# 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

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
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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