## Group Hamrle

Group Description |
Circuit simulation matrices from Jaroslav Hamrle, RIKEN, Japan. -------------------------------------------------------------------------------- Date: Tue, 24 Feb 2004 17:33:31 +0900 From: Jaroslav Hamrle <(first initial, then last name) at the domain postman dot riken dot go dot jp > To: Cleve Moler <(last name) at mathworks dotcom> Subject: Re: How to solve A\y with large ( 60MB) sparse matrices ? Hello, the matrix is coming from solving of somewhat-like very large electrical network. So, it is like a large graph, where matrix elements gives connections between graph points. I'm sending you 'spy' of three matrices of this type. Maybe the most clear is the smallest 'spy', named Amatrix-small.pdf. In this matrix, I changed colors depending on values: red ..... +1 green ... -1 blue .... different from 1 and -1, usually number between 0.01 to 100 First two upper pars of the matrix given relation between values of nodes inside the graph. More precisely, they express what is the relation between current and voltage between both sides of one resistor. The first part treats resistors, which have special boundary conditions (resistors whith connected only one end to the circuity). In this case, the sub-matrices goes by on 'diagonal'. I used quotes here, because this 'diagonal' is not exactly diagonal, by goes in the way that skip of one line is followed by skip of two raws. The second part treats relationship between both ends of the resistors. In this part, the matrix 4x4 together with four adjective '-1' goes also on 'diagonal'. Note that for large matrices, this second part is much larger that the first one. The division between first two parts and second two part is exactly in the half of the matrix. The thrid part of the matrix provide relationship, that the sum of the current in the each electrical node is zero (sum(j_i))=0. Hence, each line of this part of the matrix contains severals (in presented matrix 4x) '1' or '-1'. The fourth part of the matrix says that the voltage at each nod is the same for all resistors ending/starting at given node. Hence, this part of matrix contains in each line just once '1' and once '-1'. Hence, this part only express that that some pairs of elements in resulting vector are identical. As this part takes more that 1/4 of total matrix, maybe by some reordiring it is possible to reduce significantly number of raws/lines. In conclusion, the most of the matrix contain only '1' and '-1' elements. And half of the lines (all third and fourt part) contains only '1' and '-1' elements. More that 1/4 of lines contains just one pair of '1' and '-1'. Also, it is important to say, that the vector x (when solving A\x) contains mainly zeros. Only 1%-5% of the vector x is non-zero. (And these zeros appears only in start of the x-vector, in part corresponding to the first part of the matrix.) If you are interested, I can send you my matrix by ftp. I thank you very much for any help. Jaroslav Hamrle |
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Displaying

**all 3**collection matricesId | Name | Group | Rows | Cols | Nonzeros | Kind | Date | Download File |
---|---|---|---|---|---|---|---|---|

1201 | Hamrle3 | Hamrle | 1,447,360 | 1,447,360 | 5,514,242 | Circuit Simulation Problem | 2004 | MATLAB Rutherford Boeing Matrix Market |

1200 | Hamrle2 | Hamrle | 5,952 | 5,952 | 22,162 | Circuit Simulation Problem | 2004 | MATLAB Rutherford Boeing Matrix Market |

1199 | Hamrle1 | Hamrle | 32 | 32 | 98 | Circuit Simulation Problem | 2004 | MATLAB Rutherford Boeing Matrix Market |