1. Engineering Computation
Part 3
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2. Learning Objectives for Lecture
1. Motivate Study of Systems of Equations and particularly Systems of Linear Equations
2. Review steps of Gaussian Elimination
3. Examine how roundoff error can enter and
be magnified in Gaussian Elimination
4. Introduce Pivoting and Scaling as defenses against roundoff.
5. Consider what an engineer can do to generate well formulated problems.
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3. Systems of Equations
In Part 2 we have tried to determine the value x, satisfying f(x)=0. In this part we try to obtain the values x1,x2, xn, satisfying the system of equations:
These systems can be linear or nonlinear, but in this part we deal with linear systems:
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4. Systems of Equations
where a and b are constant coefficients, and n is the number of equations.
Many of the engineering fundamental equations are based on conservation laws. In mathematical terms, these principles lead to balance or continuity equations relating the system behavior with respect to the amount of the magnitude being modelled and the extrenal stimuli acting on the system.
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5. Systems of Equations Column 3 Row 2
Matrices are rectangular sets of elements represented by a single symbol. If the set if horizontal it is called row, and if it is vertical, it is called column.
Column 3
Row 2
Column
vector
Row vector
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6. Systems of Equations There are some special types of matrices:
Identity matrix
Symmetric matrix
Diagonal matrix
Upper triangular matrix
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7. Lower triangular matrix
Systems of Equations
Lower triangular matrix
Banded matrix
Half band width
All elements are null with the exception of thoise in a band centered around the main diagonal. This matrix has a band width of 3 and has the name of tridiagonal.
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8. Systems of Equations Linear Algebraic Equations
a11x1 + a12x2 + a13x3 + … + a1nxn = b1
a21x1 + a22x2 + a23x3 + … + a2nxn = b2
…..
an1x1 + an2x2 + an3x3 + … + anxn = bn
where all aij's and bi's are constants.
In matrix form:
n x n n x 1 n x 1
or simply [A]{x} = {b}
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9. Systems of Equations Matrix product: Matrix representation of a system
Resulting dimensions
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10. Systems of Equations
Graphic Solution: Systems of equations are hyperplanes (straight lines, planes, etc.). The solution of a system is the intersection of these hyperplanes.
Compatible and determined system. Vectors are linearly independent. Unique solution. Determinant of A is non-null.
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11. Systems of Equations
Incompatible system, Linearly dependent vectors. Null determinant of A. There is no solution.
Compatible but undetermined system. Linearly dependent vectors. Null determinant of A. There exists an infinite number of solutions.
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12. Systems of Equations
Compatible and determined system. Linearly independent vectors. Nonnull determinant of A, but close to zero. There exists a solution but it is difficult to find precisely. It is an ill conditioned system leading to numerical errors.
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13. Gauss elimination
Naive Gauss elimination method: The Gauss’ method has two phases: Forward elimination and backsustitution. In the first, the system is reduced to an upper triangular system:
First, the unknown x1 is eliminated. To this end, the first row is multiplied by -a21/a11 and added to the second row. The same is done with all other succesive rows (n-1 times) until only the first equation contains the first unknown x1.
Pivot
equation
substract
pivot
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14. Gauss elimination
This operation is repeated with all variables xi, until an upper triangular matrix is obtained.
Next, the system is solved by backsustitution.
The number of operations (FLOPS) used in the Gauss method is:
Pass 1
Pass 2
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15. Gauss elimination 1. Forward Elimination (Row Manipulation):
a. Form augmented matrix [A|b]:
b. By elementary row manipulations, reduce [A|b] to [U|b'] where U is an upper triangular matrix:
DO i = 1 to n-1
DO k = i+1 to n
Row(k) = Row(k) - (aki/aii)*Row(i)
ENDDO
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16. Gauss elimination 2. Back Substitution
Solve the upper triangular system [U]{x} = {b´}
xn = b'n / unn
DO i = n-1 to 1 by (-1)
END
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17. Gauss elimination (example)
Consider the system of equations
To 2 significant figures, the exact solution is:
