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© 2009 EMC Corporation. All rights reserved.
Data Protection: RAIDData Protection: RAID
Module 1.3
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 2
Module objectives
After completing this module, you will be able to:
Describe what is RAID and the needs it addresses
Describe the concepts upon which RAID is built
Define and compare RAID levels
Recommend the use of the common RAID levels based on performance and availability considerations
Explain factors impacting disk drive performance
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 3
Why RAID
Performance limitation of disk drive
An individual drive has a certain life expectancy – Measured in MTBF
– Example If the MTBF of a drive is 750,000 hours, and there are 100 drives in the
array, then the MTBF of the array becomes 750,000 / 100, or 7,500 hours
RAID was introduced to mitigate this problem
RAID provides: – Increase capacity
– Higher availability
– Increased performance
Redundant Array of Independent DisksRedundant Array of Inexpensive Disks
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 4
RAID Array Components
RAIDController
RAIDController
Hard Disks
Logical Array
Physical Array
RAID Array
Host
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 5
RAID Implementations
Hardware (usually a specialized disk controller card)– Controls all drives attached to it
– Array(s) appear to host operating system as a regular disk drive
– Provided with administrative software
Software – Runs as part of the operating system
– Performance is dependent on CPU workload
– Does not support all RAID levels
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 6
RAID Controller Card
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 7
RAID Levels
0 Striped array with no fault tolerance
1 Disk mirroring
Nested RAID (i.e., 1 + 0, 0 + 1, etc.)
3 Parallel access array with dedicated parity disk
4 Striped array with independent disks and a dedicated parity disk
5 Striped array with independent disks and distributed parity
6 Striped array with independent disks and dual distributed parity
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 8
Data Organization: Striping
Stripe 1
Stripe 2
Strips
Strip 1 Strip 2 Strip 3
Stripe
Strip
Stripe
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RAID 0 (Striped array with no fault tolerance)
1
95
2
106
3
117
0
Host
RAIDController
RAIDController
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 10
RAID 1 (Disk mirroring)
Block 1Block 1 Block 1Block 1Block 1Block 1Block 0Block 0Block 0Block 0
Host
Block 0Block 0 RAIDController
RAIDController
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 11
Nested RAID – 0+1 (Striping and Mirroring)
Block 3Block 3
Block 2Block 2
Block 1Block 1
Host
RAID 0
Block 0Block 0
Block 3Block 3Block 2Block 2Block 1Block 1Block 0Block 0
RAID 1
RAIDController
RAIDController
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 12
Nested RAID – 0+1 (Striping and Mirroring)
RAIDController
RAIDController
Block 3Block 3
Block 2Block 2
Block 1Block 1
RAID 0
Block 0Block 0
RAID 1
Block 3Block 3
Block 2Block 2
Block 1Block 1
Block 0Block 0
Block 3Block 3
Block 2Block 2
Block 1Block 1
Block 0Block 0
Host
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 13
Host
Nested RAID – 1+0 (Mirroring and Striping)
Block 3Block 3
Block 3Block 3
Block 1Block 1
RAID 1Block 0Block 0Block 0Block 0
Block 1Block 1
RAID 0
Block 2Block 2Block 2Block 2 RAIDController
RAIDController
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 14
Host
Nested RAID – 1+0 (Mirroring and Striping)
RAIDController
RAIDController
RAID 1
Block 0Block 0
Block 0Block 0
RAID 0
Block 2Block 2
Block 2Block 2 Block 3Block 3
Block 3Block 3
Block 1Block 1
Block 1Block 1Block 0Block 0
Block 2Block 2
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 15
RAID Redundancy: Parity
Parity Disk
1
95
3
117
0
0 1 2 34 5 6 7
4
6
1
7
18
Host
RAIDController
RAIDController
Parity calculation 4 + 6 + 1 + 7 = 18The middle drive fails:
4 + 6 + ? + 7 = 18
? = 18 – 4 – 6 – 7
? = 1
?
