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Firmware Update

idata
Esteemed Contributor III

Is there a 64 bit version of the latest firmware update?

9 REPLIES 9

idata
Esteemed Contributor III

Interesting results, redux. If I may be the devil's advocate here, your results somewhat mirror the apparent performance of the mysterious Intel 510 SSD. The 510 specs in general seem to sacrifice some 4k performance for sequential, which is just what we see in your benchmarks.

Redux, are you now worshipping at the Alter of Sequential Speed? Tsk tsk...

Looks like AS SSD Bench is not displaying the firmware number correctly, or completely, but then I guess that is nothing new.

Hmm, hmm, hmm, I wonder what those IOPs numbers are now...

idata
Esteemed Contributor III

Let me try and dig my way out of that . The pre f/w update image was taken some time ago on a drive that had just been secure erased. The post f/w image is from a used drive, so I would expect 4K and access times to be very close to the pre f/w run if the drive was fresh. Judging by other comparisons the main difference is 4K performance at queue depth 64. Not something of much practical use to me, but it's always nice to have just in case.

Below are the IOPs from that benchmark.

Completely off topic, but I have a question that is bugging me as I can't find the answer.

SATA 2.0 transfer rate = 3.0 Gbit/s minus 8b/10b encoding overhead.

SATA 3.0 transfer rate = 6.0 Gbit/s minus 8b/10b encoding plus 8 bit to one byte overhead. If 8b/10b encoding incurs an overhead of 25% for each character or 20% of overall bandwidth, what is the overhead for 8b/10b encoding plus 8 bit to one byte?

idata
Esteemed Contributor III

I can't understand your question completely, or correctly, let me state it as I see it and you correct me;

B = byte = 8 bits b = bit One character (ASCII?) (two hexadecimal digits, one per nibble) per byte.

SATA 2 Max (theoretical) transfer rate = 3Gb/s

Overhead of some kind (I'm ignorant of this) For every 8 bits data, 2 bits of overhead, thus your 8b/10b, so 25% overhead loss, or 20% overall (? I don't follow the 20% overall.)

SATA 3 Max (theoretical) transfer rate = 6Gb/s

Same 8b/10b or 25% overhead + " 8 bit to one byte overhead" (?? Lost me on 8 bit to one byte. Sounds like one byte per byte overhead? Weird, can't be right)

Please elaborate, otherwise SATA 2 Bandwidth = SATA 3 Bandwidth, which is wrong of course.

Can you link us to those figures? I'll look on my own too.

Off-On topic, never flashed firmware before, all my Intel G2's have 02HD FW. I have an unused 80GB G2, perhaps I'll try it for fun.

Currently I have it on the my Marvell 9128 SATA 3 capable mother board, with the new Marvell driver in AHCI. The AS SSD Bench was a little lower than on the ICH10R with iaStor (377 vs 407.) So no magic in that SATA 3 interface (no surprise really.) I'll move it first, of course for an FW update.

idata
Esteemed Contributor III

http://en.wikipedia.org/wiki/8b/10b_encoding 8b/10b encoding introduces overhead as 10 bits of data are required for each 8 bits of data. Both SATA 2.0 and PCI Express 2.0 express use 8b/10b. I believe Intel are moving away from this on PCI Express 3.0 and are using a technique called "scrambling" in which "a known binary polynomial is applied to a data stream in a feedback topology." This reduces overhead to 1.5%.

Wiki describes the encoding for SATA 2.0 and SATA 3.0 as:

SATA 2.0

With a native transfer rate of 3.0 Gbit/s, and taking http://en.wikipedia.org/wiki/8b/10b_encoding 8b/10b encoding into account, the maximum uncoded transfer rate is 2.4 Gbit/s (300 MB/s).

SATA 3.0

The full 3.0 standard peak throughput about 600 MB/s 10b/8b coding plus 8 bit to one byte, without the protocol, or encoding overhead was released on May 27, 2009

I can't anything that explains the difference between:

  • 8b/10b
  • 10b/8b coding plus 8 bit to one byte

Below is a more accurate way of determining the overhead impact of 8b/10b on SATA 2.0. What I try to understand is what 10b/8b coding plus 8 bit to one byte means for SATA 3.0.

EDIT: Regarding the 8b/10b overhead. I assume for large sequential file transfer's the overhead comes out at 20%. For random I assume it equates to 25%

idata
Esteemed Contributor III

Thanks redux, I need to study this, and definetly see your point regarding the SATA 3 overhead. Interesting that the SATA 2 bandwidth keeps getting smaller in reality, if all this is correct. As usual, the more I do know, the more I realize what I don't know