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VNX Expert E20-885 - What throughput do you expect to see from the LUN that was created sixth?
Can anyone assist with this question found on the E20-885 VNX Expert practice test please?
This post is more than 5 years old
4 Posts
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5075
Can anyone assist with this question found on the E20-885 VNX Expert practice test please?
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andre_rossouw
62 Posts
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November 3rd, 2014 07:00
Some things to consider:
1. They're Thick LUNs, created at 5 minute intervals. The initial data allocation will be the default [what is it?]. How will that affect data placement as tiers fill up?
2. You need to use the ROT numbers for the different drive types.
3. Remember the write penalty for the different RAID types.
Please attempt the question, and post your answer with detailed notes and calculations. We'll be able to help you if we know where you're going wrong.
DGraci
408 Posts
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October 24th, 2014 04:00
Pwarnock - what do you think the correct answer is? How did you arrive at that conclusion? Providing your understanding will help our Proven Professionals better guide you to the correct answer.
andre_rossouw
62 Posts
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November 3rd, 2014 07:00
Congratulations!
Regards,
André
Pwarnock1234
4 Posts
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November 3rd, 2014 07:00
I passed this exam last week. Thanks anyway
Anonymous
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January 5th, 2015 04:00
Hi Andre,
I have the same question. How to solve this? I know the answers for the questions you asked in this thread:
1) The initial default allocation of a thick LUN is 3GB which holds metadata and user data, and as users save additional data to the LUN, 1 GiB slices will be allocated as needed.
2) ROT numbers in this case:
SAS Flash VP (eMLC) = 3500 IOPS
SAS 10 K = 150 IOPS
NL-SAS = 90 IOPS
3) Write Penalty for R5 is 4 and R6 is 6
My calculation:
FAST VP Tier -- 3500*4 = 14000 (Excluded Parity Disks)
SAS 10 Tier -- 150*8 = 1200 (Excluded Parity Disks)
NL-SAS Tier -- 90*12 = 1080 (Excluded Parity Disks)
Total IOPS a Pool can give = 14000+1200+1080= 16280 IOPS
10 Thick LUNs are created on the Pool. So 1 LUN can deliver IOPS = 16280/10=1628
I am not getting the way to solve this question. Please help.
Regards
Shivam
andre_rossouw
62 Posts
3
January 5th, 2015 05:00
Since they're Thick LUNs, all space for them is allocated at creation time. The 10 LUNs fill the Pool, which is about 36 TB in size [calculate the correct value from the actual drive capacities], so they're each around 3.6 TB each. The default initial data placement policy [Highest First, then Auto-Tier] will fill up the Flash tier, then the SAS tier, and then the NL-SAS tier.
The Flash and NL-SAS tiers are about 6.3 TB in total [once again, calculate the correct number], so the first 3 LUNs created will fill them, and overflow to the NL-SAS tier. LUNs 4 through 10 will all be on NL-SAS drives, and will share the IOPs available from those drives.
The NL-SAS tier has 16 drives, each capable of 90 IOPs, for a total of 1440 IOPs. Those IOPs are shared across 7 LUNs [plus part of LUN 3, which I'll ignore in this rough explanation], so that's around 205 [disk] IOPs per LUN. The R/W ratio is 3:1, and this is RAID 6, so each 4 [LUN] IOPs costs us (3 x 1) + (1 x 6) disk IOPs = 9 disk IOPs. 205 / 9 = 22, meaning that we can perform 22 "groups" of 3 reads and 1 write, which is 88 LUN IOPs. The closest answer is 78 IOPs [and is what you'd get if you used exact values].
Anonymous
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January 5th, 2015 09:00
Thanks a lot Andre..
jony_pi
7 Posts
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May 18th, 2015 13:00
im not clear about this part:
so each 4 [LUN] IOPs costs us (3 x 1) + (1 x 6) disk IOPs = 9
andre_rossouw
62 Posts
1
May 19th, 2015 06:00
When doing calculations like this, I prefer to deal with I/Os in groups, where the group size is determined by the R/W ratio. As an example, when the R/W ratio is 3:1, a group is 4 I/Os - 3 reads and 1 write. For a R/W ratio of, say, 7:3, the group size would be 10 I/Os - 7 reads and 3 writes.
Once you know the group size, you can calculate the number of disk I/Os generated by those LUN I/Os. This depends, of course, on the RAID type.