Biggest problem here is that it this is (very correctly stated): sequential IO. Can you guarantee you use the disks in a strict sequential way?
If your IO pattern is not sequential but random (as probably most disks are used), the IO throughput figure drops significantly.
The 75% full thing is discussed by James Morle the last database forum. The 75% figure and potential performance drops have to do with read speed which is higher on the outer edge of a disk because of higher sector density (and lower seek time of the read/write head because movement is reduced). Apparently disk manufracturers take this in account and have optimised disks for this now. If you want to optimise this way, check if you got any benefit from it.
To calculate the sequential IO you can use the follwing:
seconds per IO=(transfer time) + (track-to-track seek time) + (avg. rotational latency)
(assuming that we have track-to-track seek time between each IO)
IOPS=1/(seconds per IO)
for random IO you can calculate
seconds per IO= (transfer time) + (random seek time) + (avg. rotational latency)
(assming we have random seek between each IO)
IOPS=1/(seconds per IO)
The transfer time, track-to-track seek time, avg. rotational latency, random seek time you can find in HDD spec.
For random seek time we can assume that we have mixture of reads and writes and take the avg. of disc spec. "avg. seek time, read" and "avg. seek time, write".
Since most IO operations done by oracle are random in nature, you should try sizing your system not to the calculated max. random access performance but to operate at around 75% of max. performance. The reason for this is that the IO response time increases as you reach the max. capacity of the disk drive and the queue length can increase greatly.
Also, take into consideration that in addition to disk drive latencies, the latencies of IO controller have to be taken into the account too.
On 9/28/06, Michael Erwin <michael.erwin@xxxxxxxxxx> wrote: > > Hi Lale, > > One would expect to find it here: > > http://h10010.www1.hp.com/wwpc/me/en/sm/WF06c/A1-1769079-1769113-1769113-1769139-1769139-16169139.html > > > Nothing really available on their site... > > However, estimate IOPS is relatively easy with > > Spindle IOPS = ((RPM/60)/3)*2 > > Or at least ~167 8k IOPS per spindle > > To figure out User Data Transfer rates, use the following: > > UDT=(Spindle Speed / 60 * Sectors Per Track * Bits per Sector * 8) / > 1,000,000 > > Bits Per Sector as defined by HP can be here: > > http://h18000.www1.hp.com/products/quickspecs/11531_div/11531_div.HTML > > So BPS=512 > > The other two number we need is the platters used in this 146GB drive > and the physical platter size (3.5", 2.5", 1", and I would assume 3.5") > to determine an areal density or at least GB/Platter density to determine > the inner & outer range of UDT. It will be a wide range considering Zone Bit > Recording (ZBR) > > BTW, the whole U320 MB/Sec transfer is a U320 SCSI transfer rate and not > the user data transfer rate for the drive itself. > > > *Michael Erwin* > Hotsos Enterprises, Ltd. > http://www.hotsos.com > Hotsos Symposium 2007 / March 4-8 > Visit www.hotsos.com<http://exchweb/bin/redir.asp?URL=http://www.hotsos.com>for curriculum and schedule details... > > ------------------------------ > *From:* oracle-l-bounce@xxxxxxxxxxxxx on behalf of lale obradovic > *Sent:* Thu 9/28/2006 12:19 PM > *To:* oracle-l > *Subject:* IO rate for U320 SCSI 15K RPM Hard Drive > > Hi All, > > Does anyone have some information of expected IO rate (IOPS and/or MBPS) > for U320 SCSI 15K RPM Hard Drive (HP 146GB U320 15K Universal HDD)? > > Thanks! > > Lale >