Database Management Systems (COP 5725) Homework 3

Database Management Systems (COP 5725) Homework 3 Instructor: Dr. Daisy Zhe Wang TAs: Yang Chen, Kun Li, Yang Peng yang, kli, ypeng@cise.uf l.edu November 26, 2013 Name: UFID: Email Address: Pledge(Must be signed according to UF Honor Code) On my honor, I have neither given nor received unauthorized aid in doing this assignment. Signature For grading use only: Question: I II III IV Total Points: 26 25 24 25 100 Score: i

COP5725, Fall 2013 Homework 3 Page 1 of 7 I. [26 points] Indexing. Make the following assumptions for questions (1) and (2): A bucket can hold two keys and a pointer. The initial database D contains one object with key 10100. Six objects with the following keys are inserted to D in the following order: 00110, 11010, 10011, 01010, 10110, and 01011. (1) [3 points] Assume that an extensible hash table is used to index the database. Show the index structure after the insertions. (2) [3 points] Assume that a linear hash table is used to index the database with the restriction that at most 80% of the hash table can be full at any time. Show the index structure after the insertions. For (1) (left) and (2) (right). i = 3 00110 2 000 001 010 011 100 101 110 111 01010 2 01011 10011 3 10100 3 10110 11010 2 i = 3 n = 5 r = 7 000 001 010 011 100 00110 01010 01011 10011 10100 10110 11010 (3) [3 points] Describe a scenario when extensible hash tables are preferred over linear hash tables. Explain your answer. Two possibilities: When the insertions are very frequent. This is because linear hash table is reorganized every time a new bucket is added. A new bucket is added frequently. When the values of the keys that the index is built on uniformly distributed keys. (i.e., the hash table is filled uniformly). For (4)-(6), consider a B+ tree whose nodes contain up to 4 keys (5 pointers). (4) [3 points] Bulkload the B+ tree with values 46, 10, 70, 49, 23, 40, 59, 29, 34, 54, 75, 30.

COP5725, Fall 2013 Homework 3 Page 2 of 7 34 54 10 23 29 30 34 40 46 49 54 59 70 75 (5) [3 points] Show the result B+ tree after inserting values 80, 24, 42. 42 24 34 54 70 10 23 24 29 30 34 40 42 46 49 54 59 70 75 80 (6) [3 points] Based on the B+ tree in (5), show the result B+ tree after deleting values 10, 40. 29 42 54 70 23 24 29 30 34 42 46 49 54 59 70 75 80 (7) [8 points] Fill in the cost table below for Alternative 1 ISAM and B+ tree indices. Assume each index takes P pages on disk, has height H, and fanout F at each internal node. Assume there are R tuples in the relation, and B tuples fit on a leaf (or overflow) page. In each case, assume infinite buffer pool size, but the buffer pool starts out empty. For each page that gets dirty, add 1 to your I/O cost since it will eventually have to be flushed to disk. For ISAM, assume that a leaf node maintains only a pointer to the beginning of an overflow list. Given the constraints of a B+ Tree/ISAM, assume whatever data you want in the tree for each case below.

COP5725, Fall 2013 Homework 3 Page 3 of 7 ISAM B+ Tree Worst-case # IOs for range query P (index consists of root with a linear string of overflow pages. Need to look at all overflow pages since they re not sorted) or H + R/B or H + (F H 1) + R/B (look at whole leaf level and all data in last leaf overflow) H + F H (range query covers the whole table) H + R/B P was not accepted here, as this would imply only 2 I/Os, given the structure of the index. Worst-case # IOs for insert P + 2 (index consist of a root with a string of overflow pages. Need to scan til the end, and add a new overflow page in the worst case, and update the previous last overflow page with a pointer) or H + R/B + 2 3H + 1 (every node needs to split, +1 for new root. Read pages we re going to split on the way down, so we don t need to read them again.) II. [25 points] Query Evaluation. Suppose we want to compute (R(a, b) S(a, c)) T (a, d) in the order indicated. We have M = 101 main memory buffers, and the number of disk blocks (pages) for R and S B(R) = B(S) = 2000. Now we decide to use one-pass or two-pass sort-merge-join algorithms to implement the query. (1) [2 points] Would you use a one- or two-pass sort-merge-join for R S? Explain. Two-pass sort-merge-join, since both operands are larger than main memory. (2) We shall use the appropriate number of passes for the second join, first dividing T into some number of sublists sorted by a, and merging them with the sorted and pipelined stream of tuples from the join R S. For what values of B(T ) should we choose for the join of T with R S: i. [3 points] A one-pass join; i.e., we read T into memory, and compare its tuples with the tuples of R S as they are generated. B(T ) 60. ii. [3 points] A two-pass join; i.e., we create sorted sublists for T and keep one buffer in memory for each sorted sublist, while we generate tuples of R S. B(T ) > 60. iii. [4 points] For cases in i. ii., what is the total number of disk I/O s (in terms of B(T ))? For i. we need 3 (2000 + 2000) = 12, 000 I/O s to perform the two-pass sortmerge-join of R and S, and B(T ) I/O s to read T in the one-pass join of (R S) T. The total # of I/O s is 12, 000 + B(T ).

