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Data Protection: RAID

Data Protection: RAID. Chapter 3. Chapter Objective. After completing this chapter, 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

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Data Protection: RAID

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  1. Data Protection: RAID Chapter 3

  2. Chapter Objective After completing this chapter, 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 Data Protection: RAID

  3. Why RAID • Due to mechanical components in a disk drive it offers limited performance • An individual drive has a certain life expectancy • Measured in Mean Time Between Failure (MTBF) • Example • If the MTBF of a drive is 750,000 hours, and there are 1000 drives in the array, then the MTBF of the array becomes 750,000 /1000, or 750 hours • RAID was introduced to mitigate this problem • RAID provides: • Increase capacity • Higher availability • Increased performance Data Protection: RAID

  4. Host RAID Array Components RAID It is a technique that combines multiple disk drives into a logical unit (RAID set) and provides protection, performance, or both. Physical Array Logical Array RAIDController Hard Disks RAID Array 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 Data Protection: RAID

  6. RAID Techniques • Three key techniques used for RAID are: • Striping • Mirroring • Parity Module 3: Data Protection - RAID

  7. Data Organization: Striping • Within each disk, a predefined no. of contiguously addressable disk blocks are defined as strips. • The set of aligned strips that spans across all the disks within a RAID set is called stripe. • The data is broken down into blocks and each block is written to a separate disk drive. • Strip size describes the no. of blocks in a strip. • Note all strips in the stripe have the same no. of blocks. • Stripe size is a multiple of strip size by the no. of HDD’s in the RAID set. Data Protection: RAID

  8. Strip Stripe=192KB Stripe Data Organization: Striping Strip 1=64KB Strip 2=64KB Strip 3=64KB Stripe 1 Stripe 2 Data Protection: RAID Strips

  9. RAID Technique – Striping Strip RAID Controller Stripe Host Module 3: Data Protection - RAID

  10. Data Organization: Mirroring • Data is stored on two different disks. • When the failed disk is replaced with a new disk, the controller copies the data from the surviving disk of the mirrored pair. • Advantages: a) provides data protection b) improves read performance Disadvantages: a) expensive b) write performance detriorates Data Protection: RAID

  11. RAID Technique – Mirroring Block 0 RAID Controller Block 0 Block 0 Host Module 3: Data Protection - RAID

  12. Data Organization: Parity • Method of protecting striped data from HDD failure without the cost of mirroring • An additional disk is added to the tripe width to hold parity. • Calculation of parity is a function of the RAID controller. • Advantages: reduces cost • Disadvantages: parity is recalculated every time there is a change in data. Data Protection: RAID

  13. RAID Technique – Parity 4 D1 6 D2 RAID Controller 1 D3 7 D4 Host 18 P Actual parity calculation is a bitwise XOR operation Module 3: Data Protection - RAID

  14. Data Recovery in Parity Technique 4 D1 6 D2 RAID Controller ? D3 7 D4 Host P Regeneration of data when Drive D3 fails: 18 4 + 6 + ? + 7 = 18 ? = 18 – 4 – 6 – 7 ? = 1 Module 3: Data Protection - RAID

  15. 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 Data Protection: RAID

  16. RAIDController Host RAID 0 0 1 5 9 2 6 10 3 7 11 Data Protection: RAID

  17. Block 1 Block 0 Block 1 Block 1 Block 0 Block 0 RAID 1 RAIDController Host Data Protection: RAID

  18. RAID 1 RAIDController RAID 0 Host Block 5 Block 4 Block 2 Block 1 Block 5 Block 4 Block 2 Block 1 Block 0 Block 3 Nested RAID – 0+1 (Striping and Mirroring) Data Protection: RAID

  19. RAID 1 RAIDController RAID 0 Host Block 2 Block 4 Block 2 Block 5 Block 1 Block 5 Block 4 Block 2 Block 1 Block 4 Block 1 Block 5 Nested RAID – 0+1 (Striping and Mirroring) Data Protection: RAID

  20. RAID 0 RAIDController RAID 1 Host Block 2 Block 5 Block 5 Block 2 Block 0 Block 4 Block 3 Block 3 Block 1 Block 1 Block 0 Block 4 Nested RAID – 1+0 (Mirroring and Striping) Data Protection: RAID

  21. RAID 0 RAIDController RAID 1 Host Block 1 Block 1 Block 4 Block 4 Block 5 Block 5 Block 2 Block 2 Block 1 Block 4 Nested RAID – 1+0 (Mirroring and Striping) Data Protection: RAID

  22. RAIDController Host The middle drive fails: RAID Redundancy: Parity 0 4 1 6 5 9 1 ? 3 7 7 11 Parity calculation 4 + 6 + 1 + 7 = 18 0 1 2 3 4 5 6 7 18 4 + 6 + ? + 7 = 18 ? = 18 – 4 – 6 – 7 ? = 1 Data Protection: RAID Parity Disk

