1	DBMS Internals How does it all work? May 3rd, 2004
2	Agenda Comments on phase 2 of the project HW 3 is out. Today: DBMS internals part 1 -- Indexing Query execution Next week: query optimization.
3	What Should a DBMS Do? Store large amounts of data Process queries efficiently Allow multiple users to access the database concurrently and safely. Provide durability of the data. How will we do all this??
4	Generic Architecture
5	Main Points to Take Away I/O model of computation We only count accesses to disk. Indexing: Basic techniques: B+-tree, hash indexes Secondary indexes. Efficient operator implementations: join Optimization: from what to how.
6	The Memory Hierarchy
7	Main Memory Fastest, most expensive Today: 512MB-2GB are common on PCs Many databases could fit in memory New industry trend: Main Memory Database E.g TimesTen Main issue is volatility
8	Secondary Storage Disks Slower, cheaper than main memory Persistent !!! Used with a main memory buffer
9	Buffer Management in a DBMS Data must be in RAM for DBMS to operate on it! Table of <frame#, pageid> pairs is maintained. LRU is not always good.
10	Buffer Manager
11	Tertiary Storage Tapes or optical disks Extremely slow: used for long term archiving only
12	The Mechanics of Disk Mechanical characteristics: Rotation speed (5400RPM) Number of platters (1-30) Number of tracks (<=10000) Number of bytes/track(10⁵)
13	Disk Access Characteristics Disk latency = time between when command is issued and when data is in memory Is not following Moore’s Law! Disk latency = seek time + rotational latency Seek time = time for the head to reach cylinder 10ms – 40ms Rotational latency = time for the sector to rotate Rotation time = 10ms Average latency = 10ms/2 Transfer time = typically 10MB/s Disks read/write one block at a time (typically 4kB)
14	The I/O Model of Computation In main memory algorithms we care about CPU time In databases time is dominated by I/O cost Assumption: cost is given only by I/O Consequence: need to redesign certain algorithms Will illustrate here with sorting
15	Sorting Illustrates the difference in algorithm design when your data is not in main memory: Problem: sort 1Gb of data with 1Mb of RAM. Arises in many places in database systems: Data requested in sorted order (ORDER BY) Needed for grouping operations First step in sort-merge join algorithm Duplicate removal Bulk loading of B+-tree indexes.
16	2-Way Merge-sort: Requires 3 Buffers Pass 1: Read a page, sort it, write it. only one buffer page is used Pass 2, 3, …, etc.: three buffer pages used.
17	Two-Way External Merge Sort Each pass we read + write each page in file. N pages in the file => the number of passes So total cost is: Improvement: start with larger runs Sort 1GB with 1MB memory in 10 passes
18	Can We Do Better ? We have more main memory Should use it to improve performance
19	Cost Model for Our Analysis B: Block size M: Size of main memory N: Number of records in the file R: Size of one record
20	External Merge-Sort Phase one: load M bytes in memory, sort Result: runs of length M/R records
21	Phase Two Merge M/B – 1 runs into a new run Result: runs have now M/R (M/B – 1) records
22	Phase Three Merge M/B – 1 runs into a new run Result: runs have now M/R (M/B – 1)² records
23	Cost of External Merge Sort Number of passes: Think differently Given B = 4KB, M = 64MB, R = 0.1KB Pass 1: runs of length M/R = 640000 Have now sorted runs of 640000 records Pass 2: runs increase by a factor of M/B – 1 = 16000 Have now sorted runs of 10,240,000,000 = 10¹⁰ records Pass 3: runs increase by a factor of M/B – 1 = 16000 Have now sorted runs of 10¹⁴ records Nobody has so much data ! Can sort everything in 2 or 3 passes !
24	Number of Passes of External Sort
25	Data Storage and Indexing
26	Representing Data Elements Relational database elements: A tuple is represented as a record
27	Record Formats: Fixed Length Information about field types same for all records in a file; stored in system catalogs. Finding i’th field requires scan of record. Note the importance of schema information!
