P2605 vs P2608

[v00d00]

Is there any particular reason the points on P2608 are less than P2605, while the amount of work done is more?

p2605_Protein in POPC 77373 3.00 4.00 1760.00 100
P2608 126010 4.00 4.00 1385.00 100

My computer frame times on both Projects.

P2605 - AMD X2 6000 @ 3.2GHz - 18mins 20secs / frame
P2608 - AMD X2 6000 @ 3.2GHz - 37mins 2secs / frame

Is this the initial push to phase out the bonus workunit (and yes, ive only just noticed this, and may be asking a dumb question )?

[_ikki_]

Maybe the benchmark machine is the reason of this amount of point

[ChasR]

P2608 is a cache intensive WU and the AMD rig you're running it on doesn't have enough cache.

On a Q6600, Ubuntu 7.04, 3.33 GHz, 2 GB ram @ 920 MHz, 4-4-4-12
p2605
TPF: 6:53.2
PPD: 3680.6

P2608:
TPF: 5:30.0
PPD: 3626.2

The benchmak machine is not a Q6600, but a the benchmark machine, a Woodcrest duallie @ 2.33, has the same 2 MB cache per core, which appears to be crucial to p2608 and p2609.

[Flathead74]

According to Intel's product brief, the 4MB L2 cache of a Woodcrest processor is shared, not discrete.

I believe this is where the performance advantage comes from, access to 4MB of L2 cache.

product brief
"4 MB shared L2 cache
• Entire L2 cache can be allocated to one core."

From the Folding@home SMP FAQ
"Before releasing any new project (series of work units), we benchmark it on a dedicated Macintosh Pro with 2 - 2.33 GHz Dual Core Xeon processors. (more specifically, 2 Woodcrest 5140 processors with 4 MB cache (each), 5 GB FBDIMM Memory (667 MHz DDR2), 1.33 GHz Bus)"

[v00d00]

Thanks for the explanation guys. I guess i need to make some more upgrades, very soon.