1. Query Optimization 3 Cost Estimation R&G, Chapters 12, 13, 14 Lecture 15

2. Administrivia Homework 2 Due Tonight –Remember you have 4 slip days for the course Homeworks 3 & 4 available later this week –3 is written assignment, deals with optimization, due before midterm 2 –4 is programming assignment, implementing query processing, due after Spring Break Midterm 2 is 3/22, 2 weeks from Thursday

3. Review: Query Processing Queries start out as SQL Database translates SQL to one or more Relational Algebra plans Plan is a tree of operations, with access path for each Access path is how each operator gets tuples –If working directly on table, can use scan, index –Some operators, like sort-merge join, or group-by, need tuples sorted –Often, operators pipelined, getting tuples that are output from earlier operators in the tree Database estimates cost for various plans, chooses least expensive

4. Today Review costs for: –Sorting –Selection –Projection –Joins Re-examine Hashing for: –Projection –Joins

5. General External Merge Sort To sort a file with N pages using B buffer pages: –Pass 0: use B buffer pages. Produce sorted runs of B pages each. –Pass 1, 2, …, etc.: merge B-1 runs. B Main memory buffers INPUT 1 INPUT B-1 OUTPUT Disk INPUT 2... * More than 3 buffer pages. How can we utilize them?

6. Cost of External Merge Sort Minimum amount of memory: 3 pages –Initial runs of 3 pages –Then 2-way merge of sorted runs (2 pages for inputs, one for outputs) –#of passes: 1 +  log 2 (N/3)  With more memory, fewer passes –With B pages, #of passes: 1 +  log (B-1) (N/B)  I/O Cost = 2N * (# of passes)

7. External Sort Example E.g., with 5 buffer pages, to sort 108 page file: –Pass 0: 22 sorted runs of 5 pages each (last run is only 3 pages) Now, do four-way (B-1) merges –Pass 1: 6 sorted runs of 20 pages each (last run is only 8 pages) –Pass 2: 2 sorted runs, 80 pages and 28 pages –Pass 3: Sorted file of 108 pages

8. Using B+ Trees for Sorting Scenario: –Table to be sorted has B+ tree index on sorting column(s). Idea: –Retrieve records in order by traversing leaf pages. Is this a good idea? Cases to consider: –B+ tree is clusteredGood idea! –B+ tree is not clusteredCould be a very bad idea! I/O Cost –Clustered tree: ~ 1.5N –Unclustered tree: 1 I/O per tuple, worst case!

9. Selections: “age < 20”, “fname = Bob”, etc No index –Do sequential scan over all tuples –Cost: N I/Os Sorted data –Do binary search –Cost: log 2 (N) I/Os Clustered B-Tree –Cost: 2 or 3 to find first record + 1 I/O for each #qualifying pages Unclustered B-Tree –Cost: 2 or 3 to find first RID + ~1 I/O for each qualifying tuple Clustered Hash Index –Cost: ~1.2 I/Os to find bucket, all tuples inside Unclustered Hash Index –Cost: ~1.2 I/Os to find bucket, + ~1 I/O for each matching tuple

10. Selection Exercise Sal > 100 –Sequential Scan: 100 I/Os –Btree on sal? Unclustered, 503 I/Os –BTree on ? Can’t use it Age = 25 –Sequential Scan: 100 I/Os –Hash on age: 1.2 I/O to get to bucket 20 matching tuples, 1 I/O for each –BTree : ~3 I/Os to find leaf + #matching pages 20 matching tuples, clustered, ~ 2 I/Os Age > 20 –Sequential Scan: 100 I/Os –Hash on age? Not with range query. –BTree : ~3 I/Os to find leaf + #matching pages (Age > 20) is 90% of pages, or ~90*1.5 = 135 I/Os

11. Selection Exercise (cont) Eid = 1000 –Sequential Scan: ~50 I/Os (avg) –Hash on eid: ~1.2 I/Os to find bucket, 1 I/O to get record Sal > 100 and age < 30 –Sequential Scan: 100 I/Os –Btree : ~ 3 I/Os to find leaf, 30% of pages match, so 30*1.5 = 45 I/Os

12. Projection Expensive when eliminating duplicates Can do this via: –Sorting: cost no more than external sort Cheaper if you project columns in initial pass, since more projected tuples fit in each page. –Hashing: build a hash table, duplicates will end up in the same bucket

13. An Alternative to Sorting: Remove duplicates with Hashing! Idea: –Many of the things we use sort for don’t exploit the order of the sorted data –e.g.: removing duplicates in DISTINCT –e.g.: finding matches in JOIN Often good enough to match all tuples with equal values Hashing does this! –And may be cheaper than sorting! (Hmmm…!) –But how to do it for data sets bigger than memory??

