1. CMSC424: Database Design Instructor: Amol Deshpande amol@cs.umd.edu

2. Query Optimization Introduction Example of a Simple Type of Query Transformation of Relational Expressions Statistics Estimation Optimization Algorithms

3. Query Optimization Why ? Many different ways of executing a given query Huge differences in cost Example: select * from person where ssn = “123” Size of person = 1GB Sequential Scan: Takes 1GB / (20MB/s) = 50s Use an index on SSN (assuming one exists): Approx 4 Random I/Os = 40ms

4. Query Optimization Equivalent relational expressions Drawn as a tree List the operations and the order

5. Query Optimization Execution plans Evaluation expressions annotated with the methods used

6. Query Optimization Steps: Generate all possible execution plans for the query Figure out the cost for each of them Choose the best Not done exactly as listed above Too many different execution plans for that Typically interleave all of these into a single efficient search algorithm

7. Query Optimization Steps: Generate all possible execution plans for the query First generate all equivalent expressions Then consider all annotations for the operations Figure out the cost for each of them Compute cost for each operation  Using the formulas discussed before  One problem: How do we know the number of result tuples for, say, Add them ! Choose the best

8. Query Optimization Introduction Example of a Simple Type of Query Transformation of Relational Expressions Statistics Estimation Optimization Algorithms

9. A Simple Case Queries with only selections on a single relation with no indexes select * from person where substr(name, 1, 1) in [‘A’, ‘B’, ‘C’] and 100ns zipcode = 94720 and 1ns date-of-birth > to_date('1978/05/31', 'yyyy/mm/dd') 1000ns Relation contains: 1,000,000 tuples CPU Costs

10. A Simple Case Possible execution plan: For each tuple Evaluate substr predicate If true,  Evaluate zipcode predicate  If true, evaluate date-of-birth predicate  If true, output the tuple 6 different possibilities How to choose one ?

11. A Simple Case Compute cost of each possibility Say, substr()  zipcode  date-of-birth Need some more information selectivity: fraction of tuples expected to pass the predicates Let selectivity(substr predicate) = 3/26 Let selectivity(zipcode predicate) = 1/100 And, selectivity(date-of-birth predicate) = 1/3 How are selectivities computed ? Must keep track of some additional information about the relations

12. A Simple Case Compute cost of each possibility Say, substr()  zipcode  date-of-birth Given that: Cost of the above plan =  1,000,000 * 100ns  + 1,000,000 * 3/26 * 1ns  + 1,000,000 * 1/26 * 1/100 * 1000ns  = approx 100.5 ms Cost of the plan: zipcode  substr()  date-of-birth: Approx 12.92 ms About a factor of 10 better.

13. A Simple Case Compute cost of each possibility Say, substr()  zipcode  date-of-birth Cost of the plan: zipcode  substr()  date-of-birth: Approx 12.92 ms About a factor of 10 better. General algorithm: Don’t need to check all n! Possibilities Sort the predicates in the increasing order by: 1 – selectivity(predicate) cost of the predicate

14. Query Optimization General case: Need: A way to enumerate all plans A way to find the cost of each plan Sub problem: Estimating the selectivities of various operations A way to search through the plans efficiently

15. Query Optimization Introduction Example of a Simple Type of Query Transformation of Relational Expressions Statistics Estimation Optimization Algorithms

16. Equivalence of Expressions Two relational expressions equivalent iff: Their result is identical on all legal databases Equivalence rules: Allow replacing one expression with another Examples: 1. 2. Selections are commutative

17. Equivalence Rules Examples: 3. 5. E 1  E 2 = E 2  E 1 7(a). If  0 only involves attributes from E 1   0  E 1  E 2 ) = (   0 (E 1 ))  E 2 And so on… Many rules of this type

18. Pictorial Depiction

19. Example Find the names of all customers with an account at a Brooklyn branch whose account balance is over $1000.  customer_name (  branch_city = “Brooklyn”  balance > 1000 (branch (account depositor)) ) Apply the rules one by one  customer_name ((  branch_city = “Brooklyn”  balance > 1000 (branch account)) depositor)  customer_name (((  branch_city = “Brooklyn” (branch)) (  balance > 1000 (account))) depositor)

20. Example

21. Equivalence of Expressions The rules give us a way to enumerate all equivalent expressions Note that the expressions don’t contain physical access methods, join methods etc… Simple Algorithm: Start with the original expression Apply all possible applicable rules to get a new set of expressions Repeat with this new set of expressions Till no new expressions are generated

