On Mon, 21 Apr 2008, Satish Balay wrote: > For eg - On intel Xeon machine with DDR2-800 - you have [othe memory bus > side]: > bandwidth = 2(banks)* 2(ddr)* 8(bytes bus) * 800 MHz/sec * = 25.6GByte/sec
My math was incorrect here.. DDR2-800 = 6.4Gb/s [its 2(ddr)* 400MHz/sec * 8bytes ] So this machine has 4 memory banks. i.e the above is: > bandwidth = 4(banks)* 2(ddr)* 400 MHz/sec* 8(bytes bus) * = 25.6GByte/sec Satish
