Chapter 1 Solution Techniques.pdf

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ECEG 6011
Advanced
Power System Analysis
Chapter1: Solution Techniques
Dr.M.Karthikeyan
Dept. of Electrical and Computer Engineering
Presented by Dr.M.Karthikeyan, AP/ECE, WSU 1

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SPARSE MATRIX
• In numerical analysis and scientific computing, a sparse
matrix or sparse array is a matrix in which most of the elements are
zero.
• A matrix is characterized as sparse, whenever a significant
percentage of it's elements are equal to zero. The admittance or
Y matrix of a power system is relatively sparse, whereas the Z of
impedance matrix of the same system is proportionately full, i.e. very
few zero elements.

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Why do use SPARSE MATRIX
• Using sparse matrices to store data that contains a large number of
zero-valued elements can both save a significant amount of memory
and speed up the processing of that data. sparse is an attribute that
you can assign to any two-dimensional MATLAB® matrix that is
composed of double or logical elements.

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Y-bus is a sparse matrix
• The Y-bus is a sparse matrix. For a large power system, more than
90% of off-diagonal elements are zero. This is due to the fact that in
power system networks each node (bus) is connected to not more than
three nodes.

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TYPES OF SPARSE MATRIX
• csc_matrix: Compressed Sparse Column format.
• csr_matrix: Compressed Sparse Row format.
• bsr_matrix: Block Sparse Row format.
• lil_matrix: List of Lists format.
• dok_matrix: Dictionary of Keys format.
• coo_matrix: COOrdinate format (aka IJV, triplet format)

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SPARSE MATRIX IN
DATA STRUCTURE
• Sparse matrix is a matrix which contains very few non-zero
elements. When a sparse matrix is represented with a 2-
dimensional array, we waste a lot of space to represent that matrix.
For example, consider a matrix of size 100 X 100 containing only 10
non-zero elements.

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ADVANTAGES OF SPARSE MATRIX
The advantages of sparse matrices are size and speed.
A 5000 by 5000 matrix in full storage requires space for 25 million
complex numbers even if only 50,000 are nonzero. The same 50,000-
nonzero element matrix, in sparse storage, would store 50,000 complex
numbers and 50,000 pairs of integer indices.

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REPRESENTATION OF
SPARSE MATRIX
• Two methods of representation:
• 1. Array Representation
• 2. Linked Lists Representation

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Sparse Matrix Techniques for
Large Scale Power System
Linear System Solution – Introduction
• A problem that occurs in many is fields is the solution of linear
systems Ax = b where A is an n by n matrix with elements aij, and x
and b are n-vectors with elements xi and bi respectively
• In power systems we are particularly interested in systems when n is
relatively large and A is sparse
• How large is changing
• A matrix is sparse if a large percentage of its elements have zero
values
• Goal is to understand the computational issues (including complexity) associated
with the solution of these systems

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Linear system – contd.,
• Sparse matrices arise in many areas, and can have domain specific
structures
• Symmetric matrices
• Structurally symmetric matrices
• Tridiagnonal matrices
• Banded matrices

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Symmetric Matrix
• A symmetric matrix in linear algebra is a square matrix that remains
unaltered when its transpose is calculated.
• That means, a matrix whose transpose is equal to the matrix itself, is
called a symmetric matrix.

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Structurally Symmetric Matrix
• A structurally symmetric sparse matrix has nonzero values with the
property that if a(i, j) 0, then a(j, i) 0 for all i and j.
• When solving a structurally symmetric system, the entire matrix must
be passed to the solver routines.

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Tridiagonal Matrix
• A tridiagonal matrix is a matrix that has non-zero elements only at
the main diagonal, diagonal below and above it.
• All other elements are zero.
• For this reason tridiagonal matrices of dimension smaller than or equal
to 3 seem meaningless.

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Banded Matrix
• A banded matrix is a matrix whose only nonzero elements lie on
diagonal bands above and/or below the main diagonal.
• The number of nonzero diagonals above the main diagonal is called
the upper bandwidth.
• The number of nonzero diagonals below the main diagonal is called
the lower bandwidth.

