Solving Scale Linear Systems (Example system continued) Lecture 14 MA/CS 471 Fall 2003.
-
Upload
aubrey-robbins -
Category
Documents
-
view
215 -
download
2
Transcript of Solving Scale Linear Systems (Example system continued) Lecture 14 MA/CS 471 Fall 2003.
Solving Scale Linear Systems (Example system continued)
Lecture 14
MA/CS 471
Fall 2003
Today
• We will discuss direct methods for a slightly larger loop current problem as introduced last time.
• Then we will look at a much larger problem and examine the memory/work requirements for direct methods.
Note• It is important that at least one loop has a
resistor that lies only on that loop (i.e. at least one resistor lying on a wire not shared by two loops).
• Otherwise the entire circuit will short circuit around the boundary of the entire circuit..
• For example:
4
3
21
6
7
+-
1
1
2 3
45
89
4
3
21
6
7
+-
1
1
2 3
45
89
Circuit Problem
Enlarged
Problem: Find the current running through each closed loop
1
4
3
2 1
5
6
7
30V +-
1
4
3
21
5
6
7
20V +-
1
4
3
21
5
6
7
10V +-
2 1
1
2 3
45
6
7
89
10
11
12
14
15
16
17
13
Shortcut to Loop Current
MatrixTo obtain the n’th row of the Matrix:1) The diagonal entry is equal to
the sum of the resistances on the n’th loop
2) There is an off diagonal entry foreach neighbor of the loop which is equal to:a) –sum(resistances) on shared wire if loop currents are in opposite direction b) sum(resistances) on shared wire if loop currents are in the same direction
1
4
3
2 1
5
6
7
30V +-
1
4
3
21
5
6
7
20V +-
1
4
3
21
5
6
7
10V +-
2 1
1
2 3
45
6
7
89
10
11
12
14
15
16
17
13
1
4
3
2 1
5
6
7
+-
1
4
3
21
5
6
7
+-
1
4
3
21
5
6
7
+-
2 1
1
2 3
45
6
7
89
10
11
12
14
15
16
17
1 7 2 0 0 3 0 1 0 0 0 0 0 0 0 0 0 0
2 2 7 1 0 0 0 0 0 0 4 0 0 0 0 0 0 0
3 0 1 12 0 0 0 0 0 0 0 0 0 0 0 6 0 0
4 0 0 0 13 7 0 0 6 0 0 0 0 0 0 0 0 0
5 3 0 0 7 14 0 0 0 4 0 0 0 0 0 0 0 0
6 0 0 0 0 0 13 0 0 0 0 0 0 0 0 0 7 6
7 1 0 0 0 0 0 7 0 0 1 3 2 0 0 0 0 0
8 0 0 0 6 0 0 0 12 1 0 0 0 0 0 0 0 0
9 0 0 0 0 4 0 0 1 7 0 0
10
11
12
13
14
15
16
17
0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 6 0 0 0 2 0 3 0
0 4 0 0 0 0 3 0 0 0 14 0 0 0 7 0 0
0 0 0 0 0 0 2 0 0 0 0 7 0 0 0 4 1
0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
0 0 0 0 0 0 0 0 0 2 0 0 1 3 0 0 0
0 0 6 0 0 0 0 0 0 0 7 0 0 0 13 0 0
0 0 0 0 0 7 0 0 0 3 0 4 0 0 0 14 0
0 0 0 0 0 6 0 0 0 0 0 1 0 0 0 0 12
13
1 7 2 0 0 3 0 1 0 0 0 0 0 0 0 0 0 0
2 2 7 1 0 0 0 0 0 0 4 0 0 0 0 0 0 0
3 0 1 12 0 0 0 0 0 0 0 0 0 0 0 6 0 0
4 0 0 0 13 7 0 0 6 0 0 0 0 0 0 0 0 0
5 3 0 0 7 14 0 0 0 4 0 0 0 0 0 0 0 0
6 0 0 0 0 0 13 0 0 0 0 0 0 0 0 0 7 6
7 1 0 0 0 0 0 7 0 0 1 3 2 0 0 0 0 0
8 0 0 0 6 0 0 0 12 1 0 0 0 0 0 0 0 0
9 0 0 0 0 4 0 0 1 7 0 0
10
11
12
13
14
15
16
17
0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 6 0 0 0 2 0 3 0
0 4 0 0 0 0 3 0 0 0 14 0 0 0 7 0 0
0 0 0 0 0 0 2 0 0 0 0 7 0 0 0 4 1
0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0
0 0 0 0 0 0 0 0 0 2 0 0 1 3 0 0 0
0 0 6 0 0 0 0 0 0 0 7 0 0 0 13 0 0
0 0 0 0 0 7 0 0 0 3 0 4 0 0 0 14 0
0 0 0 0 0 6 0 0 0 0 0 1 0 0 0 0 12
Using Sparse Matrices in Matlab
Run bigcircuit1) Create the list of non-zeros2) Convert the list to Matlab’s sparse matrix format3) Convert the sparse matrix to a full matrix (just for viewing)
Counting the Non-Zeros with nnz
• We can use Matlab’s built in nnz function to find the number of non-zeros:
i.e. there are only 58 non-zero entries out of 17x17=289 possible
Now We LU Factorize The Sparse Matrix
• Since there are only a few degrees of freedom we will use a direct method to factorize and solve the system.
