01Query Processing and Optimization-SUM25.pdf
- 2. 2 Query Processing ▪ Overview ▪ Measures of Query Cost ▪ Selection Operation ▪ Sorting ▪ Join Operation ▪ Other Operations ▪ Evaluation of Expressions
- 3. 3 Basic Steps in Query Processing 1. Parsing and translation 2. Optimization 3. Evaluation
- 4. 4 Basic Steps in Query Processing (Cont.) ▪ Parsing and translation • translate the query into its internal form. This is then translated into relational algebra. • Parser checks syntax, verifies relations ▪ Evaluation • The query-execution engine takes a query-evaluation plan, executes that plan, and returns the answers to the query.
- 5. 5 Basic Steps in Query Processing: Optimization EQUIVALENT OF EXPRESSION? ▪ Given schema instructor (id, name, dept-name, salary) Query: Write relational algebra expression to find id and salary of all instructors whose id is greater than 50501. SQL: SELECT id, salary FROM INSTRUCTOR WHERE id>50501 Answer 1: σid>50501(∏id, salary(instructor)) equivalent to Answer 2: ∏id, salary(σid>50501(instructor)) Discussion 1: Instructor is stored physically sorted order of id. Will the query processing cost of answer 1 and answer 2 be the same? How?
- 6. 6 Basic Steps in Query Processing: Optimization ▪ Each relational algebra operation can be evaluated using one of several different algorithms • Correspondingly, a relational-algebra expression can be evaluated in many ways. ▪ Annotated expression specifying detailed evaluation strategy is called an evaluation-plan. E.g.,: • Use an index on id to find instructors with id > 50501 (Answer 2), • Or perform complete relation scan and discard instructors with id <= 50501 (Answer 1)
- 7. 7 Basic Steps: Optimization (Cont.) ▪ Query Optimization: Amongst all equivalent evaluation plans choose the one with lowest cost. • Cost is estimated using statistical information from the database catalog ▪ e.g.. number of tuples in each relation, size of tuples, etc. ▪ In this chapter we study • How to measure query costs • Algorithms for evaluating relational algebra operations • How to combine algorithms for individual operations in order to evaluate a complete expression
- 8. 8 Measures of Query Cost ▪ Many factors contribute to time cost • disk access, CPU, and network communication ▪ Cost can be measured based on • response time, i.e. total elapsed time for answering query, or • total resource consumption ▪ We use total resource consumption as cost metric • Response time harder to estimate, and minimizing resource consumption is a good idea in a shared database ▪ We ignore CPU costs for simplicity • Real systems do take CPU cost into account • Network costs must be considered for parallel systems ▪ We describe how to estimate the cost of each operation • We do not include cost to writing output to disk Which cost we will consider for query evaluation plan? Explain why? Why not?
- 9. 9 Measures of Query Cost ▪ Disk cost can be estimated as: • Number of seeks * average-seek-cost • Number of blocks read * average-block-read-cost • Number of blocks written * average-block-write-cost ▪ What is number of seeks? ▪ Answer: The number of times, the disk head directly moves from one track to another track. ▪ What is number of block reads? ▪ Answer: The number of blocks need to be transferred from disk to memory to process the query. ▪ What is number of blocks written? ▪ Answer: The number of blocks need to be written to disk from memory to process the query.
