Good Day,
Today I found an article in Anandtech with the following text, about the 800Mhz bus:
It just got me confused a bit!
I do my calculations this way:
CPU at 333Mhz, needs DDR333, which gives a 8x333=2664GB/s bandwidth (it's a PC2700 DDR333) ... also for DDR400 I have 8x400=3200GB/s (it's a PC3200 DDR400). Then for PC1066 RDRAM I do believe I have to multiply like 4x1066=4264GB/s ... I do think it's because one is 32bit and the other is 64bit ... confusing since I'm not seeing all the picture ... I'm missing something here to help justify the calculations ...
I don't know exactly why I multiply by 8, but it seems the magic number to the result I want ... but then comes this 32bit, 64bit and 2x64bit stuff ... HELP!
Can anyone explain in a simple way, if possible? If not possible can you give another examples? Something that can turn the light on my head ...
Thanks
Today I found an article in Anandtech with the following text, about the 800Mhz bus:
In this case, the focus is more on balancing FSB and memory bandwidth rather than frequencies (although it is important to have frequencies that are in sync with one another in order to keep latencies as low as possible). The Pentium 4 features a 64-bit wide FSB interface, and we've already explained the frequencies this FSB can run at. Simple multiplication shows us that the 533MHz FSB can offer a maximum of 4.264GB/s of bandwidth. The 845PE chipset has a 64-bit DDR333 memory interface, offering a maximum of 2.664GB/s of memory bandwidth, and the 850E chipset has a 32-bit PC1066 RDRAM memory interface that provides at most 4.264GB/s of memory bandwidth. As you can see just by looking at the bandwidth numbers, the 850E chipset is perfectly balanced for the amount of bandwidth offered by the 533MHz FSB, which is why it is significantly faster than the 845PE.
The 800MHz FSB will offer no less than 6.4GB/s of bandwidth, which would require either a 32-bit PC1600 RDRAM memory interface (note that the PC1600 standard does not exist) or a 64-bit DDR800 memory interface (also a non-existant memory technology). Well, if you can't get faster memory, you widen the memory interface in order to increase bandwidth. Remember that bandwidth is the product of bus width and transfer rate, so if you can't improve the transfer rate, you increase the width of the bus. In the case of the upcoming 865 and 875 chipsets, Intel took the 64-bit DDR memory interface of the 845PE chipset and added a second 64-bit channel along with adding DDR400 support. A 128-bit memory interface (2 x 64-bit channels) with DDR400 memory now offers exactly 6.4GB/s of memory bandwidth, perfectly balanced with the 800MHz FSB, without using exotic memory technology or speeds that aren't readily available.
It just got me confused a bit!
I do my calculations this way:
CPU at 333Mhz, needs DDR333, which gives a 8x333=2664GB/s bandwidth (it's a PC2700 DDR333) ... also for DDR400 I have 8x400=3200GB/s (it's a PC3200 DDR400). Then for PC1066 RDRAM I do believe I have to multiply like 4x1066=4264GB/s ... I do think it's because one is 32bit and the other is 64bit ... confusing since I'm not seeing all the picture ... I'm missing something here to help justify the calculations ...
I don't know exactly why I multiply by 8, but it seems the magic number to the result I want ... but then comes this 32bit, 64bit and 2x64bit stuff ... HELP!
Can anyone explain in a simple way, if possible? If not possible can you give another examples? Something that can turn the light on my head ...
Thanks