matching CPU FSB and memory throughput

UTscott82

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Jun 1, 2007
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I'm trying to match CPU front side bus (FSB) and memory throughput to ensure I get all the CPU/memory bandwidth I'm paying for. Back in the “pre-dual channel DDR” days this was easy. If your CPU had a 133MHz FSB, you’d buy 133MHZ memory and FSB & memory throughput would be matched. I’m a little confused now…

I’ve got an Intel Core2Duo extreme quad-core with an 8-byte wide 1066MHz FSB (total FSB throughput = 8.5 Gb/s). My chipset supports PC2-4200, PC2-5300, or PC2-6400 memory in single or dual channel configurations. To match my FSB, I’d want to select memory with a 8.5Gb/s throughput. I can’t find a motherboard that supports PC2-8500 memory, so I figure I’ll need to use the dual channel configurations to achieve 8.5Gb/s throughput on the memory bus. With dual channel DDR2 I can achieve the following throughput:

Dual Channel DDR2 PC2-4200 => peak throughput = 8.4Gb/s
Dual Channel DDR2 PC2-5300 => peak throughput = 10.6Gb/s
Dual Channel DDR2 PC2-6400 => peak throughput = 12.8Gb/s

It seems the best match for the FSB 8.5Gb/s throughput would be dual channel DDR2 PC-4200 (Sure the 8.4Gb/s memory throughput would be a little slower than the 8.5Gb/s FSB, but who cares…..). However, my motherboard datasheet indicates the highest performance configuration is dual channel (2) PC2-6400 DIMMs.

Why would I want to pay extra money for money for PC2-6400 memory if I’d the performance would bottlenecked by the lower throughput FSB? It seems to me that performance would be essentially the same with all 3 memory speeds because in all cases performance will be bottlenecked by the CPU’s FSB. Does this make sense? Are the marketing folks just trying to fool into paying more for higher speed that won’t do anything to enhance my system performance?
 

joefriday

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Feb 24, 2006
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First of all, you're confusing throughput with bandwidth. Secondly, a mobo would support a higher bandwidth memory than needed for two reasons: 1. It looks good on paper 2. It helps when overclocking.