Unformatted text preview: Chapter 01.06 Propagation of Errors If a calculation is made with numbers that are not exact, then the calculation itself will have an error. How do the errors in each individual number propagate through the calculations. Let’s look at the concept via some examples. Example 1 Find the bounds for the propagation error in adding two numbers. For example if one is calculating X + Y where X = 1.5 ± 0.05 , Y = 3.4 ± 0.04 . Solution By looking at the numbers, the maximum possible value of X and Y are X = 1.55 and Y = 3.44 Hence X + Y = 1.55 + 3.44 = 4.99 is the maximum value of X + Y . The minimum possible value of X and Y are X = 1.45 and Y = 3.36 . Hence X + Y = 1.45 + 3.36 = 4.81 is the minimum value of X + Y . Hence 4.81 ≤ X + Y ≤ 4.99. One can find similar intervals of the bound for the other arithmetic operations of X − Y , X * Y , and X / Y . What if the evaluations we are making are function evaluations instead? How do we find the value of the propagation error in such cases. 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 01.06.1 01.06.2 Chapter 01.06 Example 2 The strain in an axial member of a square crosssection is given by F ∈= 2 hE where F =axial force in the member, N h = length or width of the crosssection, m E =Young’s modulus, Pa 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. Solution 72 ∈= −3 2 (4 × 10 ) (70 × 10 9 )
= 64.286 × 10 −6 = 64.286µ ∂∈ ∂∈ ∂∈ ∆ ∈= ∆F + ∆h + ∆E ∂F ∂h ∂E ∂∈ 1 =2 ∂F h E ∂∈ 2F =− 3 ∂h hE ∂∈ F =− 2 2 ∂E hE 1 2F F ∆E = 2 ∆F + 3 ∆h + 2 2 ∆E hE hE hE = 1 2 × 72 × 0 .9 + × 0.0001 9 −3 3 (4 × 10 ) (70 × 10 ) (4 × 10 ) (70 × 10 9 )
−3 2 + 72 × 1.5 × 10 9 92 (4 × 10 ) (70 × 10 )
−3 2 = 8.0357 × 10 −7 + 3.2143 × 10 −6 + 1.3776 × 10 −6 = 5.3955 × 10 −6 = 5.3955µ Hence ∈= (64.286 µ ± 5.3955µ ) implying that the axial strain, ∈ is between 58.8905µ and 69.6815µ Propagation of Errors Example 3 01.06.3 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
z = x− y ∆z = ∂z ∂z ∆x + ∆y ∂x ∂y = (1)∆x + (−1)∆y = ∆x + ∆y So the absolute relative change is ∆x + ∆y ∆z = z x− y As x and y become close to each other, the denominator becomes small and hence create large relative errors. 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%
INTRODUCTION TO NUMERICAL METHODS Topic Propagation of Errors Summary Textbook notes on how errors propagate in arithmetic and function evaluations Major All Majors of Engineering Authors Autar Kaw Last Revised May 1, 2011 Web Site http://numericalmethods.eng.usf.edu ...
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 Spring '07
 Gallivan
 Numerical Analysis, Normal Distribution, Observational error, maximum possible value, Engineering Authors Autar Kaw

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