22. Equivalence of Expressions Works, but is not feasible Consider a simple case: R1 (R2 (R3 (… Rn)))….) Just join commutativity and associativity will give us: At least: n^2 * 2^n At worst: n! * 2^n Typically the process of enumeration is combined with the search process

23. Evaluation Plans We still need to choose the join methods etc.. Option 1: Choose for each operation separately Usually okay, but sometimes the operators interact Consider joining three relations on the same attribute:  R1 a (R2 a R3) Best option for R2 join R3 might be hash-join  But if R1 is sorted on a, then sort-merge join is preferable  Because it produces the result in sorted order by a Also, we need to decide whether to use pipelining or materialization Such issues are typically taken into account when doing the optimization

24. Query Optimization Introduction Example of a Simple Type of Query Transformation of Relational Expressions Optimization Algorithms Statistics Estimation

25. Optimization Algorithms Two types: Exhaustive: That attempt to find the best plan Heuristical: That are simpler, but are not guaranteed to find the optimal plan Consider a simple case Join of the relations R1, …, Rn No selections, no projections Still very large plan space

26. Searching for the best plan Option 1: Enumerate all equivalent expressions for the original query expression Using the rules outlined earlier Estimate cost for each and choose the lowest Too expensive ! Consider finding the best join-order for r 1 r 2... r n. There are (2(n – 1))!/(n – 1)! different join orders for above expression. With n = 7, the number is 665280, with n = 10, the number is greater than 176 billion!

27. Searching for the best plan Option 2: Dynamic programming There is too much commonality between the plans Also, costs are additive  Caveat: Sort orders (also called “interesting orders”) Reduces the cost down to O(n3^n) or O(n2^n) in most cases Interesting orders increase this a little bit Considered acceptable Typically n < 10. Switch to heuristic if not acceptable

28. Left Deep Join Trees In left-deep join trees, the right-hand-side input for each join is a relation, not the result of an intermediate join Early systems only considered these types of plans Easier to pipeline

29. Heuristic Optimization Dynamic programming is expensive Use heuristics to reduce the number of choices Typically rule-based: Perform selection early (reduces the number of tuples) Perform projection early (reduces the number of attributes) Perform most restrictive selection and join operations before other similar operations. Some systems use only heuristics, others combine heuristics with partial cost-based optimization.

30. Steps in Typical Heuristic Optimization 1.Deconstruct conjunctive selections into a sequence of single selection operations (Equiv. rule 1.). 2.Move selection operations down the query tree for the earliest possible execution (Equiv. rules 2, 7a, 7b, 11). 3.Execute first those selection and join operations that will produce the smallest relations (Equiv. rule 6). 4.Replace Cartesian product operations that are followed by a selection condition by join operations (Equiv. rule 4a). 5.Deconstruct and move as far down the tree as possible lists of projection attributes, creating new projections where needed (Equiv. rules 3, 8a, 8b, 12). 6.Identify those subtrees whose operations can be pipelined, and execute them using pipelining).

31. Query Optimization Introduction Example of a Simple Type of Query Transformation of Relational Expressions Optimization Algorithms Statistics Estimation

32. Cost Estimation Cost of each operator computer as described in Chapter 13 Need statistics of input relations  E.g. number of tuples, sizes of tuples Inputs can be results of sub-expressions Need to estimate statistics of expression results To do so, we require additional statistics  E.g. number of distinct values for an attribute More on cost estimation later

33. Statistical Information for Cost Estimation n r : number of tuples in a relation r. b r : number of blocks containing tuples of r. l r : size of a tuple of r. f r : blocking factor of r — i.e., the number of tuples of r that fit into one block. V(A, r): number of distinct values that appear in r for attribute A; same as the size of  A (r). If tuples of r are stored together physically in a file, then:

34. Histograms Histogram on attribute age of relation person Equi-width histograms Equi-depth histograms

35. Selection Size Estimation  A=v (r)  n r / V(A,r) : number of records that will satisfy the selection  Equality condition on a key attribute: size estimate = 1  A  V (r) (case of  A  V (r) is symmetric) Let c denote the estimated number of tuples satisfying the condition. If min(A,r) and max(A,r) are available in catalog  c = 0 if v < min(A,r)  c = If histograms available, can refine above estimate In absence of statistical information c is assumed to be n r / 2.