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Sparse Systems
• The material presented so far applies to any arbitrary linear system
• The next step is to see what happens when we apply triangular
factorization to a sparse matrix
• For a sparse system, only nonzero elements need to be stored in the
computer since no arithmetic operations are performed on the 0’s.
• The triangularization scheme is adapted to solve sparse systems in
such a way as to preserve the sparsity as much as possible

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Inverse of a Sparse Matrix
• The inverse of a sparse matrix is NOT in general a sparse matrix
• We never (or at least very, very, very seldom) explicitly invert a
sparse matrix
• Individual columns of the inverse of a sparse matrix can be
obtained by solving x = A-1b with b set to all zeros except for a
single nonzero in the position of the desired column
• If a few desired elements of A-1 are desired (such as the
diagonal values) they can usually be computed quite efficiently
using sparse vector methods (a topic we’ll be considering soon)
• We can’t invert a singular matrix (with sparse or not)
16
Presented by Dr.M.Karthikeyan, AP/ECE, WSU

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Computer Architecture Impacts
• With modern computers the processor speed is many times faster
than the time it takes to access data in main memory
• Some instructions can be processed in parallel
• Caches are used to provide quicker access to more commonly used
data
• Caches are smaller than main memory
• Different cache levels are used with the quicker caches, like
L1, have faster speeds but smaller sizes; L1 might be 64K,
whereas the slower L2 might be 1M
• Data structures can have a significant impact on sparse matrix
computation
17

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Full Matrix versus
Sparse Matrix Storage
• Full matrices are easily stored in arrays with just one variable
needed to store each value since the value’s row and column are
implicitly available from its matrix position
• With sparse matrices two or three elements are needed to store
each value
• The zero values are not explicitly stored
• The value itself, its row number and its column number
• Storage can be reduced by storing all the elements in a
particular row or column together
• Because large matrices are often quite sparse, the total storage is
still substantially reduced
18

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Sparse Matrix Usage Can Determine the
Optimal Storage
• How a sparse matrix is used can determine the best storage scheme
to use
• Row versus column access; does structure change
• Is the matrix essentially used only once? That is, its structure and
values are assumed new each time used
• Is the matrix structure constant, with its values changed
• This would be common in the N-R power flow, in which the
structure doesn’t change each iteration, but its values do
• Is the matrix structure and values constant, with just the b vector in
Ax=b changing
• Quite common in transient stability solutions
19

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General Sparse Matrix Storage
• A general approach for storing a sparse matrix would be using three
vectors, each dimensioned to number of elements
• AA: Stores the values, usually in power system analysis as
double precision values (8 bytes)
• JR: Stores the row number; for power problems usually as an
integer (4 bytes)
• JC: Stores the column number, again as an integer
• If unsorted then both row and column are needed
• New elements could easily be added, but costly to delete
• Unordered approach doesn’t make for good computation since
elements used next computationally aren’t necessarily nearby
• Usually ordered, either by row or column
20

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Sparse Storage Example
• Assume
• Then
21
5 0 0 4
0 4 0 3
0 0 3 2
4 3 2 10

 
 

 


 
 
  
 
A
 
 
 
5 4 4 3 3 2 4 3 2 10
1 1 2 2 3 3 4 4 4 4
1 4 2 4 3 4 1 2 3 4
      


AA
JR
JC
Note, this example is a symmetric matrix, but the technique is general

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Compressed Sparse Row Storage
• If elements are ordered (as was case for previous example)
storage can be further reduced by noting we do not need to
continually store each row number
• A common method for storing sparse matrices is known as the
Compressed Sparse Row (CSR) format
• Values are stored row by row
• Has three vector arrays:
• AA: Stores the values as before
• JA: Stores the column index (done by JC in previous
example)
• IA: Stores the pointer to the index of the beginning of
each row
22

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CSR Format Example
• Assume as before
• Then
23
5 0 0 4
0 4 0 3
0 0 3 2
4 3 2 10

 
 

 


 
 
  
 
A
 
 
 
