1. Chapter 2. Data Storage
Chapter 2

2. Outline Memory hierarchy Hardware: Disks Access Times
Example - Megatron 747
Optimizations
Disk failure
RAIDs
Chapter 2

3. Hardware - Data Storage
Users
DBMS’s
Operating Systems
Hardware - Data Storage
Chapter 2

4. The Memory Hierarchy DBMS Programs, Main-memory Tertiary DBMS’s
Storage
Disk As
Virtual
Memory
File
System
Main memory
Cache
Chapter 2

5. Cache
The cache is an integrated circuit or part of the processor’s chip
Holding data or machine instructions
Copy from main-memory
If data being expelled from the cache has been modified, then the new value must be copied into the main memory.
Typical performance
Capacities up to a megabyte
Access time: 10 nanoseconds (10-8 seconds)
Moving data bet. Cache and main memory: 100 nanoseconds (10-9 seconds)
Chapter 2

6. Main Memory
Everything that happens in the computer is resident in main memory
Capacity: around 100 Mbyte to 10 Gbyte
Random access
Typical access time is nanoseconds
Chapter 2

7. Virtual Memory Is a part of disk In a 32-bit address machine
Virtual memory grows up to 232 bytes (4 Gbyte)
Data is moved between disk and main memory in entire blocks, which are also called pages in main memory
Main-memory database systems
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8. Secondary Storage (1) Slower, more capacious than main memory
Random access
magnetic, optical, magneto-optical disks
Disk read/write are done by moving a chuck of bytes called blocks (or pages)
file
buffer
Chapter 2

9. Secondary Storage (2) Accessing a block: 10-30 milliseconds
Recently, one disk unit can store data ranging from 10 to 32 Gbytes
A machine can have several disk units
Chapter 2

10. Tertiary Storage (1)
Have been developed to hold data volumes measured in terabytes
Compared with secondary storage, it offers
Higher read/write times
Larger capacities and smaller cost per byte
Not random access in general
Chapter 2

11. Tertiary Storage (2) Kinds of tertiary storage devices Capacities
Ad-hoc tape storage
Optical-disk juke boxes: CD-ROMs
Tape silo: an automated version ad-hoc tape storage
Capacities
CD: 2/3 Gbytes, 2.3 Gbytes
Tapes: 50 Gbytes
Access time: about 1000 times slower than secondary memory
Chapter 2

12. Volatile and Nonvolatile
Volatile vs. nonvolatile storage
Flush memory
A form of main memory
Nonvolatile
Becomes economical
RAM disk
A battery-backed main memory
Chapter 2

13. Access Time vs. Capacity2 1 -1 -2 -3 -4 -5 -6 -7 -8 -9 5 6 7 8 9 10 11 12 13
floppy disk
zip disk
Secondary
Main
Cache
Tertiary
X (10  seconds)
Y (10 y bytes)
Chapter 2

14. Moore’s Law
Gordon Moore observed that the followings double every 18 months
The speed of processors, i.e., the number of instructions executed per second and the ratio of the speed to cost of a processor
The cost of main memory per bit and the number of bits that can be put on one chip
The cost of disk per bit and the number of bytes that a disk can hold
Not applicable to
Main memory access time, disk access time
Chapter 2

15. Disks … A typical disk Terms: Platter, Head, Actuator Cylinder, Track
Sector, Block, Gap
A typical disk
Chapter 2

16. Disks: A Top View top view Cylinder, Track, Sector, Gap
Gaps often represents about 10% of the total tracks
A entire section cannot be used if portion of it gets destroyed
Typically a block consists of one or more sectors.
top view
Chapter 2

17. The Disk Controller Controls one or more disk drives Processor
Main
Memory
Disk
Controller
Bus
Disks
Controls one or more disk drives
controlling the mechanical actuator
selecting a surface or a sector on that surface
Transferring bits via a data bus
Chapter 2