We will use 2 decimal digit arithmetic with rounding.
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18. Gauss elimination (example)
Start with the augmented matrix:
Multiply the first row by –1/50
and add to second row.
Multiply the first row by –2/50
and add to third row:
Multiply the second row by –6/40
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19. Gauss elimination (example)
Now backsolve:
(vs , et = 2.2%)
(vs , et = 2.5%)
(vs , et = 0%)
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20. Gauss elimination (example)
Consider an alternative solution
interchanging rows:
After forward elimination, we obtain:
Now backsolve:
x3 = (vs , et = 4.4%)
x2 = (vs , et = 50%)
x1 = (vs , et = 100%)
Apparently, the order of the equations matters!
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21. Gauss elimination (example)
WHAT HAPPENED?
When we used 50 x1 + 1 x2 + 2 x3 = 1 to solve for x1, there was little change in other equations.
When we used 2 x1 + 6 x x3 = 3 to solve for x1 it made BIG changes in the other equations. Some coefficients for other equations were lost!
The second equation has little to do with x1.
It has mainly to do with x3.
As a result we obtained LARGE numbers in the table, significant roundoff error occurred and information was lost.
Things didn't go well!
If scaling factors | aji / aii | are  1 then the effect of roundoff errors is diminished.
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22. Gauss elimination (example)
Effect of diagonal dominance:
As a first approximation roots are:
xi  bi / aii
Consider the previous examples:
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23. Gauss elimination (example)
Goals: 1. Best accuracy (i.e. minimize error)
2. Parsimony (i.e. minimize effort)
Possible Problems:
A. Zero on diagonal term  ÷ by zero.
B. Many floating point operations (flops) cause numerical precision problems and propagation of errors.
C. System may be ill-conditioned: det[A]  0.
D. No solution or an infinite # of solutions: det[A] = 0.
Possible Remedies:
A. Carry more significant figures (double precision).
B. Pivot when the diagonal is close to zero.
C. Scale to reduce round-off error.
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24. Gauss elimination (pivoting)
A. Row pivoting (Partial Pivoting) -
In any good routine, at each step i, find
maxk | aki | for k = i, i+1, i+2, ..., n
Move corresponding row to pivot position.
(i) Avoids zero aii
(ii) Keeps numbers small & minimizes round-off,
(iii) Uses an equation with large | aki | to find xi
Maintains diagonal dominance.
Row pivoting does not affect the order of the variables.
Included in any good Gaussian Elimination routine.
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25. Gauss elimination (pivoting)
B. Column pivoting -
Reorder remaining variables xj
for j = i, ,n so get largest | aji |
Column pivoting changes the order of the unknowns, xi, and thus leads to complexity in the algorithm. Not usually done.
C. Complete or Full pivoting
Performing both row pivoting and column pivoting.
(If [A] is symmetric, needed to preserve symmetry.)
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26. Gauss elimination (pivoting)
How to fool pivoting:
Multiply the third equation by 100 and then performing pivoting will yield:
Forward elimination then yields (2-digit arithmetic):
Backsolution yields:
x3 = (vs , et = 4.4%)
x2 = (vs , et = 50.0%)
x1 = (vs , et = 100%)
The order of the rows is still poor!!
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27. Gauss elimination (scaling)
A. Express all equations (and variables) in comparable units so all elements of [A] are about the same size.
B. If that fails, and maxj |aij| varies widely across the rows, replace each row i by:
aij 
This makes the largest coefficient |aij| of each equation equal to 1 and the largest element of [A] equal to 1 or -1
NOTE: Routines generally do not scale automatically; scaling can cause round-off error too!
SOLUTIONS
• Don't actually scale, but use hypothetical scaling factors to determine what pivoting is necessary.
• Scale only by powers of 2: no roundoff or division required.
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28. Gauss elimination (scaling)
How to fool scaling:
A poor choice of units can undermine the value of scaling.
Begin with our original example:
If the units of x1 were expressed in µg instead of mg the matrix might read:
Scaling then yields:
Which equation is used to determine x1 ? Why bother to scale ?
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29. Gauss elimination (operation counting)
In numerical scientific calculations, the number of multiplies & divides often determines CPU time. (This represents the numerical effort!)
One floating point multiply or divide (plus any associated adds or subtracts) is called a FLOP. (The adds/subtracts use little time compared to the multiplies/divides.)