5 4 4 3 3 2 4 3 2 10
1 4 2 4 3 4 1 2 3 4
1 3 5 7
      


AA
JA
IA

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CSR Comments
• The CSR format reduces the storage requirements by taking
advantage of needing only one element per row
• The CSR format has good advantages for computation when using
cache since (as we shall see) during matrix operations we are
often sequentially going through the vectors
• An alternative approach is Compressed Sparse Column (CSC),
which identical, except storing the values by column
• It is difficult to add values.
• We’ll mostly use the linked list approach here, which makes
matrix manipulation simpler
24

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Linked Lists: Classes and Objects
• In explaining the linked list approach it is helpful to use the concepts
from object oriented programming (OOP) of classes and objects
• Approach can also be used in non-OOP programming
• OOP can be thought of as a collection of objects interacting with each
other
• Objects are instances of classes.
• Classes define the object fields and actions (methods)
• We’ll define a class called sparse matrix element, with fields of value,
column and next; each sparse matrix element is then an object of this
class
25

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Linked Lists
• A linked list is just a group of objects that represent a sequence
• We’ll used linked lists to represent a row or column of a sparse matrix
• Each linked list has a head pointer that points to the first object in the
list
• For our sparse matrices the head pointer will be a vector of the rows or
columns
26
Column a
Value a
Next a
Column b
Value b
Next b
Column c
Value c
Nil
Head

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Sparse Matrix Storage with
Linked Lists by Rows
• If we have an n by n matrix, setup a class called T sparse Element with
fields column, value and next
• Setup an n-dimensional head pointer vector that points to the first
element in each row
• Each nonzero corresponds to an object of class (type) T Sparse
Element
• We do not need to store the row number since it is given by the
object’s row
• For power system sparse matrices, which have nonzero diagonals, we
also have a header pointer vector that points to the diagonal objects
27

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Fills
• When doing a factorization of a sparse matrix some values that were
originally zero can become nonzero during the factorization process
• These new values are called “fills”
(some call them fill-ins)
• For a structurally symmetric matrix the fill occurs for both the element
and its transpose value (i.e., Aij and Aji)
• How many fills are required depends on how the matrix is ordered
• For a power system case this depends on the bus ordering
28

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Fills
• There are two key issues associated with fills
• Adding the fills
• Ordering the matrix elements (buses in our case) to reduce the
number of fills
• The amount of computation required to factor a sparse
matrix depends upon the number of non-zeros in the
original matrix, and the number of fills added
• How the matrix is ordered can have a dramatic impact on
the number of fills, and hence the required computation
• Usually a matrix cannot be ordered to totally eliminate
fills
29

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Fill Examples
4
1 2 3
1
2 3 4
5 0 0 4
0 4 0 3
0 0 3 2
4 3 2 10

 
 

 


 
 
  
 
A
10 4 3 2
4 5 0 0
3 0 4 0
2 0 0 3
  
 
 

 


 
 

 
B
No Fills Required Fills Required (matrix becomes full)
30

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Example: 7 by 7 Matrix
• Consider the 7 x 7 matrix A with the zero-nonzero
pattern shown in (a): of the 49 possible elements
there are only 31 that are nonzero
• If elimination proceeds with the given ordering, all
but two of the 18 originally zero entries, will fill in,
as seen in (b)
31

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1 2 3 4 5 6 7
1 X X X X X X
2 X X X X X
3 X X X X X
4 X X X
5 X X X X
6 X X X X X
7 X X X
(a) the original zero-
nonzero structure
1 2 3 4 5 6 7
1 X X X X X X
2 X X X F F X X
3 X X X F F X X
4 X F F X X F F
5 X F F X X X F
6 X X X F X X F
7 X X F F F X
(b) the post- elimination
zero-nonzero pattern
r
c
r
c
32
Example: 7 by 7 Matrix Structure

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• We next reorder the rows and the columns of A so as to
result in the pattern shown in (c)
• For this reordering, we obtain no fills, as shown in the
table of factors given in (d )
• In this way, we preserve the original sparsity of A
33
Example: 7 by 7 Matrix Reordering

