Index tuning-- B+tree. overview © Dennis Shasha, Philippe Bonnet 2001 B+-Tree Locking Tree Traversal –Update, Read –Insert, Delete phantom problem: need.

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Presentation transcript:

Index tuning-- B+tree

overview

© Dennis Shasha, Philippe Bonnet 2001 B+-Tree Locking Tree Traversal –Update, Read –Insert, Delete phantom problem: need for range locking ARIES KVL (implemented in DB2) Tree Traversal (next page) Lock on tuples Lock on key values Range locking: –Next key lock 4 24

© Dennis Shasha, Philippe Bonnet 2001 A B C D EF T 1 lock B+-Tree Locking

Bulk Loading of a B+ Tree If we have a large collection of records, and we want to create a B+ tree on some field, doing so by repeatedly inserting records is very slow. Bulk Loading can be done much more efficiently. Initialization: Sort all data entries, insert pointer to first (leaf) page in a new (root) page.

Bulk Loading (Contd.) Add to the root page

Bulk Loading (Contd.) Split the root and create a new root page.

Bulk Loading (Contd.) Index entries for leaf pages always entered into rightmost index page just above leaf level. When this fills up, it splits. (Split may go up right-most path to the root.)

Much faster than repeated inserts, especially when one considers locking!

Comparison: B-trees vs. static indexed sequential file Ref #1: Held & Stonebraker, “ B-Trees Re- examined ”, CACM, Feb Ref # 1 claims: - Concurrency control harder in B-Trees - B-tree consumes more space For their comparison: block = 512 bytes key = pointer = 4 bytes 4 data records per block

Example: 1 block static index 127 keys (127+1)4 = 512 Bytes -> pointers in index implicit!up to 127 blocks k1 k2 k3 k1k2k3 1 data block

Example: 1 block B-tree 63 keys 63x(4+4)+8 = 512 Bytes -> pointers needed in B-treeup to 63 blocks because index isblocks not contiguous k1 k2... k63 k1k2k3 1 data block next -

Size comparison Ref. #1 Static Index B-tree # data blocks height 2 -> > > 16, > ,130 -> 2,048, > 250, ,048 -> 15,752,961 5

Ref. #1 analysis claims For an 8,000 block file, after 32,000 inserts after 16,000 lookups  Static index saves enough accesses to allow for reorganization Ref. #1 conclusion Static index better!!

Ref #2: M. Stonebraker, “ Retrospective on a database system, ” TODS, June 1980 Ref. #2 conclusion B-trees better!! DBA does not know when to reorganize DBA does not know how full to load pages of new index Buffering –B-tree: has fixed buffer requirements –Static index: must read several overflow blocks to be efficient (large & variable size buffers needed for this)

Speaking of buffering … Is LRU a good policyfor B+tree buffers?  Of course not!  Should try to keep root in memory at all times (and perhaps some nodes from second level)

Interesting problem: For B+tree, how large should n be? … n is number of keys / node

Sample assumptions: (1) Time to read node from disk is (S+Tn) msec. (2) Once block in memory, use binary search to locate key: (a + b LOG 2 n) msec. For some constants a,b; Assume a << S (3) Assume B+tree is full, i.e., # nodes to examine is LOG n N where N = # records

 Can get: f(n) = time to find a record f(n) n opt n

 FIND n opt by f ’ (n) = 0 Answer is n opt = “ few hundred ” (see homework for details)  What happens to n opt as Disk gets faster? CPU get faster?