FLOP = FLoating point OPeration.
Examples: a * x + b
a / x – b
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30. Gauss elimination (operation counting)
Useful identities in counting FLOPS:
O(mn) means that there are terms of order mn and lower.
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31. Gauss elimination (operation counting)
Simple Example of Operation Counting:
DO i = 1 to n
Y(i) = X(i)/i – 1
ENDDO
X(i) and Y(i) are arrays whose values change when i changes. In each iteration
X(i)/i – 1
represents one FLOP because it requires one division (& one subtraction).
The DO loop extends over i from 1 to n iterations:
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32. Gauss elimination (operation counting)
Another Example of Operation Counting:
DO i = 1 to n
Y(i) = X(i) X(i) + 1
DO j = i to n
Z(j) = [ Y(j) / X(i) ] Y(j) + X(i)
ENDDO
With nested loops, always start from the innermost loop.
[Y(j)/X(i)] * Y(j) + X(i) represents 2 FLOPS
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33. Gauss elimination (operation counting)
For the outer i-loop: X(i) • X(i) + 1 represents 1 FLOP
= 3n +2n2 - n2 - n
= n2 + 2n = n2 + O(n)
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34. Gauss elimination (operation counting)
Forward Elimination:
DO k = 1 to n–1
DO i = k+1 to n
r = A(i,k)/A(k,k)
DO j = k+1 to n
A(i,j)=A(i,j) – r*A(k,j)
ENDDO
B(i) = B(i) – r*B(k)
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35. Gauss elimination (operation counting)
Operation Counting for Gaussian Elimination
Back Substitution:
X(n) = B(n)/A(n,n)
DO i = n–1 to 1 by –1
SUM = 0
DO j = i+1 to n
SUM = SUM + A(i,j)*X(j)
ENDDO
X(i) = [B(i) – SUM]/A(i,i)
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36. Gauss elimination (operation counting)
Operation Counting for Gaussian Elimination
Forward Elimination
Inner loop:
Second loop:
= (n2 + 2n) – 2(n + 1)k + k2
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37. Gauss elimination (operation counting)
Operation Counting for Gaussian Elimination
Forward Elimination (cont'd)
Outer loop =
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38. Gauss elimination (operation counting)
Operation Counting for Gaussian Elimination
Back Substitution
Inner Loop:
Outer Loop:
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39. Gauss elimination (operation counting)
Total flops = Forward Elimination + Back Substitution
= n3/3 + O (n2) + n2/ O (n)
 n3/3 + O (n2)
To convert (A,b) to (U,b') requires n3/3, plus terms of order n2 and smaller, flops.
To back solve requires:
n = n (n+1) / 2 flops;
Grand Total: the entire effort requires n3/3 + O(n2) flops altogether.
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40. Gauss-Jordan Elimination
Diagonalization by both forward and backward elimination in each column.
Perform elimination both backwards and forwards until:
Operation count for Gauss-Jordan is:
(slower than Gauss elimination)
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41. Gauss-Jordan Elimination
Example (two-digit arithmetic):
x1 = (vs , et = 6.3%)
x2 = (vs , et = 0%)
x3 = (vs , et = 2.2%)
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42. Gauss-Jordan Matrix Inversion
The solution of: [A]{x} = {b} is: {x} = [A]-1{b}
where [A]-1 is the inverse matrix of [A]
Consider: [A] [A]-1 = [ I ]
1) Create the augmented matrix: [ A | I ]
2) Apply Gauss-Jordan elimination:
==> [ I | A-1 ]
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43. Gauss-Jordan Matrix Inversion
Gauss-Jordan Matrix Inversion (with 2 digit arithmetic):
MATRIX INVERSE [A-1]
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44. Gauss-Jordan Matrix Inversion
CHECK:
[ A ] [ A ]-1 = [ I ]
[ A ]-1 { b } = { x }
Gaussian
Elimination
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45. LU decomposition
LU decomposition - The LU decomposition is a method that uses the elimination techniques to transform the matrix A in a product of triangular matrices. This is specially useful to solve systems with different vectors b, because the same decomposition of matrix A can be used to evaluate in an efficient form, by forward and backward sustitution, all cases.
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46. LU decomposition Initial system Decomposition Transformed system 1
Substitution
Transformed system 2
Backward sustitution
Forward sustitution
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47. LU decomposition
LU decomposition is very much related to Gauss method, because the upper triangular matrix is also looked for in the LU decomposition. Thus, only the lower triangular matrix is needed.
Surprisingly, during the Gauss elimination procedure, this matrix L is obtained, but one is not aware of this fact. The factors we use to get zeroes below the main diagonal are the elements of this matrix L.
Substract
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48. LU decomposition
resto
resto
