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Storage & Indexing 1

Lipyeow
November 09, 2015

Storage & Indexing 1

Storage and indexing in DBMS

Lipyeow

November 09, 2015
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  1. ICS  Data  Storage  &  Retrieval   Overview  of  Storage  &

     Indexing  (i)   Prof.    Lipyeow  Lim   Informa?on  &  Computer  Science  Department   University  of  Hawaii  at  Manoa   1   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa  
  2. Data  Storage   •  Main  Memory   –  Random  access

      –  Vola?le   •  Flash  Memory     –  Random  access   –  Random  writes  are  expensive   •  Disk   –  Random  access   –  Sequen?al  access  cheaper   •  Tapes   –  Only  sequen?al  access   –  Archiving   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   2   Cache   Main  Memory   CPU   Disk   Tapes   Op?cal   Disks   Tertiary Storage
  3. Rela?onal  Tables  on  Disk   •  Record    -­‐-­‐  a

     tuple  or  row  of  a  rela?onal   table   •  RIDs  –  record  iden?fiers  that  uniquely   iden?fy  a  record  across  memory  and   disk   •  Page  –  a  collec?on  of  records  that  is   the  unit  of  transfer  between  memory   and  disk   •  Bufferpool  –  a  piece  of  memory  used   to  cache  data  and  index  pages.   •  Buffer  Manager  –  a  component  of  a   DBMS  that  manages  the  pages  in   memory   •  Disk  Space  Manager  –  a  component  of   a  DBMS  that  manages  pages  on  disk   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   3   record page disk Bufferpool
  4. Magne?c  Disks   •  A  disk  or  plaUer  contains  mul?ple

      concentric  rings  called  tracks.     •  Tracks  of  a  fixed  diameter  of  a   spindle  of  disks  form  a  cylinder.   •  Each  track  is  divided  into  fixed  sized   sectors  (ie.  “arcs”).   •  Data  stored  in  units  of  disk  blocks   (in  mul?ples  of  sectors)   •  An  array  of  disk  heads  moves  as  a   single  unit.     •  Seek  <me:  ?me  to  move  disk  heads   over  the  required  track   •  Rota<onal  delay:  ?me  for  desired   sector  to  rotate  under  the  disk   head.   •  Transfer  <me:  ?me  to  actually   read/write  the  data   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   4   sector tracks spindle rotates Arms move over tracks
  5. Accessing  Data  on  Disk   •  Seek  <me:  ?me  to

     move   disk  heads  over  the   required  track   •  Rota<onal  delay:  ?me  for   desired  sector  to  rotate   under  the  disk  head.   –  Assume  uniform   distribu?on,  on  average   ?me  for  half  a  rota?on   •  Transfer  <me:  ?me  to   actually  read/write  the   data   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   5   sector tracks spindle rotates Arms move over tracks
  6. Example:  Barracuda  1TB  HDD  (ST31000528AS)     •  What  is

     the  average  ?me   to  read  2048  bytes  of   data  ?   =  Seek  ?me  +  rota?onal   latency  +  transfer  ?me   =  8.5  msec  +  4.16  msec  +   (  2048  /  512  )  /  63  *  (60   000  msec  /  7200  rpm  )   =  8.5  +  4.16  +  0.265     Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   6   cylinders   121601   Bytes/cylinder   16065*512   Blocks/cylinder   8029   Sectors/track   63   Heads   255   Spindle  Speed   7200  rpm   Average   Latency   4.16  msec   Random  read   seek  ?me   <  8.5  msec   Random  read   Write  ?me   <  9.5  msec  
  7. File  Organiza?ons   How  do  we  organize  records  in  a

     file  ?   •  Heap  files:  records  not  in  any  par?cular  order   –  Good  for  scans     •  Sorted  files:  records  sorted  by  par?cular  fields   –  scans  in  the  sorted  order  or  range  scans  in  the  sorted   order   •  Indexes:  Data  structures  to  organize  records  via   trees  or  hashing.       –  Like  sorted  files,  they  speed  up  searches  for  a  subset   of  records,  based  on  values  in  certain  (“search  key”)   fields   –  Updates  are  much  faster  than  in  sorted  files   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   7  
  8. Comparing  File  Organiza?ons   Consider  an  employee  table  with  search

     key  <age,sal>   •  Scans  :  fetch  all  records  in  the  file   •  Point  queries:  find  all  employees  who  are  30  years   old  (let’s  assume  there’s  only  one  such  employee)   •  Range  queries:  find  all  employees  aged  above  65.   •  Insert  a  record.   •  Delete  a  record  given  its  RID.   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   8  
  9. Analysis  of  Algorithms   •  Computa?on  model   –  CPU

