File Processing : Storage Media

2015, Spring Pusan National University Ki-Joune Li

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Major Functions of Computer

   

Computation Storage Communication Presentation STEM

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Storage of Data



Major Challenges



How to store and manage a large amount of data



Example : more than 100 peta bytes for EOS Project



How to represent sophisticated data STEM

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Modeling and Representation of Real World



Example



Building DB about Korean History Real World Computer World STEM

 

Very complicated and Depending on viewpoint Database Course : 2010 Fall semester

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Managing Large Volume of Data



Large Volume of Data



Cost for Storage Media



Not very important and negligible



Processing Time



Comparison between main memory and disk access time

 

RAM : several nanoseconds (10 -9 sec) Disk : several milliseconds (10 -3 sec)



Time is the most valuable resource



Example



Retrieving a piece of data from 100 peta bytes DB STEM

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Managing Large Volume of Data



Management of Data



Secure Management



From hacking



From any kinds of disasters



Consistency of Data



Example



Failure during a flight reservation transaction



Concurrent transaction STEM

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Goals of File Systems

STEM



To provide with 1. efficient Data Structures for storing large and complex data 2. Access Methods for rapid search 3. Query Processing Methods 4. Robust Management of Transactions

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Memory Hierarchy

STEM

 

Large Data Volume



Not be stored in main memory



But in secondary memory Memory Hierarchy Cache Memory Main Memory Secondary Memory Tertiary Memory 256 K bytes 1G bytes 100 G bytes 100 Tera bytes

Faster Cheaper

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Flash Memory

STEM

   

Non-Volatile

 

Data survives power failure, but Data can be written at a location only once, but location can be erased and written to again



Can support only a limited number of write/erase cycles.



Erasing of memory has to be done to an entire bank of memory Speed

 

Reads are roughly as fast as main memory But writes are slow (few microseconds), erase is slower Cost per unit of storage roughly similar to main memory Widely used in embedded devices such as digital cameras

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Optical Storage

  

Non-volatile :



data is read optically from a spinning disk using a laser



CD-ROM (800 MB), DVD (4.7 to 17 GB), CD-R, DVD-R



CD-RW, DVD-RW, and DVD-RAM Speed



Reads and writes are slower than with magnetic disk Juke-box systems



Large numbers of removable disks,



Few drives, and



Mechanism for automatic loading/unloading of disks



For storing large volumes of data STEM

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Tape

   

Non-volatile



Primarily Used for backup Speed



Sequential access : much slower than disk Cost



Very high capacity (40 to 300 GB tapes available)



Tape can be removed from drive



Drives are expensive Tape jukeboxes



hundreds of terabytes to even a petabyte STEM

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Data Access with Secondary Memory

STEM

Get Data Access Request

Hit Ratio r

h

= n

h

/ n

a

How to increase hit ratio ?

Get Data

If in main memory

Main Memory

Load on main memory

If not in main memory

Access to Disk

Disk

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Why Hit Ratio is so important ?



Example for(int i=0;i<1000;i++) Nbytes=read(fd,buf,100); when r

h

= 0

1000 * 10 -2 sec = 10 sec

STEM

1000 disk accesses ?

when r

h

= 1

1000 * 10 -8 sec = 10 -5 sec

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Physical Structure of Disk

200~400 sectors 512 bytes 2 * n

DF

STEM

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Disk Access Time

STEM



Disk Access Time t = t

S

+ t

R

+ t

T

, where



t S



: Seek Time Time to reposition the head over the correct track



Average seek time is 1/2 the worst case seek time



4 to 10 milliseconds on typical disks



t R

: Rotational Latency

 

Time to reposition the head over the correct sector Average rotational latency : ½ r (to find index point) + ½ r =

r



In case of 15000 rpm : r =1*60sec/15000 = 4 msec



t T

: Transfer Time



Time to transfer data from disk to main memory via channel

 

Proportional to the number of sectors to read Real transfer time is negligible

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Block-Oriented Disk Access



Example for(int i=0;i<1000;i++) Nbytes=read(fd,buf,10); 1000 times 100 times

Buffer in main memory

10 times

Number of Disk Accesses

1 block (e.g. 1024 bytes) STEM

10 bytes 1024 bytes

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Disk Block

  

Unit of Disk Access Block Size



Normally multiple of sectors



1K, 4K, 16K or 64K bytes depending on configuration Why not large block ?



Limited by the size of available main memory



Too large : unnecessary accesses of sectors



e.g. only 100 bytes, when block size is given as 64K



1 block : 128 sectors (about ½ track, ½ rotation, 2 msec)



Too wasteful STEM