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| − | [[Category: | + | [[Category:ECE]] |
| + | [[Category:QE]] | ||
| + | [[Category:CNSIP]] | ||
| + | [[Category:problem solving]] | ||
| + | [[Category:automatic control]] | ||
| + | [[Category:optimization]] | ||
| − | =QE2013_AC-3_ECE580- | + | =QE2013_AC-3_ECE580-1= |
| + | :[[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]] | ||
| + | '''(i)''' | ||
| + | <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> | ||
| + | |||
| + | |||
| + | |||
| + | '''(ii)''' | ||
| + | <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 <math> 1.0 </math> and the initial range is <math> 20 </math>, we can say | ||
| + | <math> \frac{2}{F_{N+1}} \le \frac{1.0}{20} or equivalently F_{N+1}} \ge 40 </math> | ||
| + | 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 | ||
| + | |||
| + | |||
| + | ---- | ||
| + | ---- | ||
| + | <font face="serif"></font><math>\color{blue}\text{Related Problem: }</math> | ||
| + | <math> | ||
| + | \text{Find the final uncertainty range using the Fibonacci method after 6 iterations } | ||
| + | </math> | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
| + | |||
| + | |||
| + | '''Solution:''' | ||
| + | <math> | ||
| + | \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} | ||
| + | </math> | ||
[[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | [[ QE2013 AC-3 ECE580|Back to QE2013 AC-3 ECE580]] | ||
Revision as of 14:22, 23 January 2015
QE2013_AC-3_ECE580-1
(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} $
