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Lisa Pizzo 2025-10-29 23:00:03 +01:00
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STREAM version $Revision: 5.10 $
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This system uses 8 bytes per array element.
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Array size = 10000000 (elements), Offset = 0 (elements)
Memory per array = 76.3 MiB (= 0.1 GiB).
Total memory required = 228.9 MiB (= 0.2 GiB).
Each kernel will be executed 10 times.
The *best* time for each kernel (excluding the first iteration)
will be used to compute the reported bandwidth.
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Your clock granularity/precision appears to be 1 microseconds.
Each test below will take on the order of 15540 microseconds.
(= 15540 clock ticks)
Increase the size of the arrays if this shows that
you are not getting at least 20 clock ticks per test.
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WARNING -- The above is only a rough guideline.
For best results, please be sure you know the
precision of your system timer.
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Function Best Rate MB/s Avg time Min time Max time
Copy: 14388.7 0.014571 0.011120 0.016345
Scale: 13108.2 0.014643 0.012206 0.018382
Add: 20068.4 0.014428 0.011959 0.016789
Triad: 18490.0 0.015207 0.012980 0.016955
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Solution Validates: avg error less than 1.000000e-13 on all three arrays
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FLOPS C Program (Double Precision), V2.0 18 Dec 1992
Module Error RunTime MFLOPS
(usec)
1 2.8422e-14 0.0030 4679.3328
2 2.5047e-13 0.0006 11677.4813
3 -7.6605e-15 0.0034 4957.0289
4 2.2782e-13 0.0039 3808.7061
5 3.8969e-14 0.0055 5259.0825
6 7.6050e-15 0.0054 5326.1384
7 -1.1369e-13 0.0045 2649.6869
8 1.2612e-13 0.0056 5363.6576
Iterations = 128000000
NullTime (usec) = 0.0010
MFLOPS(1) = 8091.2502
MFLOPS(2) = 3854.5260
MFLOPS(3) = 4643.6405
MFLOPS(4) = 4944.0769