Propagation of Errors
Major: All Engineering Majors
Authors: Autar Kaw, Matthew Emmons
http://numericalmethods.eng.usf.edu
Numerical Methods for STEM undergraduates
7/12/2016
http://numericalmethods.eng.usf.edu
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Propagation of Errors
In numerical methods, the calculations are not
made with exact numbers. How do these
inaccuracies propagate through the calculations?
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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
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4.81 ≤ X + Y ≤4.99.
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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
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Example 2:
The strain in an axial member of a square crosssection 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.
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Example 2:
Solution
72

3 2
9
(4  10 ) (70  10 )
 64.286  10 6
 64.286 



 
F 
h 
E
F
h
E
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Example 2:

1
 2
F h E

2F
 3
h
h E

F
 2 2
E
h E
Thus
1
2F
F
E  2 F  3 h  2 2 E
h E
hE
h E

1
2  72

0
.
9

 0.0001
3 2
9
3 3
9
(4 10 ) (70 10 )
(4 10 ) (70 10 )

Hence
72
9

1
.
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
10
(4 103 ) 2 (70 109 ) 2
 5.3955
 (64.286   5.3955 )
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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
x  y
z

z
x y
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Example 3:
For example if
x  2  0.001
y  2.003  0.001
0.001  0.001
z

z
| 2  2.003 |
= 0.6667
= 66.67%
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