Package com.imsl.test.example.math

Class SuperLUEx1

java.lang.Object

com.imsl.test.example.math.SuperLUEx1

public class SuperLUEx1
extends Object

SuperLU Example 1: Computes the LU factorization of a sparse matrix.

The LU Factorization of the sparse \(6 \times 6\) matrix $$ A=\begin{pmatrix} 10.0 & 0.0 & 0.0 & 0.0 & 0.0 & 0.0 \\ 0.0 & 10.0 & -3.0 & -1.0 & 0.0 & 0.0 \\ 0.0 & 0.0 & 15.0 & 0.0 & 0.0 & 0.0 \\ -2.0 & 0.0 & 0.0 & 10.0 & -1.0 & 0.0 \\ -1.0 & 0.0 & 0.0 & -5.0 & 1.0 & -3.0 \\ -1.0 & -2.0 & 0.0 & 0.0 & 0.0 & 6.0 \end{pmatrix} $$

is computed. The sparse coordinate form for \(A\) is given by a series of row, column, and value triplets:

row	column	value
\(0\)	\(0\)	\(10.0\)
\(1\)	\(1\)	\(10.0\)
\(1\)	\(2\)	\(-3.0\)
\(1\)	\(3\)	\(-1.0\)
\(2\)	\(2\)	\(15.0\)
\(3\)	\(0\)	\(-2.0\)
\(3\)	\(3\)	\(10.0\)
\(3\)	\(4\)	\(-1.0\)
\(4\)	\(0\)	\(-1.0\)
\(4\)	\(3\)	\(-5.0\)
\(4\)	\(4\)	\(1.0\)
\(4\)	\(5\)	\(-3.0\)
\(5\)	\(0\)	\(-1.0\)
\(5\)	\(1\)	\(-2.0\)
\(5\)	\(5\)	\(6.0\)

Let \(y^T = (1.0, 2.0, 3.0, 4.0, 5.0, 6.0)\). Then we have \(b_1 := Ay = {(10.0, 7.0, 45.0, 33.0, -34.0, 31.0)}^T\) and \(b_2 := A^Ty = {(-9.0,8.0,39.0, 13.0,1.0, 21.0)}^T \).

In the example, the The LU factorization of \(A\) is used to solve the sparse linear system $$\begin{array}{cc} Ax &=& b_1 \\ A^Tx &=& b_2 \end{array}$$

with iterative refinement. The reciprocal pivot growth factor and the reciprocal condition number are also computed. Note that by construction \(x=y\) is the solution to this system.

See Also:

• Code
• Output

Constructor Summary

Constructors

Constructor	Description
SuperLUEx1()

Method Summary

Modifier and Type	Method	Description
static void	main(String[] args)

Methods inherited from class java.lang.Object

clone, equals, finalize, getClass, hashCode, notify, notifyAll, toString, wait, wait, wait

Constructor Details

SuperLUEx1

public SuperLUEx1()

Method Details

main

public static void main(String[] args)
                 throws Exception

Throws:

Exception