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49. LU decomposition (Complexity)
Basic Approach
Consider [A]{x} = {b}
a) Gauss-type "decomposition" of [A] into [L][U] n3/3 flops
[A]{x} = {b} becomes [L] [U]{x} = {b}; let [U]{x}  {d}
b) First solve [L] {d} = {b} for {d} by forward subst n2/2 flops
c) Then solve [U]{x} = {d} for {x} by back substitution n2/2 flops
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50. LU Decompostion: notation
[A] = [L] + [U0]
[A] = [L0] + [U]
[A] = [L0] + [U0] + [D]
[A] = [L1] [U]
[A] = [L] [U1]
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51. LU decomposition LU Decomposition Variations
Doolittle [L1][U] General [A]
Crout [L][U1] General [A]
Cholesky [L][L] T Pos. Def. Symmetric [A]
Cholesky works only for Positive Definite Symmetric matrices
Doolittle versus Crout:
• Doolittle just stores Gaussian elimination factors where Crout uses a different series of calculations (see C&C ).
• Both decompose [A] into [L] and [U] in n3/3 FLOPS
• Different location of diagonal of 1's
• Crout uses each element of [A] only once so the same array can be used for [A] and [L\U] saving computer memory!
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52. Definition of a matrix inverse:
LU decomposition
Matrix Inversion
Definition of a matrix inverse:
[A] [A]-1 = [ I ]
==> [A] {x} = {b}
[A]-1 {b} = {x}
First Rule: Don’t do it.
(numerically unstable calculation)
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53. LU decomposition Matrix Inversion If you really must --
1) Gaussian elimination: [A | I ] –> [U | B'] ==> A-1
2) Gauss-Jordan: [A | I ] ==> [I | A-1 ]
Inversion will take
n3 + O(n2)
flops if one is careful about where zeros are (taking advantage of the sparseness of the matrix)
Naive applications (without optimization) take 4n3/3 + O(n2) flops. For example, LU decomposition requires n3/3 + O(n2) flops. Back solving twice with n unit vectors ei:
2 n (n2/2) = n3 flops.
Altogether: n3/3 + n3 = 4n3/3 + O(n2) flops
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54. FLOP Counts for Linear Algebraic Equations
Summary
FLOP Counts for Linear Algebraic Equations, [A]{x} = {b}
Gaussian Elimination (1 r.h.s) n3/3 + O (n2)
Gauss-Jordan (1 r.h.s) n3/2 + O (n2)
LU decomposition n3/3 + O (n2)
Each extra LU right-hand-side n2
Cholesky decomposition (symmetric A) n3/6 + O (n2)
Inversion (naive Gauss-Jordan) 4n3/3 +O (n2)
Inversion (optimal Gauss-Jordan) n3 + O (n2)
Solution by Cramer's Rule n!
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55. Errors in Solutions to Systems of Linear Equations
System of Equations
Errors in Solutions to Systems of Linear Equations
Objective: Solve [A]{x} = {b}
Problem:
Round-off errors may accumulate and even be exaggerated by the solution procedure. Errors are often exaggerated if the system is ill-conditioned
Possible remedies to minimize this effect:
1. Partial or complete pivoting
2. Work in double precision
3. Transform the problem into an equivalent system of linear equations by scaling or equilibrating
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56. Errors in Solutions to Systems of Linear Equations
Ill-conditioning
A system of equations is singular if det|A| = 0
If a system of equations is nearly singular it is ill-conditioned.
Systems which are ill-conditioned are extremely sensitive to small changes in coefficients of [A] and {b}. These systems are inherently sensitive to round-off errors.
Question:
Can we develop a means for detecting these situations?
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57. Errors in Solutions to Systems of Linear Equations
Ill-conditioning of [A]{x} = {b}:
Consider the graphical interpretation for a 2-equation system:
We can plot the two linear equations on a graph of x1 vs. x2. x1 x2
b2/a22
b1/a11
b1/a12
a11x1+ a12x2 = b1
a21x1+ a22x2 = b2
b2/a21
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58. Errors in Solutions to Systems of Linear Equations
Ill-conditioning of [A]{x} = {b}:
Consider the graphical interpretation for a 2-equation system:
We can plot the two linear equations on a graph of x1 vs. x2. x1 x1 x2 x2
Uncertainty
in x2
Uncertainty
in x2
Well-conditioned
Ill-conditioned
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59. Errors in Solutions to Systems of Linear Equations
Ways to detect ill-conditioning:
1. Calculate {x}, make small change in [A] or {b} and determine change in solution {x}.
2. After forward elimination, examine diagonal of upper triangular matrix. If aii << ajj, i.e. there is a relatively small value on diagonal, then this may indicate ill-conditioning.
3. Compare {x}single with {x}double
4. Estimate "condition number" for A.
Substituting the calculated {x} into [A]{x} and checking this against {b} will not always work!!!
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60. Errors in Solutions to Systems of Linear Equations
Ways to detect ill-conditioning:
If det|A| = 0 the matrix is singular