     comparison  opera?on   –  General:  most  expensive   opera?on   •  Worst-­‐case   –  How  bad  can  it  get  ?   •  Average-­‐case   –  Assump?on  about   probabili?es   •  Analysis:  count  the  number   of  some  opera?on  w.r.t.   some  input  size   •  Asympto?cs:  Big  “O”   –  Constants  don’t  maUer   –  500n+10000  =  O(n)   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   9   SELECT  *   FROM  Employees  E   WHERE  E.age=30   For each tuple t in Employees { if (t.age==30) { output t } } What  is  the  worse  case  number  of  output   tuples  ?   Assume  input  size  :   n    tuples   What  is  the  worse  case  running  ?me  in  the   number  of  comparisons  ?  
  10. Search  Algorithms  on  Sorted  Data   Lipyeow  Lim  -­‐-­‐  University

     of  Hawaii  at  Manoa   10   SELECT  *   FROM  Employees  E   WHERE  E.age=30   For each tuple t in Employees { if (t.age==30) { output t } elsif ( t.age > 30 ) { exit } } Tuples  are   sorted  by  age   What  is  the  worse  case   running  ?me  in  the   number  of  comparisons  ?   Shortcircuited Linear Search (lo, hi) = (0,n-1) mid = lo+(hi-lo)/2 While(hi>lo && E[mid].age!=30) { if (E[mid].age < 30) { lo=mid } else { hi=mid } mid = lo+(hi-lo)/2 } Output all satisfying tuples around E[mid] Binary Search
  11. Analysis  of  Binary  Search   •  Number  tuples  searched  per

     itera?on  =  n,  n/2,  n/4,  ...  1   •  Hence  the  number  of  itera?ons  =  O(  log  n  )   •  Therefore  number  of  comparisons  =  O(log  n)   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   11   (lo, hi) = (0,n-1) mid = lo + (hi-lo)/2 While(hi>lo && E[mid].age!=30) { if (E[mid].age < 30) { lo=mid } else { hi=mid } mid = lo + (hi-lo)/2 } Output all satisfying tuples around E[mid] 16, 19, 19, 20, 26, 30, 30, 31, 36 16, 19, 19, 20, 26, 30, 30, 31, 36 2  cmp   2  cmp   16, 19, 19, 20, 30, 31, 31, 31, 36 16, 19, 19, 20, 30, 31, 31, 31, 36 What  is  the  worse  case?  
  12. Analysis  of  DBMS  Algorithms   Lipyeow  Lim  -­‐-­‐  University  of

     Hawaii  at  Manoa   12   SELECT  *   FROM  Employees   WHERE  age=30   for each page p of Employees table { if (p not in bufferpool) { Fetch p from disk } for each tuple t in page p { if (t.age==40) { output t } } } What  is  the  most  expensive   opera?on  ?   Worst  case  running  ?me  =     +  <me  to  fetch  all  pages  of   Employees    from  disk   +  <me  to  compare  age     +  <me  to  output  result   Table Scan How  would  you  es?mate  these   ?mes  ?   What  is  the  worst  case  number   of  disk  access  ?  
  13. Analysis  Model   •  B  :  number  of  data  pages

      •  R  :  number  of  records  per  page   •  D  :  average  ?me  to  read/write  a  disk  page   – From  previous  calcula?ons,  if  a  page  is  2K  bytes,  D   is  about  13  milliseconds   •  C  :  average  ?me  to  process  a  record   – For  the  1  Ghz  processors  we  have  today,  assuming   it  takes  100  cyles,  C  is  about  100  nanoseconds   Lipyeow  Lim  -­‐-­‐  University  of  Hawaii  at  Manoa   13  
  14. Table  Scans  on  Heap  Files   Lipyeow  Lim  -­‐-­‐  University

     of  Hawaii  at  Manoa   14   SELECT  *   FROM  Employees   for each page p of Employees table { if (p not in bufferpool) { Fetch p from disk } for each tuple t in page p { output t if (t.age==30) { output t } if (t.age>20 && t.age<30) { output t } } } SELECT  *   FROM  Employees   WHERE  age  >  20  and  age  <  30   SELECT  *   FROM  Employees   WHERE  age=30   O(B)  pages  get  fetched  +   O(B*R)  tuples  processed