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SAN 6Gb SAS question

So my SAN controller (Dell SC8000) has 2x quad port 6Gb cards in it for the back end.

Back End
6Gb gives a theoretical maximum of 600MB/second

So with 8 ports it would handle in theory 600x8=4800MB/sec

Front end FC

2x Dual port 8Gb FC cards, each FC port can give 800MB/sec, so each card can give 1600MB/sec
So 4 ports would in theory give a maximum of 1600x2=3200MB/sec

I just want to confirm that is correct?

Cheers

Scott Gorcester

I don't believe that is an accurate calculation. It does not aggregate the bandwidth in that manner, the Dell website lists the Total IOPS at 300k and the maximum read bandwidth of 18GB sec and maximum write b/w at 10GB sec

================================================

Available Storage Interfaces/Protocols:
FC 8Gbit,16Gbit; iSCSI 1Gbit,10Gbit

Total Max Raw HDD Capacity:
1.5PB

Total Max Operational HDD Capacity:
1PB, Supports RAID 0, 5, 50, 6, 60, RAID 1, 10 and RAID 10 DM (dual mirror).

Total Max Raw Flash Capacity:
1.5PB

Total Max Operational Flash Capacity:
1PB

Maximum IOPS:
300K

Max Read Bandwidth:
>18GB/sec

Max Write Bandwidth:
>10GB/sec

Hard Drive Type:
960, less the number of SSD’s, in a wide variety of options ranging from as IO-intensive as 450GB 15Krpm to as capacity-intensive as 4TB 7.2krpm

Solid State Drive Type:
936@1.6TB each (read-intensive) 960@200GB or 400GB each (write-intensive)

Power Requirement:
Power: Dual, redundant 750W 80 Plus Platinum-certified power supplies
Maximum Power: 425W

Andrew Hancock (VMware vExpert PRO / EE Fellow/British Beekeeper)

If you really want to know how many IOPS for your workload test it for VMware - TEST IT!

Here is a list of applications you can use to test!

When we are producing results for clients, we use the following applications, build spreadsheets, and average the results

For a quick rough, result run CrystalDiskMark in a VM, and see what figures you get, I think you'll be surprised if you get anything like the following:-

But it's not about numbers, but what IOPS do you need for your applications ?

1.IO Meter
http://www.iometer.org/

2. CrystalDiskMark
http://crystalmark.info/software/CrystalDiskMark/index-e.html

3. HD Tach
http://www.majorgeeks.com/files/details/hdtach.html

4. Intel NAS Toolkit
http://www.intel.com/products/server/storage/NAS_Perf_Toolkit.htm

5. VDBench
http://sourceforge.net/projects/vdbench

6. VMware Fling IO Blazer
http://labs.vmware.com/flings/ioblazer

7. Jet Stress
http://www.microsoft.com/en-us/download/details.aspx?id=36849

8. SQLIO
http://www.microsoft.com/en-gb/download/details.aspx?id=20163

9. HDPARM
http://linux.about.com/od/commands/l/blcmdl8_hdparm.htm

10. dd for (Linix, Unix and Windows)
http://en.wikipedia.org/wiki/Dd_%28Unix%29

11. ATTO
http://www.attotech.com/disk-benchmark/

12. Bonnie++
http://www.coker.com.au/bonnie++/

13. FIO
http://www.linux.com/learn/tutorials/442451-inspecting-disk-io-performance-with-fio

14. hIOmon
http://www.hyperio.com/productsAndServices.htm

15. DiskTester
http://diglloydtools.com/manual/disktester-iops.html

piedthepiper

ASKER

Max Read Bandwidth:
>18GB/sec

Max Write Bandwidth:
>10GB/sec

Is that the max b/w for the front end or back end?

piedthepiper

ASKER

I agree testing it would be the best way, I am just trying to understand the theory a bit better esp the way Dell do it

SOLUTION

Andrew Hancock (VMware vExpert PRO / EE Fellow/British Beekeeper)

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Scott Gorcester

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David

real-world, not even close to that. You are only as fast as the weakest link.

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piedthepiper

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ASKER CERTIFIED SOLUTION

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piedthepiper

ASKER

Ah ok that makes sense, I have access to that portal but the guides aren't there. Which sucks, I am emailing them to see if I can get them.

You are correct, the top enclosure has one link going to each controllers sas card and the bottom enclosure has one link going to each controllers sas cards. So 4 in total 2 to each controller.

So even though each controller has 2 connections per loop, only one connection will be in use at any given time?

So it is patched in as 4x6Gb x 4 wide = 96Gb = 9600MB/sec in total for that loop.

But only one connection to each controller will be active for that loop so:

2x6Gb x 4 wide = 48Gb in total, and that means each controller can do 24Gb (2.4GB/sec) per loop?

SOLUTION

David

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piedthepiper

ASKER

Cheers guys plenty of food for thought!