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Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

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Thank you for the explanation.

You replied that the solution uses 15 SAS and 13 NL-SAS drives  , one of each type is for HS.

(4+1)R5 for SAS and (4+2) R6 for NL-SAS

After that the calculations were based on 10 SAS drives and 12 NL-SAS , should it not be 14 SAS instead ( 15-1 HS)

Q1.  I am not sure why you 8 x 536( formatted capacity) =4288 GB, how did you get 8 ? Should it not be 14 SAS  x 536 ?

14 x 536 =  7504 * [(n-1)/n]  , therefore usable = 7003 GB.

Q2.  Similarly, how did you 8 x 1834 , how did you get 8 ? Should it not be 12 NL-SAS x 1834 ?

12 x 1834 = 22008 * [(n-2)/n], usable = 18339 GB

Q3. I know that a 600 GB drive has a formatted capacity of 537 GB ( found this in the OE 31.5 best practices), so I use a factor of 1.11732.

If I take 2TB/1.11732 that should give me 1790 for the formatted capacity instead of 1834, is this alright ?

Thanks once again.

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Re: E20-324 VideoILT Module 5 - Scenario 1 & 2 solutions missing

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As I said in my previous answer, I ignored the 4 system drives. That leaves 10 SAS drives, which would normally be configured as 2x 4+1 R5. In 1 4+1 R5 group, only the equivalent of 4 drives hold user data, while the remaining drive holds parity information. For 2x 4+1 groups, that gives 8 data drives, hence the 8x 536 GB.

The same is true of the NL-SAS drives. My answer used 2x 4+2 R6, so there are also 8 data drives, giving the answer you can find above.

I looked at a VNX system to get the drive capacities I used, so I must assume they are correct. EMC has a document that lists drive types and capacities, and it's better to use that than to try to calculate usable capacity based on the manufacturer's published capacity.

In the examples you showed, n refers to the number of drives in the RAID Group, not the number of drives in the VNX. For a 4+2 R6 group, the usable capacity with 2 TB drives is 6 * 1834 * ((6-2) / 6) = 1834 * 4 = 7336 GB. Two of those give you 14672 GB - the answer I showed above.

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