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CSE 551 Computational Methods 2018/2019 Fall Chapter 7 Systems of Linear Equations. Outline. Naive Gaussian Elimination Gaussian Elimination with Scaled Partial Pivoting Tridiagonal and Banded Systems Matix Factorizations. References.
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CSE 551 Computational Methods 2018/2019 Fall Chapter 7 Systems of Linear Equations
Outline Naive Gaussian Elimination Gaussian Elimination with Scaled Partial Pivoting Tridiagonal and Banded Systems Matix Factorizations
References • W. Cheney, D Kincaid, Numerical Mathematics and Computing, 6ed, • Chapter 7 • Chapter 8
Naive Gaussian Elimination A Larger Numerical Example Algorithm Pseudocode Testing the Pseudocode Residual and Error Vectors
Naive Gaussian Elimination • fundamental problems in many scientific and engineering applications is to • solve an algebraic linear system Ax = b • for the unknown vector x • the coefficient matrix A • and right-hand side vector b • known • arise – various applications; • approximating nonlinear equations by linear equations • or differential equations by algebraic equations • cornerstone of many numerical methods • the efficient and accurate solution of linear • systems • The system of linear algebraic equations Ax = b • may or may not have a solution • if it has a solution, it may or may not be unique.
Gaussian elimination – standard method for solving the linear system • When the system has no solution • other approaches, e.g., linear least squares • assume that: • the coefficient matrix A is n × n and • invertible (nonsingular) • pure mathematics - solution of Ax = b : x =A−1b • A−1:inverse matrix. • most applications - solve the system directly for the unknown vector x
many applications, daunting task: • to solve accurately extremely large systems - thousands or millions of unknowns • Some questions: • How to store such large systems in the computer? • Are computed answers correct? • What is the precision of the computed results? • Can the algorithm fail? • How long will it take to compute answers? • What is the asymptotic operation count of the algorithm? • Will the algorithm be unstable for certain systems? • Can instability be controlled by pivoting? (Permuting the order of the rows of the matrix is called pivoting.) • Which strategy of pivoting should be used? • Whether the matrix is ill-conditioned? • whether the answers are accurate?
Gaussian elimination transforms a linear system into an upper triangular form • easier to solve • process equivalent to finding the factorization • A = LU • L: unit lower triangular matrix • U: upper triangular matrix • This • useful solving many linear systems • the same coefficient matrix • different right-hand sides
When A - special structure: • symmetric, positive definite, triangular, banded, block, or sparse • Gaussian elimination with partial pivoting • modified or rewritten specifically for th system.
develop a good program for solving a system of n • linear equations in n unknowns: • In compact form:
A Larger Numerical Example • simplest form of Gaussian elimination - naive • not suitable for automatic computation • essential modifications • e.g., naive Gaussian elimination:
first step • elimination procedure: • certain multiples of the first equation • subtracted from the second, third, and fourth equations • eliminate x1 from these equations. • to create 0’s as coefficients for each x1 below the first • 12, 3, −6 stand • subtract 2 times the first equation from the second • multiplier: 12/6 . • subtract 12 times the first equation from the third • multiplier:3/6 • subtract −1 times the first equation from the fourth.
the first equation not altered pivot equation • Systems (2) and (3) are equivalent: • Any solution of (2) is also a solution of (3)
second step • ignore the first equation and the first column of coefficients • system of three equations with three unknowns same process repeated: • the top equation in the smaller system current pivot equation • subtracting 3 times the second equation from the third. • multiplier: −12/(-4) • subtract −1/2 times the second equation from the fourth
final step: • subtracting 2 times the third equation from the fourth: • system is in upper triangular form • equivalent to System (2)
completes first phase (forward elimination) • The second phase (back substitution) • solve System (5) for the unknowns starting at the • bottom • from the fourth equation: • Putting x4 = 1 in the third equation: • The solution:
Algorithm • write System (1) in matrix-vector form • ai j: n × n square array, or matrix • unknowns xi • right-hand side elements bi:n×1 arrays, or vectors • or
Operations between equations correspond to operations between rows in this notation • naive Gaussian elimination algorithm for the general system, • n equations and n unknowns • original data are overwritten
forward elimination phase • n − 1 principal steps • first step: first equation to produce n − 1 • zeros as coefficients for each x1 in all other equations i, i >1 • subtracting appropriate multiples of the first equation from the others • first equation - first pivot equation • a11 as the first pivot element
remaining equations (2 i n): • ← : replacement • content of the memory location allocated to ai jreplaced by ai j− (ai 1/a1 1)a1 j, and so on. • (ai1/a11) multipliers • new coefficient of x1 in the ith equation:0 • ai1 − (ai1/a11)a11 = 0.
