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## Query Processing and Optimization

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**General Overview**• Relational model - SQL • Formal & commercial query languages • Functional Dependencies • Normalization • Physical Design • Indexing • Query Processing and Optimization**Review: QP & O**SQL Query Query Processor Parser Query Optimizer Algebraic Expression Execution plan Evaluator Data: result of the query**Review: QP & O**Query Optimizer Algebraic Representation Query Rewriter Algebraic Representation Data Stats Plan Generator Query Execution Plan**Review-Plan Generation**Metadata: DBMS maintains statistics about each relation, attribute and index. Plan generation: • Generate many alternative plans • We saw many for selections, joins • Estimate cost for each and choose best Plans examined: Selection (exact match): Linear, binary, PI, SI Range: PI, SI Joins: NLJ, BNLJ, INLJ, SMJ, HJ**Review-Plan Generation**Depends upon a cost model For any query, must know its estimated cardinality its estimated cost (in # of I/Os) E.g.: A = K (R ) cardinality SC(A, R) cost: depends on the plan, attribute: Linear Scan bR /2 bR Binary Search log2(bR) log2(bR)+sc(A, R)/fR -1 PI Scan HTi +1 HTi +sc(A, R) / fR**Cost Model**How do we predict the cost of a plan? Ans: Cost model • For each plan operator and each algorithm we have a cost formula • Inputs to formulas depend on relations, attributes • Database maintains statistics about relations for this (Metadata)**Metadata**• Given a relation r, DBMS likely maintains the following metadata: • Size (# of tuples) nr • Size (# of blocks) br • Block size (#tuples) fr (typicallybr =nr / fr ) • Tuple size (in bytes) sr • Attribute Variance (for each attribute r, # of different values) V(att, r) • Selection Cardinality (for each attribute in r, expected size of a selection: att = K (r ) ) SC(att, r)**Cardinality of Joins in General**Assume join: R S • If R, S have no common attributes: nr*ns • If R,S have attribute A in common: (take min) • If R, S have attribute A in common and: • A is a candidate key for R: ≤ ns • A is candidate key in R and candidate key in S : ≤ min(nr, ns) • A is a key for R, foreign key for S: = ns**Join Operation**• Size and plans for join operation • Running example: depositor customer Metadata: ncustomer = 10,000 ndepositor = 5000 fcustomer = 25 fdepositor = 50 bcustomer= 400 bdepositor= 100 V(cname, depositor) = 2500 (each customer has on average 2 accts) cname in depositor is foreign key depositor(cname, acct_no) customer(cname, cstreet, ccity)**Nested-Loop Join**Query: R S Algorithm 1: Nested Loop Join Idea: t1 u1 Blocks of... t2 u2 t3 u3 R S results Compare: (t1, u1), (t1, u2), (t1, u3) ..... Then: GET NEXT BLOCK OF S Repeat: for EVERY tuple of R**Join Algorithms**Query: R S Algorithm 2: Block Nested Loop Join Idea: t1 u1 Blocks of... t2 u2 t3 u3 R S results Compare: (t1, u1), (t1, u2), (t1, u3) (t2, u1), (t2, u2), (t2, u3) (t3, u1), (t3, u2), (t3, u3) Then: GET NEXT BLOCK OF S Repeat: for EVERY BLOCK of R**Block Nested-Loop Join**• Block Nested Loop Join for each block BRofR dofor each block BSof S do for each tuple trin BR do for each tuple usin Bsdo beginCheck if (tr,us) satisfy the join condition if they do (“match”), add tr• usto the result.**Block Nested-Loop Join (Cont.)**Cost: • Worst case estimate: br bs + br block accesses. • Improvements to nested loop and block nested loop algorithms for a buffer with M blocks: • In block nested-loop, use M — 2 disk blocks as blocking unit for outer relations, where M = memory size in blocks; use remaining two blocks to buffer inner relation and output • Cost = br / (M-2) bs + br • If equi-join attribute forms a key on inner relation, stop inner loop on first match • Scan inner loop forward and backward alternately, to make use of the blocks remaining in buffer (with LRU replacement)**Join Algorithms**Query: R S Algorithm 3: Indexed Nested Loop Join Idea: t1 Blocks of... t2 t3 R S results (fill w/ blocks of S or index blocks) For each tuple ti of R if ti.A = K (A is the attribute R,S have in common) then use the index to compute att = K (S ) Demands: index on A for S**Indexed Nested-Loop Join**Indexed Nested Loop Join • For each tuple tRin the outer relation R, use the index to look up tuples in S that satisfy the join condition with tuple tR. • Worst case: buffer has space for only one page of R, and, for each tuple in R, we perform an index lookup on s. • Cost of the join: br + nr c • Where c is the cost of traversing the index and fetching all matching s tuples for one tuple from r • c can be estimated as cost of a single selection on s using the join condition. • If indices are available on join attributes of both R and S,use the relation with fewer tuples as the outer relation.