Revision as of 17:17, 26 January 2013

QE2012_AC-3_ECE580-1

Part 1,2,3,4,5

(i)
Solution:

   The reduction factor is (1 − ρ1)(1 − ρ2)(1 − ρ3)...(1 − ρN − 1)
  Since 
  $  1- \rho_{N-1} = \frac{F_{2}}{F_{3}} = \frac{2}{3},  $
  we have 
  $  1- \rho_{N-2} = \frac{F_{3}}{F_{4}} $     and so on.

   Then, we have
  $  (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}} $
  Therefore, the reduction factor is
  $ \frac{2}{F_{N+1}} $

Solution 2:

The uncertainty interval is reduced by $ (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}} $

(ii)
Solution:

      Final Range: 1.0; Initial Range: 20.

      $  \frac{2}{F_{N+1}} \le \frac{1.0}{20} $, or $  F_{N+1} \ge 40 $ 

      So, N + 1 = 9

      Therefore, the minimal iterations is N-1 or 7.

Solution 2:

Since the final range is $ 1.0 $ and the initial range is $ 20 $, we can say $ \frac{2}{F_{N+1}} \le \frac{1.0}{20} or equivalently F_{N+1}} \ge 40 $ From the inequality, we get $ F_{N+1} \ge 40 , so N+1=9 $. Therefore the minimum number of iteration is N-1=7

$ \color{blue}\text{Related Problem: For function } $

        $ f\left( x_{1},x_{2} \right) =\frac{1}{3} x_{1}^{3} + \frac{1}{3} x_{2}^{3} -x_{1}x_{2} $

$ \color{blue} \text{Find point(s) that satisfy FONC and check if they are strict local minimizers.} $

$ \color{blue}\text{Solution:} $

$ \text{Applying FONC gives } \nabla f\left ( x \right )=\begin{bmatrix} x_{1}^{2}-x_{2}\\ x_{2}^{2}-x_{1} \end{bmatrix}=0 $

        $ \Rightarrow x^{\left ( 1 \right )}=\begin{bmatrix} 0\\ 0 \end{bmatrix} \text{ and }x^{\left ( 2 \right )}=\begin{bmatrix} 1\\ 1 \end{bmatrix} $

$ \text{The Hessian matrix: } F\left ( x \right )=\begin{bmatrix} 2x_{1} & -1\\ -1 & 2x_{2} \end{bmatrix} $

        $ \text{The matrix } F\left ( x^{\left ( 1 \right )} \right )=\begin{bmatrix} 0 & -1\\ -1 & 0 \end{bmatrix} \text{ is indefinite. The point is not a minimizer.} $

        $ \text{The matrix } F\left ( x^{\left ( 2\right )} \right )=\begin{bmatrix} 0 & -1\\ -1 & 0 \end{bmatrix} \text{ is positive definite. } $

$ \therefore x^{\left ( 2 \right )}=\begin{bmatrix} 1\\ 1 \end{bmatrix} \text{ satisfies SOSC to be a strict local minimizer.} $

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