18. Disk Storage Characteristics (as of 1999)
Rotation speed of the disk assembly
5400 RPM (one rotation every 11 milliseconds)
Number of platters per unit
Typical disk drive: 5 platters (10 surfaces)
Floppy/zip disk: 1 platter (2 surfaces)
Number of tracks per surface
Have as many as 10,000 tracks
3.5 inch diskette : 40 tracks
Number of bytes per track
Common disk: 105 or more bytes
3.5 inch diskette: 150K
Chapter 2

19. Megatron 747 Disk (1) Characteristics Diameters of tracks
Have 4 platters (8 surfaces)
8192 (213) tracks per surface
On average 256 (28) sectors per track
512 (29) bytes per sector
Diameters of tracks
outermost track is 3.5 inches
innermost track is 1.5 inches
Track consists of two parts
gap: 10 %
data: 90%
Chapter 2

20. Megatron 747 Disk (2) The capacity of the disk
8 surfaces * 8192 tracks * 256 sectors * 512 bytes = 8G bytes
A single track on average
256 sectors * 512 bytes = 128K bytes = 1 Mbits
A cylinder is of 1 Mbytes on average
If a block is 4096 bytes (212)
A block uses 8 sectors (= 4096 bytes / 512 bytes)
A track consists of 32 blocks (= 256 sectors / 8)
Chapter 2

21. Megatron 747 Disk (3)
If each track had the same number (i.e. 256) of sectors, then the density of bits around the tracks would be greater
Length of the outermost track
0.9 * 3.5 *  ≒ 9.9 inch
1 megabit / 9.9 ≒ 100,000 bits per inch
Length of the innermost track
0.9 * 1.5 *  ≒ 4.2 inch
1 megabit /4.2 ≒ 250,000 bits per inch
Each track in Megatron 747 has the different numbers of sectors
outer: 320 sectors
middle: 250 sectors
inner: 192 sectors
The outermost track
1,801,800 bit / 9.9 ≒ 182,000 bpi
The innermost track
47,880 bit / 4.2 ≒ 114,000 bpi
Chapter 2

22. The Latency of The Disk Disk access time I want block X block x
in memory
disk access time
Disk access time
seek time
rotational delay
transfer time
others
Chapter 2

23. Seek Time
The time to position the head assembly at the proper cylinder
0(zero): already to be at the proper cylinder
Otherwise: move to be at the proper cylinder
In range
3 or 20x x 1
Max
Cylinders Traveled
Time
Chapter 2

24. Rotational latency Time
The time for disk to rotate the first of the sectors containing the block
One rotation takes 10 ms, so rotational latency on average 5 ms.
Head Here
Block I Want
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25. Transfer Time/Other delays
the time to read/writes the data on the appropriate disk surface
10 Mbytes per second
Other delays (here, those are neglected)
taken by the processor and disk controller
due to contention for the disk controller
other delays due to contention
Chapter 2

26. Modifying Blocks Not possible to modify a block on disk directly
Sequence of procedures
Read block (time: rt)
Modify in memory (time: mt)
Write block (time: wt)
Verify (time: vt) if appropriate
Total time
rt + mt + wt + vt
Chapter 2

27. Example 2.3 (1)
Let us examine the time to read a 4096-byte block from the Megatron 747 disk
Characteristic
4 platters (8 surfaces), 1 surface = 8192 tracks
1 track = 256 sectors, 1 sector = 512 bytes
Disk rotates at 3840 RPM, one rotation = 1/64 of a second
To move the head assembly
1ms (to start and stop)+ 1ms for every 500 cylinders
Heads move one track in ms
To move heads from innermost to outermost track
1 + (8192 / 500) = 17.4 ms
Chapter 2

28. Example 2.3 (2) Minimum time (the best case)
No seek time, no rotational latency, only transfer time
Note: 1 track = 256 sectors, 1 sector = 512 bytes
4096 bytes / 512 bytes = 8 sectors (including 7 gap)
gaps/sectors occupy 10%/90% of track
A track has 256 gaps and 256 sectors
36 * 7/ * 8/256 = degrees
(11.109/360)/64 = 4.8e-4 seconds = 0.5 ms
Chapter 2

