QE2012_AC-3_ECE580-1

Part 1,2,3,4,5

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}} $

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

Back to QE2012 AC-3 ECE580