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Converting IOPS to disks required

Posted on 2012-08-20
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Hi

We are running HP servers. One of our application vendors states that their new application we'll be using has an IOPS footprint of 0.3. IOPS per user.

We'll have about 500 users per server.

What's the best method to find out what the best server/ disk/ RAID architecture is? Do HP provide any sort of calculators or knowledge on their website (I can't see anything myself)?
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Question by:richlionel

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Expert Comment

The best is a function of budget, unless, of course, you are the federal government :(
So how many TB of redundant data do you need, and what is the budget for the STORAGE + controller(s).
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Accepted Solution

Here's an example of how you'd calculate the number of disk drives needed (I posted this about 5 minutes ago at http://www.experts-exchange.com/Hardware/Servers/Q_27833963.html):

Let's say you have an environment that's generating 5000 IOPS, 66% reads, 34% writes, highly random:
First step is to calculate the workload at the disks:
RAID 5 (write penalty is 4):
(5000 x .34) x 4 = 6800 IOPS
5000 x .66 = 3300 IOPS
6800 + 3300 = 10,100 IOPS
Assume 15K SAS drives @ 200 IOPS per drive. You need:
10,100/200 = 51 drives to handle the workload
Assume 7200 rpm SATA drives @ 80 IOPS per drive:
10,100/80 = 127 drives

Now for RAID 1/0 (write penalty is 2)
(5000 x .34) x 2 = 3400 IOPS
5000 x .66 = 3300 IOPS
3400 + 3300 = 10,100 IOPS
Assume 15K SAS drives @ 200 IOPS per drive. You need:
6,700/200 = 34 drives to handle the workload
Assume 7200 rpm SATA drives @ 80 IOPS per drive:
6,700/80 = 84 drives

Now for RAID 5 sequential (write penalty is 1.25)

You can use these formulas to work out ratios for performance calculations. Note that you can consider a 64KB write or larger to be sequential.

(5000 x .34) x 1.25 = 2125 IOPS
5000 x .66 = 3300 IOPS
2125 + 3300 = 5425 IOPS
Assume 15K SAS drives @ 200 IOPS per drive. You need:
5425/200 = 28 drives to handle the workload
Assume 7200 rpm SATA drives @ 80 IOPS per drive:
5425/80 = 68 drives

Let's do SSDs for fun:

(5000 x .34) x 4 = 6800 IOPS
5000 x .66 = 3300 IOPS
6800 + 3300 = 10,100 IOPS
Assume  SSD drives @ 3000 IOPS per drive. You need:
10,100/3,000 = 4 drives to handle the workload

... and all your performance problems go away. Now all you need is space. Enter EMC FAST, NetApp Flash Cache, Compellent, etc etc

So as you can see in some cases RAID 5 is most economical, in others, RAID 1/0.

Remember that this a rule-of-thumb sizing method.
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LVL 46

Expert Comment

It doesn't really work that way.  You are pretty much asking the equivalent of trying to convert percent CPU utilization to number of Mhz needed.

Myersd gives as good a number as anybody else can, but real world, the problem is much more complicated.

Example, give me \$500.00 and I can give you a subsystem that will give you 100,000 IOPs using a pair of SSDs and software RAID1. I doubt that will meet your needs, because you probably want terabytes of data, not gigabytes.  Nevertheless, you can get 100,000 IOPS with the right technology.
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LVL 30

Expert Comment

True, but it works well as a rule-of-thumb for performance sizing - and after all, you have to start somewhere. It was easier in the days of yore with 9GB FC drives where performance tended to take care of itself as long as you got the space right. Now of course, you have to work out performance requirements since space requirements are so easily satisfied.

"you want how much for a disk?? But I can buy one from Newegg fr \$99!"
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LVL 55

Expert Comment

0.3*500=150, if that's all the IOPS you need then a pair of 15K SAS disks covers it.

HP do provide calculators for applications, e.g. http://h71019.www7.hp.com/ActiveAnswers/cache/70593-0-0-0-121.html for MS Exchange 2010; they don't provide a generic one based on IOPS as far as I know.
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