==> the determinant may be an indicator of conditioning
If det|A| is near zero is the matrix ill-conditioned?
Consider:
After scaling:
==> det|A| will provide an estimate of conditioning if it is normalized by the "magnitude" of the matrix.
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61. Norms Norms and the Condition Number
We need a quantitative measure of ill-conditioning.
This measure will then directly reflect the possible magnitude of round off effects.
To do this we need to understand norms:
Norm: Scalar measure of the magnitude of a matrix or vector ("how big" a vector is).
Not to be confused with the dimension of a matrix.
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62. Vector Norms Vector Norms: Scalar measure of the magnitude of a vector
Here are some vector norms for n x 1 vectors {x} with typical elements xi.
Each is in the general form of a p norm defined by the general relationship:
1. Sum of the magnitudes:
2. Magnitude of largest element:
(infinity norm)
3. Length or Euclidean norm:
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63. Norms Vector Norms Required Properties of vector norm:
1. ||x||  0 and ||x|| = 0 if and only if [x]=0
2 ||kx|| = k ||x|| where k is any positive scalar
3. ||x+y||  ||x|| + ||y|| Triangle Inequality
For the Euclidean vector norm we also have
4. ||x•y||  ||x|| ||y||
because the dot product or inner product property satisfies:
||xy|| = ||x||•||y|| |cos()|  ||x|| • ||y||.
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64. Matrix Norms
Matrix Norms: Scalar measure of the magnitude of a matrix.
Matrix norms corresponding to vector norms above are defined by the general relationship:
1. Largest column sum: (column sum norm)
2. Largest row sum:
(row sum norm)
(infinity norm)
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65. Matrix norms 3. Spectral norm: ||A|| 2 = (µmax)1/2
where µmax is the largest eigenvalue of [A]T[A]
If [A] is symmetric, (µmax)1/2 = max , is the largest eigenvalue of [A].
(Note: this is not the same as the Euclidean or Frobenius norm, seldom used:
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66. Matrix norms Matrix Norms
For matrix norms to be useful we require that
0. || Ax ||  || A || ||x ||
General properties of any matrix norm:
1. || A ||  0 and || A || = 0 iff [A] = 0
2. || k A || = k || A || where k is any positive scalar
3. || A + B ||  || A || + || B || "Triangle Inequality"
4. || A B ||  || A || || B ||
Why are norms important?
Norms permit us to express the accuracy of the solution {x} in terms of || x ||
Norms allow us to bound the magnitude of the product [ A ] {x} and the associated errors.
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67. Error Analysis
Forward and backward error analysis can estimate the effect of truncation and roundoff errors on the precision of a result. The two approaches are alternative views:
Forward (a priori) error analysis tries to trace the accumulation of error through each process of the algorithm, comparing the calculated and exact values at every stage.
Backward (a posteriori) error analysis views the final solution as the exact solution to a perturbed problem. One can consider how different the perturbed problem is from the original problem.
Here we use the condition number of a matrix [A] to specify the amount by which relative errors in [A] and/or {b} due to input, truncation, and rounding can be amplified by the linear system in the computation of {x}.
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68. Error Analysis
Backward Error Analysis of [A]{x} = {b} for errors in {b}
Suppose the coefficients {b} are not precisely represented. What might be the effect on the calculated value for {x + dx}?
Lemma: [A]{x} = {b} yields ||A|| ||x||  ||b|| or
Now an error in {b} yields a corresponding error in {x}:
[A ]{x + dx} = {b + db}
[A]{x} + [A]{ dx} = {b} + {db}
Subtracting [A]{x} = {b} yields:
[A]{dx} = {db} ––> {dx} = [A]-1{db}
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69. Error Analysis
Backward Error Analysis of [A]{x} = {b} for errors in {b}
Taking norms we have:
And using the lemma:
we then have :
Define the condition number as
k = cond [A]  ||A-1|| ||A||  1
If k  1 or k is small, the system is well-conditioned
If k >> 1, system is ill conditioned.
1 = || I || = || A-1A ||  || A-1 || || A || = k = Cond(A)
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70. Error Analysis
Backward Error Analysis of [A]{x} = {b} for errors in [A]
If the coefficients in [A] are not precisely represented, what might be effect on the calculated value of {x+ dx}?
[A + dA ]{x + dx} = {b}
[A]{x} + [A]{ dx} + [dA]{x+dx} = {b}
Subtracting [A]{x} = {b} yields:
[A]{ dx} = – [dA]{x+dx} or
{dx} = – [A]-1 [dA] {x+dx}
Taking norms and multiplying by || A || / || A || yields :
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71. E. T. S. I. Caminos, Canales y Puertos
Linfield & Penny, 1999