29. Example 2.3 (3) Maximum time (the worst case)
full seek time and rotational latency, plus transfer time
full seek time: 17.4 ms
full rotational time: 1/64 of a second = 15.6 ms
transfer time: 0.5 ms
= 33.5 ms
Chapter 2

30. Example 2.3 (4) Average Time Transfer time: 0.5 ms
Average rotational time: half of the full rotation = 7.8 ms
Average seek time
average distance traveled = 1/3 of the disk = 2730 cylinders
/500 = 6.5ms
= 14.8 ms
4096
2048
8192
Average
travel
Starting track
Chapter 2

31. RAM model vs. I/O model computation
Dominance of I/O cost
Remember, in-memory operations take the same time as one disk I/O
Should minimize the number of block accesses
Data Structure vs. File Processing
Chapter 2

32. Using Secondary Storage Effectively
In general database
Whole databases are much too large to fit in main memory
Key parts of databases are buffered in main memory
Disk I/O’s occur frequently
Main memory sorts (such as “Quick sort”) are inadequate
Chapter 2

33. Merge Sort Step List 1 List 2 Output start 1, 3, 4, 9 2, 5, 7, 8 none1)
3, 4, 9 1 2)
5, 7, 8
1,2 3)
4, 9
1,2,3 4) 9
1,2,3,4 5)
7, 8
1,2,3,4,5 6) 8
1,2,3,4,5,7 7)
1,2,3,4,5,7,8 8)
1,2,3,4,5,7,8,9
Chapter 2

34. Two-Phase, Multiway Merge-Sort (1)
Sort main-memory-sized pieces of the data
Fill all available main memory with blocks
Sort the records in main memory
Write the sorted records
Chapter 2

35. Two-Phase, Multiway Merge-Sort (2)
Merge all the sorted sublists into a single sorted list
Find the smallest key among the first remaining elements of all the lists
Move the smallest element to the first available position of the output block
If output block is full, write it to disk and reinitialize the same buffer
Repeat until all input blocks become exhausted.
Chapter 2

36. Main-memory Organization
Pointers
to first
unchosen
record
Input buffers, one for each sorted list
Select
smallest
for output
Output
buffers
Chapter 2

37. Merge Sort Example (1) Assumption
10,000,000 tuples, 1 tuple = 100 bytes
So, 1 Gbyte data
50 Mbytes memory available
4096 byte blocks, so each block contains 40 records
Total # of blocks: 250,000
# of blocks in main memory: 12,800 (= 50*220 / 212)
Number of sublists
19 sublists (12,800 blocks) + 1 sublists (6,800 blocks)
Each block read or write: 15 ms
Chapter 2

38. Merge Sort Example (2) Computation First phase Second phase
Read each of the 250,000 blocks once
Write 250,000 new blocks
Total time
(250,000 * 15 ms) * 2 = 7500 seconds = 125 minutes
Second phase
Similar with the first phase
Total time: 125 minutes
Chapter 2

39. Improving the Access Time of Secondary Storage
Place blocks on the same cylinder
Divide the data among several small disks
Mirroring disks
Use a disk-scheduling algorithm
Prefetch blocks to main memory in anticipation of their later use
Chapter 2

40. Organizing Data by Cylinders
Use several adjacent cylinders
Read all the blocks on a single track or on a cylinder consecutively
Neglect all but the first seek time and the first rotational latency
Chapter 2

41. Example 2.9 (1) Recall examples 2.3 and 2.7
Original data may be stored on consecutive cylinders
Total # of cylinders: 1000 (= 1Gbytes / 1M bytes)
Main memory can hold 50 cylinders (i.e. 50M)
To read 50 cylinder data into main memory
6.5 ms for average seek time
49 ms for 49 one-cylinder seeks (1 ms each)
6.4 seconds for transfer of 12,800 blocks
(12,800 * 0.5 ms) / 1000 = 6.4 seconds
So, ,400 = ms
Chapter 2

42. Example 2.9 (2) First phase Second phase Read
((6.5 ms + 49 ms seconds) * 20 times) = 2.15 minutes
Write: The same as reading
Total time: 4.3 minutes
Second phase
Still takes about 125 minutes (WHY ?)
Chapter 2

