CS222: Principles of Database Management
Fall 2010
Professor Chen Li
Department of Computer Science
University of California, Irvine
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Topic 1: Data Storage and RecordOriented File Systems
• Data Storage
– Storage hierarchy
– Disks
• Record-oriented file systems
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Storage hierarchy
CPU
...
cache
Memory
Controller
...
Disk/tape
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Storage Media
• Cache: inside/outside CPU
– CPU: becoming faster and faster (>=3 GHz now)
• Main Memory
–
–
–
–
–
costs $100/Mbyte -- reduces every year
‘volatile’ -- does not survive system failures
random I/O very fast
data can be processed by CPU directly
capacity limited to orders of magnitude lower than what database
needs.
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Storage Media: secondary storage
• Disks (floppy disks, hard disks, CD)
–
–
–
–
Cheap, and price reduces each year
Non-volatile (except when disk crashes)
Random I/O slow
Data needs to be transferred to memory to be processed by CPU
• Tape
– Cheaper but slower than disks.
– Sequential I/O devices.
– Handy for backups, sometimes for archival.
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Databases and Storage Devices
•
•
•
•
Due to capacity, cost, volatility factors, DBs usually stored in disks.
Data brought to main memory for processing from disks
There are many ways to interface memory with disk resident data
E.g., virtual memory:
– VM size limited to max address generated by CPU
– Existing VM does not support durability
• File system provides a more powerful mapping between memory and
disk storage
• A bunch of tricks used ensure that high latency of secondary storage
does not impact application response time and system throughput
– access disks asynchronously with active applications
– prefetch data before application needs it
– intelligent caching techniques
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Disk Storages -- Outline
•
•
•
•
•
Disk mechanics
Access times (random, sequential)
Examples
Optimization
Other topics
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Disk mechanics
…
Terms: Spindle, Platters, Magnetic surfaces,
Disk head, Disk controller, …
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Top Views
Tracks
Sectors
Gaps
Cylinders
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Characteristics
•
•
•
•
•
•
Diameter:
Cylinders:
Surfaces:
Tracks/Cyl:
Sector Size:
Capacity:
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1 inch -- 15 inches
100 -- 2000
1 (CDs) -- many
2 (floppies) -- 30
512B -- 50K
360 KB (old floppy) -- >=200GB
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“Block”
• Corresponds to 1 or multiple sectors
• Its address consists of:
–
–
–
–
Physical device # (in case of multi disks)
Cylinder #
Surface #
Sector #
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Random disk access time
block x
in memory
I want
block X
Time = Seek Time + Rotational Delay + Transfer Time + Other
time 1
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time 2
time 3
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time 4
12
Time 1: seek time
3 or 5x
Time
x
1
N
Cylinders Traveled
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Average Random Seek Time
N
N
 
S=
i=1
SeekTime(Track i  Track j)
j=1
ji
N(N-1)
• Assumptions:
– Each track has the same probability to be accessed.
– Each track has the probability to jump to another track.
• Typical S value: 10 ms – 50 ms
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Time 2: Rotational Delay
Initial Head
Block Wanted
• Average delay:
– R = 1/2 revolution
– If disk speed 3600 RPM, then R = 8.33 ms
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Complication
May have to wait for start of track
before we can read desired block
Track Start
Head Here
Block We Want
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Time 3: Transfer time
• Transfer time: block size/transfer rate
• Typical transfer rate:1  3 MB/sec
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Time 4: Other Delays
• CPU time to issue I/O
• Contention for controller
• Contention for bus, memory, etc.
Typical value: “0”
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Sequential disk access
• Reading “Next” block
• Additional time = Block size/transfer rate
• Other time negligible:
– skip gaps
– once in a while, next cylinder
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Random I/O vs Sequential I/O
• Average sequential IO time much smaller than random IO
time
– Random I/O:  20 ms (most time on the initial delay)
– Sequential I/O:  1 ms.
• When designing a structure, try to use sequential
IOs.
– Data layout on disk becomes critical
– Do not just look at the number of IOs
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Modify blocks
•
•
•
•
Read block
Modify in memory
Write block
Verify
– Optional
– If so, the access time needs to add:
full rotation + block size/transfer rate
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Example 1
Disk Specs:
• 3.5 in diameter
• 3600 RPM
• 1 surface
• Usable capacity: 16 MB = 224
• # of cylinders: 128 = 27
• 1 block = 1 sector = 1 KB
• 10% overhead between blocks (gaps)
• seek time:
– average = 25 ms.
– adjacent cyl = 5 ms.
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Cylinder
• bytes/cyl = 224/27 = 217 = 128 KB
• blocks/cyl = 128 KB / 1 KB = 128
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Track
...
One track
• Speed:
– 3600 RPM  60 revolutions / sec  16.66 ms/rev
• In each revolution:
–
–
–
–
Time over useful data: 16.66 * 0.9=14.99 ms
Time over gaps: 16.66 * 0.1 = 1.66 ms
Transfer time 1 block = 14.99/128 = 0.117 ms
Trans. time 1 block + gap = 16.66/128 = 0.13ms
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Bandwidths
• Burst bandwidth:
– No time on gaps (10%)
– 1 KB in 0.117 ms.
BB=1KB / 0.117ms = 8.54 KB/ms = 8.33MB/sec
• Sustained bandwidth:
– Including time on gaps
– 128 KB in 16.66 ms.
SB=128KB /16.66ms = 7.68 KB/ms = 7.50 MB/sec
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Time of random block access
• Time to read one random block T1
• T1 = seek time + rotational delay + Transfer time
–
–
–
–
–
Assume we do not have to wait for track start
Seek time = 25ms
Rotational delay = 16.66ms /2 = 8.33 ms
Transfer time = .117 ms
Total = 25 ms + 8.33 ms + .117 ms= 33.45 ms
• Most of the time is on “seek time” and “rotational
delay”!
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Larger blocks?
1
2
3
4
...
1 block
• Suppose OS deals with 4 KB blocks
• We need to include the time of reading 1 block (without gap)
and 3 blocks (with gaps)
• T4 = 25ms + (16.66ms/2) + (.117) x 1 + (.130) * 3 = 33.83 ms
• Compare to T1 = 33.45 ms – not much difference
– That’s why we want to use sequential IOs!
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Reading a track
• TT = Time to read a full track (start at any block)
• TT = 25ms
(seek time)
+ (0.13ms / 2) (rotational delay, half of a block)
+ 16.66 ms
(transfer time)
= 41.73 ms
• The time could be a bit less by ignoring the last gap.
• Question: what if we need to wait for the start of a
track?
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