28	Record Header
29	Variable Length Records
30	Records With Repeating Fields
31	Storing Records in Blocks Blocks have fixed size (typically 4k)
32	Storage and Indexing How do we store efficiently large amounts of data? The appropriate storage depends on what kind of accesses we expect to have to the data. We consider: primary storage of the data additional indexes (very very important).
33	Cost Model for Our Analysis As a good approximation, we ignore CPU costs: B: The number of data pages R: Number of records per page D: (Average) time to read or write disk page Measuring number of page I/O’s ignores gains of pre-fetching blocks of pages; thus, even I/O cost is only approximated. Average-case analysis; based on several simplistic assumptions.
34	File Organizations and Assumptions Heap Files: Equality selection on key; exactly one match. Insert always at end of file. Sorted Files: Files compacted after deletions. Selections on sort field(s). Hashed Files: No overflow buckets, 80% page occupancy. Single record insert and delete.
35	Cost of Operations
36	Indexes An index on a file speeds up selections on the search key fields for the index. Any subset of the fields of a relation can be the search key for an index on the relation. Search key is not the same as key (minimal set of fields that uniquely identify a record in a relation). An index contains a collection of data entries, and supports efficient retrieval of all data entries with a given key value k.
37	Index Classification Primary/secondary Clustered/unclustered Dense/sparse B+ tree / Hash table / …
38	Primary Index File is sorted on the index attribute Dense index: sequence of (key,pointer) pairs
39	Primary Index Sparse index
40	Primary Index with Duplicate Keys Dense index:
41	Primary Index with Duplicate Keys Sparse index: pointer to lowest search key in each block: Search for 20
42	Primary Index with Duplicate Keys Better: pointer to lowest new search key in each block: Search for 20 Search for 15 ? 35 ?
43	Secondary Indexes To index other attributes than primary key Always dense (why ?)
44	Clustered/Unclustered Primary indexes = usually clustered Secondary indexes = usually unclustered
45	Clustered vs. Unclustered Index
46	Secondary Indexes Applications: index other attributes than primary key index unsorted files (heap files) index clustered data
47	Applications of Secondary Indexes Clustered data Company(name, city), Product(pid, maker)
48	Composite Search Keys Composite Search Keys: Search on a combination of fields. Equality query: Every field value is equal to a constant value. E.g. wrt <sal,age> index: age=20 and sal =75 Range query: Some field value is not a constant. E.g.: age =20; or age=20 and sal > 10
49	B+ Trees Search trees Idea in B Trees: make 1 node = 1 block Idea in B+ Trees: Make leaves into a linked list (range queries are easier)
50	B+ Trees Basics Parameter d = the degree Each node has >= d and <= 2d keys (except root) Each leaf has >=d and <= 2d keys:
51	B+ Tree Example
52	B+ Tree Design How large d ? Example: Key size = 4 bytes Pointer size = 8 bytes Block size = 4096 byes 2d x 4 + (2d+1) x 8 <= 4096 d = 170
53	Searching a B+ Tree Exact key values: Start at the root Proceed down, to the leaf Range queries: As above Then sequential traversal
54	B+ Trees in Practice Typical order: 100. Typical fill-factor: 67%. average fanout = 133 Typical capacities: Height 4: 133⁴ = 312,900,700 records Height 3: 133³ = 2,352,637 records Can often hold top levels in buffer pool: Level 1 = 1 page = 8 Kbytes Level 2 = 133 pages = 1 Mbyte Level 3 = 17,689 pages = 133 MBytes
55	Hash Tables Secondary storage hash tables are much like main memory ones Recall basics: There are n buckets A hash function f(k) maps a key k to {0, 1, …, n-1} Store in bucket f(k) a pointer to record with key k Secondary storage: bucket = block, use overflow blocks when needed
56	Hash Table Example Assume 1 bucket (block) stores 2 keys + pointers h(e)=0 h(b)=h(f)=1 h(g)=2 h(a)=h(c)=3
57	Searching in a Hash Table Search for a: Compute h(a)=3 Read bucket 3 1 disk access
58	Insertion in Hash Table Place in right bucket, if space E.g. h(d)=2
59	Insertion in Hash Table Create overflow block, if no space E.g. h(k)=1 More over- flow blocks may be needed
60	Hash Table Performance Excellent, if no overflow blocks Degrades considerably when number of keys exceeds the number of buckets (I.e. many overflow blocks). Typically, we assume that a hash-lookup takes 1.2 I/Os.
