From mboxrd@z Thu Jan 1 00:00:00 1970 From: jim owens Subject: Re: MD performance options: More =?windows-1252?Q?CPU=92s_or_?= =?windows-1252?Q?more_Hz=92s=3F?= Date: Wed, 04 Nov 2009 10:41:44 -0500 Message-ID: <4AF1A0B8.2080807@hp.com> References: <66781b10911040149q165edf1s94a86f179f9af9fc@mail.gmail.com> <4AF18A39.6010404@tmr.com> <66781b10911040634mcdc3c70jec3410c09003fb6c@mail.gmail.com> Mime-Version: 1.0 Content-Type: text/plain; charset=windows-1252; format=flowed Content-Transfer-Encoding: QUOTED-PRINTABLE Return-path: In-Reply-To: <66781b10911040634mcdc3c70jec3410c09003fb6c@mail.gmail.com> Sender: linux-raid-owner@vger.kernel.org To: mark delfman Cc: Bill Davidsen , Linux RAID Mailing List List-Id: linux-raid.ids mark delfman wrote: > Thank you Bill... maybe a silly question but doesn=92t the fact we ca= n > achieve 1.6GBsec on MD RAID0 mean we have drive bandwidth available t= o > MD to use? To the block device (sas drives in this case) it does not > know if the data is been xor=92d or just striped, its just blocks of > data. So wouldn=92t this mean that we know that MD RAID6 =91could=92 > achieve up to 1.6GB if the software could send it out quicker? >=20 > Hence my logic of =91deal with MD software aspects faster =3D faster = writes=92 Are your tests and assumptions valid... remember that raid6 will put 1.6GB of data on the drives for each 800MB of data from the application in minimum 4 stripe mode (2 data stripes, 1 P stripe, 1 Q stripe). So the proper comparison of raid6 software overhead is to run the same application writing a fixed large amount of data to 4 drives at raid0 and the same 4 drives at raid6. overhead =3D raid0_time * 2 - raid6_time; Of course if you add more data stripes, the ratio of user data to disk data gets better than 1/2. jim -- To unsubscribe from this list: send the line "unsubscribe linux-raid" i= n the body of a message to majordomo@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html