- 10. 10 Magnetic Hard Disk Mechanism Schematic diagram of magnetic disk drive Photo of magnetic disk drive
- 11. 11 Measures of Query Cost ▪ For simplicity we just use the number of block transfers from disk and the number of seeks as the cost measures • tT – time to transfer one block ▪ Assuming for simplicity that write cost is same as read cost • tS – time for one seek • Cost for b block transfers plus S seeks b * tT + S * tS ▪ tS and tT depend on where data is stored; with 4 KB blocks: • High end magnetic disk: tS = 4 msec and tT =0.1 msec • SSD: tS = 20-90 microsec and tT = 2-10 microsec for 4KB
- 12. 12 Selection Operation ● File scan ● Algorithm A1 (linear search). Scan each file block and test all records to see whether they satisfy the selection condition. ● Cost estimate = br block transfers + 1 seek 4 br denotes number of blocks containing records from relation r ● If selection is on a key attribute, can stop on finding record 4 cost = (br /2) block transfers + 1 seek ● Linear search can be applied regardless of 4 Selection condition or 4 Ordering of records in the file, or 4 Availability of indices ● Note: binary search generally does not make sense since data is not stored consecutively ● except when there is an index available, ● and binary search requires more seeks than index search
- 13. 13 Selections Using Indices ● Index scan – search algorithms that use an index ● Selection condition must be on search- key of index. ● A2 (primary index, equality on key). Retrieve a single record that satisfies the corresponding equality condition ● Let hi denote the height of the index tree ● Cost = (hi + 1) * (tT + tS) ● A5 (primary index, comparison). (Relation is sorted on A) 4 For A V(r) use index to find first tuple v and scan relation sequentially from there 4 For AV (r) just scan relation sequentially till first tuple > v; do not use index ID Name Dept_name tot_Cred 25001Fatema ECE 120 25002Sajid Math 100 25003Abid Accounting 120 25004Fatema ECE 50 25005Sajid Math 120 25006Shafique Accounting 100 25007Abid ECE 120 25008Fatema Math 50 25009Sajid ECE 120 25010Shafique Math 100 ROOT LEAF hi
- 14. 14 Selections Using Indices ● Index scan – search algorithms that use an index ● Selection condition must be on search- key of index. ● A3 (primary index, equality on nonkey) Retrieve multiple records. ● Records will be on consecutive blocks 4 Let b = number of blocks containing matching records ● Cost = hi * (tT + tS) + tS + tT * b ID Name Dept_name tot_Cred 25003Abid Accounting 120 25007Abid ECE 120 25001Fatema ECE 120 25004Fatema ECE 50 25008Fatema Math 50 25002Sajid Math 100 25005Sajid Math 120 25009Sajid ECE 120 25006Shafique Accounting 100 25010Shafique Math 100 ROOT LEAF hi
- 15. 15 Selections Using Indices ● A4 (secondary index, equality on nonkey). ● Retrieve a single record if the search-key is a candidate key 4 Cost = (hi + 1) * (tT + tS) ● Retrieve multiple records if search-key is not a candidate key 4 each of n matching records may be on a different block 4 Cost = (hi + n) * (tT + tS) – Can be very expensive! ID Name Dept_name tot_Cred 25001Fatema ECE 120 25002Sajid Math 100 25003Abid Accounting 120 25004Fatema ECE 50 25005Sajid Math 120 25006Shafique Accounting 100 25007Abid ECE 120 25008Fatema Math 50 25009Sajid ECE 120 25010Shafique Math 100 ROOT LEAF hi bucket
- 16. 16 Selections Using Indices ● A6 (secondary index, comparison). 4 For A V(r) use index to find first index entry v and scan index sequentially from there, to find pointers to records. 4 For AV (r) just scan leaf pages of index finding pointers to records, till first entry > v 4 In either case, retrieve records that are pointed to – requires an I/O for each record – Linear file scan may be cheaper ID Name Dept_name tot_Cred 25001Fatema ECE 120 25002Sajid Math 100 25003Abid Accounting 120 25004Fatema ECE 50 25005Sajid Math 120 25006Shafique Accounting 100 25007Abid ECE 120 25008Fatema Math 50 25009Sajid ECE 120 25010Shafique Math 100
- 17. 17 Implementation of Complex Selections ● Conjunction: 1 2. . . n(r) ● A7 (conjunctive selection using one index). ● Select a combination of i and algorithms A1 through A7 that results in the least cost for i (r). ● Test other conditions on tuple after fetching it into memory buffer.
