Comments on: Storage Basics – Part III: RAID

By: Vlad

Vlad — Fri, 02 Sep 2011 21:40:30 +0000

Hi Josh,
Thank you for the great article series!!!
Not an expert in the filed but the formula for calculation of the total IOPS for a RAID5 array, given the read/write workload percentage, looks to me more like
I=(n*i*r) + (n*i*w)/4 = (n*i)(r + w/f). Am I wrong?

Regards,
Vlad

By: Welcome to vSphere-land! » Storage Links

Welcome to vSphere-land! » Storage Links — Fri, 08 Jul 2011 15:12:34 +0000

[...] Storage Basics – Part I: An Introduction (VM Today) Storage Basics – Part II: IOPS (VM Today) Storage Basics – Part III: RAID (VM Today) Storage Basics – Part IV: Interface (VM Today) Storage Basics – Part V: Controllers, [...]

By: Sam

Sam — Tue, 29 Mar 2011 12:43:51 +0000

I enjoyed reading this article and I learned a lot. I loved the “over simplification” as I am learning hardware after being in software industry for too long and I could not find anything else that even comes close. Thank you so much for writing this. Your time, effort and energy is well spent here.

By: VCAP-DCA Study notes – 1.1 Implement and manage complex storage | www.vExperienced.co.uk

Mon, 21 Mar 2011 09:40:48 +0000

[...] Common RAID types: 0, 1, 5, 6, 10. Wikipedia do a good summary of the basic RAID types if you’re not familiar with them. Scott Lowe has a good article about RAID in storage arrays, as does Josh Townsend over at VMtoday. [...]

By: Objective 1.1 – Implement and Manage Complex Storage Solutions | The world of Marc O'Polo – Blog

Thu, 27 Jan 2011 18:28:28 +0000

[...] Storage Basics – Part III: RAID [...]

By: Storage, it actually needs to be understood! | Technical Trickery

Storage, it actually needs to be understood! | Technical Trickery — Tue, 17 Aug 2010 03:20:19 +0000

[...] http://vmtoday.com/2009/12/storage-basics-part-ii-iops/ http://vmtoday.com/2010/01/storage-basics-part-iii-raid/ http://vmtoday.com/2010/01/storage-basics-part-iv-interface/ [...]

By: Joshua Townsend

Joshua Townsend — Sat, 24 Jul 2010 21:00:26 +0000

Karthik – Thanks for the comment. RAID5 is indeed distributed parity – no single disk holds all parity data. Parity bits are read only during rebuilds and so they do not impose a penalty during normal read operations. This means that for reads, your calculation of a RAID5 set with 5 15k disks (5*180=900 IOPS) is correct.

Your formula for writes, however, is not correct. For writes, the actual work of writing parity imposes a penalty – 4 IOPS are consumed to write data and parity for each I/O operation requested by your server/OS/application/workload – the “factor of 4″ I wrote about. This penalty takes away from the IOPS available to your real workload (we call this overhead). In reality, your 5-disk RAID5 could write a total of 900 IOPS (5*180=900) before queuing. However, your workload can’t take advantage of all 900 IOPS because most of the IOPS go to reading-modifying-writing the parity bits. Put another way, the disks are busy reading, modifying and writing parity bits instead of writing your workload’s “real” data. Your workload only gets the left-over IOPS after the parity writing overhead has been consumed.

Summing it up, your formula for calculating RAID5 read IOPS is (Number of Disks)*(IOPS per disk)=READ IOPS. Your forumla for calculating RAID5 write IOPS available to your server/OS/application/workload is (Number of Disks)*(IOPS per disk)/4=WRITE IOPS.

Keep in mind this is all theoretical – many arrays use cache, coalescing, and other tricks to minimize the RAID5 write penalty.

Hope this helps,
Josh

By: karthik

karthik — Fri, 23 Jul 2010 12:07:55 +0000

Hi,

In the above IOPS calculation begining (RAID5) its been said for 5 disk you take it as four reason being one disk for Parity.

But actually RAID 5 doesnt have a separate disk for just holding the Parity. It takes a junk of each disk in RAID 5. So in the case the calculation should be 5*180=900 IOPS (ideally) and not (5-1)*180=720.

Am I right?

By: cmdln.org (a sysadmin blog) » Blog Archive » Analyzing I/O performance in Linux

Mon, 17 May 2010 15:41:26 +0000

[...] http://vmtoday.com/2010/01/storage-basics-part-iii-raid/ [...]

By: Storage Basics – Part V: Controllers, Cache and Coalescing | VMtoday

Storage Basics – Part V: Controllers, Cache and Coalescing | VMtoday — Wed, 24 Mar 2010 12:41:28 +0000

[...] I introduced IOPS and the math that goes into calculating the IOPS capacity of a disk array. In Part III we considered a RAID implementation’s impact on performance and availability. And most [...]

By: uberVU - social comments

uberVU - social comments — Wed, 03 Feb 2010 07:14:36 +0000

Social comments and analytics for this post…

This post was mentioned on Twitter by joshuatownsend: New VMtoday.com post: Storage Basics – Part III: RAID http://cli.gs/Hyt2m #vmware…

By: Joshua Townsend

Joshua Townsend — Thu, 07 Jan 2010 16:32:42 +0000

Good catch – I have updated the post.

By: Matthew Shuter

Matthew Shuter — Thu, 07 Jan 2010 16:08:20 +0000

Raid 0 portion, you have “This is basic aggregation with now redundancy”
now should be no

By: Sumesh

Sumesh — Thu, 07 Jan 2010 07:20:31 +0000

Excelent one dude,i was lookign for this

By: Tweets that mention Storage Basics – Part III: RAID | VMtoday -- Topsy.com

Tweets that mention Storage Basics – Part III: RAID | VMtoday -- Topsy.com — Thu, 07 Jan 2010 07:18:39 +0000

[...] This post was mentioned on Twitter by joshuatownsend, WillHamann. WillHamann said: RT @joshuatownsend: New VMtoday.com post: Storage Basics – Part III: RAID http://cli.gs/Hyt2m #vmware [...]