# Young Modulus Example The pairs (1,1), (2,2), (4,3) represent strain (millistrains) and stress (ksi) measurements. Estimate Young modulus using the three.

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Young Modulus Example The pairs (1,1), (2,2), (4,3) represent strain (millistrains) and stress (ksi) measurements. Estimate Young modulus using the three commonly used error norms. Estimate the error in Young modulus using cross validation Our model is y=Ex, where x is strain and y is stress

Best fits Data provides y(1)=1, y(2)=2, y(4)=3. With y=Ex then the First determine range of reasonable moduli: Then the errors are Sum of absolute errors – Must take minimum value of E=0.75 Sum of square of errors – E=17/21=0.81 For max error – E=5/6=0.83

Sketch of fits Average absolute error has linear objective so it ends at the extreme range. Max equated the two largest errors and rms is, as expected, in between.

Cross validation for average absolute error Cross validation gives us error in prediction not coefficients If at point x the error in y is e, we will consider it equivalent to an error in Young modulus of e/x y(1)=1, y(2)=2, y(4)=3. If we leave out the first point, Error at x=1 is ¼, which is also error in E. If we leave out second point Error at x=2 is ½, so error in E is again ¼. If we omit the third point, obviously E=1, error at point is 1, error in modulus is ¼. So average cross validation error is ¼.

Cross validation for average absolute error Cross validation gives us error in prediction not coefficients If at point x the error in y is e, we will consider it equivalent to an error in Young modulus of e/x y(1)=1, y(2)=2, y(4)=3. If we leave out the first point, Error at x=1 is ¼, which is also error in E. If we leave out second point Error at x=2 is ½, so error in E is again ¼. If we omit the third point, obviously E=1, error at point is 1, error in modulus is ¼. So average cross validation error is ¼.

Cross validation for max error y(1)=1, y(2)=2, y(4)=3. When we omit first point, need e 2 =-e 3, E=5/6, error at point one is 1/6. When we omit second point, e 1 =-e 3, E=4/5, error at point 2 is 2/5, error in E is 1/5. When we omit third point obviously E=1, error at point 3 is 1, error in E is ¼. Maximum error is ¼. Note that the average absolute error is 0.206. When we minimized average absolute error was 0.25. How come? I leave the rms case to you.

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