1. Fiber Optics Design and Solving Symmetric Banded Systems
Linda Kaufman
William Paterson University

2. Dispersion Compensation fibers
Signal degradation and Restoration
Transmission (~80km)
Dispersion Compensation
Amplification
EDFA
High Speed Optical Communication Segment

3. Fiber Optics Design and Solving symmetric Banded systems
Linda Kaufman
William Paterson University

5. Marcuse model for fiber wrapped around spool
For a radially symmetric fiber
where r is the radius from the center of the fiber, R is the radius of the spool which leads to a matrix problem of the structure.
b’s and c’s are 0(r/R)

6. Objective: Create a algorithm to factor a symmetric indefinite banded matrix A
An n x n matrix A is symmetric if
ajk = akj
A matrix is indefinite if any of these hold
a. eigenvalues are not necessarily all positive or all negative
b. One cannot factor A into LLT
is indefinite- determinant is negative
-5 2
A has band width 2m+1 if
ajk = 0 for |k-j| >m
m=2
x x x
x x x x 0
x x x x x
x x x x

7. Uses
(1)Solve a system of equations Ax=b
If A = MDMT , D is block diagonal, one solves
Mz=b
Dy =z
MTx =y
(2)Find the inertia of a system- the number of positive and negative eigenvalues of a matrix
If A = MDMT, The inertia of A is the inertia of D.
Given a matrix B, the inertia of a matrix A = B-cI
is the number of eigenvalues greater, less than and equal to c.

8. Competition
Ignore symmetry- use Gaussian elimination- does not give inertia info- 0(nm2) time-(n is size of matrix, m is bandwidth)
Band reduction to tridiagonal (Schwarz,Bischof, Lang, Sun, Kaufman) followed by Bunch for tridiagonal-0(n2m)
Snap Back-Irony and Toledo-Cerfacs-2004, 0(nm2) time generally faster than GE but twice space
Bunch – Kaufman for general matrices and hope bandwidth does not grow as Jones and Patrick noticed

9. Difficulties Consider the linear system .0001x + y = 1.0 x + y = 2.0
Gaussian elimination- 3 digit arithmetic
10000y = 10000
Giving y = 1, x =0
But true solution is about x =1.001, y =.999
If interchange first and second rows and columns before Gaussian elimination get
x + y = 2.0
y = = 1.0
So x = 1.0- a bit better

10. Gaussian elimination with partial pivoting to prevent division by zero and unbounded elemental growth
Unsymmetric pivoting yields
0 x x x f
0 x x f f
Eliminating first column yields
5 diagonal becomes 7 diagonal
In general 2k+1 diagonal becomes 3k+1 diagonal

11. Symmetric pivoting But pivoting tends to upset the band structure!!
But to maintain symmetry one must pivot rows and columns simultaneously
What if matrix is ?
1 0
Interchanging first row and column does not help
If matrix is
Pivot it to
And use top 2 x 2 as a “pivot”
But pivoting tends to upset the band structure!!

12. Bunch -Kaufman for symmetric indefinite non banded
Partition A as
Where D is either 1 x 1 or 2 x 2
and B’ = B – Y D-1 YT
Choice of dimension of D depends on magnitude of a11 versus other elements
If D is 2 x 2, det(D)<0

13. 2 x 2 vs 1 x 1 for nonbanded symmetric system x x x

14. Bandwidth spread with Bunch-Kaufman on banded matrix

15. Banded algorithm based on B-K
1) Let c = |ar 1 | = max in abs. in col. 1
2) If |a11 | >= w c, use a 1 x 1 pivot. Here w is a scalar to balance element growth, like 1/3
Else
3)Let f= max element in abs. in column r
4) If w c*c <= |a11 | f, use a 1 x 1 pivot
5)interchange the rth and second rows and columns of A
6) perform a 2 by 2 pivot
7) fix it up if elements stick out beyond the original band width
Never pivot with 1 x 1

16. Fix up algorithm Worst case r =m+1, what happens in pivoting
x x x x x x
x x x x x x x a b c d
x x x x x x x x
x x x x x x x x x
x x x x x x x x x x
x x x x x x x
x x x x x x x x x x x
a x x x x x x x x x x
b x x x x x x x x x x
c x x x x x x x x x
d x x x x x x x x
x x x x x x x x
x x x x x x x
x x x x x x

17. Partition A as If we don’t remove elements outside the band,
Reset B’ = B – Y D-1 YT
Let Z = D-1 YT = x x x x x x x x x
p q r s x x x x x .
Then B’ looks like
x x x x x x bp cp dp
x x x x x x x cq dq
x x x x x x x cr dr
x x x x x as bs cs ds x
x x x x x x x x x x x x
x x x as x x x x x x x x
bp x x bs x x x x x x x x
cp cq x cs x x x x x x x x
dp dq dr ds x x x x x x x x
x x x x x x x x
x x x x x x x
x x x x x x
If we don’t remove elements outside the band,
the bandwidth could explode to the full matrix.