- 18. 18 Algorithms for Complex Selections ● Disjunction:1 2 . . . n (r). ● A10 (disjunctive selection by union of identifiers). ● Applicable if all conditions have available indices. 4 Otherwise use linear scan. ● Use corresponding index for each condition, and take union of all the obtained sets of record pointers. ● Then fetch records from file
- 19. 19 Sorting ▪ We may build an index on the relation, and then use the index to read the relation in sorted order. May lead to one disk block access for each tuple. Discussion: The case when it May lead to one disk block access for each tuple. ▪ For relations that fit in memory, techniques like quicksort can be used. ▪ For relations that don’t fit in memory, quicksort is not applicable. Why? external sort-merge is a good choice in this case.
- 20. 20 Example: External Sorting Using Sort-Merge g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19
- 21. 21 External Sort-Merge (Run Creation) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Run Creation Let M denote memory size (in pages). Here M = 3 1. Create sorted runs. Let i be 0 initially. Repeatedly do the following till the end of the relation: (a) Read M blocks of relation into memory (b) Sort the in-memory blocks (c) Write sorted data to run Ri; increment i. Let the final value of i be N (Here N = ?) Here N = 4 2. Merge the runs (next slide)….. Memory R1 R2 R3 R4
- 22. 22 External Sort-Merge (Sort-Merge) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Merge pass1 Input Buffer R1 R2 R3 R4 Merge the runs (2-way merge). Here N > M. M=3, N=4 Use 2 blocks of memory to buffer input runs, and 1 block to buffer output. Read the first block of each run into its buffer page repeat Select the first record (in sort order) among all buffer pages Write the record to the output buffer. If the output buffer is full write it to disk. Delete the record from its input buffer page. If the buffer page becomes empty then read the next block (if any) of the run into the buffer. until all input buffer pages are empty: a 19 b 14 Output Buffer a 19
- 23. 23 External Sort-Merge (Sort Merge) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Merge pass1 Memory R1 R2 R3 R4 Merge the runs (2-way merge). Here N > M. M=3, N=2 Use 2 blocks of memory to buffer input runs, and 1 block to buffer output. Read the first block of each run into its buffer page repeat Select the first record (in sort order) among all buffer pages Write the record to the output buffer. If the output buffer is full write it to disk. Delete the record from its input buffer page. If the buffer page becomes empty then read the next block (if any) of the run into the buffer. until all input buffer pages are empty: b 14 d 31 b 14
- 24. 24 External Sort-Merge (Algorithm Analysis) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Merge pass1 Memory R1 R2 R3 R4 ▪ If N M, several merge passes are required. ▪ Here, it is 2 ▪ In each pass, contiguous groups of M - 1 runs are merged. ▪ A pass reduces the number of runs by a factor of M -1, and creates runs longer by the same factor. E.g. If M=11, and there are 90 runs, one pass reduces the number of runs to 9, each 10 times the size of the initial runs ▪ Repeated passes are performed till all runs have been merged into one. ▪ N = 90, M = 11 ▪ After merge, number of runs = 90/(11- 1)=9 b 14 d 31 b 14 Merge pass2
- 25. 25 External Sort-Merge (Algorithm Analysis) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Merge pass1 Memory R1 R2 R3 R4 b 14 d 31 b 14 Merge pass2 Problem: Find the number of runs, number of passes, number of blocks transfer and number of seeks for a relation with br blocks. Memory size is M, number of runs is N and N > M. ▪ Cost analysis (Total number of passes) ▪ Total number of blocks of the relation is br ▪ Size of the memory is M ▪ Total number of runs = (br/M) ▪ No of blocks = 100 ▪ Memory size = 11 ▪ Number of runs = 100/11 = 10
