RK20 power flow sparse - University of Iceland · Lecture 20 Power Engineering - Egill Benedikt Hreinsson 1 Sparse Matrices in Power Flow Calculations

1Lecture 20 Power Engineering - Egill Benedikt Hreinsson

Sparse Matrices in Power Flow Calculations


Any matrix that has a high proportion of its elements equal to zero is a sparse matrix.

•Most large matrices that arise in engineering applications are sparse.

Sparse Matrices?


The Sparsity of a Matrix - A definition

The sparsity of a matrix

Number of elements Total number of elements

=⋅0 100 %

A sparse matrix: A matrix, where most of the elements are zero, but a few are different from zero

A full matrix: A matrix, where all the elements are non-zero

=

4Lecture 20 Power Engineering - Egill Benedikt HreinssonWhy and What Are Sparse Matrices in Power Systems?

• Matrices in power systems are often very large and very sparse!• A large matrix: is for instance of dimension 100x100,

1000x1000 or 10000x10000!• A sparse matrix: A matrix where, for instance, 99% of the

elements are zero, but 1% are different from zero.• Examples of these matrices are the Ybus matrix and the Jacobi-

matrix, which both are used in power flow calculations.


Ybus Matrix Composition• The diagonal elements of the Ybus

matrix consist of the sum of the (series and shunt) admittances connected to the bus in question

• These elements are always present!

• The off-diagonal elements of the Ybusmatrix consist of the negative value of the series admittances connecting the 2 buses in question. These elements will be zero when no direct connection is between buses. (The Ybus matrix is very sparse!)

• These elements are present only when there is a link (a line or transformer)

11 12 1

21 22 1

1 2

n

n

n n nn

y y yy y y

y y y

⎡ ⎤⎢ ⎥⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

busY

ikik iky y eγ=

We review the rule for building the Ybus matrix:


The Sparsity of a Matrix: Example 1:

An example: Assume, in an electrical power system, each bus is on the average connected to 1.5 other buses. Further assume that we have a 100 bus system. We get a 100 by 100 matrix with 100 · 100 elements.

With the above assumption in the Ybus matrix there are 100 diagonal elements ≠ 0 and 150 elements above the diagonal and 150 elements below the diagonal ≠ 0. The sparsity of this matrix will therefore be:

Sparsity = 100 100 150 150 100

100 100

960

⋅ − − −⋅

= , %

We can say that the matrix is 4% full.

Sparsity


The Sparsity of a Matrix: Example 2:

Similarly for a 1000 bus system we get a matrix with 1000 · 1000 elements. We have for the same assumption of how each bus is interconnected for the Ybus matrix 1000 diagonal elements ≠ 0 and 1500 elements above the diagonal and 1500 elements below the diagonal ≠ 0. The sparsity of this matrix will be:

Sparsity = 1000 1000 1500 1500 1000

1000 1000

996

⋅ − − −⋅

= , %

We can perhaps say that the matrix is 0,4% full.

Sparsity


The Sparsity of the Jacobi-Matrix The J1 sub-matrix :

The J2 sub-matrix :

The J3 sub-matrix

The J4 sub-matrix :

Conclusion: If Ybus is a sparse matrix, the Jacobi matrix is also sparseThe elements of Ybus are also in the Jacobi matrices

( )1,

sinn

ii k ik i k ik

k k ii

PV V y δ δ γ

δ = ≠

∂= − ⋅ − −

∂ ∑

( )sinii k ik i k ik

k

PV V y δ δ γ

δ∂

= ⋅ − −∂

i k≠

( ) ( )1,

2 cos cosn

ii ii ii j ij i j ij

j j ii

PV y V y

Vγ δ δ γ

= ≠

∂= ⋅ − + ⋅ − −

∂ ∑

( )cosii ik i k ik

k

PV y

Vδ δ γ

∂= ⋅ − −

∂k i≠

( )1,

cosn

ii j ij i j ij

j j ii

QV V y δ δ γ

δ = ≠

∂= ⋅ − −

∂ ∑

( )cosii k ik i k ik

k

QV V y δ δ γ

δ∂

= − ⋅ − −∂

k i≠

( ) ( )1,

2 sin sinn

ii ii ii j ij i j ij

j j ii

QV y V y

Vγ δ δ γ

= ≠

∂= ⋅ − + ⋅ − −

∂ ∑

( )sinii ik i k ik

k

QV y

Vδ δ γ

∂= ⋅ − −

∂ k i≠


Random Sparse MatricesA Random Sparse Matrix

Sparse matrices can be of different nature


IEEE TEST systems• IEEE test systems are standardized power systems to test

research theories in power system analysis• We will look at different sparse Ybus matrices for different