61	Where are we? File organizations: sorted, hashed, heaps. Indexes: hash index, B+-tree Indexes can be clustered or not. Data can be stored in the index or not. Hence, when we access a relation, we can either scan or go through an index: Called an access path.
62	Current Issues in Indexing Multi-dimensional indexing: how do we index regions in space? Document collections? Multi-dimensional sales data How do we support nearest neighbor queries? Indexing is still a hot and unsolved problem!
63	Multidimensional Indexes Applications: geographical databases, data cubes. Types of queries: partial match (give only a subset of the dimensions) range queries nearest neighbor Where am I? (DB or not DB?) Conventional indexes don’t work well here.
64	Indexing Techniques Hash like structures: Grid files Partitioned indexing functions Tree like structures: Multiple key indexes kd-trees Quad trees R-trees
65	Grid Files
66	Partitioned Hash Functions A hash function produces k bits identifying the bucket. The bits are partitioned among the different attributes. Example: Age produces the first 3 bits of the bucket number. Salary produces the last 3 bits. Supports partial matches, but is useless for range queries.
67	Tree Based Indexing Techniques
68	Multiple Key Indexes
69	KD Trees
70	Quad Trees
71	R-Trees
72	Query Execution
73	Query Execution Plans
74	The Leaves of the Plan: Scans Table scan: iterate through the records of the relation. Index scan: go to the index, from there get the records in the file (when would this be better?) Sorted scan: produce the relation in order. Implementation depends on relation size.
75	How do we combine Operations? The iterator model. Each operation is implemented by 3 functions: Open: sets up the data structures and performs initializations GetNext: returns the the next tuple of the result. Close: ends the operations. Cleans up the data structures. Enables pipelining! Contrast with data-driven materialize model. Sometimes it’s the same (e.g., sorted scan).
76	Implementing Relational Operations We will consider how to implement: Selection ( ) Selects a subset of rows from relation. Projection ( ) Deletes unwanted columns from relation. Join ( ) Allows us to combine two relations. Set-difference Tuples in reln. 1, but not in reln. 2. Union Tuples in reln. 1 and in reln. 2. Aggregation (SUM, MIN, etc.) and GROUP BY
77	Schema for Examples Purchase: Each tuple is 40 bytes long, 100 tuples per page, 1000 pages (i.e., 100,000 tuples, 4MB for the entire relation). Person: Each tuple is 50 bytes long, 80 tuples per page, 500 pages (i.e, 40,000 tuples, 2MB for the entire relation).
78	Simple Selections Of the form With no index, unsorted: Must essentially scan the whole relation; cost is M (#pages in R). With an index on selection attribute: Use index to find qualifying data entries, then retrieve corresponding data records. (Hash index useful only for equality selections.) Result size estimation: (Size of R) * reduction factor. More on this later.
79	Using an Index for Selections Cost depends on #qualifying tuples, and clustering. Cost of finding qualifying data entries (typically small) plus cost of retrieving records. In example, assuming uniform distribution of phones, about 54% of tuples qualify (500 pages, 50000 tuples). With a clustered index, cost is little more than 500 I/Os; if unclustered, up to 50000 I/Os! Important refinement for unclustered indexes: 1. Find sort the rid’s of the qualifying data entries. 2. Fetch rids in order. This ensures that each data page is looked at just once (though # of such pages likely to be higher than with clustering).
80	Two Approaches to General Selections First approach: Find the most selective access path, retrieve tuples using it, and apply any remaining terms that don’t match the index: Most selective access path: An index or file scan that we estimate will require the fewest page I/Os. Consider city=“seattle AND phone<“543%” : A hash index on city can be used; then, phone<“543%” must be checked for each retrieved tuple. Similarly, a b-tree index on phone could be used; city=“seattle” must then be checked.