- 26. 26 External Sort-Merge (Algorithm Analysis) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Merge pass1 Memory R1 R2 R3 R4 b 14 d 31 b 14 Merge pass2 ▪ Cost analysis (Total number of passes) ▪ Total number of blocks of the relation is br ▪ Size of the memory is M ▪ Total number of runs = (br/M) ▪ After merge pass 1, number of runs = (br/M) /(M-1) ▪ After merge pass 2, number of runs = (br/M) /(M-1) (M-1) = (br/M) /(M-1) 2 Total number of pass = log (M–1)(br/M). log (3–1)(12/3). = 2
- 27. 27 External Sort-Merge (Algorithm Analysis) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Merge pass1 Memory R1 R2 R3 R4 b 14 d 31 b 14 Merge pass2 Cost analysis (Total number of block transfer) Block transfers for initial run creation as well as in each pass is br + br = 2br for final pass, we don’t count write cost since the output of an operation may be sent to the parent operation Block transfer for run creation = br for read+ br for write = 2 br Block transfers for merging = 2 br * log (M–1)(br/M) BT for run and merge (B_RM) = 2 br + 2 br * log (M–1)(br/M) Final merge, no write. So Net block transfer = B_RM - br = 2 br + 2 br * log (M–1)(br/M) - br = br * (2 * log (M–1)(br/M) +1)
- 28. 28 External Sort-Merge (Algorithm Analysis) g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19 Merge pass1 Memory R1 R2 R3 R4 b 14 d 31 b 14 Merge pass2 Cost analysis (Total number of seek) During run generation: one seek to read each run and one seek to write each run 2 br / M During the merge phase Need 2 br seeks for each merge pass except the final one which does not require a write Total number of seeks: = seek for run creation + seek for merge = 2 br / M + 2br logM–1(br / M) - br = 2 br / M + br ( 2 logM–1(br / M) - 1)
- 29. 29 Problem to Solve Question: The size of a relation is 200 blocks and the memory size is 5 blocks. a. Find the number of initial sorted runs and explain the run creation method.
- 30. 30 Problem to Solve Question: The size of a relation is 200 blocks and the memory size is 5 blocks. b. Find the number of runs created after first merge pass and explain the first merge- pass . Number or runs = 200/5 = 40 Input buffer = 4 Output buffer =1 Number of runs after merge pass 1 = 40/4 = 10
- 31. 31 Problem to Solve Question: The size of a relation is 200 blocks and the memory size is 5 blocks. c. Find the number of block transfer in final merge pass and explain why the final sorted run is not saved in disk. Number or runs = 200/5 = 40 Input buffer = 4 Output buffer =1 Number of runs after merge pass 1 = 40/4 = 10 Merge 1 > 40 to 10 Merge 2 > 10 to 3 Merge 3 > 3 to 1 Number of merge pass =
- 32. 32 Problem to Solve Question: The size of a relation is 200 blocks and the memory size is 5 blocks. c. Find the number of block transfer in final merge pass and explain why the final sorted run is not saved in disk. Merge 1 > 40 to 10 Merge 2 > 10 to 3 Merge 3 > 3 to 1 Number of merge pass = 3 Number of block transfer in final merge pass = br = 200 blocks
- 33. 33 Join Operation ▪ Several different algorithms to implement joins • Nested-loop join • Block nested-loop join ▪ Choice based on cost estimate ▪ Examples use the following information • Number of records of student: 20,000 takes: 40,000 • Number of blocks of student: 2000 takes: 400
- 34. 34 Nested-Loop Join To compute the theta join r ⨝ s for each tuple tr in r do begin for each tuple ts in s do begin test pair (tr,ts) to see if they satisfy the join condition if they do, add tr • ts to the result. end end r is the outer relation and s the inner relation of the join. Expensive since it examines every pair of tuples in the two relations. Cost Analysis (Block transfer) Case 1: (worst case), nr = total number of tuples in r br = number of blocks in r bs= number of blocks in s if there is enough memory only to hold one block of each relation, the estimated block transfers is nr bs + br Number of seek = nr + br B1(r) B2(r) B3(r) …….. Br(r) B1(s) B2(s) …….. Bs(s) M1 M2 Iteration 1 r s
