=\begin{bmatrix}
-\\
-  -1
- \end{bmatrix}
-</math>
-<br>
 '''Problem 3. (20pts) For the system of linear equations,<math> Ax=b </math> where

Difference between revisions of "QE2012 AC-3 ECE580" - Rhea

Revision as of 05:31, 26 January 2013

AC - 3 August 2012 QE

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