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4 5 1 6 7 3 2
4 X X X
5 X X X X
1 X X X X X X
6 X X X X X
7 X X X
3 X X X X X
2 X X X X X
(c) the reordered system
4 5 1 6 7 3 2
4 X X X
5 X X X X
1 X X X X X X
6 X X X X X
7 X X X
3 X X X X X
2 X X X X X
(d) the post- elimination
reordered system
r
c
r
c
34
Example: 7 by 7 Matrix Reordered Structure

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• For structurally symmetric matrices we can gain good
insights into the problem by examining the graph-
theoretic interpretation of the triangularization process
• This assumption involves essentially no loss in
generality since if Aij  0 but Aji = 0 we simply treat Aji
as a nonzero element with the value 0; in this way, we
ensure that A has a symmetric structure
• We term a matrix as structurally symmetric whenever it
has a symmetric zero-nonzero pattern
Fills for Structurally Symmetric
Matrices
35

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• We make use of graph theoretic notions to develop a practical
reordering scheme for sparse systems
• We associate a graph G with the zero-nonzero structure of the n
by n matrix A
• We construct the graph G associated with the matrix A as
follows:
i. G has n nodes corresponding to the dimension n of the
square matrix: node i represents both the column i and the
row i of A;
ii. a branch (k, j) connecting nodes k and j exists if and only if
the element Ajk (and, by structural symmetry, Akj) is nonzero;
the self loop corresponding to Akk is not represented
36
Graph Associated with A

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• We consider the simple linear resistive network
with the topology shown below
 1
s
 2
s
 4
s  3
s
37
Example: Linear Resistive Network

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• The corresponding loop impedance matrix has the
following zero - nonzero pattern:
1 2 3 4
1 X X 0 X
2 X X X 0
3 0 X X X
4 X 0 X X
r
c
38

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• The associated 4-node graph of the matrix is
• We next illustrate the graphical interpretation of the
elimination process with another example
1 2
3 4
39

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Example: 5 by 5 System
• Suppose that A has the zero-nonzero pattern
1 2 3 4 5
1 X X X X
2 X X X
3 X X X
4 X X X X
5 X X X
r
c
40

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• Then, the associated graph G is
1 2
3
4
5
41

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• The graph-theoretic interpretation of the elimination of
the node (bus) j is as follows
• The deletion of the node j involves all its incident
branches (k, j) and ( j, k) connected to j, kj
• In the pre-elimination graph of the eliminated node j,
the elimination of the branches ( j, k) and (l, j) results
in the addition of the new branch (k, l), if one does not
already exist
Graph-Theoretic Interpretation
42

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• We eliminate the Bus (Node) 1 variable with the
resulting zero-nonzero pattern as shown the array
bordered by the broken
lines with on the new graph
new G1
1 2 3 4 5
1 X X X X
2 X X X F F
3 X X X
4 X F X X X
5 X F X X
r
c
43

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Graph G1
• We obtain the graph G1 from G by removing Bus 1 with
the new added branches (2, 4) and (2, 5) corresponding
to the fills
5 2
4 3
new branch
44

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• The elimination of Bus 2 results in the
submatrix shown below
3 4 5
3 X X F
4 X X X
5 F X X
r
c
45

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with the corresponding graph G2
• The elimination of Bus 3 yields
5
4 3
4 5
4 X X
5 X X
r
c
46

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with the corresponding graph G3
• Finally, upon Bus 4 we have
• and the corresponding G4 is simply the point
5
5 X
5
4
5
r
c
47

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• We next examine how we may reorder the rows and
columns of A to preserve its sparsity, i.e., to
minimize the number of fills
• Eventually we’ll introduce an algorithm to try to
minimize the fills
• This is motivated by revisiting the graph G
1 2
3
4
5
Reording the Rows/Columns
48

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• To minimize the number of fills, i.e., the number of
new branches in G, we eliminate first the node
which upon deletion introduces the least number of
new branches
• This is node 5 and upon deletion no new branches
are added and the resulting graph G1 is
1
4
2
3
49
Reording Motivating Example