43. Using Multiple Disks in place of One
Use several disks with their independent heads
Transfer data at a higher rate
Roughly speaking, total time could be divided by the number of disks
Chapter 2

44. Example 2.10 (1)
Replace one 747 by four 737’s which have one platter and two surfaces
Assumption
Divide the given records among the four disks
Occupy 1000 adjacent cylinders on each disk
Fill ¼ of main memory each disk
Recall previous examples
Average seek time and rotational latency: 0
Number of full memory blocks: 12,800
¼ memory size: 3,200 blocks
Chapter 2

45. Example 2.10 (2) Computation First phase
Transfer time: 3200 * 0.5 ms = 1.6 seconds
Read: (6.5 ms + 49 ms seconds) * 20 = 33 sec.
Write: similar with reading
Total time: about 1 minute
Chapter 2

46. Example 2.10 (3) Second phase
Apply delicate techniques (?) to reduce disk I/O time
Start comparisons among the 20 lists as soon as the first element of the block appears in main memory
Use four output buffers …
Total time: about 1 hours (?)
Chapter 2

47. Mirroring Disks Two or more disks hold identical copies of data
Survive a head crash by either disk
If we make n copies of a disk, we can read any n blocks in parallel.
Using mirror disks does not speed up writing, but neither does it slow writing down (to some extent)
Chapter 2

48. Scheduling Requests by the Elevator Algorithm
Disk controller choose which of several requests to execute first, to increase throughput
Elevator Algorithm
Proceed in the same direction until the next cylinder with blocks to access is encountered
When no requests ahead in direction of travel, reverse direction
Chapter 2

49. Example 2.11 Cylinder of Request First time available 1000 3000 7000
3000
7000
2000 20
8000 30
5000 40
Cylinder of Request
Time completed
1000
8.3
3000
21.6
7000
38.9
8000
50.2
5000
65.5
2000
80.8
Cylinder of Request
Time completed
1000
8.3
3000
21.6
7000
38.9
2000
58.2
8000
79.5
5000
94.8
Arrival times for six block-access requests
Finishing times for block accesses using the elevator algorithm
Finishing times for block accesses using the first-come-first-served algorithm
Chapter 2

50. Prefetching Data on Track- or Cylinder-sized Chunks
Can we predict the order in which blocks will be requested from disk ?
For example,
Devote two block buffers to each list when merged (when there is plenty of memory)
When a buffer is exhausted, switch to the other buffer for the same list
Chapter 2

51. Single Buffering Single buffering Computation Read B1  Buffer
Process Data in Buffer
Read B2  Buffer
Process Data in Buffer ...
Computation
P = time to process/block
R = time to read in 1 block
n = # of blocks
Single buffer time = n(P+R)
Chapter 2

52. Single Buffering vs. Double Buffering
done
process A C
process B
done
Memory:
Disk: A A B C D G E F
Chapter 2

53. Double Buffering Computation P = processing time/block
R = IO time/block
n = # of blocks
Double buffering time: R + nP
Single buffering time: n(R+P)
Chapter 2

54. Prefetching Combine prefetching with the cylinder-based strategy
Store the sorted sublists on whole, consecutive cylinders
Read whole tracks or cylinders whenever we need some records from a given list
Chapter 2

55. Example 2.14 (1) Consider the second phase of the sort
Have in main memory two track-sized buffers
A track: 128KB
Total space requirement: 128KB * 20 lists * 2 = 5 Mbyte
Read all the blocks on 1000 cylinders (8000 tracks)
Computation
average seek time : 6.5 ms
the time for disk to rotate once: 15.6 ms
total time (for reading): ( ) * 8000 = 2.95 minutes
Chapter 2

56. Example 2.14 (2)
Have in main memory two cylinder-sized buffers per sorted sublist
1 cylinder = 8 tracks = 128K * 8 = 1M
Use 40 buffers of a megabyte each
50 megabytes available main memory
Need only do a seek once per cylinder
Read all the block on 1000 cylinders (8000 tracks)
Total time (for reading)
( * 15.6) * 1000 cylinders) = 2.19 minutes
Chapter 2

