1. CS522 Advanced database Systems
1/26/2018
CS522 Advanced database Systems
4. Disks
Huiping Guo
Department of Computer Science
California State University, Los Angeles

2. Review File organizations Index Data entry Index entry Classification
1/26/2018
Review
File organizations
Index
Data entry
Format
Structure
Index entry
Classification
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3. Disks and files DBMS stores information on hard disks
This has major implication for DBMS design!
READ: transfer data from disk to main memory (RAM)
WRITE: transfer data from RAM to disk
Both are high cost operation, relative to in-memory operations.
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4. Disks and files (Cont.) Why not store everything in main memory?
Costs too much
Main memory is volatile
Typical storage hierarchy
RAM for currently used data
Disk for the main database (secondary storage)
Random access
Tapes are for archiving older versions of the data (tertiary storage)
Sequential access
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5. Disks Secondary storage Advantages over tapes
Random access vs. sequential
Data is stored and retrieved in units called pages
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6. Components of a disk sector block Cylinder Spindle Disk head Tracks
Arm assembly
Platters
sector
Arm movement
Cylinder
surface
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7. Components of a disk Data blocks Tracks Platters Surface
Data is stored on disk in units called disk blocks
A data block is a contiguous sequence of bytes and is the unit in which data is written to a disk and read from a disk
Tracks
Blocks are arranged in concentric rings called tracks
Platters
Tracks are on one or more platters
Surface
Tracks can be recorded on one or both surfaces of a platter (single-sided platter or double-sided platter)
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8. Components of a disk Cylinder Sectors Disk head
The set of all tracks is shaped like a cylinder
A cylinder contains one track per platter surface
Sectors
Each track is divided into arcs, called sectors
Sector size is a characteristic of the disk
A block contains multiple sectors
Disk head
There is a disk head for each surface
An array of disk heads moves as a unit
when one head is positioned over a block, the other heads are in identical positions with respect to their platters
To read a block, a disk head must be positional on top of the block
At most one disk head is allowed to read/write at a time
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9. Components of a disk sector block Cylinder Spindle Disk head Tracks
Arm assembly
Platters
sector
Arm movement
Cylinder
surface
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10. Summary of disk components
Platter, surface, head
Cylinder, track
Block, sector
A block stores a data page
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11. Disk access time Seek time Rotational delay Transfer time
moving arms to position disk head on track
Rotational delay
Waiting for block to rotate under head
Transfer time
Actually moving data to/from disk surface
Transfer time =
Block size/Average transfer rate
Access time =
seek time + rotational delay + transfer time
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12. Reduce I/O cost Seek time and rotational delay dominate.
Seek time varies from about 1 to 20msec
Rotational delay varies from 0 to 10msec
Transfer rate is about 1msec per 4KB page
Key to lower I/O cost
reduce seek/rotation delays!
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13. Arranging Pages on Disk
`Next’ block concept:
blocks on same track, followed by
blocks on same cylinder, followed by
blocks on adjacent cylinder
To minimize seek and rotational delay
Blocks in a file should be arranged sequentially on disk (by `next’),
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14. Exercise 1 A disk with What’s the capacity of a track, the disk?
a sector size of 512 bytes,
2000 tracks per surface,
50 sector per track,
five double-sided platters
Average seek time 10 msec
What’s the capacity of a track, the disk?
How many cylinders?
If the disk platters rotate at 5400 revolutions per min, what is the maximum rotational delay?
If one track of data can be transferred per revolution, what is the transfer rate?
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15. Answers Capacity of a track = 512x50 = 25K
Capacity of a disk = 2000x5x2x25K = 500,000K
2000 Cylinders
Maximum rotational delay
= (1/5400) x 60 = seconds
Transfer rate = 25K/0.011 = 2,250K/Second
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16. Exercise 2 The same disk specification from ex1 Block size is 1024byte
A file containing 100,000 records of 100 bytes each
No record is allowed to span two blocks
How many records fit onto a block?
What time is required to read a file containing 100,000 of 100 bytes each sequentially?
What time is required to read a file containing 100,000 of 100 bytes each in a random order?
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17. Answers A block holds 1024/100 = 10 records A track has 25 blocks
The file needs 100,000/(10x25) = 400 tracks (40 cylinders)
One track of data can be transferred in sec
So it takes x 400 = 4.4 sec to transfer 400 tracks of data.
This access seeks the track 40 times, so seek time is 40x0.01=0.4 sec.
The total access time = = 4.8 sec.