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• The structure of G1 is such that any one of the
remaining nodes may be chosen as the next node to be
eliminated since each of the 4 remaining nodes
introduces a new branch after its elimination
• We arbitrarily pick node 1 and we obtain the graph G2
• We continue with the next three choices arbitrary,
resulting in no new fills
2
3
4
new branch
50
Reordering Motivating Example

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• We may relabel the original graph in such a way that
the label of the node refers to the order in which it is
eliminated
• Thus we renumber the nodes as shown below
1
2 3
4
5
51

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• Clearly, relabeling the nodes corresponds to reordering
the rows and columns of A
• For the reordered system, the zero-nonzero pattern of A
is
1 2 3 4 5
1 X X X
2 X X X X
3 X X X
4 X X X
5 X X X X
r
c
52

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and of its table of factors has the zero-nonzero
structure
Compared to the original ordering scheme, the new
ordering scheme has saved us 4 fill-ins
1 2 3 4 5
1 X X X
2 X X X X
3 X X X F
4 X X X
5 X X F X X
r
c
53

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• The associated graph of the structurally symmetric
matrix A is useful in gaining insights into the
factorization process
• We make the following observations
• If A is originally structurally symmetric, then it
remains so in all the steps of the factorization;
• A good ordering scheme is independent of the
values of the elements of A and depends only on
its the zero-nonzero pattern
54
General Findings

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Sparse Vector Methods
• Sparse vector methods are useful for cases in solving Ax=b in which
• A is sparse
• b is sparse
• only certain elements of x are needed
• In these right circumstances sparse vector methods can result in
extremely fast solutions!
• A common example is to find selected elements of the inverse of A,
such as diagonal elements.
55

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Sparse Vector Methods Introduced
• Assume we are solving Ax = b with A factored so we solve LUx = b
by first doing the forward substitution to solve Ly = b and then the
backward substitution to solve Ux = y
• A key insight: In the solution of Ly = b if b is sparse then only certain
columns of L are required, and y is often sparse
y1
.
.
.
yn
.
.
.
x
.
.
.
x
.
.
.
x
=
56

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Fast Forward Substitution
• If b is sparse, then the fast forward (FF) substitution takes advantage
of the fact that we only need certain columns of L
• We define {FF} as the set of columns of L needed for the solution of
Ly = b; this is equal to the nonzero elements of y
• In general the solution of Ux = y will NOT result in x being a sparse
vector
• However, oftentimes only certain elements of x are desired
• E.g., the sensitivity of the flows on certain lines to a change in generation at a
single bus; or a diagonal of A-1
57

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Fast Backward Substitution
• In the case in which only certain elements of x are
desired, then we only need to use certain rows in U
below the desired elements of x; define these columns as
{FB}
• This is known as a fast backward substitution (FB),
which is used to replace the standard backward
substitution .
.
.
x
.
.
.
x
.
.
.
x
.
.
y1
y2
.
.
.
yn
=
58

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Factorization Paths
• We observe that
• {FF} depends on the sparsity structures of L and b
• {FB} depends on the sparsity structures of U and x
• The idea of the factorization path provides a systematic way to
construct these sets
• A factorization path is an ordered set of nodes associated with the
structure of the matrix
• For FF the factorization path provides an ordered list of the columns
of L
• For FB the factorization path provides an ordered list of the rows of U
59

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Gaussian Elimination
• The best known and most widely used method for
solving linear systems of algebraic equations is
attributed to Gauss
• Gaussian elimination avoids having to explicitly
determine the inverse of A, which is O(n3)
• Gaussian elimination can be readily applied to sparse
matrices
• Gaussian elimination leverages the fact that scaling a
linear equation does not change its solution, nor does
adding on linear equation to another
1 2 1 2
2 4 10 2 5
x x x x
    
60

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Gaussian Elimination, cont.
• Gaussian elimination is the elementary procedure in
which we use the first equation to eliminate the first
variable from the last n-1 equations, then we use the
new second equation to eliminate the second variable
from the last n-2 equations, and so on
• After performing n-1 such eliminations we end up with
a triangular system which is easily solved in a
backward direction
61