57. Block Size Selection Big block  amortize I/O cost
Big block  read in more useless stuff and takes longer to read
As memory prices drop, blocks get bigger…
Chapter 2

58. Disk Failures Intermittent failure Media decay Write failure
An attempt to read or write a sector is unsuccessful, but with repeated tries we are able to read or write successfully.
Media decay
A bit or bits are permanently corrupted, and the sector becomes unreadable.
Write failure
We can neither write successfully nor can we retrieve the previously written sector.
Disk Crash
When a disk becomes unreadable permanently
Chapter 2

59. Checksums (1)
Each section has additional bits, called the checksum, to check reading or writing operations
(w, s)
w: the data that is read
s: a status bit
A simple form of checksum: parity
Chapter 2

60. Checksums (2) Example 1 (even parity) Example 2 (even parity)
The sequence of bits in a sector :
The parity bit is 1
Data becomes
Example 2 (even parity)
The sequence of bits in a sector :
The parity bit is 0
Data becomes
Chapter 2

61. Checksums (3)
Possible that we cannot detect an error if more than one bit of the sector may be corrupted
If we use n independent bits as a checksum, then the chance of missing an error is only 1/2n (WHY ?)
Chapter 2

62. Stable Storage (1) How to correct errors ?
Stable storage is a technique for organizing a disk so that media decays or failed writes do not result in permanent loss.
The general idea is that sectors are paired, and each pair represents one sector-contents X
As the left (XL) and right (XR) copies
Chapter 2

63. Stable Storage (2) Writing policy
Write the value of X into XL
if status is good, write the value
if status is bad, repeat writing
If fails after a number of times, a media failure in the sector
Repeat above scheme for XR
Reading policy (to obtain the value of X)
Read XL
if status bad is returned, repeat reading
if status good is returned, take that value as X
If can’t read XL , repeat above with XR
Chapter 2

64. Recovery from Disk Crashes
Disk crash is fatal in mission-critical applications
RAID (redundant arrays of independent disks)
Here, we talk levels 5, 6, and 7
These RAID schemes also handle failures discussed previously
Chapter 2

65. The Failure Model of Disks
Mean time to failure represents the length of time by which 50% of a population of disks will have failed catastrophically.
For modern disks, it is about 10 years
Fraction
surviving
Time
Chapter 2

66. RAID Level 1 To protect against data loss
Use mirroring disks
The only way data can be lost is if there is a second disk crash while the first crash is being repaired.
Chapter 2

67. How often will a data loss occur?
Assume
The process of replacing the failed disk
take 3 hours, 1/8 day, 1/2920 year
A failure rate of 5% per year
Probability that the mirror disk will fail during copying
(1/20) * (1/2920) = 1/58,400
Mean time to a failure involving data loss
One of the two disks will fail once in 5 years on the average
5 * 58,400 = 292,000 years
Chapter 2

68. RAID Level 4 (1)
Use one redundant disks no matter how many data disks there are
In the redundant disk, the ith block consists of parity checks for the ith blocks of all the data disks
Use modulo-2 sum: an even parity
disk1:
disk2:
disk3:
disk4:
Data disks
Redundant disk
Chapter 2

69. The Algebra of Modulo-2 Sums
The commutative law
x  y = y  x
The associative law
x  (y  z) = (x  y)  z
The all-0 vector of the appropriate length is the identity for 
x  Ō = Ō  x = x
 is its own inverse
x  x = Ō
If x  y = z, y = x  z
Chapter 2

70. RAID Level 4: Reading (2) Read disks normally.
We could read the redundant disk !
Example
read disk 2, 3, and 4, and get the contents of disk 1 using modulo-2 sum.
disk2 :
disk3 :
disk4 :
disk1 :
Chapter 2

