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![Ibrahim Omar Habiballah Sep. 2013
If the first pivot is moved to the last axis, i.e.,
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then the table of factors (TOF) will be sparse, and for this particular
matrix structure, the TOF will have same sparse structure but with
different nonzero (fill-ins) values.
References:
[1] G.T. Heydt, "Computer Analysis Methods for Power System",
Macmillan, New York, 1986.](https://image.slidesharecdn.com/04programming-2-190410143727/85/04-programming-2-10-320.jpg)
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- 1. Ibrahim Omar HabiballahSep. 2013 2.3 Triangular Factorization This programming technique is necessary to solve a simultaneous set of n algebraic linear equations, which can be written in a matrix form as A x = b where A is an n x n square matrix, x and b are n x 1 vectors. There are several triangular factorization techniques for different applications. Such techniques are ease to use with sparsity programming. Two of these factorization techniques will be presented. 2.3.1) LU Factorization Let A be factored into two matrices, A = LU1 or A = L1U where L is lower left triangular, L1 is lower left triangular with a unit diagonal, U is upper right triangular, and U1 is upper right triangular with a unit diagonal. Consider the L1U factorization (similar steps can be applied for LU1). Then L1U x = b Let U x = w (this is known as backward substitution), then L1w = b (this is known as forward substitution). where w is an n-vector. The factorized entries of L1U are stored in the same matrix (say J), i.e., there is no need for new storage areas. Techniques that result in the destruction of the original data and replacement of the factorized data are known as "in situ" methods.
- 2. Ibrahim Omar HabiballahSep. 2013 The new factorized entries are found from the following two equations, 1,....,3,2,1 )( 1 1 −= − = ∑ − = rc u ulj l cc c q qcrqrc rc (2) nrrculju r q qcrqrcrc ,....,1, 1 1 +=−= ∑ − = (3) The forward substitution equation is nrwlbw r q qrqrr ,....,3,2,1 1 1 =−= ∑ − = (4) The backward substitution equation is 1,....,2,1, 1 −−= − = ∑ += nnnr u xuw x rr n rq qrqr r (5) Notice that the summations in Eqs. (2-5) are equal to zero when the lower index of the summation is greater than the upper index. A drawback of this method is that if the vector b is changed, the forward/backward substitution process must be repeated (unlike the direct inversion of the matrix). In most of power system applications the inversion of a matrix is not needed. The LU factorization, therefore, is favored for such applications.
- 3. Ibrahim Omar HabiballahSep. 2013 Example (6): Use L1U-factorization to solve the following linear equations, ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 1 0 3 1 4000 0102 5010 1001 x Solution: This problem requires three steps, 1) Replacing the entries of the matrix by the table of factors (TOF). The first row new entries, using Equation (3), are u11 =j11 - 0 = 1, u12 = j12 - 0 = 0, u13 = j13 - 0 = 0, u14 = j14 - 0 = 1 (Note that the sums are zero because the upper index of summation is zero). The second row new entries, using Eqs. (2) & (3), are 0 1 )00()0( 11 21 21 = − = − = u j l , u22 = j22 – l21u12 =1 – 0 = 1, u23 = j23 – l21u13 =0 – 0 = 0, u24 = j24 – l21u14 =5 – 0 = 5 The third row new entries, using Eqs. (2) & (3), are 2 1 )02()0( 11 31 31 = − = − = u j l 0 1 )00()( 22 123132 32 = − = − = u ulj l u33 = j33 – l31u13 – l32u23 = 1 – 0 – 0 = 1, u34 = j34 – l31u14 – l32u24 = 0 – 2 – 0 = -2
- 4. Ibrahim Omar HabiballahSep. 2013 The fourth row new entries, using Eqs. (2) & (3), are ,0 1 )00()0( 11 41 41 = − = − = u j l ,0 1 )00()( 22 124142 42 = − = − = u ulj l 0 1 )00()( 33 2342134143 43 = − = −− = u ululj l u44 = j44 – l41u14 – l42u24 – l43u34 = 4 – 0 – 0 - 0 = 4 Therefore, the completed table of factors (TOF), which contains both L1 and U superimposed, is ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ − 4000 2102 5010 1001 Where: ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ ⎛ = 1000 0102 0010 0001 1L & ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − = 4000 2100 5010 1001 U 2) Using forward substitution to find the vector w according to Eq. (4). Then w1 = b1 - 0 = 1 - 0 = 1, w2 = b2 - l21w1 = 3 - 0 = 3, w3 = b3 - l31w1 - l32w2 = 0 - 2 x 1 - 0 = -2, and w4 = b4 - l41w1 - l42w2 – l43w3 = 1 - 0 - 0 - 0 = 1
