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HPLinpack 1.0a -- High-Performance Linpack benchmark -- January 20, 2004<br />
Written by A. Petitet and R. Clint Whaley, Innovative Computing Labs., UTK<br />
============================================================================<br />
<br />
An explanation of the input/output parameters follows:<br />
T/V : Wall time / encoded variant.<br />
N : The order of the coefficient matrix A.<br />
NB : The partitioning blocking factor.<br />
P : The number of process rows.<br />
Q : The number of process columns.<br />
Time : Time in seconds to solve the linear system.<br />
Gflops : Rate of execution for solving the linear system.<br />
<br />
The following parameter values will be used:<br />
<br />
N : 19360<br />
NB : 220<br />
PMAP : Row-major process mapping<br />
P : 2<br />
Q : 2<br />
PFACT : Left<br />
NBMIN : 4<br />
NDIV : 2<br />
RFACT : Right<br />
BCAST : 1ringM<br />
DEPTH : 0<br />
SWAP : Mix (threshold = 64)<br />
L1 : transposed form<br />
U : transposed form<br />
EQUIL : yes<br />
ALIGN : 8 double precision words<br />
<br />
----------------------------------------------------------------------------<br />
<br />
- The matrix A is randomly generated for each test.<br />
- The following scaled residual checks will be computed:<br />
1) ||Ax-b||_oo / ( eps * ||A||_1 * N )<br />
2) ||Ax-b||_oo / ( eps * ||A||_1 * ||x||_1 )<br />
3) ||Ax-b||_oo / ( eps * ||A||_oo * ||x||_oo )<br />
- The relative machine precision (eps) is taken to be 1.110223e-16<br />
- Computational tests pass if scaled residuals are less than 16.0<br />
<br />
============================================================================<br />
T/V N NB P Q Time Gflops<br />
----------------------------------------------------------------------------<br />
WR01R2L4 19360 220 2 2 101.40 4.771e+01<br />
----------------------------------------------------------------------------<br />
||Ax-b||_oo / ( eps * ||A||_1 * N ) = 0.0041180 ...... PASSED<br />
||Ax-b||_oo / ( eps * ||A||_1 * ||x||_1 ) = 0.0062675 ...... PASSED<br />
||Ax-b||_oo / ( eps * ||A||_oo * ||x||_oo ) = 0.0011940 ...... PASSED<br />
============================================================================<br />
<br />
Finished 1 tests with the following results:<br />
1 tests completed and passed residual checks,<br />
0 tests completed and failed residual checks,<br />
0 tests skipped because of illegal input values.<br />
----------------------------------------------------------------------------<br />
<br />
End of Tests.<br />
============================================================================<br />
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