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Numerical Analysis
CC413
Propagation of Errors
Propagation of Errors
In numerical methods, the calculations
are not made with exact numbers. How
do these inaccuracies propagate through
the calculations?
2
Source of errors




Rounding
Truncation
Inherent
Mistakes
Underflow and Overflow
Numbers occurring in calculations
that have a magnitude less than 2 1023 .(1+2 -52) result in underflow
and are generally set to zero.
 Numbers greater than 2 1024 .(2-2 -52)
result in overflow.

Significant Figures

Number of significant figures indicates
precision. Significant digits of a number
are those that can be used with confidence,
e.g., the number of certain digits plus one
estimated digit.
53,800 How many significant figures?
5.38 x 104
3
5.380 x 104
4
5.3800 x 104
5
Significant Figures

Zeros are sometimes used to locate the decimal point not
significant figures.
0.00001753
4
0.0001753
4
0.001753
4
 N = 3.141592653589793232 then N =
0.14159265358979232 x 10+1
t= 4
N = 0.1415 x 10+1
Error Definitions



Numerical error - use of approximations
to represent exact mathematical
operations and quantities
true value = approximation + error
Error = true value - approximation
•

Error = |N – n| Absolut error
Relative error = error / true value
•
R.E. = |N-n|/|N| = |N-n|/|n|
true error
t 
 100
true value
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.
Example 2:
The strain in an axial member of a square cross-section is
given by
Given
F
 2
h E
F  72  0.9 N
h  4  0.1 mm
E  70  1.5 GPa
Find the maximum possible error in the measured strain.
Example 2:
Solution
72

3 2
9
(4  10 ) (70  10 )
 64.286  10 6
 64.286 
10
http://numericalmethods.eng.usf.edu
Example
Consider a problem where the true answer is
7.91712. If you report the value as 7.92, answer
the following questions.
1. What is the true error?
2. What is the relative error?
Example

Determine the absolute and relative
errors when approximating p by p∗
when
Solution


This example shows that the same relative error,
0.3333×10−1, occurs for widely varying absolute errors.
the absolute error can be misleading and the relative
error more meaningful, because the relative error takes
into consideration the size of the value.
Error Definitions cont.
Round off error – Symmetric
rounding originate from the fact that
computers retain only a fixed
number of significant figures: y =
0.73248261 to be 0.7325
 Error = y-round(y) = 0.00001739
and relative error = error /y = 0.000024

Error Definitions cont.
 Truncation errors – Chopping
errors that result from using an
approximation in place of an
exact mathematical procedure: y
= 0.73248261 to be 0.7324
 Error = 0.0008261 and relative
error = 0.00011
Example
Determine the five-digit (a) chopping and (b) rounding
values of the irrational number π.
 Solution The number π has an infinite decimal expansion
of the form
π = 3.14159265. . . .
 Written in normalized decimal form, we have
π = 0.314159265 . . . × 10.
(a) The floating-point form of π using five-digit chopping is
f l(π) = 0.31415 × 10 = 3.1415.
(b) The sixth digit of the decimal expansion of π is a 9, so
the floating-point form of π using five-digit rounding is
f l(π) = (0.31415 + 0.00001) × 10 = 3.1416.

Example


Suppose that x = 57 and y = 13. Use five-digit
chopping for calculating x + y, x − y, x × y, and x ÷
y.
Solution: Note that

Solution (Cont.)


Use absolute value.
Computations are repeated until stopping criterion is
satisfied.
 a  s

Pre-specified % tolerance
based on the knowledge of
your solution
If the following criterion is met
 s  (0.5 10 (2-n) )%
you can be sure that the result is correct to at least n
significant figures.
Round-off Errors (Error Bounds Analysis)
z    0.a1a2  an an1    e ,
a1  0
  1 (sign )   base e  exponent

  (0.a1a2  an )    e

fl ( z )  

e


[(
0
.
a
a

a
)

(
0
.
00
...
01
)
]


1 2
n 



 

 n
0  a n 1 

2
Round down

2
 a n 1  
Round up
fl(z) is the rounded value of z
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Chopping Errors (Error Bounds Analysis)
Suppose the mantissa can only support n digits.
z  0.a1a2 an an 1an 2    e ,
a1  0
fl( z )  0.a1a2 an    e
Thus the absolute and relative chopping errors are
e
en
z  fl( z )  (0.00
...
0
a
a

