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Linear Equations. PSCI 702 September 21, 2005. Numerical Errors. Round off Error Truncation Error. Linear Algebraic Equations. Linear Algebraic Equations. In Matrix Format Solution:. Cramer’s Rule. M ij is the determinant of the Matrix A with the ith row and jth column removed.

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## Linear Equations

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**Linear Equations**PSCI 702 September 21, 2005**Numerical Errors**• Round off Error • Truncation Error**Linear Algebraic Equations**• In Matrix Format • Solution:**Cramer’s Rule**• Mij is the determinant of the Matrix A with the ith row and jth column removed. • (-1)ij is called the cofactor of element aij.**Cramer’s Rule**• 3n2 operations for the determinant. • 3n3 operations for every unknown. • Unstable for large Matrices. • Large error propagation. • Good for small Matrices (n<20).**Gaussian Elimination**• Divide each row by the leading element. • Subtract row 1 from all other rows. • Move to the second row an continue the process.**Gaussian Elimination**• Back substitution**Gaussian Elimination**• Division by zero: May occur in the forward elimination steps. • Round-off error: Prone to round-off errors.**Gaussian Elimination**Consider the system of equations: Use five significant figures with chopping = At the end of Forward Elimination =**Gaussian Elimination**Back Substitution**Gaussian Elimination**Compare the calculated values with the exact solution**Improvements**Increase the number of significant digits Decreases round off error Does not avoid division by zero Gaussian Elimination with Partial Pivoting Avoids division by zero Reduces round off error**Partial Pivoting**Gaussian Elimination with partial pivoting applies row switching to normal Gaussian Elimination. How? At the beginning of the kth step of forward elimination, find the maximum of If the maximum of the values is In the pth row, then switch rows p and k.**Partial Pivoting**What does it Mean? Gaussian Elimination with Partial Pivoting ensures that each step of Forward Elimination is performed with the pivoting element |akk| having the largest absolute value.**Partial Pivoting: Example**Consider the system of equations In matrix form = Solve using Gaussian Elimination with Partial Pivoting using five significant digits with chopping**Partial Pivoting: Example**Forward Elimination: Step 1 Examining the values of the first column |10|, |-3|, and |5| or 10, 3, and 5 The largest absolute value is 10, which means, to follow the rules of Partial Pivoting, we switch row1 with row1. Performing Forward Elimination**Partial Pivoting: Example**Forward Elimination: Step 2 Examining the values of the first column |-0.001| and |2.5| or 0.0001 and 2.5 The largest absolute value is 2.5, so row 2 is switched with row 3 Performing the row swap**Partial Pivoting: Example**Forward Elimination: Step 2 Performing the Forward Elimination results in:**Partial Pivoting: Example**Back Substitution Solving the equations through back substitution**Partial Pivoting: Example**Compare the calculated and exact solution The fact that they are equal is coincidence, but it does illustrate the advantage of Partial Pivoting**Gauss Jordan Elimination**• Start with the following system of Matrices. • Divide each row by the leading element. • Subtract row 1 from all other rows. • In addition to subtracting the line whose diagonal term has been made unity from all those bellow it, also subtract from the equations above it as well.**Matrix Factorization**• Assume A can be written as A=VU where V and U are triangular Matrices.**Matrix Factorization**Proof If solving a set of linear equations If Then Multiply by Which gives Remember which leads to Now, if then Now, let Which ends with and (1) (2)**Matrix Factorization**How can this be used? Given Decompose into and Then solve for And then solve for**Matrix Factorization**How is this better or faster than Gauss Elimination? Let’s look at computational time. n = number of equations To decompose [A], time is proportional to To solve and time is proportional to**Matrix Factorization**Therefore, total computational time for LU Decomposition is proportional to or Gauss Elimination computation time is proportional to How is this better?**Matrix Factorization**What about a situation where the [C] vector changes? In VU factorization, VU decomposition of [A] is independent of the [C] vector, therefore it only needs to be done once. Let m = the number of times the [C] vector changes The computational times are proportional to VU factorization = Gauss Elimination= Consider a 100 equation set with 50 right hand side vectors VU factorization = Gauss Elimination =**Matrix Factorization**Another Advantage Finding the Inverse of a Matrix VU Factorization Gauss Elimination For large values of n**Matrix Factorization**Method: Decompose [A] to [V] and [U] [U] is the same as the coefficient matrix at the end of the forward elimination step. [V] is obtained using the multipliers that were used in the forward elimination process**Matrix Factorization**Finding the [U] matrix Using the Forward Elimination Procedure of Gauss Elimination**Matrix Factorization**Finding the [U] matrix Using the Forward Elimination Procedure of Gauss Elimination**Matrix Factorization**Finding the [V] matrix Using the multipliers used during the Forward Elimination Procedure From the first step of forward elimination From the second step of forward elimination**Matrix Factorization**Does ?**Matrix Factorization**Example: Solving simultaneous linear equations using VU factorization Solve the following set of linear equations using VU Factorization Using the procedure for finding the [V] and [U] matrices**Matrix Factorization**Example: Solving simultaneous linear equations using VU factorization Complete the forward substitution to solve for**Matrix Factorization**Example: Solving simultaneous linear equations using VU factorization Set Solve for The 3 equations become**Matrix Factorization**Example: Solving simultaneous linear equations using VU factorization From the 3rd equation Substituting in a3 and using the second equation**Matrix Factorization**Example: Solving simultaneous linear equations using VU factorization Substituting in a3 and a2 using the first equation Hence the Solution Vector is:**Gauss Method**An iterative method. • Basic Procedure: • Algebraically solve each linear equation for xi • Assume an initial guess solution array • Solve for each xi and repeat • Use absolute relative approximate error after each iteration to check if error is within a prespecified tolerance.**Gauss Method**Why? The Gauss Method allows the user to control round-off error. Elimination methods such as Gaussian Elimination and VU Factorization are prone to round-off error. Also: If the physics of the problem is understood, a close initial guess can be made, decreasing the number of iterations needed.**Gauss Method**Algorithm A set of n equations and n unknowns: If: the diagonal elements are non-zero Rewrite each equation solving for the corresponding unknown ex: First equation, solve for x1 Second equation, solve for x2 . . . . . .**Gauss Method**Algorithm Rewriting each equation From Equation 1 From equation 2 From equation n-1 From equation n**Gauss Method**Algorithm General Form of each equation**Gauss Method**Gauss Algorithm General Form for any row ‘i’ How or where can this equation be used?

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