- 5. Ibrahim Omar HabiballahSep. 2013 Therefore, the w vector is ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − = 1 2 3 1 w 3) Using backward substitution to find the vector x according to Eq. (5). Then ,25.0 4 010 44 4 4 = − = − = u w x ,5.1 1 25.0)2(2 33 4343 3 = −−− = − = x u xuw x ,75.1 1 25.0503 22 4243232 2 = −− = −− = x u xuxuw x and ,75.0 1 25.01001 11 4143132121 1 = −−− = −−− = x u xuxuxuw x Therefore, the x vector (which is the final solution) is ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − = 25.0 5.1 75.1 75.0 x Checkout the results, ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ +++ +−++ +++ +++ = ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ − ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ 25.0*4000 0)5.1(*1075.0*2 25.0*5075.1*10 25.0*10075.0*1 25.0 5.1 75.1 75.0 * 4000 0102 5010 1001 ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = 1 0 3 1
- 6. Ibrahim Omar HabiballahSep. 2013 2.3.2) Chipley-Coleman Matrix Inversion This method allows inversion of a nonsingular square matrix in situ (i.e., the inverted matrix will replace the original matrix). The advantage of this method is that if the b vector is changed, the inversion matrix is never recalculated. Another advantage is its ease way in programming and its connection to power engineering problems. The algorithm to invert, say, an n x n matrix is: a- Let the pivot axis, p, be axis 1. (a pivot is defined as a nonzero entry in any row/column of a matrix which is located at or will be assigned as a diagonal element). b- Kron reduces all elements outside the pivot axis. pjpi A AA AA old pp old pj old ipold ij new ij ≠≠−= ; (6) c- Replace the pivot position by its negative inverse. old pp new pp A A 1− = (7) d- Reduce the elements in the pivot axis, but outside the p, p position. piAAA new pp old ip new ip ≠= (8) pjAAA new pp old pj new pj ≠= (9) e- Repeat steps (b) through (d) for p = 2, 3, …, n. The result is –A-1 . Therefore, reversal of the sign is A-1 . Example (7): Invert A using Chipley-Coleman method.
- 7. Ibrahim Omar HabiballahSep. 2013 ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ = 0.30.05.0 0.00.30.1 5.00.10.4 A Solution: Let p = 1, according to Eq. (6) ,75.2 0.4 0.10.1 0.3 11 1221 2222 =−=−= x A AA AA old oldold oldnew Similarly .9375.2,125.0 333223 =−== newnewnew AAA According to Eqs. (7-9), ,25.0 0.4 11 11 11 −= − = − = old new A A 25.0)25.0(0.111122112 −=−=== xAAAA newoldnewnew , and 125.0)25.0(5.011133113 −=−=== xAAAA newoldnewnew Therefore, for p = 1, the modified matrix is ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− −−− 9375.2125.0125.0 125.075.225.0 125.025.025.0 Repeating this process for p = 2, 3, the –A-1 is ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎣ ⎡ −− −− − 34109.001550.004651.0 01550.036434.009302.0 04651.009302.027906.0
- 8. Ibrahim Omar HabiballahSep. 2013 A subroutine INVERT for inverting a symmetry matrix is given below. SUBROUTINE INVERT (A,N) DIMENSION A(N,N) DO 1 IP=1,N DO 2 IR=1,N IF(IR.EQ.IP) GOTO 2 DO 3 IC=1,N IF(IC.EQ.IP) GOTO 3 A(IR,IC)=A(IR,IC)-(A(IR,IP)*A(IP,IC)/A(IP,IP)) 3 CONTINUE 2 CONTINUE A(IP,IP)=-1.0/A(IP,IP) DO 4 I=1,N IF(I.EQ.IP) GOTO 4 A(I,IP)=A(I,IP)*A(IP,IP) A(IP,I)=A(IP,I)*A(IP,IP) 4 CONTINUE 1 CONTINUE DO 5 IR=1,N DO 5 IC=1,N 5 A(IR,IC)=-A(IR,IC) RETURN END Note that this subroutine stores the A matrix in full. Considerable savings of memory and time are possible using sparsity programming and optimal ordering techniques. 2.4 Optimal Ordering It is a procedure by which the 'fill-ins' (generating new nonzeros) during the factorization or inversion process is minimized. Optimal ordering refers to renumbering the matrix axes so that fill-ins are as minimum as possible.
- 9. Ibrahim Omar HabiballahSep. 2013 The process of optimal ordering results in sparse (with slightly higher fill- ins than the original sparse matrix) table of factors and inverted matrices. However, if the original matrix is dense (not sparse) then the application of optimal ordering becomes ineffective. To illustrate the objective of optimal ordering, consider the following matrix with 'x' representing the nonzeros. ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ xx xx xx xx xx xx xx xx xxxxxxxxx If this matrix is factorized without ordering the axes of the matrix, the table of factors (TOF) will be dense (notice that there may by few nonzeros, but this could be due to cancellation between some of the numbers during the factorization process), ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ xxxxxxxx xxxxxxxxx xxxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxxx xxxxxxxxx
- 10. Ibrahim Omar HabiballahSep. 2013 If the first pivot is moved to the last axis, i.e., ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ xxxxxxxxx xx xx xx xx xx xx xx xx then the table of factors (TOF) will be sparse, and for this particular matrix structure, the TOF will have same sparse structure but with different nonzero (fill-ins) values. References: [1] G.T. Heydt, "Computer Analysis Methods for Power System", Macmillan, New York, 1986.