更新时间:2022-06-17 15:36:26
这是几件事的组合:
这是我机器上的数字:Core i7 920 @ 3.5 GHz(4核)
Here's the numbers on my machine: Core i7 920 @ 3.5 GHz (4 cores)
>> a = rand(10000);
>> b = rand(10000);
>> tic;a*b;toc
Elapsed time is 52.624931 seconds.
任务管理器显示4个CPU使用核心.
Task Manager shows 4 cores of CPU usage.
现在进行一些数学运算
Number of multiplies = 10000^3 = 1,000,000,000,000 = 10^12
Max multiplies in 53 secs =
(3.5 GHz) * (4 cores) * (2 mul/cycle via SSE) * (52.6 secs) = 1.47 * 10^12
因此Matlab可以在最大可能的CPU吞吐量上达到1 / 1.47 = 68%
的效率.
So Matlab is achieving about 1 / 1.47 = 68%
efficiency of the maximum possible CPU throughput.
我没有发现异常.