IEEE test systems along with their one-line diagram:– 14 bus– 30 bus– 57 bus– 118 bus– 300 bus

• These test systems can be found at:– http://www.ee.washington.edu/research/pstca/– (http://www.pserc.cornell.edu/matpower/)


IEEE 14 BUS TEST CASE

0 5 10 15

0

5

10

15nz = 54

Use the SPY function in Matlab



Use the “SPY” function in Matlab to obtain the graphical representation of the sparse matrix

0 5 10 15 20 25 30

0

5

10

15

20

25

30

nz = 112



Sparsity 93%

0 10 20 30 40 50

0

10

20

30

40

50

nz = 213

Use the SPY function in Matlab


IEEE 118 bus test system

(Ybus)

0 20 40 60 80 100

0

20

40

60

80

100

nz = 476


IEEE 300 bus test system(Ybus)

0 50 100 150 200 250 300

0

50

100

150

200

250

300

nz = 1118

See details for the one-line diagrams for the 300 bus system on the following slides


300 bus IEEE test system (1) one line diagram


300 bus IEEE test system (2)

one line diagram


300 bus IEEE test system (3) one line diagram


Sparsity structure of a 108 bus Icelandic system Ybus matrix

0 10 20 30 40 50 60 70 80 90 100

0

10

20

30

40

50

60

70

80

90

100

nz = 346


Solutions to the power system problem with sparse matrices

• For large systems (and power systems are large) we need to exploit the sparsity of the system matrices

• Therefore solutions to power flow problem with Newton’s method involves solution to a set of linear equations rather than direct inversion of matrices

• We use Gauss elimination as a standard set of solving large linear systems of equations where we exploit simultaneously the sparsity structure of the system matrix (Jacobi)


Sparse Matrices and Solutions to Sets of Linear Equations in Power Systems

1,1 1,2 1, 1 1

2,1 2,2 2, 2 2

,1 ,2 ,

n

n

n n n n n n

a a a x ba a a x b

a a a x b

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⋅ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

To solve: Find a vector of unknown x-es. Other quantities are known or the a-s and b-s are known


Matrix Notation in Linear Equations

1,1 1,2 1, 1 1

2,1 2,2 2, 2 2

,1 ,2 ,

n

n

n n n n n n

a a a x ba a a x b

a a a x b

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥= = =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

A x b

A is a known matrixx is a vector with unknown variables. b is a known vector.

⋅ =A x b


Direct methods

Gaussian eliminationGauss-Jordan eliminationLU factorizationCholesky factorization

Iterative methods

Jacobi iterationsGauss-Seidel iteration

Solution methods for linear equations based on numerical analysis


Solutions to a Set of Linear Equations

• If we can transform the matrix to a triangular form, it is possible to solve the equations

• “Elementary Operations” can be carried out without changing the solution

• Examples of “Elementary Operation”:– Multiply rows (equations) with a constant– Add rows (equations) to each other

• This is the Gauss elimination method


⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

⋅

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

nnnn

n

n

b

bb

x

xx

a

aaaaa

2

1

2

1

,

,22,2

,12,11,1

00

0

Solutions to Linear Equations in Power Systems. A Special Case (triangular form)

Assume for a moment that a matrix has the special triangular form case -- The matrix, A is triangular with all zeros below the diagonal:

Solution method: Back substitution


Back Substitution

1,111,1 −−−−− =+ nnnnnnn bxaxa

,

nn

n n

bxa

=

1,1

,111

−−

−−−

−=

nn

nnnnn a

xabx

1,1

,133,122,111

,

1,

2,2

11,2,222

axaxaxab

x

a

xabx

axaxab

x

nn

kk

n

kjjjkk

k

nn

nnnnnnnn

−−−−=

−=

−−=

∑+=

−−

−−−−−−

…

1,1 1,2 1, 1 1

2,2 2, 2 2

,

0

0 0

n

n

n n n n

a a a x ba a x b

a x b

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥⋅ =⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎣ ⎦ ⎣ ⎦⎣ ⎦

nnnn bxa =,

2,211,222,2 −−−−−−−− =++ nnnnnnnnnn bxaxaxa

The last line

The second last line

The 3rd line from the end


Solution method

• How do we get any matrix to a triangular form?• Transform the matrix by Gauss elimination to a lower

triangular form matrix• Solve such a matrix by back substitution

• We have already checked out back substitution. Let us look at Gauss elimination!!


