1.7 KiB
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4 de Outubro 2023 - #CP
Ex. 2
a) Limitações vetoriais
A -> consecutive elements in a row -> consecutive access in the vector C -> same element B -> consecutive elements in a collumn
Não vai ser vetorizável.
b) Enable vectorization
result of change cycles to i, k , j : A -> same element C -> consecutive elements in a row -> consecutive access in the vector B -> consecutive elements in a row -> consecutive access in the vector
i k j 0 0 1
Vai ser vetorizável.
128b 8B -> 64b 2 elements
Without vectorization:
!
With vectorization:
!
Estimated: ( n^3 / 2 )* 8
c)Measure and analyze results
| N | Version | Time | CPI | #I |
|---|---|---|---|---|
| 512 | base_v() | 0.492484818 | 0.91 | 1113554887 |
| 512 | vect() | 0.081604350 | 2.88 | 578275097 |
[!note]- Commands run module load gcc/9.3.0 gcc -O2 -ftree-vectorize -msse4 mmult.c srun --partition=cpar perf stat -e cycles,instructions ./a.out
d) Vectorization fine-tuning
Ganhos de 4 vezes mais.
Ex. 3
a) Peak Performance
2 operações em FP 4 elementos de cada vez em cada ciclo 2.5 GHz -> 2.5 billion cycles per second
conclusion: 20 GFlop/s
b)
peak with vectorization: 20 GFlop/s peak without vectorization: 5 GFlop/s memory bandwith limitation: 20 GFlop/s real achievable performance: measured performance:
c)
memory bandwith limitation
| GFlop/s | Flop/Byte |
|---|---|
| 0.125 | |
| 0.25 | |
| 0.5 | 10 |
| 1 | 20 |
| 2 | |
| 4 | |
| 8 |