)



(
0
.
a
a

)


n 1 n  2

 n 1 n 2 
n zeroes
(0.00...0an 1an 2 )    e
z  fl( z )

z
(0.a1a2 an an 1an 2 )    e
Suppose ß = 10 (base 10), what are the values of ai such
that the errors are the largest?
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Chopping Errors (Error Bounds Analysis)
Because
0.an1an2an3   1
z  fl( z )  0.an 1an  2    en   en
z  fl( z)
  en
z  fl( z )
0.00...0an 1an  2    e

z
0.a1a2  an an 1an  2    e
 en

0.a1a2  an an 1an  2    e
 en

e
0.100000
a
a



 n 1 n  2
n digits
 en
 en
e  n ( e 1)
1 n






0.1   e  1   e
z  fl ( z )
z
  1 n
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Round-off Errors (Error Bounds Analysis)
Absolute error of fl(z)
When rounding down
z  fl( z )
z  fl( z )
an 1
   0.000an 1an  2 an 3    e
   0.an 1an  2 an 3    en
 0.an 1an  2 an 3    en

1
  (.an 1 )  
2
2
z  fl ( z )
1 en
 
2
z  fl ( z )
1 en
 
2
Similarly, when rounding up
i.e., when

2
 an 1  
23
Round-off Errors (Error Bounds Analysis)
Relative error of fl(z)
1 en
z  fl ( z ) 

2
z  fl ( z )
1  n

z
2 z  e
1
 n

because z  (.a1a2 )    e
2 (.a1a2 ) 
1  n

because (.a1 )   (0.1) 
2 (.1) 
1  n

z  fl ( z )
1 1 n
2  1


z
2
24
Summary of Error Bounds Analysis
Chopping Errors
Round-off errors
Absolute
z  fl( z)  
1 en
z  fl( z )  
2
Relative
z  fl( z )
  1n
z
β
n
e n
z  fl( z ) 1 1n
 
z
2
base
# of significant digits or # of digits in the mantissa
Regardless of chopping or round-off is used to round the
numbers, the absolute errors may increase as the
numbers grow in magnitude but the relative errors are
bounded by the same magnitude.
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Numerical Stability


Rounding errors may accumulate and
propagate unstably in a bad algorithm.
Can be proven that for Gaussian
elimination the accumulated error is
bounded
Ill-conditioned problem







If small error in data input produce large
error in solution
When |f’(x)| is large
1.0 x +3.5 y = 8
2.1 x + 7.0 y = 16.1
Sol is (1.0, 2.0) if Changed to
2.12x + 7.0 y = 16.1
Sol is (0.8333, 2.048)
Ill-conditioned system of
equations

A small deviation in the entries of A
matrix, causes a large deviation in the
solution.
x1  1
2   x1   3 

 1

 x   1
0.48 0.99  x   1.47

 2  

 2  
2   x1   3 
 1
0.49 0.99  x   1.47 

 2  

28
 x1  3
 x   0 
 2  
Machine Epsilon
Relative chopping error
z  fl( z )
z
Relative round-off error
z  fl( z )
z
eps is known as the
machine epsilon – the
smallest number such that
epsilon = 1;
while (1 + epsilon > 1)
epsilon = epsilon / 2;
epsilon = epsilon * 2;
1 + eps > 1
  1n  epsChopping

1 1n
  epsRound-off
2
Algorithm to compute machine epsilon
29
Exercise
Discuss to what extent
(a + b)c = ac + bc
is violated in machine arithmetic.
30
Error Propagation
Let xfl refer to the floating point representation of the real number x.
 Since computer has fixed word length, there is a difference between x and xfl
(round-off error)
and we would like to estimate the error in the calculation of f(x) :

f ( x fl )  f ( x)  f ( x fl )
• Both x and f(x) are unknown.
• If xfl is close to x, then we can use first order Taylor expansion and compute:
f ( x)  f ( x fl )  f ( x fl )( x  x fl )
f ( x fl )  f ( x fl ) * x
Result: If f’(xfl) and x are known, then we can estimate the error using this formula