Barcelona Opteron 2350(B1) arrived

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Originally Posted by kyosen

CPU-Z latest beta screenshot
Now it can treat each core clock...we can select the core with mouse right-click.
And it also shows Core VID:)
http://www.oohashi.jp/c-board/file/C...core_clock.png
Thanks to Franck, as usual:up:

3DMark06 with GeForce8800GT on WinXP x64
2350(B1)-2.2G=220x10、GeForce8800GT,
3DMark Score, CPU Score = 9973, 3212
http://www.oohashi.jp/c-board/file/C...core_clock.png
In previous test on WinXP 32bit: 10148, 3267
http://www.oohashi.jp/c-board/file/3....2G-220x10.png
So, I couldn't find no merit for WinXP x64 about 3DMark06, as expected.

NorthBridge multiplier again
I rebooted after changing register in this time,
and I could confrim that NB clock was down on BIOS display.
http://www.oohashi.jp/c-board/file/O...NB1.0_BIOS.jpg
And SuperPI4M time at same Core clock and different NB clock:

NB-1.8G: 3m48.406s, http://www.oohashi.jp/c-board/file/S...-1.8_WinXP.png
NB-1.6G: 3m48.016s, http://www.oohashi.jp/c-board/file/S...-1.6_WinXP.png
NB-1.4G: 3m50.500s, http://www.oohashi.jp/c-board/file/S...-1.4_WinXP.png
NB-1.2G: 3m53.532s, http://www.oohashi.jp/c-board/file/S...-1.2_WinXP.png
NB-1.0G: 3m58.359s, http://www.oohashi.jp/c-board/file/S...-1.0_WinXP.png

NB>=1.6G looks enough for single thread program with dual DDR2-667.
There may be difference between NB-1.6G and NB-1.8G in case of DDR2-800 setting.

BTW, I've learned reboot(warm reset) is needed for changing HT Link multiplier too.
Yeah, tictac and macci was/is right, as usual:yepp:

Folding@Home
Quick test on 2350(B1)-2.0G, DDR2-667.
Screenshot at 7% steps:
http://www.oohashi.jp/c-board/file/F...0_step-35k.png

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Thanks for the SMP tests :clap: now at least I have an idea of its performance.

Something seems weird about that SMP test.... my Opteron is only 6 minutes slower than that. I would think it would be better. I guess my opteron is running 800MHz faster, but still I guess I was expecting it to be faster than that.

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Originally Posted by SparkyJJO

Something seems weird about that SMP test.... my Opteron is only 6 minutes slower than that. I would think it would be better. I guess my opteron is running 800MHz faster, but still I guess I was expecting it to be faster than that.

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Just quick guesstimation:
my result: 15m24s=924s for each 5000 steps with 2.0G x4 cores K10 Optreon
your score: 15m24s + 6m = 1284s with 2.8G x2 cores K8 Optreon
So, 1284/x * 2.8/2.0 /y = 924
...here x is efficiency of increased cores x2->x4, and y is performance gain per core.
for example, if y is ~1.05, x is ~1.85 from the formula above...
...yeah x should be within 2.0 in this case.
In my experience for SuperPI, y=~1.05 is feasible, at least on my board and current BIOS, so far.
I don't know usual efficiency x for Folding@Home, but 1.85 looks feasible too...

are you saying that, from this comparison, that the performance gain from K8 to K10 is about... 5%?

2.0ghz Barcelona gets 924s
2.8ghz K8 Opteron gets 1284s

2.0 x 8 = 16,000mhz
2.8 x 4 = 11,200mhz

16,000/924 = 17.316
11,200/1284 = 8.723

17.316/8.723 = 1.985 speedup factor.

edit - i'm assuming a dual quad and a dual dual here, but half the numbers (1 proc each) and you get the same... unless there's a number out somewhere in which case my bad ;)

Does the SMP client work one work unit across all cores or is it one per core? I assume it is one per core in this calculation. One across all cores will be different numbers.

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Originally Posted by JohannesRS

are you saying that, from this comparison, that the performance gain from K8 to K10 is about... 5%?

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~5% performance gain from K8 to K10 is based on my own results about SuperPI4M run.
I don't know whether the gain is same or not for both SuperPI and F@H, at this moment.
My intention of previous post is just suggestion of rough estimation formula.
Under assumption of that formula,
*if x = 2.0(ideal scaling), y < 1.0, i.e. gain from K8 to K10 is negative...it's not feasible.
*if x = ~1.95, y = 1.0, i.e. no gain from K8 to K10...it's not feasible too.
*if x = ~1.90, y = ~1.02
*if x = ~1.85, y = ~1.05
*if x = ~1.8, y = ~1.08
*if x = ~1.6, y = ~1.2
...
x=1.8~1.85 looks feasible for me as result of multi-thread program,
then, y=1.08~1.05...and it's not inconsistent with my SuperPI1M&4M results.

