Propagation of Errors - Numerical Methods - Lecture Slides, Slides for Mathematical Methods for Numerical Analysis and Optimization. Ankit Institute of Technology and Science
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pankarithi17 April 2013

Propagation of Errors - Numerical Methods - Lecture Slides, Slides for Mathematical Methods for Numerical Analysis and Optimization. Ankit Institute of Technology and Science

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The main points are: Propagation of Errors, Numerical Methods, Maximum Possible Value, Errors in Formulas, Axial Member, Formula for Error Propagation, Subtraction of Numbers, Objective Function, Vector Operator, Initial...
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Propagation of Errors

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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 4.81 ≤ X + Y ≤4.99.

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Propagation of Errors In Formulas

f nn XXXXX ,,.......,,, 1321 − f

n n

n n

X X fX

X fX

X fX

X ff

∂ ∂

+∆ ∂ ∂

++∆ ∂ ∂

+∆ ∂ ∂

≈∆ − −

1 1

2 2

1 1

.......

If is a function of several variables then the maximum possible value of the error in is

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Example 2:

The strain in an axial member of a square cross- section is given by Given Find the maximum possible error in the measured strain.

Eh F 2∈=

N9.072 ±=F mm1.04 ±=h GPa5.170 ±=E

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Example 2:

)1070()104( 72

923 ×× ∈= −

610286.64 −×= µ286.64=

E E

h h

F F

∆ ∂ ∈∂

+∆ ∂ ∈∂

+∆ ∂ ∈∂

∈=∆

Solution

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Example 2:

EhF 2 1

= ∂ ∈∂

Eh F

h 3 2 −=

∂ ∈∂

22 Eh F

E −=

∂ ∈∂

E Eh

Fh Eh FF

Eh E ∆+∆+∆=∆ 2232

21

9 2923

933923

105.1 )1070()104(

72

0001.0 )1070()104(

7229.0 )1070()104(

1

×× ××

+

× ××

× +×

×× =

−−

µ3955.5=

Thus

Hence )3955.5286.64( µµ ±∈=

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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 Then So the relative change is

yxz −=

y y zx

x zz

∂ ∂ +∆

∂ ∂

=∆

yx ∆−+∆= )1()1( yx ∆+∆=

yx yx

z z

∆+∆ =

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Example 3:

For example if 001.02 ±=x

001.0003.2 ±=y

|003.22| 001.0001.0

+ =

z z

= 0.6667 = 66.67%

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