- 35. 35 Nested-Loop Join To compute the theta join r ⨝ s for each tuple tr in r do begin for each tuple ts in s do begin test pair (tr,ts) to see if they satisfy the join condition if they do, add tr • ts to the result. end end r is the outer relation and s the inner relation of the join. Expensive since it examines every pair of tuples in the two relations. Cost Analysis (Block transfer) Case 1: (worst case), nr = total number of tuples in r br = number of blocks in r bs= number of blocks in s if there is enough memory only to hold one block of each relation, the estimated block transfers is nr bs + br Number of seek = nr + br B1 B2 B3 …….. Br B1 B2 …….. Bs M1 M2 Iteration 1
- 36. 36 Nested-Loop Join To compute the theta join r ⨝ s for each tuple tr in r do begin for each tuple ts in s do begin test pair (tr,ts) to see if they satisfy the join condition if they do, add tr • ts to the result. end end r is the outer relation and s the inner relation of the join. Expensive since it examines every pair of tuples in the two relations. Cost Analysis (Block transfer) Case 1: (worst case), nr = total number of tuples in r br = number of blocks in r bs= number of blocks in s if there is enough memory only to hold one block of each relation, the estimated block transfers is nr bs + br Number of seek = nr + br B1 B2 B3 …….. Br B1 B2 …….. Bs M1 M2 Iteration 1
- 37. 37 Nested-Loop Join for each tuple tr in r do begin for each tuple ts in s do begin test pair (tr,ts) to see if they satisfy the join condition if they do, add tr • ts to the result. end end Cost Analysis (Block transfer) Case 2 (Best Case): the smaller relation fits entirely in memory, use that as the inner relation Br = number of blocks in r Bs= number of blocks in s the estimated block transfers is bs + br Number of seek = 2 B1 B2 B3 …….. Br B1 B2 …….. Bs M1 M2 M3 …… Mn
- 38. 38 Nested-Loop Join for each tuple tr in student do begin for each tuple ts in takes do begin test pair (tr,ts) to see if they satisfy the join condition if they do, add tr • ts to the result. end end Problem : The relation schema are Student (id, name, cgpa, tot-cred, street, city, NID) Takes (id, course-id, semester, year) a. Write the SQL and algebra to find id, name, course id and semester b. Find the worst case number of block transfer and number of seek Relation No. of records No. of blocks student 20000 2000 takes 40000 400 Cost Analysis (Block transfer) Case 1: (worst case), Br = number of blocks in r Bs= number of blocks in s if there is enough memory only to hold one block of each relation, the estimated block transfers is nr bs + br Number of seek = nr + br B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 1
- 39. 39 Nested-Loop Join for each tuple tr in student do begin for each tuple ts in takes do begin test pair (tr,ts) to see if they satisfy the join condition if they do, add tr • ts to the result. end end Relation No. of records No. of blocks student 20000 2000 takes 40000 400 Solution br = 2000 bs= 400 memory only holds one block of each relation, the estimated block transfers is nr bs + br = 20000 * 400 + 2000 Number of seek = nr + br = 20000+2000 B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 1 Problem : The relation schema are Student (id, name, cgpa, tot-cred, street, city, NID) Takes (id, course-id, semester, year) a. Write the SQL and algebra to find id, name, course id and semester b. Find the worst case number of block transfer and number of seek
- 40. 40 Nested-Loop Join for each tuple tr in student do begin for each tuple ts in takes do begin test pair (tr,ts) to see if they satisfy the join condition if they do, add tr • ts to the result. end end Problem: Find the best case number of block transfer and number of seek Relation No. of records No. of blocks student 20000 2000 takes 40000 400 Solution br = 2000 bs= 400 memory only holds one block of each relation, the estimated block transfers is bs + br = 400 + 2000 Number of seek = 2 B1 B2 B3 …….. B2000 B1 B2 …… B400 M1 M2 M3 …… Mn Discussion: Write a discussion on nested loop join considering cost.