The thing is, for multithreaded programs, you can't fairly compare the K8 vs K10 because there isn't any quad cores K8. The only way to do a fair comparison would be to disable two of the cores when testing.

Quote:

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Originally Posted by kyosen

Just quick guesstimation:
my result: 15m24s=924s for each 5000 steps with 2.0G x4 cores K10 Optreon
your score: 15m24s + 6m = 1284s with 2.8G x2 cores K8 Optreon
So, 1284/x * 2.8/2.0 /y = 924
...here x is efficiency of increased cores x2->x4, and y is performance gain per core.
for example, if y is ~1.05, x is ~1.85 from the formula above...
...yeah x should be within 2.0 in this case.
In my experience for SuperPI, y=~1.05 is feasible, at least on my board and current BIOS, so far.
I don't know usual efficiency x for Folding@Home, but 1.85 looks feasible too...

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FYI, this is about on par with Kentsfield at 2.0GHz.

I pull down about 14min flat per 1% on a 2653 WU on my octacore Clovertown server (running 2 instances - one per socket).

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Originally Posted by STEvil

Does the SMP client work one work unit across all cores or is it one per core? I assume it is one per core in this calculation. One across all cores will be different numbers.

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The SMP client spawns 4 total threads that actively do calculations. If you have less than 4 cores, 2 threads are shared on a core.

If you have MORE than 4 cores (dual socket quad cores), then you have to run multiple instances of the client to load up all cores.

Here is a SS from my server:
http://img474.imageshack.us/img474/6...3223xe1.th.jpg

Dual G0 Xeon E5310 @ 2.0GHz.

On a single client on this machine I normally get about 13.8-14.2 min per 1&#37;. When you load up 2 clients, the time increases to 14.4 min per 1% per client. I can only assume that is b/c both processors share the same FSB and memory.

Which part of the core does the SMP client use most (what does it depend most on)?
How does it scale with frequency?

*edit* nvm,

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Originally Posted by KTE

Which part of the core does the SMP client use most (what does it depend most on)?
How does it scale with frequency?

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P2653
X3210 @ 2.00Ghz : 13:15
X3210 @ 2.96Ghz : 9:30

Core 2 quad (Kentsfield) 2.96 GHz: 1 SMP instance 2653: 8:5x
Core 2 quad (Kentsfield) 2.96 Ghz: 2 SMP 2653: 15:45

OS: tweaked WinXP pro
IP-35E, 370 fsb, 2 GB Adata 1110 MHz CL5, performance level 5

If I could get 450+ fsb it would run even faster, but this mobo is not a good clocker.

Concerning unusualy high speedup in 64bit Cinebench, notice that with 4 threads, total L2 cache available to application is 4x0.5MB =2MB, while single thread only has 0.5MB available to it.

It is known that sometimes it is possible to get more than 100&#37; (n*100%) speedup because of bigger aggregate cache available then application is running several threads.

Note that Core2 has significantly lower speedup because of shared nature of L2.

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Originally Posted by JVguest

Core 2 quad (Kentsfield) 2.96 GHz: 1 SMP instance 2653: 8:5x
Core 2 quad (Kentsfield) 2.96 Ghz: 2 SMP 2653: 15:45

OS: tweaked WinXP pro
IP-35E, 370 fsb, 2 GB Adata 1110 MHz CL5, performance level 5

If I could get 450+ fsb it would run even faster, but this mobo is not a good clocker.

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I think I'll have to investigate what impact FSB and memory has on SMP results. Because you are getting about a 7&#37; increase in performance from me with a faster FSB and memory speed on the same project.

i know L3 cache is being recognized, but is it being utilized?

Well about the ganged and unganged memory, I asked my professor about it, so hopefully he'll give a good response.

Excerpt from BKDG For AMD Family 10h Processors Page 60.

2.8 DRAM Controllers (DCTs)

The DCTs support DDR2 DIMMs or DDR3 DIMMs. Products may be configurable between DDR2 and DDR3 operation.

A DRAM channel is the group of the DRAM interface pins that connect to one series of DIMMs. The processor supports two DDR channels. The processor includes two DCTs. Each DCT controls one 64-bit DDR DIMM channel.
For DDR products, DCT0 controls channel A DDR pins and DCT1 controls channel B DDR pins. However, the processor may be configured: (1) to behave as a single dual-channel DCT; this is called ganged mode; or

(2) to behave as two single-channel DCTs; this is called unganged mode.
A logical DIMM is either one 64-bit DIMM (as in unganged mode) or two identical DIMMs in parallel to create a 128-bit interface (as in ganged mode). See section 1.5.2 [Supported Feature Variations] on page 20 for information about supported package/DRAM configurations.