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 16
Host
RAIDController
RAIDController
Block 1Block 1
Block 2Block 2
Block 3Block 3
P 0 1 2 3
Block 0Block 0Block 3Block 3Block 2Block 2Block 1Block 1Block 0Block 0
ParityGenerated
RAID 3 (Parallel access array with dedicated parity disk) (Bit-interleaved Parity)
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 17
RAID 4 - Striped array with independent disks and a dedicated parity disk(Block-interleaved Parity)
RAIDController
RAIDController
P 0 1 2 3
Block 0Block 0
Block 0Block 0
Block 4Block 4
Block 1Block 1
Block 5Block 5
Block 2Block 2
Block 6Block 6
Block 3Block 3
Block 7Block 7
P 0 1 2 3P 0 1 2 3
P 4 5 6 7P 4 5 6 7
ParityGenerated
Block 0Block 0
P 0 1 2 3P 0 1 2 3
Host
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 18
Host
Block 0Block 0
P 0 1 2 3P 0 1 2 3
Block 7Block 7
RAIDController
RAIDController
P 0 1 2 3
Block 0Block 4Block 0
Block 1Block 1
Block 5Block 5
Block 2Block 2
Block 6Block 6
Block 3Block 3
ParityGenerated
Block 0Block 0
P 0 1 2 3P 0 1 2 3
Block 4Block 4
P 4 5 6 7P 4 5 6 7P 4 5 6 7P 4 5 6 7
Block 4Block 4
P 4 5 6 7
Block 4ParityGenerated
RAID 5 (Striped array with independent disks and distributed parity)
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 19
RAID 6 – Dual Parity RAID
Two disk failures in a RAID set leads to data unavailability and data loss in single-parity schemes, such as RAID-3, 4, and 5
Increasing number of drives in an array and increasing drive capacity leads to a higher probability of two disks failing in a RAID set
RAID-6 protects against two disk failures by maintaining two parities– Horizontal parity which is the same as RAID-5 parity– Diagonal parity is calculated by taking diagonal sets of data blocks
from the RAID set members
Even-Odd, and Reed-Solomon are two commonly used algorithms for calculating parity in RAID-6
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 20
RAID 6 (P+Q)
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 21
RAID 6-DP (Diagonal Parity)
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 22
RAIDMin
DisksStorage
Efficiency %Cost Read Performance Write Performance
0 2 100 Low
Very good for both random and sequential
readVery good
1 2 50 HighGood
Better than a single disk
GoodSlower than a single
disk, as every write must be committed to two
disks
3 3
(n-1)*100/nwhere n= number of
disksModerate
Good for random reads and very good for sequential reads
Poor to fair for small random writesGood for large,
sequential writes
5 3
(n-1)*100/nwhere n= number of
disksModerate
Very good for random reads
Good for sequential reads
Fair for random writeSlower due to parity
overhead Fair to good for
sequential writes
6 4
(n-2)*100/nwhere n= number of
disks
Moderate but more
than RAID 5
Very good for random reads
Good for sequential reads
Good for small, random writes
(has write penalty)
1+0and0+1
4 50 High Very good Good
RAID Comparison
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 23
Small (less than element size) write on RAID 3 & 5 Ep = E1 + E2 + E3 + E4 (XOR operations)
If parity is valid, then: Ep new = Ep old – E4 old + E4 new (XOR operations)– 2 disk reads and 2 disk writes
Parity Vs Mirroring– Reading, calculating and writing parity segment introduces penalty to every write operation– Parity RAID penalty manifests due to slower cache flushes– Increased load in writes can cause contention and can cause slower read response times
Ep new
RAID Controller
2 XOR
Ep new Ep old E4 old E4 new
+-= E4 oldEp old E4 new
RAID Impacts on Performance
P0 D1 D2 D3 D4
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 24
RAID Penalty Exercise
Total IOPS at peak workload is 1200
Read/Write ratio 2:1
Calculate IOPS requirement at peak activity for– RAID 1/0
– RAID 5
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 25
Solution
Total IOPS = 1200
Read / Writ ratio 2:1
For RAID 1/0:
(1200x2/3) + (1200x(1/3)x2) = 800 + 800 = 1600 IOPS
For RAID 5:
(1200x2/3) + (1200x(1/3)x4) = 800 + 1600 = 2400 IOPS
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 26
RAIDController
RAIDController
Hot Spares
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 27
Module Summary
Key points covered in this module:
What RAID is and the needs it addresses
The concepts upon which RAID is built
Some commonly implemented RAID levels
© 2009 EMC Corporation. All rights reserved. Data Protection: RAID - 28
Check Your Knowledge
What is a RAID array?
What benefits do RAID arrays provide?
What methods can be used to provide higher data availability in a RAID array?
What is the primary difference between RAID 3 and RAID 5?
What is advantage of using RAID 6?
What is a hot spare?