not alter first equation • not alter coefficients x1 • a multiplier times 0 subtracted from 0 is still 0 • ignore the first row and the first column • repeat the process on the smaller system. • With the second equation - pivot equation: • compute for each remaining equation (3 i n)
prior to the kth step, the system: • a wedge of 0 coefficients created • first k equations processed - now fixed • kth equation - pivot equation • select multipliers - create 0’s coefficients xibelow the akkcoefficient.
compute for each remaining equation • (k + 1i n) • assume that all the divisors in this algorithm are nonzero
Pseudocode • pseudocode for forward elimination • coefficient array: (ai j ) • right-hand side of the system of equations: (bi ) • solution computed: (xi )
multiplier aik/akknot depend on j • moved outside the j loop • the new values in column k will be 0-theoretically, when j = k:
The location where the 0 created good place to store multiplie :
At the beginning of the back substitution phase, • the linear system: • the ai j’s and bi’s not original ones from System (6) • altered by the elimination process
The back substitution starts by solving the nth equation for xn: • using the (n − 1)th solve for xn−1: • continue upward, recovering each xi:
crucial computation: • a triply nested for-loop containing a replacement operation:
expect all quantities to be infected with roundoff error • Such a roundoff error • in ak jmultiplied by the factor (aik/akk) • large: pivot element |akk| small relative to |aik|. • conclude, tentatively, that • small pivot elements lead to • large multipliers and to worse roundoff errors.
Testing the Pseudocode • One way to test a procedure: • set up an artificial problem • solution is known beforehand • test problem - parameter • changed to vary the difficulty. • example: • Fixing n:
coefficients all equal to 1 • try to recover known coefficients • values of p(t) at integers t = 1+i for i = 1, 2, . . . , n • coefficients: x1, x2, . . . , xn: • used formula sum of a geometric series on the right-hand side: • linear system
Example • write a pseudocode for a specific test case that solves the system of Equation (8) for various values of n. • Solution: use Naive Gauss • calling program: n = 4, 5, 6, 7, 8, 9, 10 • for testing
run on a machine approximately seven decimal digits of accuracy • solution with complete precision until n reached 9, • then, the computed solution worthless • one component exhibited a relative error of 16,120%!
The coefficient matrix for this linear system • example of a well-known illconditioned • matrix the Vandermonde matrix • cannot be solved accurately - naive Gaussian elimination • amazing • the trouble happens so suddenly! n: 9 • roundoff error - computing xipropagated and magnified throughout the back substitution phase • most computed values for xiworthless • Insert intermediate print statements • to see what is going on hereRes
Residual and Error Vectors • linear system Ax = b - true solution x • and a computed solution • define: • relationship between the error vector and the residual vector:
using different computers solve the same linear system, Ax = b • algorithm and precision not known • two solutions xaand xbtotally different! • How determine which, if either, computed solution is correct?
check the solutions substituting them into the original system, • computing the residual vectors: ra= Axa− b and rb= Axb− b • computed solutions not exact • some roundoff errors • accept the solution with the smaller residual vector. • However, if we knew • the exact solution x, then we would just compare the computed solutions with the exact • solution, which is the same as computing the error vectors ea= x −xaand eb= x −xb • computed solution - smaller error vector • most assuredly be the better answer
exact solution not known • accept • the computed solution - smaller residual vector • may not be the best computed solution • original problem sensitive to roundoff errors • illconditioned. • the question: • whether a computed solution to a linear system is a good solution is extremely difficult
Gaussian Elimination with Scaled Partial Pivoting Naive Gaussian Elimination Can Fail Partial Pivoting and Complete Partial Pivoting Gaussian Elimination with Scaled Partial Pivoting A Larger Numerical Example Pseudocode Long Operation Count Numerical Stability Scaling
Naive Gaussian Elimination Can Fail • naive Gaussian elimination algorith - unsatisfactory: • the algorithm fails if a11 = 0. • If fails for some values of the data • untrustworthy - values near the failing values: • ε small number different from 0
naive algorithm works • after forward elimination: • back substitution: • ε−1 large - computer – fixed word length • small ε, (2 − ε−1),(1 − ε−1) computed as −ε−1
8-digit decimal machine with a 16-digit accumulator, when ε =10−9 • ε−1 = 109 • subtract -interpret the number: • ε−1−2computed: 0.09999 99998 00000 0 × 1010 rounded to 0.10000 000×1010 = ε−1
for values of ε sufficiently close to 0 • computer calculates x2:1 x1: 0 • correct solution: • relative error x1: 100%
naive Gaussian elimination algorithm works well on Systems (1) and (2) • if the equations are first permuted: x2 = 1, x1 = 2 - x2=1
second systems: • forward elimination and back substitution: • not to rearrange the equations • select a different pivot row • difficulty in (2): • not ε small but small relative to other coefficients in the same row
