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I´m writing my master thesis on working with Intel Cache Allocation Technology.
In the specs of my CPU (https://ark.intel.com/de/products/92981/Intel-Xeon-Processor-E5-2630-v4-25M-Cache-2_20-GHz Intel® Xeon® Processor E5-2630 v4 (25M Cache, 2.20 GHz) Produktspezifikationen ) there is a number of 68.3GB/s maximum memory throughput.
I calculated the throughput on my own (clock * bus-width = 2.2GHz * (4 * 64 Bit)) and the result is 70.4 GB/s .. or 65.6 GiB/s
So why is there a difference? Maybe someone got confused by GibiByte and Gigabyte?
thanks,
Norbert Schramm
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Hello stoertebeker:
In regard to your question, I just wanted to let you know that we will review this case in order to confirm and to provide the most accurate response related to this matter.
As soon as I get any updates I will post all the details on this thread.
Any further questions, please let me know.
Regards
Alberto
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Hello stoertebeker:
I just wanted to let you know that we are still working on this case to gather the information you requested.
As soon get any information, I will post all the details on this thread.
Any questions, please let me know.
Regards
Alberto
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Is there anything new?
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The memory BW is dependent on the speed of DDR4, not on the speed of the core itself. For E5-2630 v4, max DDR4 speed is 2133 MT/s. So, the total BW is 4 channels * 64 bits per channel * 2.133 GHz then divide by 8 and you have what you see on ARK, 68.256GB/s
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