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in Storage - 30 Sep, 2016
by mordi - no comments
VCAP – Design: Storage calculation – Part 1

I will like to start a series of posts regrading VCAP design storage calculation and performance. we will start with the basics and build on it to more advanced topics.

Raid levels:

RAID – Redundant Array of In Depended Disks , RAID technology is a combination of in depended disks to form redundancy or create a better performance.

There are many type of RAID levels but recently vendors start to use a few common raid levels:

 

RAID 0:

raid0

RAID 0 technical details :

Method: Block level striping , data is writing across all drives in the stripe set. Using RAID 0 will increase performance since we utilizing all drives from the RAID set at the same time.

Fault Tolerance:  RAID0 DO NOT offer any fault tolerance, failure in one of the drives will cause data loss in the array.

Efficiency: 100% , all disks in the array are been used.

Design consideration:

  • Good for application that do not need data protection
  • Minimum 2 drives
  • Good performance in sequential read and write
  • Good performance for random I/O.
  • Low cost

 

RAID 1:

raid1

RAID 1 technical details :

Method: Block level mirroring, data is mirrored to additional disk /disks

Fault Tolerance: RAID 1 provide fault tolerance by using the mirror technique, if a disk fails you can recover the data from the mirrored disk/disks.

Efficiency: 50% , because we have additional full copy of the data.

Design consideration:

  • Basic protection
  • Minimum 2 drives
  • Good performance in sequential read
  • Poor write performance (all writes need to be mirrored)
  • High cost

 

RAID 5:

raid5

RAID 5 technical details :

Method: Block level striping with distributed single parity , each disk in the RAID set include a parity information

Fault Tolerance: RAID 5 provide fault tolerance by using the distributed parity technique , if a single disk in the RAID set fails you can recover the data from the parity that exist on the other drive in the RAID.

Efficiency: [(n-1)/n]*100 , n = number of drives. example: using 10 X 500GB drives in RAID 5:

calculate efficiency:  [(10-1)/10] *100 = 90%

now that we know the efficiency we can calculable the capacity in the raid  : (500GB X 10) *.90 = 4.5TB

Design consideration:

  • Good for application that need data protection
  • Minimum 3 drives
  • Good performance in sequential reads
  • Good performance for random I/O.
  • Moderate cost

 

RAID 6:

raid6

RAID 6 technical details :

Method: Block level striping with distributed dual parity , each disk in the RAID set includes  dual parity information

Fault Tolerance: RAID 6 provide fault tolerance by using the distributed dual parity technique , if a two disk in the RAID set fails you can recover the data from the parity that exist on the other drive in the RAID.

Efficiency: [(n-2)/n]*100 , n = number of drives. example: using 10 X 500GB drives in RAID 6:

calculate efficiency:  [(10-2)/10] *100 = 80%

now that we know the efficiency we can calculable the capacity in the raid  : (500GB X 10) *.80 = 4TB

Design consideration:

  • Good for application that need data protection
  • Minimum 4 drives
  • Good performance in sequential reads
  • Good performance for random I/O.
  • High cost

 

RAID 10:

raid10

RAID 10 technical details :

Method: combined method of RAID 1 mirroring  and RAID 0 striping , data is striping across the mirrored sets

Fault Tolerance: RAID 10 provide fault tolerance by using the mirror technique , if a disk in the mirrored RAID set fails you can recover the data from the mirror disk, for every mirror in the stripe you can loose a disk.

Efficiency: 50%

Design consideration:

  • Good for application that need data protection
  • Minimum 4 drives
  • Good performance in sequential reads and writes
  • Good performance for random I/O.
  • High cost

 

Thanks for reading

Mordi.

 

 

 

 

 

 

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