**Example of Nested-Loop Join Costs**Query: depositor customer (cname, acct_no) (cname, ccity, cstreet) Metadata: customer: ncustomer = 10,000 fcustomer = 25 bcustomer = 400 depositor: ndepositor = 5000 fdepositor = 50 bdepositor = 100 V (cname, depositor) = 2500 i a primary index on cname (dense) for customer (fi = 20) Minimal buffer**Plan generation for Joins**Alternative 1: Block Nested Loop 1a: customer = OUTER relation depositor = INNER relation cost: bcustomer + bdepositor * bcustomer = 400 +(400 *100) = 40,400 1b: customer = INNER relation depositor = OUTER relation cost: bdepositor + bdepositor * bcustomer = 100 +(400 *100) = 40,100**Plan generation for Joins**Alternative 2: Indexed Nested Loop We have index on cname for customer. Depositor is the outer relation Cost: bdepositor + ndepositor * c = 100 +(5000 *c ) , c is the cost of evaluating a selection cname=K using index. What is c? Primary index on cname, cname a key for customer c = HTi +1**Plan generation for Joins**What is HTi ? cname a key for customer. V(cname, customer) = 10,000 fi = 20, i is dense LBi = 10,000/20 = 500 HTi ~ logfi(LBi) + 1 = log20 500 + 1 = 4 Cost of index nested loop is: = 100 + (5000 * (4+1)) = 25,100 BA (cheaper than NLJ)**pR**pS Another Join Strategy Query: R S Algorithm: Merge Join Idea: suppose R, S are both sorted on A (A is the common attribute) A A 2 2 3 5 1 2 3 4 ... ... Compare: (1, 2) advance pR (2, 2) match, advance pS add to result (2, 2) match, advance pS add to result (2, 3) advance pR (3, 3) match, advance pS add to result (3, 5) advance pR (4, 5) read next block of R**Merge-Join**GIVEN R, S both sorted on A • Initialization • Reserve blocks of R, S into buffer reserving one block for result • Pr= 1, Ps =1 • Join (assuming no duplicate values on A in R) WHILE !EOF( R) && !EOF(S) DO if BR[Pr].A == BS[Ps].A then output to result; Ps++ else if BR[Pr].A < BS[Ps].A then Pr++ else (same for Ps) if Pr or Ps point past end of block, read next block and set Pr(Ps) to 1**Merge-Join (Cont.)**• Each block needs to be read only once (assuming all tuples for any given value of the join attributes fit in memory) • Thus number of block accesses for merge-join is bR + bS • But.... What if one/both of R,S not sorted on A? Ans: May be worth sorting first and then perform merge join (Sort-Merge Join) Cost: bR + bS + sortR + sortS**External Sorting**Not the same as internal sorting Internal sorting: minimize CPU (count comparisons) best: quicksort, mergesort, .... External sorting: minimize disk accesses (what we ‘re sorting doesn’t fit in memory!) best: external merge sort WHEN used? 1) SORT-MERGE join 2) ORDER BY queries 3) SELECT DISTINCT (duplicate elimination)**d**e g m p r 31 16 24 3 2 16 External Sorting Idea: 1. Sort fragments of file in memory using internal sort (runs). Store runs on disk. 2. Merge runs. E.g.: a b c 19 14 33 a d g 19 31 24 sort a a b c d d d e g m p r g a d c b e r d m p d a 14 19 14 33 7 21 31 16 24 3 2 16 24 19 31 33 14 16 16 21 3 2 7 14 merge sort b c e 14 33 16 merge sort a d d 14 7 21 d m r 21 3 16 sort merge a d p 14 7 2**External Sorting (cont.)**Algorithm Let M = size of buffer (in blocks) 1. Sort runs of size M blocks each (except for last) and store. Use internal sort on each run. 2. Merge M-1 runs at a time into 1 and store. Merge for all runs. 3. if step 2 results in more than 1 run, goto step 2. Run m-1 Output Run 1 Run 2 ........ Run 3**External Sorting (cont.)**Cost: 2 bR * (logM-1(bR / M) + 1) Intuition: Step 1: create runs every block read and written once cost 2 bR I/Os Step 2: Merge every merge iteration requires reading and writing entire file (2 bR I/Os) Total: logM-1(bR / M) Iteration # --------------- 1 2 3 ..... Runs Left to Merge ----------------------------**What if we need to sort?**Query: depositor customer Merge-sort Join Sorting depositor: bdepositor = 100 Sort depositor = 2 * 100 * (log2(100 / 3) + 1) = 1400 Similarly, for customer we get 7200 I/Os. Total: 100 + 400 + 1400 + 7200 = 9100 I/O’s! Still beats BNLJ (40K), INLJ (25K) Why not use SMJ always? Ans: 1) Sometimes inner relation can fit in memory 2) Sometimes index is small 3) SMJ only work for natural joins, “equijoins”**Hybrid Merge Join**• hybrid merge-join: If one relation is sorted, and the other has a secondary B+-tree index on the join attribute • Merge the sorted relation with the leaf entries of the B+-tree . • Sort the result on the addresses of the unsorted relation’s tuples • Scan the unsorted relation in physical address order and merge with previous result, to replace addresses by the actual tuples • Sequential scan more efficient than random lookup