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18. Answers (cont.) 3. for any block,
access time = seek time + rotational delay + transfer time
Seek time = 10 msec
Rotational delay = 0.011/2 = 6 msec
Transfer time = 1k/(2,250K/sec) = msec
The access time for a block of data is msec
The file contains 100,000/10 blocks, so the access time is
164.4 sec.
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19. RAID
Disks are potential bottlenecks for system performance and storage system reliability
Disk performance has been improved slowly
Disks have much higher failure rates
Disk Array
An arrangement of several disks that gives abstraction of a single, large disk
Goals
Increase performance and reliability
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20. Two main techniques Improve performance: data striping
Data is partitioned
size of a partition is called the striping unit
Partitions are distributed over several disks.
Improve reliability: Redundancy
More disks => more failures
Redundant information allows reconstruction of data if a disk fails.
Redundancy level
RAID
Redundant Array of Independent Disks
Disk arrays that implement the two techniques
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21. Structure of a DBMS DBMS 4. Disks CS522_S16 File and Access Methods
Buffer Manager
Disk Space Manager
Recovery
Manager
Plan Executor
Parser
Operator Evaluator
Optimizer
Query evaluation
engine
Transaction
Lock
Concurrency control
DBMS
query
Results
Database
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22. Disk Space Management Lowest layer of DBMS Manage space on disk How?
Supports the concepts of a page as a unit of data
Provides commands to allocate or deallocate a page and read/write a page
Manage space on disk
Keep track of which disk blocks are in use
Keep track of which pages are on which disk blocks
How?
Maintain a list of free blocks
Or maintain a bit map with one bit for each disk block
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23. Buffer Manager Buffer manager is a software layer responsible for
Bring pages from disk to main memory as needed
Managing available main memory
Buffer pool
The main memory is partitioned into a collection of pages, called buffer pool.
The main memory page in the buffer pool are called frames
One frame holds one data page
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24. Buffer Manager Page Requests from Higher Levels DB
1/26/2018
Buffer Manager DB
MAIN MEMORY
DISK
disk page
free frame
Page Requests from Higher Levels
BUFFER POOL
choice of frame dictated
by replacement policy
P318
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25. Buffer manager (BM) and higher level codes(HLC)
HLC needs a page
Asks BM for the page
BM brings the page into a frame if it is not in the buffer pool
HLC doesn’t need a page
Asks BM to release the frame
The frame can be reused
HLC also needs to tell BM whether the page has been modified
BM ensures that any modification is propagated to the copy of the page on disk
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26. Information buffer manager keeps
Table of <frame#, pageid> pairs
For each frame, keep two variables
pin_count
The number of current users
Initially, the pin_count for every frame is set to 0
dirty
Whether the page has been modified since it was brought into the buffer pool
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27. When a Page is Requested ...
BM checks the buffer pool to see of some frame contains the requested page
If it is in pool
Increase its pin_count (pinning the page)
If requested page is not in pool:
Choose a frame for replacement, increase its pin_count (pinning the page)
If the frame is dirty, write it to disk
Read requested page into chosen frame
Return the address of the frame containing the requested page to the requestor
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28. Choose a frame for replacement
Candidate frames for replacement
Free frames
Frames with pin_count = 0
No candidate frames
Wait until some page is released
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29. Buffer Replacement Policy: LRU
Least-recently-used (LRU)
Use a queue of pointers to frames with pin_count 0
A frame is added to the end of the queue when it becomes a candidate for replacement
The page chosen for replacement is the one in the frame at the head of the queue
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30. Buffer Replacement Policy: Clock
Frames are arranged in a circle, like clock face
“current” variable (1 – N)is like clock hand moving across the face
Each frame also has an associated reference bit, which is turned on the page pin_count goes to 0
The current frame is considered for replacement
If the frame is not chosen for replacement, current++, next frame is considered
If the frame has pin_count>0, current++
If the frame has the referenced bit turned on (pin_count=0) , turn it off, current++
Recently referenced page is less likely to be replaced
If the frame has pin_count=0, and its reference bit turned off, it’s chosen for replacement
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31. Buffer Replacement Policy
Other policies
Most-Recently-Used (MRU)
FIFO
Random
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32. DBMS vs. OS File System
OS does disk space & buffer mgmt: why not let OS manage these tasks?
Differences in OS support: portability issues
Some limitations, e.g., files can’t span disks.
Buffer management in DBMS requires ability to:
pin a page in buffer pool, force a page to disk (important for implementing CC & recovery),
adjust replacement policy, and pre-fetch pages based on access patterns in typical DB operations.
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