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Old 09-23-07   #71 (permalink)
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Thanks for taking the time to write such a great guide.
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Old 09-26-07   #72 (permalink)
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ok, i think i got it now. thank u

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Old 09-26-07   #73 (permalink)
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Make sure and Rep+ if you feel he deserves it... I sure do myself!
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Old 11-01-07   #74 (permalink)
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Why isn't this stickied? REP+
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Old 11-13-07   #75 (permalink)
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BUMP, as I feel some of the newcomers should be reading this.

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Old 12-24-07   #76 (permalink)
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this article doesn't discuss dual channel ddr2. On a system running, say a q6600 at 266 FSB, the ram will be running at ddr2-533. However, isn't it running dual channel, giving the equivalent bandwidth of ddr2 1066 single channel to the fsb? i'm a little confused because my dual channel ddr2-800 setup is showing a bandwidth of about 5.9 gb/sec in sandra, but is that only testing one channel so that the actual bandwidth is double that?
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Old 01-01-08   #77 (permalink)
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Quote:
Originally Posted by Ddejean
Hi everyone

Great guide, but I've got one question about memory bandwidth.
It seems to me that you are not considering double data transfer speed when using two memory sticks in your ratio calculations.
Quote:
Originally Posted by papa_smurf View Post
this article doesn't discuss dual channel ddr2. On a system running, say a q6600 at 266 FSB, the ram will be running at ddr2-533. However, isn't it running dual channel, giving the equivalent bandwidth of ddr2 1066 single channel to the fsb? i'm a little confused because my dual channel ddr2-800 setup is showing a bandwidth of about 5.9 gb/sec in sandra, but is that only testing one channel so that the actual bandwidth is double that?
Dual channel affects the bandwidth of the memory, not the speed that it runs at. There's a difference between the two. Think of it as adding another lane on a highway. More traffic can flow but it won't be going any faster than traffic going in the other direction.

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Last edited by t4ct1c47 : 01-01-08 at 10:15 PM
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Old 01-03-08   #78 (permalink)
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i understand that, but when matching fsb bandwidth, the actual bandwidth of the memory configuration, not just the speed is what needs to be considered. Am i correct in saying this?
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Old 01-23-08   #79 (permalink)
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Quote:
Originally Posted by pauldovi View Post
[B](Version 1.10)

The NBCC can be calculated by dividing your CPU current multiplier by its default multiplier and then multiplier the sum by your FSB.

For Example:

E6600 @ 500Mhz and a 7 multiplier:

(9 / 7) x 500 = 642Mhz NBCC

So it can be seen that lowering your multiplier, even though offering addition headway for FSB on the CPU, will increase the NBCC, reduce NB stability and thus cause the overall system stability to decrease.
According to your formulae, you are dividing the Default Multiplier by the current set multiplier, but your wording describes it in reverse. Which way is correct? For a Q6700 with 10 max, current set to 8 with 400 FSB
Is it (10/8) * 400 = 500NBCC, or (8/10) * 400 = 320NBCC? I am assuming it is the former, in which case you should change your wording to match your formula. Confusing to a noob like myself.

This is a great guide! I agree with others that a greater understanding of the NB strap would be very helpful. I think my lack of knowledge there is preventing me from getting to a rock solid 3.6 OC of my 2.66 Q6700, instead of the steady 3.2 I am at.

I using 1600MHz DDR3 and getting 16:8, which now seems kind of good, since it follows your logic of the mem being at twice the CPU FSB (800 to 400) So is a 2:1 DRAM:FSB ratio good news?

Last edited by redeyemike : 01-24-08 at 12:06 AM Reason: Need to ask explicit question
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Old 02-05-08   #80 (permalink)
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There is an article on Anandtech, in their review of the QX9650, which discusses the NB strap and memory latency in detail and I think adds something meaningful to this discussion (specifically the relative importance of Trd.) I'm a noob so maybe I'm missing the point but I think that the Anandtech article might shed some light on what's going on behind the scenes of NB straps and memory performance.

http://www.anandtech.com/cpuchipsets...spx?i=3184&p=1
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