71. RAID Level 4: Writing (3)
When a block is written, we need to change the redundant disk
Naïve approach
N-1 reads of blocks not being rewritten
One write of new block
Rewrite new redundant disk
In total, N+1 disk I/O’s
There is a better way to do that !
Chapter 2

72. Writing Example (4) When disk 2 changes from 10101010 to 11001100
disk2 :
disk4 :
Modulo-2 sum of old and new bits of disk 2
Modulo-2 sum of old redundant disk and modulo-2 sum of disk 2’s
Chapter 2

73. RAID Level 4: Failure Recovery (5)
Recomputing any missing data is simple, and does not depend on which disk (data or redundant) is failed.
Chapter 2

74. RAID Level 5
We could treat each disk as the redundant disk for some of the blocks
That is, do not have to treat one disk as the redundant disk and the others as data disks
When there are n+1 disks (disk 0 – disk n)
If (i mod n+1) = j, then we can treat the ith cylinder of disk j as redundant
Chapter 2

75. Example 2.21 (1) How redundant blocks compute for 4 disks (n=3)?
redundant for block 4, 8, 12, …
Disk 1
redundant for block 1, 5, 9, …
Disk 2
redundant for block 2, 6, 10, …
Disk 3
redundant for block 3, 7, 11, …
Chapter 2

76. Example 2.21 (2)
The reading and writing load for each disk is the same
If all blocks are equally likely to be written
each disk has a 1/4 chance
If not
each disk has a 1/3 chance
Each of four disks is involved in ½ of the writes
1/4 + 3/4 * 1/3 = 1/2
Chapter 2

77. RAID Level 6 (1)
To handle with any number of disk crashes – data or redundant
Here, focused on a simple example, where two simultaneous crashes are correctable and the strategy is based on a simple error-correcting code, Hamming code
Consider a system with seven disks
data disks: disk 1-4
redundant disks: disk 5-7
Chapter 2

78. RAID Level 6 (2) The relationship between data and redundant disks
Note
every possible column of three 0’s and 1’s, except for the all-0 column
the columns for the redundant disk have a singe 1
the columns for the data disks each have at least two 1’s
DATA
Redundant
Disk Number 1 2 3 4 5 6 7
Chapter 2

79. RAID Level 6 (3)
DATA
Redundant
Disk Number 1 2 3 4 5 6 7
The disks with 1 in a row are treated as if they were the entire set of disks in a RAID level 4 scheme.
The bits of disk 5
are the modulo-2 sum of bits of disk 1,2, and 3
The bits of disk 6
are the modulo-2 sum of bits of disk 1,2, and 4
The bits of disk 7
are the modulo-2 sum of bits of disk 1,3, and 4
Chapter 2

80. RAID Level 6 – Read/Write
Reading: Just read data from any data disk normally
Writing
Need to recalculate several redundant disks
Chapter 2

81. A Writing Example (1) Writing Disk 2 is changed to be 0000111
Corresponding redundant disks
disk 5 and 6
Using modulo-2 sum
between old and new disk 2
between modulo-2 sum of disk 2’s and disk 5
between modulo-2 sum of disk 2’s and disk 6
Disk
Contents 1 2 3 4 5 6 7
Chapter 2

82. A Writing Example (2) Disk Contents 1 11110000 2 00001111 3 00111000 4 5 6 7
(old disk 2)
(new disk 2)
(modulo-2 sum )
(modulo-2 sum)
(disk 5)
(new disk 5)
(modulo-2 sum)
(disk 6)
(new disk 6)
Chapter 2

83. RAID Level 6 – Failure Recovery
Assume that disk a and b fails simultaneously
Find a row r in which the columns of a and b are different
For example, a has 0 in row r, b has 1 in row r
Compute the correct b by taking the modulo-2 sum of corresponding bits from all the disks other than b that have 1 in row r.
Then, compute the correct a
Chapter 2

84. A Recovery Example Pick the second row Disk 2: Disk 5:
modulo-2 sum of disks 1, 4, and 6
Disk 5:
modulo-2 sum of disks 1, 2, and 3
Disk
Contents 1 2
???????? 3 4 5 6 7
Chapter 2