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[[Category:QE2013 AC-3 ECE580]][[Category:QE2013 AC-3 ECE580]][[Category:QE2013 AC-3 ECE580]][[Category:QE2013 AC-3 ECE580]][[Category:QE2013 AC-3 ECE580]][[Category:QE2013 AC-3 ECE580]]
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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]]
  
=QE2013_AC-3_ECE580-3=
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=QE2013_AC-3_ECE580-1=
  
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:[[QE2013_AC-3_ECE580-1|Part 1]],[[QE2013_AC-3_ECE580-2|2]],[[QE2013_AC-3_ECE580-3|3]],[[QE2013_AC-3_ECE580-4|4]],[[QE2013_AC-3_ECE580-5|5]]
  
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'''(i)'''
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<br> '''Solution: '''
  
Put your content here . . .
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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>
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  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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'''(ii)'''
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<br> '''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, <span class="texhtml">''N'' + 1 = 9</span>
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      Therefore, the minimal iterations is N-1 or 7.
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<br>
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'''Solution 2:'''
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Since the final range is <math> 1.0 </math> and the initial range is <math> 20 </math>, we can say
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<math> \frac{2}{F_{N+1}} \le \frac{1.0}{20}  or equivalently F_{N+1}} \ge 40 </math>
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From the inequality, we get <math> F_{N+1} \ge 40 , so N+1=9 </math>. Therefore the minimum number of iteration is N-1=7
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----
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----
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<font face="serif"></font><math>\color{blue}\text{Related Problem: }</math>
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<math>
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\text{Find the final uncertainty range using the Fibonacci method after 6 iterations }
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</math>
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<math>
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\text{Assume the last step has the form: }
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1-\rho_{N-1}=\frac{F_2}{F_3}=\frac{2}{3}
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\text{The initial range is 10}
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</math>
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'''Solution:'''
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<math>
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\text{The reduction factor is } \frac{F_{2}}{F_{7+1}} =\frac{1}{34}
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\text{, So the final range is }
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10 \frac{F_{2}}{F_{7+1}}= \frac{5}{17}
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</math>
  
  
 
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Revision as of 14:22, 23 January 2015


QE2013_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: } $ $ \text{Find the final uncertainty range using the Fibonacci method after 6 iterations } $ $ \text{Assume the last step has the form: } 1-\rho_{N-1}=\frac{F_2}{F_3}=\frac{2}{3} \text{The initial range is 10} $


Solution: $ \text{The reduction factor is } \frac{F_{2}}{F_{7+1}} =\frac{1}{34} \text{, So the final range is } 10 \frac{F_{2}}{F_{7+1}}= \frac{5}{17} $


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