Propagation of Errors by 9c01PDQ

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									                 Propagation of Errors



                   Major: All Engineering Majors

             Authors: Autar Kaw, Matthew Emmons

         http://numericalmethods.eng.usf.edu
             Transforming Numerical Methods Education for STEM
                             Undergraduates

10/15/2012                http://numericalmethods.eng.usf.edu    1
     Propagation of Errors



http://numericalmethods.eng.usf.edu
            Propagation of Errors
    In numerical methods, the calculations are not
    made with exact numbers. How do these
    inaccuracies propagate through the calculations?




3                                   http://numericalmethods.eng.usf.edu
                          Example 1:
    Find the bounds for the propagation in adding two numbers. For example
    if one is calculating X +Y where
                        X = 1.5 ± 0.05
                        Y = 3.4 ± 0.04
    Solution
    Maximum possible value of X = 1.55 and Y = 3.44

    Maximum possible value of X + Y = 1.55 + 3.44 = 4.99

    Minimum possible value of X = 1.45 and Y = 3.36.

    Minimum possible value of X + Y = 1.45 + 3.36 = 4.81

    Hence
                     4.81 ≤ X + Y ≤4.99.



4                                             http://numericalmethods.eng.usf.edu
       Propagation of Errors In Formulas

    If f is a function of several variables X 1 , X 2 , X 3 ,......., X n1 , X n
    then the maximum possible value of the error in f is


           f          f                   f              f
     f       X 1       X 2  .......        X n1       X n
          X 1        X 2                 X n1          X n




5                                                    http://numericalmethods.eng.usf.edu
                      Example 2:
    The strain in an axial member of a square cross-
    section is given by
          F
        2
         h E
    Given
        F  72  0.9 N
        h  4  0.1 mm
        E  70  1.5 GPa

    Find the maximum possible error in the measured
    strain.

6                                    http://numericalmethods.eng.usf.edu
                 Example 2:
    Solution
                     72
      
         (4  10 3 ) 2 (70  10 9 )
        64.286  10 6
        64.286

                     
         F     h     E
         F      h      E

7                                  http://numericalmethods.eng.usf.edu
                         Example 2:
          1                   2F                    F
           2                    3                    2 2
       F h E                h   hE                E   h E
    Thus
            1      2F     F
       E  2 F  3 h  2 2 E
           h E    hE     h E
                          1                        2  72
                      3 2
                                     0.9        3 3
                                                                0.0001
                 (4 10 ) (70 10 )
                                 9
                                            (4 10 ) (70 10 )
                                                            9


                             72
                                            1.5 109
                   (4 103 ) 2 (70 109 ) 2
             5.3955
    Hence
            (64.286  5.3955 )
8                                                    http://numericalmethods.eng.usf.edu
                                Example 3:
    Subtraction of numbers that are nearly equal can create unwanted
    inaccuracies. Using the formula for error propagation, show that this is true.

    Solution
    Let
        z  x y
    Then       z      z
       z        x     y
               x      y
            (1)x  (1)y
            x  y
    So the relative change is
      z   x  y
         
       z    x y
9                                                  http://numericalmethods.eng.usf.edu
                       Example 3:
     For example if
        x  2  0.001
        y  2.003  0.001

      z   0.001  0.001
         
       z    | 2  2.003 |
          = 0.6667
          = 66.67%


10                            http://numericalmethods.eng.usf.edu
          Additional Resources
For all resources on this topic such as digital audiovisual
lectures, primers, textbook chapters, multiple-choice
tests, worksheets in MATLAB, MATHEMATICA, MathCad
and MAPLE, blogs, related physical problems, please
visit

http://numericalmethods.eng.usf.edu/topics/propagatio
n_of_errors.html
         THE END


http://numericalmethods.eng.usf.edu

								
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