14. General Idea Two phases: –Partition: use a hash function h to split tuples into partitions on disk. Key property: all matches live in the same partition. –ReHash: for each partition on disk, build a main- memory hash table using a hash function h2

15. Two Phases Partition: Rehash: Partitions Hash table for partition R i (<= B pages) B main memory buffers Disk Result hash fn h2 B main memory buffers Disk Original Relation OUTPUT 2 INPUT 1 hash function h B-1 Partitions 1 2 B-1...

16. Duplicate Elimination using Hashing read one bucket at a time for each group of identical tuples, output one Partitions Hash table for partition R i (<= B pages) B main memory buffers Disk Result hash fn h2 B main memory buffers Disk Original Relation OUTPUT 2 INPUT 1 hash function h B-1 Partitions 1 2 B-1...

17. Hashing in Detail Two phases, two hash functions First pass: partition into (B-1) buckets E.g., B = 5 pages, h(x) is two low order bit 1 MemoryInput File Output Files 65149126

18. Memory, I/O costs Requirement If we can hash in two passes -> cost is 4N How big of a table can we hash in two passes? –B-1 “partitions” result from Phase 0 –Each should be no more than B pages in size –Answer: B(B-1). Said differently: We can hash a table of size N pages in about space –Note: assumes hash function distributes records evenly! Have a bigger table? Recursive partitioning!

19. How does this compare with external sorting?

20. Memory Requirement for External Sorting How big of a table can we sort in two passes? –Each “sorted run” after Phase 0 is of size B –Can merge up to B-1 sorted runs in Phase 1 –Answer: B(B-1). Said differently: We can sort a table of size N pages in about space Have a bigger table? Additional merge passes!

21. So which is better ?? Based on our simple analysis: –Same memory requirement for 2 passes –Same IO cost Digging deeper … Sorting pros: –Great if input already sorted (or almost sorted) –Great if need output to be sorted anyway –Not sensitive to “data skew” or “bad” hash functions Hashing pros: –Highly parallelizable (will discuss later in semester) –Can exploit extra memory to reduce # IOs (stay tuned…)

22. Nested Loops Joins R, with M pages, joins S, with N Pages Nested Loops –Simple nested loops Insanely inefficient M + P R *M*n –Paged nested loops – only 3 pages of memory M + M*N –Blocked nested loops – B pages of memory M +  M/(B-2)  * N If M fits in memory (B-2), cost only M + N –Index nested loops M + P R *M* index cost Only good in M very small

23. Sort-Merge Join Simple case: –sort both tables on join column –Merge –Cost: external sort cost + merge cost 2M*(1 +  log (B-1) (M/B)  ) + 2N*(1 +  log (B-1) (N/B)  ) + M + N Optimized Case: –If we have enough memory, do final merge and join in same pass. This avoids final write pass from sort, and read pass from merge –Can we merge on 2 nd pass? Only in #runs from 1 st pass < B –#runs for R is  M/B . #runs for S is  N/B . Total #runs ~~ (M+N)/B –Can merge on 2 nd pass if M+N/B < B, or M+N < B 2 –Cost: 3(M+N)

24. Hash Join Partitions of R & S Input buffer for Si Hash table for partition Ri (B-2 pages) B main memory buffers Disk Output buffer Disk Join Result hash fn h2 B main memory buffers Disk Original Relation OUTPUT 2 INPUT 1 hash function h B-1 Partitions 1 2 B-1...

25. Cost of Hash Join Partitioning phase: read+write both relations  2(|R|+|S|) I/Os Matching phase: read+write both relations  |R|+|S| I/Os Total cost of 2-pass hash join = 3(|R|+|S|) Q: what is cost of 2-pass merge-sort join? Q: how much memory needed for 2-pass sort join? Q: how much memory needed for 2-pass hash join?

26. Have B memory buffers Want to hash relation of size N An important optimization to hashing N # passes B B2B2 1 2 If B < N < B 2, will have unused memory … cost N 3N

27. 1 Hybrid Hashing Idea: keep one of the hash buckets in memory! B main memory buffers Disk Original Relation OUTPUT 3 INPUT 2 h B-k Partitions 2 3 B-k... h3 k-buffer hashtable Q: how do we choose the value of k?

28. Cost reduction due to hybrid hashing Now: N # passes B B2B2 1 2 cost N 3N

29. Summary: Hashing vs. Sorting Sorting pros: –Good if input already sorted, or need output sorted –Not sensitive to data skew or bad hash functions Hashing pros: –Often cheaper due to hybrid hashing –For join: # passes depends on size of smaller relation –Highly parallelizable

30. Summary Several alternative evaluation algorithms for each operator. Query evaluated by converting to a tree of operators and evaluating the operators in the tree. Must understand query optimization in order to fully understand the performance impact of a given database design (relations, indexes) on a workload (set of queries). Two parts to optimizing a query: – Consider a set of alternative plans. Must prune search space; typically, left-deep plans only. – Must estimate cost of each plan that is considered. Must estimate size of result and cost for each plan node. Key issues: Statistics, indexes, operator implementations.