81	Intersection of Rids Second approach Get sets of rids of data records using each matching index. Then intersect these sets of rids. Retrieve the records and apply any remaining terms.
82	Implementing Projection Two parts: (1) remove unwanted attributes, (2) remove duplicates from the result. Refinements to duplicate removal: If an index on a relation contains all wanted attributes, then we can do an index-only scan. If the index contains a subset of the wanted attributes, you can remove duplicates locally.
83	Equality Joins With One Join Column R S is a common operation. The cross product is too large. Hence, performing R S and then a selection is too inefficient. Assume: M pages in R, p_R tuples per page, N pages in S, p_S tuples per page. In our examples, R is Person and S is Purchase. Cost metric: # of I/Os. We will ignore output costs.
84	Discussion How would you implement join?
85	Simple Nested Loops Join For each tuple in the outer relation R, we scan the entire inner relation S. Cost: M + (p_R * M) * N = 1000 + 1001000500 I/Os: 140 hours! Page-oriented Nested Loops join: For each page of R, get each page of S, and write out matching pairs of tuples <r, s>, where r is in R-page and S is in S-page. Cost: M + MN = 1000 + 1000500 (1.4 hours)
86	Index Nested Loops Join If there is an index on the join column of one relation (say S), can make it the inner. Cost: M + ( (Mp_R) cost of finding matching S tuples) For each R tuple, cost of probing S index is about 1.2 for hash index, 2-4 for B+ tree. Cost of then finding S tuples depends on clustering. Clustered index: 1 I/O (typical), unclustered: up to 1 I/O per matching S tuple.
87	Examples of Index Nested Loops Hash-index on name of Person (as inner): Scan Purchase: 1000 page I/Os, 1001000 tuples. For each Person tuple: 1.2 I/Os to get data entry in index, plus 1 I/O to get (the exactly one) matching Person tuple. Total: 220,000 I/Os. (36 minutes) Hash-index on buyer of Purchase (as inner): Scan Person: 500 page I/Os, 80500 tuples. For each Person tuple: 1.2 I/Os to find index page with data entries, plus cost of retrieving matching Purchase tuples. Assuming uniform distribution, 2.5 purchases per buyer (100,000 / 40,000). Cost of retrieving them is 1 or 2.5 I/Os depending on clustering.
88	Block Nested Loops Join Use one page as an input buffer for scanning the inner S, one page as the output buffer, and use all remaining pages to hold ``block’’ of outer R. For each matching tuple r in R-block, s in S-page, add <r, s> to result. Then read next R-block, scan S, etc.
89	Sort-Merge Join (R S) Sort R and S on the join column, then scan them to do a ``merge’’ on the join column. Advance scan of R until current R-tuple >= current S tuple, then advance scan of S until current S-tuple >= current R tuple; do this until current R tuple = current S tuple. At this point, all R tuples with same value and all S tuples with same value match; output <r, s> for all pairs of such tuples. Then resume scanning R and S.
90	Cost of Sort-Merge Join R is scanned once; each S group is scanned once per matching R tuple. Cost: M log M + N log N + (M+N) The cost of scanning, M+N, could be M*N (unlikely!) With 35, 100 or 300 buffer pages, both Person and Purchase can be sorted in 2 passes; total: 7500. (75 seconds).
91	Hash-Join Partition both relations using hash fn h: R tuples in partition i will only match S tuples in partition i.
92	Cost of Hash-Join In partitioning phase, read+write both relations; 2(M+N). In matching phase, read both relations; M+N I/Os. In our running example, this is a total of 4500 I/Os. (45 seconds!) Sort-Merge Join vs. Hash Join: Given a minimum amount of memory both have a cost of 3(M+N) I/Os. Hash Join superior on this count if relation sizes differ greatly. Also, Hash Join shown to be highly parallelizable. Sort-Merge less sensitive to data skew; result is sorted.
93	Double Pipelined Join (Tukwila) Hash Join Partially pipelined: no output until inner read Asymmetric (inner vs. outer) — optimization requires source behavior knowledge
94	Discussion How would you build a query optimizer?