- 41. 41 Block Nested-Loop Join Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation. for each block Br of r do begin for each block Bs of s do begin for each tuple tr in Br do begin for each tuple ts in Bs do begin Check if (tr,ts) satisfy the join condition if they do, add tr • ts to the result. end end end end B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 1
- 42. 42 Block Nested-Loop Join B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 1 Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation. for each block Br of r do begin for each block Bs of s do begin for each tuple tr in Br do begin for each tuple ts in Bs do begin Check if (tr,ts) satisfy the join condition if they do, add tr • ts to the result. end end end end
- 43. 43 Block Nested-Loop Join B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 1 Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation. for each block Br of r do begin for each block Bs of s do begin for each tuple tr in Br do begin for each tuple ts in Bs do begin Check if (tr,ts) satisfy the join condition if they do, add tr • ts to the result. end end end end
- 44. 44 Block Nested-Loop Join B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 2 Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation. for each block Br of r do begin for each block Bs of s do begin for each tuple tr in Br do begin for each tuple ts in Bs do begin Check if (tr,ts) satisfy the join condition if they do, add tr • ts to the result. end end end end
- 45. 45 Block Nested-Loop Join B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 2 Cost Analysis (Block transfer) Case 2 (Worst Case): the estimated block transfers is br * bs + br Number of seek = bs + br Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation. for each block Br of r do begin for each block Bs of s do begin for each tuple tr in Br do begin for each tuple ts in Bs do begin Check if (tr,ts) satisfy the join condition if they do, add tr • ts to the result. end end end end
- 46. 46 Block Nested-Loop Join B1 B2 B3 …….. B2000 B1 B2 …….. B400 M1 M2 Iteration 2 Cost Analysis (Block transfer) Case 2 (Best Case): the smaller relation fits entirely in memory, use that as the inner relation the estimated block transfers is bs + br Number of seek = 2 Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation. for each block Br of r do begin for each block Bs of s do begin for each tuple tr in Br do begin for each tuple ts in Bs do begin Check if (tr,ts) satisfy the join condition if they do, add tr • ts to the result. end end end end
- 47. 47 Evaluation of Expressions So far: we have seen algorithms for individual operations Alternatives for evaluating an entire expression tree Materialization: generate results of an expression whose inputs are relations or are already computed, materialize (store) it on disk. Repeat. Pipelining: pass on tuples to parent operations even as an operation is being executed We study above alternatives in more detail name((building = ‘Watson’(department)) ⨝ instructor)
- 48. 48 Materialization Materialized evaluation: evaluate one operation at a time, starting at the lowest-level. Use intermediate results materialized into temporary relations to evaluate next-level operations. The department and tinstructor schemas are given below: instructor (id, name, dept-name, salary) Department (dept-name, building, budget) i. Write relational algebra to find the names of all instructors of all departments in ‘Watson’ building. ii. Show the query expression tree (operation tree) for the query. name((building = ‘Watson’(department)) ⨝ instructor) Relational algebra Query expression tree
- 49. 49 Materialization Materialized evaluation: evaluate one operation at a time, starting at the lowest-level. Use intermediate results materialized into temporary relations to evaluate next-level operations. Problem 12: The student and takes schemas are given below: student (id, name, CGPA, year-admit) Takes (id, course-id, semester, year) i. Write relational algebra to find the names and CGPA of all students of spring 2019. ii. Show the query expression tree (operation tree) for the query. name((building = ‘Watson’(department)) ⨝ instructor) Relational algebra Query expression tree
- 50. 50 Materialization (Cont.) Materialized evaluation is always applicable Cost of writing results to disk and reading them back can be quite high Our cost formulas for operations ignore cost of writing results to disk, so Overall cost = Sum of costs of individual operations + cost of writing intermediate results to disk Write Cost Write Cost