72. E. T. S. I. Caminos, Canales y Puertos
Linfield & Penny, 1999

73. Error Analysis Estimate of Loss of Significance:
Consider the possible impact of errors [dA] on the precision of {x}.
implies that if
Or, taking log of both sides: s > p - log10()
log10() is the loss in decimal precision; i.e., we start with p decimal figures and end-up with s decimal figures.
It is not always necessary to find [A]-1 to estimate k = cond[A]. Instead, use an estimate based upon iteration of inverse matrix using LU decomposition.
E. T. S. I. Caminos, Canales y Puertos

74. Iterative Solution Methods
Impetus for Iterative Schemes:
1. May be more rapid if coefficient matrix is "sparse"
2. May be more economical with respect to memory
3. May also be applied to solve nonlinear systems
Disadvantages:
1. May not converge or may converge slowly
2. Not appropriate for all systems
Error bounds apply to solutions obtained by direct and iterative methods because they address the specification of [dA] and {db}.
E. T. S. I. Caminos, Canales y Puertos

75. Iterative Solution Methods
Basic Mechanics:
Starting with:
a11x1 + a12x2 + a13x a1nxn = b1
a21x1 + a22x2 + a23x a2nxn = b2
a31x1 + a32x2 + a33x a3nxn = b3
: :
an1x1 + an2x2 + an3x annxn = bn
Solve each equation for one variable:
x1 = [b1 – (a12x2 + a13x a1nxn )} / a11
x2 = [b2 – (a21x1 + a23x a2nxn )} / a22
x3 = [b3 – (a31x1 + a32x a3nxn )} / a33 :
xn = [bn – (an1x2 + an2x an,n-1xn-1 )} / ann
E. T. S. I. Caminos, Canales y Puertos

76. Iterative Solution Methods
Start with initial estimate of {x}0.
Substitute into the right-hand side of all the equations.
Generate new approximation {x}1.
This is a multivariate one-point iteration: {x}j+1 = {g({x}j)}
Repeat process until the maximum number of iterations reached or until:
|| xj+1 – xj ||  d + e || xj+1 ||
E. T. S. I. Caminos, Canales y Puertos

77. Convergence To solve [A]{x} = {b}
Separate [A] into: [A] = [Lo] + [D] + [Uo]
[D] = diagonal (aii)
[Lo] = lower triangular with 0's on diagonal
[Uo] = upper triangular with 0's on diagonal
Rewrite system:
[A]{x} = ( [Lo] + [D] + [Uo] ){x} = {b}
[D]{x} + ( [Lo] + [Uo] ){x} = {b}
Iterate:
[D]{x}j+1 = {b} – ( [Lo]+[Uo] ) {x}j
{x}j+1 = [D]-1{b} – [D]-1 ( [Lo]+[Uo] ) {x}j
Iterations converge if:
|| [D]-1 ( [Lo] + [Uo] ) || < 1
(sufficient if equations are diagonally dominant)
E. T. S. I. Caminos, Canales y Puertos