  23. RAIDController ParityGenerated Host Block 1 Block 2 Block 3 Block 0 Block P 0 1 2 3 RAID 3 Data Protection: RAID

  24. P 0 1 2 3 P 0 1 2 3 Block 0 Block 0 Block 3 Block 1 Block 2 P 4 5 6 7 Block 5 Block 4 Block 6 Block 7 RAIDController Block 0 Block 0 ParityGenerated Host P 0 1 2 3 RAID 4 Data Protection: RAID

  25. P 0 1 2 3 P 0 1 2 3 Block 0 Block 0 Block 3 Block 1 Block 2 ParityGenerated P 4 5 6 7 P 4 5 6 7 Block 5 Block 6 Block 4 Block 4 Block 7 RAIDController Block 0 Block 4 Block 0 Block 4 P 4 5 6 7 ParityGenerated Host P 0 1 2 3 RAID 5 Data Protection: RAID

  26. 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 Data Protection: RAID

  27. RAID Comparison Data Protection: RAID

  28. Suitable RAID Levels for Different Applications • RAID 1+0 • Suitable for applications with small, random, and write intensive (writes typically greater than 30%) I/O profile • Example: OLTP, RDBMS – Temp space • RAID 3 • Large, sequential read and write • Example: data backup and multimedia streaming • RAID 5 and 6 • Small, random workload (writes typically less than 30%) • Example: email, RDBMS – Data entry Module 3: Data Protection - RAID

  29. RAID Impacts on Performance RAID Controller Cpnew Cpold C4 old C4 new = - + • In RAID 5, every write (update) to a disk manifests as four I/O operations (2 disk reads and 2 disk writes) • In RAID 6, every write (update) to a disk manifests as six I/O operations (3 disk reads and 3 disk writes) • In RAID 1, every write manifests as two I/O operations (2 disk writes) 2 3 4 1 A1 A2 A3 A4 AP B1 B2 B3 BP B4 C1 C2 CP C3 C4 Module 3: Data Protection - RAID

  30. D4 D2 D1 P0 D3 RAID Impacts on Performance RAID Controller Ep new Ep old E4 old E4 new • Small (less than element size) write on RAID 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 = - + 2 XOR Data Protection: RAID

  31. 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 Data Protection: RAID

  32. Solution: RAID Penalty • For RAID 1/0, the disk load (read + write) = (1200 x 2/3) + (1200 x (1/3) x 2) = 800 + 800 = 1600 IOPS • For RAID 5, the disk load (read + write) =(1200 x 2/3) + (1200 x (1/3) x 4) = 800 + 1600 = 2400 IOPS Module 3: Data Protection - RAID

  33. Hot Spares • It refers to a spare HDD in RAID array that temporarily replaces a failed HDD. • If parity RAID is used, then the data is rebuilt onto the hot spare from the parity and the data on the surviving HDDs in the RAID set. • If mirroring is used, then the data from the surviving mirror is used to copy the data. Data Protection: RAID

  34. Hot Spares • When failed HDD is replaced with new HDD, one of foll. takes place: • The hot spare replaces the new HDD permanently. • When a new HDD is added, data from the hot spare is copied to it. • A hot spare can be configured as automatic or user-initiated. Data Protection: RAID

  35. Hot Spare Failed disk RAIDController Replace failed disk Hot spare Module 3: Data Protection - RAID

  36. Chapter Summary Key points covered in this chapter: • What RAID is and the needs it addresses • The concepts upon which RAID is built • Some commonly implemented RAID levels Data Protection: RAID

  37. Exercise 1: RAID • A company is planning to reconfigure storage for their accounting application for high availability • Current configuration and challenges • Application performs 15% random writes and 85% random reads • Currently deployed with five disk RAID 0 configuration • Each disk has an advertised formatted capacity of 200 GB • Total size of accounting application’s data is 730 GB which is unlikely to change over 6 months • Approaching end of financial year, buying even one disk is not possible • Task • Recommend a RAID level that the company can use to restructure their environment fulfilling their needs • Justify your choice based on cost, performance, and availability Module 3: Data Protection - RAID

  38. Exercise 2: RAID • A company (same as discussed in exercise 1) is now planning to reconfigure storage for their database application for HA • Current configuration and challenges • The application performs 40% writes and 60% reads • Currently deployed on six disk RAID 0 configuration with advertised capacity of each disk being 200 GB • Size of the database is 900 GB and amount of data is likely to change by 30% over the next 6 months • It is a new financial year and the company has an increased budget • Task • Recommend a suitable RAID level to fulfill company’s needs • Estimate the cost of the new solution (200GB disk costs $1000) • Justify your choice based on cost, performance, and availability Module 3: Data Protection - RAID

  39. 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? Data Protection: RAID

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