TRIANGULAR FACTORIZATION

By Syed Zulqadar Hassan

CIIT Abbottabad Campus

TRIANGULAR FACTORIZATION

• It involves three steps:

• Step 1 Triangular Factorization

• Step 2 Forward Substitution

• Step 3 Back Substitution

𝑌 𝑏𝑢𝑠=[𝑌 11 𝑌 12 𝑌 13 𝑌 14

𝑌 21 𝑌 22 𝑌 23 𝑌 24

𝑌 31 𝑌 32 𝑌 33 𝑌 34

𝑌 41 𝑌 42 𝑌 43 𝑌 44]

The coefficients , , and are eliminated from the first column of the original coefficient matrix of Ybus by dividing it by all the other elements of column are zero and we have the following 1st row

𝑌11

𝑌11

=¿1𝑌12

𝑌 11

𝑌13

𝑌 11

𝑌14

𝑌 11

Multiply row 1 by and subtract it from row 2

∗ (𝑌 22−𝑌 21𝑌12

𝑌 11)=¿𝑌 22

(1 )𝑌 23(1 )𝑌 24

(1 )

As diagonal elements are 1 so divide row 2 by

❑𝑌 22

(1) =¿1❑𝑌 22

(1)❑𝑌 22

(1)𝑌 23

(1)

𝑌 22(1 )

𝑌 24(1)

𝑌 22(1 )

Similarly multiply row 2 by and subtract it from row 2 & As diagonal elements are 1 so divide row 3 by

(𝑌 33( 1) −

𝑌 32( 1)𝑌 23

(1 )

𝑌 22(1 ) )=𝑌 3 3

( 2 )𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑌 33( 2)1 𝑌 34

(2 )

𝑌 33(2 )

∗∗

As diagonal elements are 1 so divide row 3 by

(𝑌 44( 2) −

𝑌 43(2 )𝑌 34

( 2 )

𝑌 33( 2) )=𝑌 44

( 3 )𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑌 44( 3)

1

∗ ∗∗

𝑌 𝑏𝑢𝑠=[𝑌 11 𝑌 12 𝑌 13 𝑌 14

𝑌 21 𝑌 22 𝑌 23 𝑌 24

𝑌 31 𝑌 32 𝑌 33 𝑌 34

𝑌 41 𝑌 42 𝑌 43 𝑌 44]

As we have 1st same because we have to made lower triangular matrix

𝑌 11 𝑌 2 1 𝑌 4 1𝑌 3 1

∗ ∗ ∗ ∗ ∗ ∗

(𝑌 22−𝑌 21𝑌12

𝑌 11)=¿𝑌 22

(1 )

Similarly for 2nd column we have

𝑌 32(1 ) 𝑌 42

(1 )

Similarly for 3rd column we have

(𝑌 33( 1) −

𝑌 32( 1)𝑌 23

(1 )

𝑌 22(1 ) )=¿𝑌 33

(2 )𝑌 43(2 )

(𝑌 44( 2) −

𝑌 43(2 )𝑌 34

( 2 )

𝑌 33( 2) )=¿

Similarly for 4th column we have

𝑌 44(3 )

TRIANGULAR FACTORIZATION Count…

• So we can write

¿𝑌 𝑏𝑢𝑠• We Know that

𝑉=𝐼

L𝑈

𝑌 𝑏𝑢𝑠

we may replace the product UV by a new voltage vector such that so we have

𝐿𝑈𝑉 𝐼¿𝑉 ′𝑉 ′

¿

• Finally we get

TRIANGULAR FACTORIZATION Count…

[𝑌 11 ∗ ∗ ∗

𝑌 21 𝑌 22(1 ) ∗ ∗

𝑌 31 𝑌 32(1 ) 𝑌 33

( 2) ∗

𝑌 41 𝑌 42(1 ) 𝑌 43

(2 ) 𝑌 44( 3) ] [

𝑉 ′1

𝑉 ′2

𝑉 ′3

𝑉 ′4

]=[ 𝐼1

𝐼2

𝐼3

𝐼 4]

[ 1𝑌 12

𝑌 11

𝑌 13

𝑌 11

𝑌 14

𝑌 11

∗ 1𝑌 23

( 1)

𝑌 22( 1)

𝑌 24( 1)

𝑌 22(1 )

∗ ∗ 1𝑌 34

( 2 )

𝑌 33(2 )

∗ ∗ ∗ 1

] [𝑉 1

𝑉 2

𝑉 3

𝑉 4]=[𝑉

′1

𝑉 ′2

𝑉 ′3

𝑉 ′4

]

ADVANTAGES

• Solution of a linear system by triangular factorization and subsequent forward and back substitution is very popular because of the many advantages of the method:

• Efficiency

• Ability to preserve sparsity of the matrix

Sparsity• The fraction of zero elements (non-zero elements) in a matrix is called

the sparsity (density).

Sparsity in Power System

𝑅𝑠=Number   of   Nonzero   Elements   in   the   Table   of   FactorsNumber   of   Nonzero   Elements   in   the   Original  Matrix

• Let us analyze the requirements for a 1000 node/2000 branch circuit.

• For this network, the admittance matrix Y will have approximately 5000 nonzero elements. The table of factors for this matrix will have 5000Rs nonzero elements.

• If Rs = 2.5, then 12,500 nonzero elements need to be stored.

• The degree to which the triangular factorization procedure preserves the sparsity properties of the matrix is quantified with the Sparsity Preservation Index, defined with:

Sparsity in Power System Count…• The sparsity preservation index also impacts the efficiency of the method.

• This becomes obvious by considering the fact that the forward and back substitutions require as many multiply-adds as the number of non-zeros entries in the table of factors.

• If Rs = 2.5, then only 12,500 multiply-adds are required in the forward and back substitution, a small number compared with the required multiply-adds for the operation inverse of Y.

• The inverse of Y have 10,000,000 Multiply-adds while Factorization have 900,000 Multiply-adds.

References

• Power System Modeling, Analysis and Control By A. P. Sakis Meliopoulos (Page 18)

• https://en.wikipedia.org/wiki/Sparse_matrix

• “Triangular Factorization Method for Power Flow Analysis” by Y.Okamoto Published in journal “Electrical Engineering in Japan” Vol 96, No 1, January 1976, pp 31-35