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E20-554 Practice Exam Question: Video surveillance
Hello everybody,
I have some trouble understanding the video surveillance question in E20-554.
Your customer is moving from a Restaurant/Bar business model to a Restaurant/Casino business model. You are helping architect the upgrade of a current 4-node X200 48TB Isilon cluster, with 65% utilized for their video surveillance operations.
They wish to achieve the following:
--Upgrade from 1280x720 resolution, with a 15 days retention policy, to 2048x1536 resolution with a 30 days retention policy.
What is the minimum number of nodes of the same type will you need to add in order to meet the required changes and not exceed 70% raw capacity?
a) 6 nodes
b) 10 nodes
c) 14 nodes
d) 18 nodes
Current use of storage: 4 x 48 TB = 192 TB
65% of storage utilized for video data: 192 TB x 0,65 = 124,8 TB
>> With 1280x720 and 15 days retention policy they have ~124,8 TB video data.
1280x720 = 921.600 pixel
2048x1536 = 3.145.728 pixel
3.145.728 / 921.600 = 3,41333
>> 2048x1536 needs 3,4133x more storage than 1280x720
>> 30 days retention policy needs 2x more storage than 15 days
Future use: 124,8 TB x 3,4133 x 2 = 851,95968 TB video data
Not exceed 70% raw capacity: 851,95968 / 70 x 100 = 1.217,085 TB
>> The new storage must have at least 1.217 TB capacity to fulfill the requirements.
The old storage has already 192 TB. So we need only 1.025,085 TB new storage.
1.025,085 TB / 48 TB = 21,36 nodes
>> We have to add at least 22 new nodes of the same typ!
But the right answer of this question is b) 10 nodes (what means in total we have a storage with 14 nodes).
Am I wrong with my calculation?
jeffspencer1
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December 10th, 2013 09:00
The 48TB configuration relates to the cluster size, rather than the node type. I guess this is a bit of "tribal knowledge". The X200 only supports up to 3TB drives today.
48TB / 4 nodes = 12TB per node...
So, taking some of the "rules of thumb" from Infrastructure Solutions, we can assume that the data consumption is:
48TB * .65 = 31.2TB (3 nodes ~= 60%, 4 nodes ~= 65%, etc.)
31.2TB * .65 = 20.3TB (of the usable capacity, this is utilized)
Then for the video changes... 2x increase in retention, and let's use your 3.4133x value for the resolution change.
20.3TB * 2 * 3.4133 = 138.6TB
138.6TB / 12TB = 11.6 nodes just for data, rounded up to a whole node count is twelve. From there add another 2 nodes for protection and a bit of headroom. Since we already have four nodes, adding another ten makes sense for a "minimum" configuration.
Not the most "scientific" approach, but get us to the answer that is marked correct.
philippspohr
107 Posts
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December 11th, 2013 00:00
Hi Jeff,
thanks a lot for your explanation.
That is the main mistake I did. In other examples there is written the Node storage, not the total cluster storage.
Other facts and rule of thumbs - e.g. that in a 4-Node-Cluster only 65% of the raw capacity is usable - isn't currently wired enough into my brain. If the protection level is not mentioned, can I assume it is N+1 or N+2:1?
The next steps you did are clear to me. In future the customer needs ~138.6 TB storage only for his video data.
Is that another rule of thumb?
Or did you have in mind the table "Data Protection Overhead" which says that there is round about 10% protection overhead at a 12-Node-Cluster (9% with N+1 and 11% with N+2:1)?
Than 138.6 / 12 = 11.5 = ~12 Nodes rounded up
12 * 1.1 = 13.2 = ~14 Nodes rounded up
minus 4 Nodes the customer already owns = 10 Nodes needed.
Am I right?