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2002
E20-554 Practice Exam Question
Hi All,
I think I figure out how to find the answer to this question but I wanted to verify my method with someone else.
Click on the calculator icon. Your customer is looking for a storage solution that will be able to store seven million three MB files which are written and seldom accessed. Read and write operations are both completed by a web based application, which requires 1.3 Gbps throughput. The customer's network has not been upgraded in many years, so the network interfaces are 1Gbps.
Which cluster configuration would best meet the customer's requirements?
I used the following rule below and doing the math I picked the 4 X200 since that will give me 1200MBps. The other options give me more than I need and for the S200 the price would be greater as compare to the 4 X200 nodes.
The rule of thumb to repeat is: 7, 5, 3, 2 (700 MBps X400, 500 MBps S200, 300 MBps X200, and 200 MBps
NL400).
jeffspencer1
3 Posts
0
December 9th, 2013 09:00
Responding generally to all comments so far...
1.3Gbps (little b, meaning gigabits per second) is 166MB/s. Any current Isilon storage node has been benchmarked to accommodate for that with the right IO characteristics, especially in aggregate. When considering the capacity math:
7000000 * 3MB = 21000000 MB
21000000/1024 ~= 20507 GB
20507/1024 ~= 20.1 TB
With only 20TB of logical data, extrapolating capacity per node type certainly achieves a comfortable utilization rate with the X200 platform. Any other SATA based product type will be much more capacity than needed for this data set, and the S200 series will provide much more performance than needed.
The X200 will support the capacity and performance requirements. The next consideration I would have is whether Isilon is the best fit of the portfolio at your disposal for this use case... which is outside the scope of this question.
atully
8 Posts
0
October 25th, 2013 06:00
Hi,
I have also struggled with trying to understand the answer of 4 x X200 nodes.
My thinking on it is as follows.
Discount the NL nodes despite the reference to "seldom accessed" (archive) since the text indicates NL nodes for opportunities with a capacity requirement greater than 100TB.
Regarding throughput - I have assumed equal split between read and write even though the text of the question indicates mostly writes "seldom accessed". This gives me a write throughput requirement of 650Mbps and the same for read.
The X200 nodes give 4 x 190 = 760 Mbps write throughput and 4 x 270 = 1080Mbps. These figures appear to be the closest match to my assumed read / write throughput requirements.
I discount the X400 nodes despite better throughput performance because this provide too much capacity than is needed and probably cost a lot more.
I discount the S200 nodes partly on price, rack space (twice that of the X200 nodes) and because they are more generally suited for higher performance / even transaction workloads.
Would love to here other thoughts on this.
Note: you mention the rule of thumb in your post - the figures you quoted are based on a 5-node cluster read throughput. So 4 X200 read only is 4 x 270 = 1080 Mbps a good bit away from assuming a read only requirement of 1300Mbps (1.3Gbps).
philippspohr
107 Posts
0
December 3rd, 2013 01:00
Hi,
I am confused with this question too. I tried to find the right solution from the side of capacity but I failed. Maybe I am a bit slow today - or there is a mistake in the question?
7.000.000 files á 3 Megabyte
= 7.000.000 x 3 MB
= 21.000.000 MB
= ~20.507 GB
= ~20 PB
But it is not possible to reach a capacity of 20 PB with any of the four given answers. In the following calculation there is no protection or something similar, just the maximum raw capacity:
A) 3x NL400 = 3 x 36 x 4TB = 432 TB
B) 3x X400 = 3 x 36 x 4TB = 432 TB
C) 4x X200 = 4 x 12 x 3TB = 144 TB
D) 5x S200 = 5 x 24 x 900MB = 105 TB
Where is my mistake?
philippspohr
107 Posts
0
December 10th, 2013 00:00
Hi Jeff (master of Isilon ILT and VILT),
thank you for the solution. That was my fail. I jumped over the TB...
But anyway: If there is a value of 1-3 Gbps throughput need in the room, the first look should be at the needed capacity because nearly any node has a better throughput? And aggregated with at least three nodes in any case?
But just for my better understanding: How did you calculate the value of 166 MB/s?
If you are in the area of communications (= throughput, not data storage) I would use the metric definition:
1.5 Gbps = 1,500 Mbits = 1,500,000 Kbits = 1,500,000,000 Bits
/8 = 187,500,000 Byte/s = 187,500 KByte/s = 187.5 MByte/s
If you are in the area of storage I would use the binary definition:
1.5 Gbps = 1,500 Megabit/s = 1,500,000 Kilobit/s = 1,500,000,000 bit/s
/8 = 187,500,000 Byte/s (so far so good) = 183,105.46875 KByte/s = 178.81 MByte/s
I am confused because it seems that everybody uses the notation (e.g. big vs. small letters; binary vs. metric calculation) in a different way. Additionally there is sometimes a distinction between US, British and European notation...
BTW: Can you have a look at my other calculation-question? Maybe I have a similar misunderstanding in calculating storage needs.
Thank you.
jeffspencer1
3 Posts
0
December 10th, 2013 09:00
Oops, I used 1024 rather than 1000 for the Gbit/s to Mbit/s conversion, as you stated, the metric definition. I blame being raised in the States.
Therefore, the real number:
1.3 * 1000 = 1300 Mbit/s
1300 / 8 = 162.5 MB/s
For capacity consumption the most specific use is KiB, MiB, GiB, TiB, PiB, etc., and I am used to Mbit/s and MB/s for bandwidth and throughput. Hopefully more of us will align to one standard as time goes on.