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Example 1
• Solve the equations using Gauss Elimination Method.
3x1 + 12x3 = 15 3 0 12 x1 = 15
6x2 + 12x3 = 15 0 6 12 x2 15
3x1 + 10x2 + 16x3 = 24 or 3 10 16 x3 24

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Triangular Decomposition
• In this example, we have:
• triangularized the original matrix by
Gaussian elimination using column elimination
• then, we used back substitution to solve the
triangularized system
• The following slides present a general scheme for
triangular decomposition by Gaussian elimination
• The assumption is that A is a nonsingular matrix
(hence its inverse exists)
• Gaussian elimination also requires the diagonals to be
nonzero; this can be achieved through ordering
• If b is zero then we have a trivial solution x = 0
63

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• We form the matrix Aa using A and b with
and show the steps of the triangularization scheme
 
 
 
 
 
 
 
 
 
 
 
 
 
11 12 1 1
21 22 2 2
31 32 3 3
1 2



    

n
n
n
n n nn n
a a a b
a a a b
a a a b
a a a b
   
  
   
   

 
a  
A A b

64

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Triangular Decomposition, Step 1
• Step 1: normalize the first equation

1
(1)
1
11
1
(1)
1
11
= 2 ...
=
j
j
a
a j , ,n
a
b
b
a
65

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• Step 2: a) eliminate x1 from row 2:
• Step 2: b) normalize the second equation
66
,

(1) (1)
2 2 21 1 2 ...
j j j
a = a a a j = , ,n

(1) (1)
2 2 21 1
b = b a b
,
(1)
2
(2)
2 (1)
22
3 ...
j
j
a
a = j = , ,n
a
(1)
2
(2)
2 (1)
22
b
b =
a

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67
and we end up at the end of step 2 with
 
 
 
 
 
 
 
 
 
 
 
 
(1 ) (1 ) (1 ) (1 )
1 2 1 3 1 1
(2 ) (2 ) (2 )
2 3 2 2
3 1 3 2 3 3 3 3
1 2 3
1
1



     

n
n
n
n n n n n n
a a a b
0 a a b
a a a a b
a a a a b
67

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68
• Step 3: a) eliminate x1 and x2 from row 3:
• Step 3: b) normalize the third equation:
2
 
( ) (1) (1) (2)
3 3 32 2 3 ...
j j j
a a a a j = , ,n

(1) (1)
3 3 31 1 2
j j j
a = a a a j = , ... ,n

(1) (1)
3 3 31 1
b = b a b

(2) (1) (1) (2)
3 3 32 2
b = b a b
(2)
3
(3)
3 (2)
33
4 ...
j
j
a
a = j = , ,n
a
(2)
3
(3)
3 (2)
33
b
b =
a
68

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69
and we have the system at the end of step 3
 
 
 
 
 
 
 
 
 
 
 
(1) (1) (1) (1) (1)
12 13 14 1 1
(2) (2) (2) (2)
21 24 2 2
(3) (3) (3)
34 3 3
41 42 43 44 4 4
1 2 3 4
1
1
1




      

n
n
n
n
n n n n nn n
a a a a b
0 a a a b
0 0 a a b
a a a a a b
a a a a a b
69

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70
Triangular Decomposition, Step k
• In general, we have for step k:
a) eliminate from row k:
b) normalize the kth equation:
 

1 1
1 ...
(m) (m ) (m ) (m)
kj kj km mj
a = a a a j = m + , ,n
, , ,
1 2 1
... k
x x x 
 

1 1
1 2 ... 1
(m) (m ) (m ) (m)
k k km m
b = b a b m = , , , k -


1
1
= = 1 ...
(k )
kj
(k)
kj (k )
kk
j
a
a k + , ,n
a


1
1
=
(k )
k
(k)
k (k )
kk
b
b
a
70

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71
Triangular Decomposition:
Upper Triangular Matrix
• and proceed in this manner until we obtain the upper
triangular matrix (the nth derived system):
 
 
 
 
 
 
 
 
 
 
 
(1) (1) (1) (1) (1)
12 13 14 1n 1
(2)
(2) (2) (2)
21 24 2 2
(3)
(3) (3)
34 3 3
(4) (4)
4 4
1
1
1
1
1




