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Gaussian Elimination

Gaussian Elimination. Industrial Engineering Majors Author(s): Autar Kaw http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates. Naïve Gauss Elimination http://numericalmethods.eng.usf.edu. Naïve Gaussian Elimination.

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Gaussian Elimination

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  1. Gaussian Elimination Industrial Engineering Majors Author(s): Autar Kaw http://numericalmethods.eng.usf.edu Transforming Numerical Methods Education for STEM Undergraduates

  2. Naïve Gauss Eliminationhttp://numericalmethods.eng.usf.edu

  3. Naïve Gaussian Elimination A method to solve simultaneous linear equations of the form [A][X]=[C] Two steps 1. Forward Elimination 2. Back Substitution

  4. Forward Elimination The goal of forward elimination is to transform the coefficient matrix into an upper triangular matrix

  5. Forward Elimination A set of n equations and n unknowns . . . . . . (n-1) steps of forward elimination

  6. Forward Elimination Step 1 For Equation 2, divide Equation 1 by and multiply by .

  7. Forward Elimination Subtract the result from Equation 2. − _________________________________________________ or

  8. Forward Elimination Repeat this procedure for the remaining equations to reduce the set of equations as . . . . . . . . . End of Step 1

  9. Forward Elimination Step 2 Repeat the same procedure for the 3rd term of Equation 3. . . . . . . End of Step 2

  10. Forward Elimination At the end of (n-1) Forward Elimination steps, the system of equations will look like . . . . . . End of Step (n-1)

  11. Matrix Form at End of Forward Elimination

  12. Back Substitution Solve each equation starting from the last equation Example of a system of 3 equations

  13. Back Substitution Starting Eqns . . . . . .

  14. Back Substitution Start with the last equation because it has only one unknown

  15. Back Substitution

  16. THE END http://numericalmethods.eng.usf.edu

  17. Naïve Gauss EliminationExamplehttp://numericalmethods.eng.usf.edu

  18. Example: Production Optimization To find the number of toys a company should manufacture per day to optimally use their injection-molding machine and the assembly line, one needs to solve the following set of equations. The unknowns are the number of toys for boys, x1, number of toys for girls, x2, and the number of unisexual toys, x3. Find the values of x1, x2,and x3 using Naïve Gauss Elimination.

  19. Example: Production Optimization Forward Elimination: Step 1 Yields

  20. Example: Production Optimization Forward Elimination: Step 1 Yields

  21. Example: Production Optimization Forward Elimination: Step 2 Yields This is now ready for Back Substitution

  22. Example: Production Optimization Back Substitution: Solve for x3 using the third equation

  23. Example: Production Optimization Back Substitution: Solve for x2 using the second equation

  24. Example: Production Optimization Back Substitution:Solve for x1 using the first equation

  25. Example: Production Optimization Solution: The solution vector is 1440 toys for boys should be produced 1512 toys for girls should be produced 36 unisexual toys should be produced

  26. THE END http://numericalmethods.eng.usf.edu

  27. Naïve Gauss EliminationPitfallshttp://numericalmethods.eng.usf.edu

  28. Pitfall#1. Division by zero

  29. Is division by zero an issue here?

  30. Is division by zero an issue here? YES Division by zero is a possibility at any step of forward elimination

  31. Pitfall#2. Large Round-off Errors Exact Solution

  32. Pitfall#2. Large Round-off Errors Solve it on a computer using 6 significant digits with chopping

  33. Pitfall#2. Large Round-off Errors Solve it on a computer using 5 significant digits with chopping Is there a way to reduce the round off error?

  34. Avoiding Pitfalls • Increase the number of significant digits • Decreases round-off error • Does not avoid division by zero

  35. Avoiding Pitfalls • Gaussian Elimination with Partial Pivoting • Avoids division by zero • Reduces round off error

  36. THE END http://numericalmethods.eng.usf.edu

  37. Gauss Elimination with Partial Pivotinghttp://numericalmethods.eng.usf.edu

  38. Pitfalls of Naïve Gauss Elimination • Possible division by zero • Large round-off errors

  39. Avoiding Pitfalls • Increase the number of significant digits • Decreases round-off error • Does not avoid division by zero

  40. Avoiding Pitfalls • Gaussian Elimination with Partial Pivoting • Avoids division by zero • Reduces round off error

  41. What is Different About Partial Pivoting? At the beginning of the kth step of forward elimination, find the maximum of If the maximum of the values is in the p th row, then switch rows p and k.

  42. Matrix Form at Beginning of 2nd Step of Forward Elimination

  43. Example (2nd step of FE) Which two rows would you switch?

  44. Example (2nd step of FE) Switched Rows

  45. Gaussian Elimination with Partial Pivoting A method to solve simultaneous linear equations of the form [A][X]=[C] Two steps 1. Forward Elimination 2. Back Substitution

  46. Forward Elimination Same as naïve Gauss elimination method except that we switch rows before each of the (n-1) steps of forward elimination.

  47. Example: Matrix Form at Beginning of 2nd Step of Forward Elimination

  48. Matrix Form at End of Forward Elimination

  49. Back Substitution Starting Eqns . . . . . .

  50. Back Substitution

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