You have a server named Server1 that runs Windows Server 2012.
You plan to create a storage pool that will containa new volume.
You need to create a new 600-GB volume by using thin provisioning. The new volume must use the parity
layout.
What is the minimum number of 256-GB disks requiredfor the storage pool?
A.
2
B.
3
C.
4
D.
5
I think is C
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Parity – This is a striped set with distributed parity by striping data and parity information across multiple disks, similar to RAID 5. It increases reliability with reduced capacity. This configuration requires at least (!)three(!) disks to protect data from a single disk failure, and cannot be used in a failover cluster.
http://blogs.technet.com/b/yungchou/archive/2012/08/31/windows-server-2012-storage-virtualization-explained.aspx
The answer is B. (!)Thin(!) provisioned means ONLY virtual limits (MAX 64 TB within 2012 R2)
(!)Thick(!) provisioned the answer would be C.
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yeah in a raid 5 you will need 3 to make a 600GB Volume and 1 for Partiy
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in RAID 3 you have data disks and one for parity, but in raid 5 the parity is distributed in data disks. This question is not explained very well. The minimum number od disks for parity is 3, in this question they say one disk is 256GB in raid 3 we need 4 disks 3 for data 3*256=768GB and one for parity. I think MS for parity in Space layout use distributed parity than the answer B is correct. 3 disks.
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I think “thin provisioning” is the magic word.
Normally 1 Disk should be enough to fake a 600GB drive.
But with the other word “parity” the correct answer should be A -> 2 disks
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You need at least 3 physical disks to make parity.
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I think B is the correct answer as it it required to have parity. in order to have parity you need to create RAID5.
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sorry, correction B is correct ! you are right makhan.
already found it on a MS page, for parity theres a min. of 3 disks required
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good anwswer is: C, 4 disk have = 768 GB, 3 disk have = 512 GB
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Yes, but remember with thin provisioning you can specify whatever size you want.
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Confirmation? C seems correct.
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What’s correct now?
If I want to have at least 600 GB for data I obviously need 4 disk since always one block in one read block is used as a parity block.
But, if I want to have a storage pool with that size, 3 disks are enough. How is that question meant??
I am confused …
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Answer is ‘B’, 3 disks. With ‘thin provisioning’ you do not need the full 600GB to be available, so the question really is ‘how many disks do you need to create a parity pool’.
RESILIENCY TYPE …. MIN DISKS
Simple (no resiliency) … 1 disk
Two-way mirror … 2 disks
Three-way mirror … 5 disks
Single Parity … 3 disks *** <- Answer
Dual Parity … 7 disks
http://social.technet.microsoft.com/wiki/contents/articles/11382.storage-spaces-frequently-asked-questions-faq.aspx#What_types_of_storage_arrays_can_I_use_with_Storage_Spaces
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At first I thought it was c, but I can confirm its b. I just did it in my lab. There is also a message saying that when you think provision, you can make the vdisk larger than what you actually have. Freaking MS with their tricky tests. no wonder people use dumps
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B.3 LAN right
Single Parity … 3 disks *** <- Answer
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http://blogs.technet.com/b/yungchou/archive/2012/08/31/windows-server-2012-storage-virtualization-explained.aspx
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Check this website;
http://www.ibeast.com/content/tools/raidcalc/raidcalc.asp
calculator says you need 4 drives with 256GB to make 709GB free space.
Wth 3 disks you will have 473GB space.
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I think they are asking parity space. Not raid.so sea1731 user is not correct.
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I am thinking parity layout uses the same striping as Raid5, but it is not Raid5
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I am thinking parity layout uses the same striping method as Raid5. This is the method how they put the parity data by spanning it, just like how Raid5 puts its data. It is not Raid5. The storing of parity data is use to recover a disk failure.
Therefore, I believe the answer is ‘B’, since so many users say the same thing.
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this is the final answer
if i am not sure 100% i will not comment
• The keywords here are provisioning and parity
• Parity requires at least three disks, its LOGIC, so if one disk fail the other one can take over, otherwise why we will need parity
• Using provisioning, you can make the vdisk larger than what you actually have, so you can use only two 256-GB disks to create a 600-GB volume
• The right answer will be three because you will need at least three disks to use parity
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3 disks only give you 512GB storage with parity (RAID 5) so correct answer is C:
http://www.raid-calculator.com/default.aspx
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The answer is B. I confirmed this with my 70-410 Exam Ref. On page 54 it says, “Parity requires the pool to contain at least three physical disks and provides fault tolerance by striping parity information along with data.”
B is the correct answer.
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It doesn’t say that it has to be RAID-5.
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“What is the minimum number of 256-GB disks requiredfor the storage pool?”
The keyword is “minimum”. We definitely need 4 disks for the 600gb parity layout, as some of you already pointed out. But we need the minimum for the parity layout, as the disk array is thin provisioned. Answer is 3 disks. B
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Search in google.com for Raidcalc
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That is still misleading because eople then argue about the intent of the question as the question does not specify what type of RAID. With RAID 0, you could say 3 disks are fine for parity, but if you argue that the question is referring to RAID 5, you could say 5 is the requirement. To me, the answer is given away the by “THIN PROVISIONING”, meaning it does not need to be fixed at the minimum 600GB and can therefore get away with 3 disks in either RAID configuration.
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I mean 4 not 5.
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3 disks for DATA. (To allow up to 600gb)
1 disk for PARITY. (Parity is not shared between disks in raid 5)
=C. 4 disks total.
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I’ve tried the setup.
When I have 3 disks of 256 GB, then it’s not possible to create a 600 GB parity disk. Even before the option to create a fixed or thin provisioned disk, the wizard tells me that there isn’t enough space in the storagepool.
When I have 4 disks of 256 GB, then it IS possible to create the 600 GB disk.
The correct answer is: C.
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my tests says otherwise. I added 3 disks, created a storage pool.
I then created a volume with Parity, Thin Provisioning and was able to set the capacity to 600.
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I can confirm this. I did the same test in my lab, and it is indeed possible to create a 600GB virtual disk. The key to this question is “thin provisioning”.
As with many of these questions, this one is also poorly worded. As some have already pointed out, it is correct that you will never be able to use the entire 600GB, as you will lose the capacity of 1 disk due to the parity layout. This will leave you with a little less than 500GB of usable space.
However, this is not what they are asking with this question. We simply want to know whether it is possible to create the 600GB THIN PROVISIONED virtual disk on the specified disk layout.
The answer is “yes”, so “B. 3” is correct.
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Thin provisioning doesn’t make something from nothing, it just provisions enough to meet the needs of the volume at a given time, which isn’t necessarily always the full capacity; so to parity the physical pool you still need 4, so the answer is C.
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The answer I got from the sample questions:
It takes 3 discs (minimum) in order to create a storage pool array with parity. If this array were using fixed provisioning, this would not be enough given the 256MB
capacity (since only 2/3rds of 256 X 3 – less than 600 – could be used as actual data with the rest being parity bits), but since this array uses thin provisioning, a
600GB volume could technically be set up on a 20GB disc and it would still show as 600GB. (So, essentially, the question really becomes how many drives it takes
in a storage pool to create a parity array.)
Correct Answer: C
References:
http://technet.microsoft.com/en-us/library/hh831391.aspx
http://www.ibeast.com/content/tools/RaidCalc/RaidCalc.asp
http://www.raid-calculator.com/default.aspx
https://www.icc-usa.com/raid-calculator
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