• We can use the built in LU factorization of the matrix…
• i.e. find two matrices L & U such that A=LU where L is logically lower triangle and U is logically upper triangle..
LU Factorization of Loop Current Matrix
Solving The System
• Now we have factorized the system into A=LU we can solve in three stages.
1) Build the source vector, v
2) Solve y = L\v
3) Solve for the currents I = U\y
1
4
3
2 1
5
6
7
+-
1
4
3
21
5
6
7
+-
1
4
3
21
5
6
7
+-
2 1
1
2 3
45
6
7
89
10
11
12
14
15
16
17
13
30V
20V
10V
Solving for the loop currents using an LU factorization
1
4
3
2 1
5
6
7
+-
1
4
3
21
5
6
7
+-
1
4
3
21
5
6
7
+-
2 1
1
2 3
45
6
7
89
10
11
12
14
15
16
17
13
30V
20V
10V
Solving for the loop currents using an LU factorization
Let’s Renumber
• Matlab has a built in routine symrcm which takes a symmetric matrix and returns a permutation array so that if we use this to permute the unknowns (by column and row swaps) the bandwidth of the matrix may be reduced…
Renumbering with symrcm
How Were The Loops Renumbered
• We can examine the permutation matrix:
• i.e. the old 13 cell becomes the new 1 cell
• 6->2, 14->3…
1
4
3
2 1
5
6
7
+-
1
4
3
21
5
6
7
+-
1
4
3
21
5
6
7
+-
2 1
1
2 3
45
6
7
89
10
11
12
14
15
16
17
13
1
4
3
2 1
5
6
7
+-
1
4
3
21
5
6
7
+-
1
4
3
21
5
6
7
+-
2 1
13
12 10
1514
2
11
1716
7
9
8
3
6
4
5
1
Let’s Figure Out the Sequence of
Shells in symrcm
1
4
3
2 1
5
6
7
+-
1
4
3
21
5
6
7
+-
1
4
3
21
5
6
7
+-
2 1
13
12 10
1514
2
11
1716
7
9
8
3
6
4
5
1
Level 1
Level 2
Level 3
Level 4
Level 5
Level 6
Let’s Try LU Factorization of the Reordered Matrix
Effect of Reordering
Before
After
Notes• We could have also used Cholesky
factorization (since the loop current matrix is symmetric).
• Just by reordering unknowns we have changed the amount of fill in the L,U factors of the matrix.
• Changing the ordering of cells will not change the answer (beyond round off).
• However, we have reduced the amount of work required in the backsolves.
Solving The Reordered System
Comparing The Results• We can compare the results from solving using the
original matrix and using the reordered matrix:No reordering With reordering
OK – Let’s Get Serious And Look At A Large Circuit Case
We can construct a random circuit:
The Sparsity Pattern of a Loop Circuit Matrix for a Random Circuit (with 1000 closed loops)
Notes
• For a more realistic circuit (i.e. with less random interconnects) the fill in the L & U matrices will be reduced more after reordering using RCM
Lab Task• Construct a more realistic “random” circuit for an arbitrary N (i.e.
design a process which randomly grows a circuit). Use an interconnect of say 4 per cell on average.
• Time how long the LU factorization takes for N=10,100,1000,1e4,1e5,1e6,1e7,1e8and plot a graph of time v. N
• Calculate the number of non-zeros (i.e. how much memory is taken by the L & U matrices). Plot this as a function of N.
• Solve with a random source vector for N=10,100,…,1e6 and time how long the backsolves take and plot a graph of time v. N
• Perform a polynomial fit of the timings (estimate the polynomial growth rate with N)…
• USE SPARSE MATRICES!. You can use Cholesky if you wish.