78. Iterative Solution Methods – the Jacobi Method
E. T. S. I. Caminos, Canales y Puertos

79. Iterative Solution Methods -- Gauss-Seidel
In most cases using the newest values within the right-hand side equations will provide better estimates of the next value. If this is done, then we are using the Gauss-Seidel Method:
( [Lo]+[D] ){x}j+1 = {b} – [Uo] {x}j
or explicitly:
If this is done, then we are using the Gauss-Seidel Method
E. T. S. I. Caminos, Canales y Puertos

80. Iterative Solution Methods -- Gauss-Seidel
If either method is going to converge,
Gauss-Seidel will converge
faster than Jacobi.
Why use Jacobi at all?
Because you can separate the n-equations into n independent tasks, it is very well suited computers with parallel processors.
E. T. S. I. Caminos, Canales y Puertos

81. Convergence of Iterative Solution Methods
Rewrite given system: [A]{x} = { [B] + [E] } {x} = {b}
where [B] is diagonal, or triangular so we can solve [B]{y} = {g} quickly. Thus,
[B] {x}j+1= {b}– [E] {x}j
which is effectively: {x}j+1 = [B]-1 ({b} – [E] {x}j )
True solution {x}c satisfies: {x}c = [B]-1 ({b} – [E] {x}c)
Subtracting yields: {x}c – {x}j+1= – [B]-1 [E] [{x}c – {x}j]
So ||{x}c – {x}j+1 ||  ||[B]-1 [E]|| ||{x}c – {x}j ||
Iterations converge linearly if || [B]-1 [E] || < 1 => || ([D] + [Lo])-1 [Uo] || < 1 For Gauss-Seidel
=> || [D] -1 ([Lo] + [Uo]) || < 1 For Jacobi
E. T. S. I. Caminos, Canales y Puertos

82. Convergence of Iterative Solution Methods
Iterative methods will not converge for all systems of equations, nor for all possible rearrangements.
If the system is diagonally dominant,
i.e., | aii | > | aij | where i  j then
with all < 1.0, i.e., small slopes.
E. T. S. I. Caminos, Canales y Puertos

83. Convergence of Iterative Solution Methods
A sufficient condition for convergence exists:
Notes:
1. If the above does not hold, still may converge.
2. This looks similar to infinity norm of [A]
E. T. S. I. Caminos, Canales y Puertos

84. Improving Rate of Convergence of G-S Iteration
Relaxation Schemes:
where < l  2.0
(Usually the value of l is close to 1)
Underrelaxation ( 0.0 < l < 1.0 )
More weight is placed on the previous value. Often used to:
- make non-convergent system convergent or
- to expedite convergence by damping out oscillations.
Overrelaxation ( 1.0 < l  2.0 )
More weight is placed on the new value. Assumes that the new value is heading in right direction, and hence pushes new value close to true solution.
The choice of l is highly problem-dependent and is empirical, so relaxation is usually only used for often repeated calculations of a particular class.
E. T. S. I. Caminos, Canales y Puertos

85. Why Iterative Solutions?
We often need to solve [A]{x} = {b} where n = 1000's
• Description of a building or airframe,
• Finite-Difference approximations to PDE's. Most of A's elements will be zero; a finite-difference approximation to Laplace's equation will have five aij0 in each row of A.
Direct method (Gaussian elimination)
• Requires n3/3 flops (say n = 5000; n3/3 = 4 x 1010 flops)
• Fills in many of n2-5n zero elements of A
Iterative methods (Jacobi or Gauss-Seidel)
• Never store [A] (say n = 5000; [A] would need 4n2 = 100 Mb)
• Only need to compute [A-B] {x}; and to solve [B]{xt+1} = {b}
E. T. S. I. Caminos, Canales y Puertos

86. Why Iterative Solutions?
• Effort:
Suppose [B] is diagonal,
solving [B] {v} = {b} n flops
Computing [A-B] x 4n flops
For m iterations 5mn flops
For n = m = 5000, 5mn = 1.25x108
At worst O(n2).
E. T. S. I. Caminos, Canales y Puertos