      

n
n
n
(n)
n
a a a a b
0 a a a b
0 0 a a b
0 0 0 a b
0 0 0 0 b
71

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72
• Note that in the scheme presented, unlike in the
first example, we triangularly decompose the
system by eliminating row-wise rather than
column-wise
• In successive rows we eliminate (reduce to 0) each
element to the left of the diagonal rather than those
below the diagonal
• Either could be used, but row-wise operations will
work better for power system sparse matrices
72

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73
Example 2
• Solve for x in the system using OTF
• The three elimination steps are given on the next
slides; for simplicity, we have appended the r.h.s.
vector to the matrix
• First step is set the diagonal element of row 1 to 1 (i.e.,
normalize it)
1
2
3
4
2 3 1 0 20
6 5 0 2 45
2 5 6 6 3
4 2 2 3 30
x
x
x
x
  
   
 
   
  
 
   

  
 
   
 
   

   
 
73

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74
Example 2, cont.
• Eliminate x1 by subtracting row 1 from all the rows
below it
1
1
2
multiply row by

1 2
3
multiply row by
and add to row

1 4
4
multiply row by
and add to row
1 6
2
multiply row by
and add to row
74

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75
Example 2, cont.
• Eliminate x2 by subtracting row 2 from all the
rows below it
1
2
4
multiply row by
2 8
3
multiply row by
and add to row
2 4
4
multiply row by
and add to row




 
 
 
 
 
 
 
 
 
3 1
1 0 10
2 2
0 1
0 0 1 2 7
0 0 1 1 5
3 1 15
4 2 4
75

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76
Example 2, cont.
• Elimination of x3 from row 3 and 4
3
4
subtract row
fromrow




 
 
 
 
 
 
 
 
 
3 1
1 0 10
2 2
0 1
0 0 1 2 7
0 0 0 1 2
3 1 15
4 2 4
76

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77
Example 2, cont.
• Then, we solve for x by “going backwards”, i.e.,
using back substitution:
    
2 3 4 2
3 1 15
7
4 2 4
x x x x
 
4
2
x
   
3 4 3
2 7 3
x x x
    
1 2 3 1
3 1
10 1
2 2
x x x x
77

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78
Solving for X
• To compute x we perform back substitution
   
   
 

  

1 1
1 1 1
1
=
= = 1 2 ... 1

n n
n- n- n- ,n n
n
k k
k k k j j
j=k+
x b a x
x b a x k n , n , ,
 

n
n n
x b
78

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79
Upper Triangular Matrix
• The triangular decomposition scheme applied to the
matrix A results in the upper triangular matrix U with
the elements
• The following theorem is important in the development
of the sparse computational scheme







1 =
(i)
ij ij
i j
u = a j > i
0 j < i
79

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80
LU Decomposition Theorem
• Any nonsingular matrix A has the following
factorization:
where U could be the upper triangular matrix
previously developed (with 1’s on its diagonals) and L
is a lower triangular matrix defined by
A = LU






1
=
(j )
i j
ij
a
0

j i

j i

80

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81
LU Decomposition Application
• As a result of this theorem we can rewrite
• Can also be set so U has non unity diagonals
• Once A has been factored, we can solve for x by
first solving for y, a process known as forward
substitution, then solving for x in a process known
as back substitution
• In the previous example we can think of L as a
record of the forward operations preformed on b.
Define
Then
Ax = LUx = b
y = Ux
Ly = b
81

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82
• A =LU Where
L = 1 0 0 U = U11 U12 U13
L21 1 0 0 U22 U23
L31 L32 1 0 0 U33
LU = U11 U12 U13
L21U11 L21U12+U22 L21U13+U23
L31U11 L31U12+L32U22 L31U13+L32U23+U33
LU Decomposition Application

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83
Symmetric Matrix Factorization
• There are also some computational benefits from factoring symmetric
matrices. However, since symmetric matrices are not common in
power applications, we will not consider them in-depth
• However, topologically symmetric sparse matrices are quite common,
so those will be our main focus
83

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