Gaussian Elimination (1st step)

1,1 1,2 1,3 1,

2,1 2,2 2,3 2,

3,1 3,2 3,3

,1 ,2 ,3 ,

n

n

n n n n n

a a a aa a a aa a a

a a a a

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

The original matrix:

1,1 1,2 1,3 1,

2,1 2,1 2,12,2 1,2 2,3 1,3 2, 1,

1,1 1,1 1,1

3,1 3,1 3,13,2 1,2 3,3 1,3 3, 1,

1,1 1,1 1,1

,1 ,1 ,1,2 1,2 ,3 1,3 , 1,

1,1 1,1 1,1

0

0

0

n

n n

n n

n n nn n n n n

a a a aa a a

a a a a a aa a aa a a

a a a a a aa a a

a a aa a a a a a

a a a

⎡⎢⎢ − ⋅ − ⋅ − ⋅⎢⎢⎢ − ⋅ − ⋅ − ⋅⎢⎢⎢⎢⎢ − ⋅ − ⋅ − ⋅⎢⎣

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥

⎢ ⎥⎦

Multiply the 1st rowwith a2,1/a1,1 and subtract from the 2nd row

Multiply the 1st rowwith a3,1/a1,1 and subtract from the 3rd row

Multiply the 1st row with an,1/a1,1 and subtract from the last row

2,1

1,1

aa


Gaussian Elimination (2nd step)

1,1 1,2 1,3 1,(1) (1) (1)2,2 2,3 2,

(1) (1) (1)1,2 1,3 1,

(1) (1) (1),2 ,3 ,

0

00

n

n

n n n n

n n n n

a a a aa a a

a a aa a a− − −

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

The matrix after the 1st step:

The 1st column is with zerosexcept at the top row


Gaussian Elimination (2nd step)The matrix after the 1st step:

1,1 1,2 1,3 1,(1) (1) (1)2,2 2,3 2,

(1) (1)3,2 3,2(1) (1) (1) (1)

3,3 2,3 3, 2,(1) (1)2,2 2,2

(1) (1),2 ,2(1) (1) (1) (1)

,3 2,3 , 2,(1) (1)2,2 2,2

0

0 0

0 0

n

n

n n

n nn n n n

a a a aa a a

a aa a a a

a a

a aa a a a

a a

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− ⋅ − ⋅⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− ⋅ − ⋅⎜ ⎟⎜ ⎟⎝ ⎠

Multiply the 2nd row with

and subtract from the 3rd row

Multiply the 2nd row with

and subtract from the last row

1,2 1,2 1,3 1,(1) (1) (1)2,2 2,3 2,

(1) (1) (1)1,2 1,3 1,

(1) (1) (1),2 ,3 ,

0

00

n

n

n n n n

n n n n

a a a aa a a

a a aa a a− − −

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

)1(2,2

)1(2,3

aa

)1(2,2

)1(2,

aan

Now the 2nd row plays the same role as the 1st row before:


1,2 1,2 1,3 1,(1) (1) (1)2,2 2,3 2,

(2) (2)1,3 1,

(2) (2),3 ,

0

0 00 0

n

n

n n n

n n n

a a a aa a a

a aa a− −

⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

Gaussian Elimination (3rd step)

The matrix after the 2nd step:

The 1st column with zeros except at the top

The 2nd column with zeros, except in the 2 top rows


Gauss-Elimination Transforms the Matrix to the Following Form :

This matrix is U or an “upper” or “unit upper”triangular matrix. The elements are calculated with iteration, where each step is an “elementary operation”. It is possible to solve the equation with a simple “back substitution”. It is found by dividing each row with the diagonal element

1 112 1

2 22

10 1

0 0 1

n

n

n n

x ba ax ba

x b

′′ ′ ⎡ ⎤ ⎡ ⎤⎛ ⎞⎜ ⎟ ⎢ ⎥ ⎢ ⎥′′⎜ ⎟ ⎢ ⎥ ⎢ ⎥=⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ⎢ ⎥ ⎢ ⎥⎜ ⎟ ′⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎣ ⎦ ⎣ ⎦

⋅ =U x G


Gaussian Elimination

• After n-1 steps the matrix has become triangular!

• Computer time grows fast with the matrix size (increasing n)

• Computer time grows in proportion to n3, if the matrix is not sparse

• We can obtain a “1” on the diagonal by dividing by the diagonal element in each step

Documents

RK20 power flow sparse - University of Iceland · Lecture 20 Power Engineering - Egill Benedikt Hreinsson 1 Sparse Matrices in Power Flow Calculations