- 51. 51 Materialization (Cont.) Problem 13: The cost of (building = ‘Watson’(department) is 1 seek and 20 block transfer. The cost of ((building = ‘Watson’ (department)) ⨝ instructor) is 10 seek and 100 block transfer. The cost of name (…….) is 1 seek and 50 block transfer. Write cost for (building = ‘Watson’(department) is 1 seek and 10 block transfer Write cost for ⨝ is 1 seek and 50 block transfer Find overall cost of the query. Write Cost Write Cost
- 52. 52 Materialization (Cont.) Problem 13: The cost of (building = ‘Watson’(department) is 1 seek and 20 block transfer. The cost of ((building = ‘Watson’ (department)) ⨝ instructor) is 10 seek and 100 block transfer. The cost of name (…….) is 1 seek and 50 block transfer. Write cost for (building = ‘Watson’(department) is 1 seek and 10 block transfer Write cost for ⨝ is 1 seek and 50 block transfer Find overall cost of the query. Write Cost Write Cost Overall cost = Sum of costs of individual operations + cost of writing intermediate results to disk = ((1+10+1) seek + (20+100+50) blocks) + ((1+1) seek + (10+50) blocks) = (12Seek+170Blocks) + (2 seek + 60blocks)
- 53. 53 Materialization (Cont.) ▪ Materialized evaluation is always applicable ▪ Cost of writing results to disk and reading them back can be quite high ▪ Our cost formulas for operations ignore cost of writing results to disk, so ▪ Overall cost = Sum of costs of individual operations + cost of writing intermediate results to disk ▪ Double buffering: use two output buffers for each operation, when one is full write it to disk while the other is getting filled • Allows overlap of disk writes with computation and reduces execution time Write (X) Buffer 1 Buffer 2 DB Double Buffering
- 54. 54 Materialization (Cont.) ▪ Materialized evaluation is always applicable ▪ Cost of writing results to disk and reading them back can be quite high ▪ Our cost formulas for operations ignore cost of writing results to disk, so ▪ Overall cost = Sum of costs of individual operations + cost of writing intermediate results to disk ▪ Double buffering: use two output buffers for each operation, when one is full write it to disk while the other is getting filled • Allows overlap of disk writes with computation and reduces execution time Write (Y) Buffer 1 Buffer 2 DB If Buffer 1 full
- 55. 55 Pipelining ▪ Pipelined evaluation: evaluate several operations simultaneously, passing the results of one operation on to the next. ▪ E.g., in previous expression tree, don’t store result of ▪ instead, pass tuples directly to the join.. Similarly, don’t store result of join, pass tuples directly to projection. ▪ Much cheaper than materialization: no need to store a temporary relation to disk. ) ( " Watson " department building= Write Cost Write Cost Discussion: Explain how the above query is executed using pipelining and compare the cost with materialization × × name((building = ‘Watson’(department)) ⨝ instructor
- 56. 56 Pipelining ▪ Pipelining may not always be possible – e.g., sort, hash- join. ▪ For pipelining to be effective, use evaluation algorithms that generate output tuples even as tuples are received for inputs to the operation. ▪ can be executed in two ways: demand driven and producer driven Discussion: Explain why pipelining may not be possible in case of database sorting. How much it is possible?
- 57. 57 Example: External Sorting Using Sort-Merge g a d 31 c 33 b 14 e 16 r 16 d 21 m 3 p 2 d 7 a 14 a 14 a 19 b 14 c 33 d 7 d 21 d 31 e 16 g 24 m 3 p 2 r 16 a 19 b 14 c 33 d 31 e 16 g 24 a 14 d 7 d 21 m 3 p 2 r 16 a 19 d 31 g 24 b 14 c 33 e 16 d 21 m 3 r 16 a 14 d 7 p 2 initial relation create runs merge pass–1 merge pass–2 runs runs sorted output 24 19
- 58. 58 Pipelining (Cont.) ▪ In demand driven or lazy evaluation • system repeatedly requests next tuple from top level operation • Each operation requests next tuple from children operations as required, in order to output its next tuple • In between calls, operation has to maintain “state” so it knows what to return next • Alternative name: pull model of pipelining Output-1 Tuple 1 Tuple 1 Tuple 1
- 59. 59 Pipelining (Cont.) ▪ In producer-driven or eager pipelining ▪ Operators produce tuples eagerly and pass them up to their parents ▪ Buffer maintained between operators, child puts tuples in buffer, parent removes tuples from buffer ▪ if buffer is full, child waits till there is space in the buffer, and then generates more tuples ▪ System schedules operations that have space in output buffer and can process more input tuples ▪ Alternative name: push model of pipelining Output-1 Tuple 1 Tuple 1 Tuple 1 Discussion: Give a comparison between pull and push model of pipelining