36. Size Estimation of Complex Selections The selectivity of a condition  i is the probability that a tuple in the relation r satisfies  i. If s i is the number of satisfying tuples in r, the selectivity of  i is given by s i /n r. Conjunction:   1   2 ...   n ( r). Assuming independence, estimate of tuples in the result is: Disjunction:   1   2 ...   n (r). Estimated number of tuples: Negation:   (r). Estimated number of tuples: n r – size(   (r))

37. Join Operation: Running Example Running example: depositor customer Catalog information for join examples: n customer = 10,000. f customer = 25, which implies that b customer =10000/25 = 400. n depositor = 5000. f depositor = 50, which implies that b depositor = 5000/50 = 100. V(customer_name, depositor) = 2500, which implies that, on average, each customer has two accounts. Also assume that customer_name in depositor is a foreign key on customer. V(customer_name, customer) = 10000 (primary key!)

38. Estimation of the Size of Joins The Cartesian product r x s contains n r.n s tuples; each tuple occupies s r + s s bytes. If R  S = , then r s is the same as r x s. If R  S is a key for R, then a tuple of s will join with at most one tuple from r therefore, the number of tuples in r s is no greater than the number of tuples in s. If R  S in S is a foreign key in S referencing R, then the number of tuples in r s is exactly the same as the number of tuples in s.  The case for R  S being a foreign key referencing S is symmetric. In the example query depositor customer, customer_name in depositor is a foreign key of customer hence, the result has exactly n depositor tuples, which is 5000

39. Estimation of the Size of Joins If R  S = {A} is not a key for R or S. If we assume that every tuple t in R produces tuples in R S, the number of tuples in R S is estimated to be: If the reverse is true, the estimate obtained will be: The lower of these two estimates is probably the more accurate one. Can improve on above if histograms are available Use formula similar to above, for each cell of histograms on the two relations

40. Estimation of the Size of Joins (Cont.) Compute the size estimates for depositor customer without using information about foreign keys: V(customer_name, depositor) = 2500, and V(customer_name, customer) = 10000 The two estimates are 5000 * 10000/2500 - 20,000 and 5000 * 10000/10000 = 5000 We choose the lower estimate, which in this case, is the same as our earlier computation using foreign keys.

41. Size Estimation for Other Operations Projection: estimated size of  A (r) = V(A,r) Aggregation : estimated size of A g F (r) = V(A,r) Set operations For unions/intersections of selections on the same relation: rewrite and use size estimate for selections  E.g.   1 (r)    2 (r) can be rewritten as   1   2 (r) For operations on different relations:  estimated size of r  s = size of r + size of s.  estimated size of r  s = minimum size of r and size of s.  estimated size of r – s = r.  All the three estimates may be quite inaccurate, but provide upper bounds on the sizes.

42. Size Estimation (Cont.) Outer join: Estimated size of r s = size of r s + size of r  Case of right outer join is symmetric Estimated size of r s = size of r s + size of r + size of s

43. Estimation of Number of Distinct Values Selections:   (r) If  forces A to take a specified value: V(A,   (r)) = 1.  e.g., A = 3 If  forces A to take on one of a specified set of values: V(A,   (r)) = number of specified values.  (e.g., (A = 1 V A = 3 V A = 4 )), If the selection condition  is of the form A op r estimated V(A,   (r)) = V(A.r) * s  where s is the selectivity of the selection. In all the other cases: use approximate estimate of min(V(A,r), n  (r) ) More accurate estimate can be got using probability theory, but this one works fine generally

44. Estimation of Distinct Values (Cont.) Joins: r s If all attributes in A are from r estimated V(A, r s) = min (V(A,r), n r s ) If A contains attributes A1 from r and A2 from s, then estimated V(A,r s) = min(V(A1,r)*V(A2 – A1,s), V(A1 – A2,r)*V(A2,s), n r s ) More accurate estimate can be got using probability theory, but this one works fine generally

45. Estimation of Distinct Values (Cont.) Estimation of distinct values are straightforward for projections. They are the same in  A (r) as in r. The same holds for grouping attributes of aggregation. For aggregated values For min(A) and max(A), the number of distinct values can be estimated as min(V(A,r), V(G,r)) where G denotes grouping attributes For other aggregates, assume all values are distinct, and use V(G,r)

46. Query Optimization Introduction Example of a Simple Type of Query Transformation of Relational Expressions Optimization Algorithms Statistics Estimation Summary

47. Query Optimization Integral component of query processing Why ? One of the most complex pieces of code in a database system Active area of research E.g. XML Query Optimization ? What if you don’t know anything about the statistics Better statistics Etc …