For DDR products, when the DCTs are in ganged mode, as specified by [The DRAM Controller Select Low Register] F2x110[DctGangEn], then each logical DIMM is two channels wide. Each physical DIMM of a 2-channel logical DIMM is required to be the same size and use the same timing parameters. Both DCTs must be programmed with the same information (see section 2.8.1 [DCT Configuration Registers] on page 61). When the DCTs are in 64-bit mode, a logical DIMM is equivalent to a 64-bit physical DIMM and each channel is controlled by a different DCT.

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Originally Posted by tictac

:D

http://img229.imageshack.us/img229/6...oreeditzo9.png

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One Question, I have two Barcelonas in my system, is it correct that register C0010065 is for second CPU?

In my system the first cpu has VCore 1.15V and second has 1.1V at default, the register shows the number 40 and 44 when made rdmsr.

Hi, guys! Is it now confirmed that the ram clock is derived only from the external clock ("Ext.CLK", "System Clock", etc.), and not from the NB clock? (I think the IMC is part of the NB, so it would seem logical to me.)

ps., BTW, I'm sorry, but just why do you keep on writing FSB and "bus speed", in relation to the external clock? This is only a clock, not a bus, expecially not a Front Side Bus, like at Intel. (HT bus clock is derived from it by a multiplier, as we know.)

Another Quote from AMD's k10 dev guide

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2.6 The Northbridge (NB)
Each processor includes a single Northbridge that provides the interface to the local CPU core(s), the interface to system memory, the interface to other processors, and the interface to system IO devices. The NB includes all power planes except VDD; see section 2.4.1 [Processor Power Planes And Voltage Control] on page 25 for more information.
The NB of each node is responsible for routing transactions sourced from CPU cores and links to the appropriate CPU core, cache, DRAM, or link. See section 2.9.3 [Access Type Determination] on page 107 for more information.

2.6.1 Northbridge (NB) Architecture

Major NB blocks are: System Request Interface (SRI), Memory Controller (MCT), DRAM Controllers (DCTs), L3 cache, and Cross Bar (XBAR). SRI interfaces with the CPU core(s). MCT maintains cache coherency and interfaces with the DCTs; MCT maintains a queue of incoming requests called MCQ. XBAR is a switch that routes packets between SRI, MCT, and the links.
The MCT operates on physical addresses. Before passing transactions to the DCTs, the MCT converts physical addresses into normalized addresses that correspond to the values programmed into [The DRAM CS Base Address Registers] F2x[1, 0][5C:40]. Normalized addresses include only address bits within the DCTs’ range.
The normalized address varies based on DCT interleave and hoisting settings in [The DRAM Controller Select Low Register] F2x110 and [The DRAM Controller Select High Register] F2x114 as well as node interleaving based on [The DRAM Base/Limit Registers] F1x[1, 0][7C:40].

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Code:

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Core 0 ----
          |
Core 1 ----               
          |---- SRI ---- XBAR ---- cHT
Core 2 ----                |
          |              MCT --- L3
Core 3 ----                |
                          ---
                          |  |
                        DCT0 DCT1
                          |  |
                      LDIMM0 LDIMM1

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So SRI, XBAR, MCT, L3 DCT0 and DCT1 all run on NB speed?
Or do the DCT's run on ram clock?

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Originally Posted by indiana_74

One Question, I have two Barcelonas in my system, is it correct that register C0010065 is for second CPU?

In my system the first cpu has VCore 1.15V and second has 1.1V at default, the register shows the number 40 and 44 when made rdmsr.

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I have to answer myself and to tell that i am wrong :(

With the step postet by tictac i only changed the vcore of the first core, to change the second, third till eight&#180;s core i have to switch the cores in Cyrstalcpuid and then reload with MSR Editor.

I made a regdump with cpu-z and i wonder about the msr code of core 1-3

Now I inserted second CPU and made a new regdump.

Second CPU runs with 1.10V and the MSR Code is

40 = 1.15V
48 = 1.10V

Now I changed every register to 38 = 1.20V and beginn a stresstest at 2.0GHz with my Barcelona 2344

kyosen

Could you please test a sisoft sandra 2007 benchmark on a 32 bit system to obtain such kind of results:
http://www.expreview.com/img/news/07...6047070_rs.jpg

I'd like to know if the low memory score is strange or is due to sandra.
thanks

sorry.. indiana i cant get back to you.. i am viewing this from opera mini mobile phone.. each processor have it own MSR.. 64 stand for p-state0.. 65 p-state1.. and so on..

yeah you did it right.. change cpu id from crystal spui than open MSR editor for each processor... :up: :up:

other p-state than 0 used for power saving CnQ implementation..

Maybe I just missed it, but how is the memory clock calculated? Is there a calculateing difference between Barcelona and Phenom?