18. Because of structure eliminating 1 element gets rid of column
x x x x x x
x x x x x x x cq dq
x x x x x x x cr dr
x x x x x x bt ct dt
x x x x x x x x x x
x x x x x x x x x x x x
x x bt x x x x x x x x
cq cr ct x x x x x x x x
dq dr dt x x x x x x x x
x x x x x x x x
x x x x x x x
x x x x x x

19. Continue with eliminating another element
x x x x x x
x x x x x x x
x x x x x x x x u
x x x x x x x x m
x x x x x x x x x x
x x x x x x x x x x x x
x x x x x x x x x x x
x x x x x x x x x x
u m x x x x x x x x
x x x x x x x x
x x x x x x x
x x x x x x
Now eliminate u to restore bandwidth

20. In practice one would just use the elements in the first part of Z to determine the Givens/stabilized elementary transformations and never bother to actually form the bulge. Thus there is never any need to generate the elements outside the original bandwidth.

21. Alternative Formulation
Partition A as
Where D is 2 x 2
Let Z = D-1 YT
Reset B’ = QT(B – Y Z)Q= QTB Q-HG
Where H=QTY and G =ZQ
Therefore do “retraction” followed by rank 2 correction.
Recall H looks like h1 h2
0 h3
Where the “0” has length r-1 and the whole H has length m+r-1

22. Alternative-2 Q is created such that G =ZQ has the form G =
where 0 has r-3 elements and G5 is a multiple of G3
1)Explicitly use 0’s in rank-2 correction
Thus correction has form below where numbers indicate the rank of the correction
(r-3 rows)
(m-r+2 rows)
(r-1 rows)
2) Store Q info in place of 0’s and G3 to reduce space to (2m+1)n
3) Reduce solve time for each 2 x 2 from 4m+6r mults to 4m+2r- In worst case, 3mn vs. 5mn

23. Properties
Space- (2m+1)n-in order to save transformations compared to (3m+1)n for unsymmetric G.E.
Never pivots for positive definite or 1 x 1
Decrease operations by not applying second column of pivot when these will be undone
Operation count depends on types of transformations
Elementary –between m2n/2 and 5 m2n/4
Compared with between m2n and 2m2n for G.E.

24. Elemental growth Let a = max element of matrix
For 1 x 1, Ajk = Ajk – A1k Aj1 /A11 taking norms element growth is a(1+1/w)
For 2 x 2 if no retraction, it turns out that element growth after 1 step is a(1 + (3+w)/(1-w))
For 2 x 2 if retraction, it turns out that element growth is (4+8/(1-w))a
2 steps of 1 x 1 = 1 step of retraction when w=1/3, giving growth bound of 4n-1

25. Comparison using Atlas Blas on Sun
Problem 1 2 3 4
Time-retraction (ms)elem/orthog
98/98
161/240
296/520
220/220
Time-Snapback
elem/orthog
140/140
550/590
1640/
2200
1290/1370
Time-Lapack-tf2/trf
195/136
395/235
Error-retraction
3e-15
/3e-15
1.6e-13 /1.3e-13
3e-13/
9e-14
2e-11
/5e-12
Error-snapback
5e-15
/5e-15
4e-11
1e-13
2.2e-4
4e-12
3e-5
Error Lapack-tf2
7e-15
1e-15
6e-15
1e-12

26. Retraction vs. Lapack

27. Elementary Orthogonal Time 150 160 2 x2’s 204 179 error 3.04e-11
Elementary vs. orthogonal on 2 random matrices, n=1000, m=100
Elementary
Orthogonal
Time
150
160
2 x2’s
204
179
error
3.04e-11
1.10e-11
growth
146.4
1292.1
221
178
1.00e-12
3e-12
130.5
1762.8
I thought orthogonal would be more accurate, have less elemental growth, and take much more time-but these assumptions were wrong for random matrices

28. Future
Could do pivoting with 1 by 1 at price of space and time- needs to be implemented
Adaptations for small bandwidth as in optical fiber code.
Condition estimator