如何计算CPU的理论峰值性能 [英] How to compute the theoretical peak performance of CPU
问题描述
这是我的 cat / proc / cpuinfo
输出:
...
processor : 15
vendor_id : GenuineIntel
cpu family : 6
model : 26
model name : Intel(R) Xeon(R) CPU E5520 @ 2.27GHz
stepping : 5
cpu MHz : 1600.000
cache size : 8192 KB
physical id : 1
siblings : 8
core id : 3
cpu cores : 4
apicid : 23
fpu : yes
fpu_exception : yes
cpuid level : 11
wp : yes
flags : fpu vme de pse tsc msr pae mce cx8 apic ...
bogomips : 4533.56
clflush size : 64
cache_alignment : 64
address sizes : 40 bits physical, 48 bits virtual
power management :
这台机器有两个CPU,每个CPU带有4个核,超线程功能,因此处理器总数为16(2 CPU * 4核心* 2超线程)。这些处理器具有相同的输出,为了保持清洁,我只显示最后一个的信息,并在标志行中省略部分标志。
This machine has two CPUs, each with 4 cores with hyperthreading capability, so the total processor number is 16(2 CPU * 4 core * 2 hyperthreading). These processors have same output, to keep clean, I just show the last one's info and omit part of flags in the flags line.
所以我该如何计算峰值性能这台机器的GFlops?
让我知道是否应该提供更多信息。
So how do I calculate the peak performance of this machine in terms of GFlops? Let me know if more info should be supplied.
谢谢。
推荐答案
您可以查看英特尔出口规范 。
图表中的GFLOP通常称为单个芯片的峰值。
它显示E5520的36.256 Gflop / s。
You can check the Intel export spec. The GFLOP in the chart is usually referred as the peak of a single chip. It shows 36.256 Gflop/s for E5520.
该单芯片具有4个带有SSE的物理内核。
因此,该GFLOP也可以计算为:
2.26GHz * 2(mul,add)* 2(SIMD双精度)* 4(物理核)= 36.2。
This single chip has 4 physical cores with SSE. So this GFLOP can also be calculated as: 2.26GHz*2(mul,add)*2(SIMD double precision)*4(physical core) = 36.2.
您的系统有两个CPU,因此峰值为36.2 * 2 = 72.4 GFLOP / S。
You system has two CPUs, so your peak is 36.2*2 = 72.4 GFLOP/S.
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