COP5725, Fall 2013 Homework 3 Page 4 of 7 For ii. we need 2B(T ) disk I/O s to sort B(T ) into sublists; 12,000 disk I/O s to join R S; B(T ) to read the sorted lists of T. The total number of disk I/O s is 12, 000 + 3B(T ). (3) [4 points] Consider the query (R(a, b) S(a, c)) T (c, d), i.e., the second join is based on attribute c instead of a. How would you choose the join algorithms? Provide a new cost estimation if your choices differ from (2). We need to re-sort the intermediate result R S based on the attribute c. New cost is 12, 000 + 3B(T ) + 2B(R S), where the 2B(R S) term comes from writing out the sublists of R S and read them in again while joining (R S) T. For (4)-(6), you are given M memory blocks and a relation R. (4) [3 points] Describe a two-pass hash-based algorithm for duplicate elimination, δ(r). (Hint: review the aggregation algorithm with grouping). Hash R into M 1 buckets based on all attributes. Perform δ on each bucket in isolation, using M memory blocks. (5) [3 points] What is the largest relation your algorithm can handle given M blocks of main memory? M(M 1). (6) [3 points] What is the number of disk I/O s of your algorithm? B(R) for reading R and hashing; B(R) for writing out the buckets; B(R) for reading the buckets and do the actual duplication. 3B(R) in total. III. [24 points] Query Optimization. Consider the following database schema: Employees(eid: integer, ename: string, sal: integer, title: string, age: integer) Suppose that the following indexes, all using Alternative (2) for data entries, exist: a hash index on eid, a B+ tree index on sal, a hash index on age, and a clustered B+ tree index on (age, sal). Each Employees record is 100 bytes long, and you can assume that each index data entry is 20 bytes long. The Employees relation contains 10,000 pages. (1) Consider each of the following selection conditions and, assuming that the reduction factor (RF) for each term that matches an index is 0.1, compute the cost of the most selective access path for retrieving all Employees tuples that satisfy the condition (in terms of the number of I/O s):

COP5725, Fall 2013 Homework 3 Page 5 of 7 i. [4 points] age=25. The clustered B+ tree index would be the best option here, with a cost of 2 (lookup) + 10000 0.1 (data pages) + 10000 0.2 (index pages) 0.1 = 1202. Although the hash index has a less lookup time, the potential number of record lookups (10000 0.1 20 tuples per page = 20000) renders the clustered index more efficient. ii. [4 points] sal>200 AND age>30 AND title= CFO. Here an age condition is present, so the clustered B+ tree index on (age, sal) can be used. The cost is 2 + 10, 000 0.2 0.1 (all index pages needs to be fetched satisfying age>30) + 10, 000 0.1 0.1 (data pages) = 302. Consider the following relational schema and SQL query: Emp(eid, did, sal, hobby) Dept(did, dname, floor, phone) Finance(did, budget, sales, expences) SELECT D.dname, F.budget FROM Emp E, Dept D, Finance F WHERE E.did = D.did AND D.did = F.did AND D.floor = 1 AND E.sal >= 59000 AND E.hobby = yodelling; (2) [5 points] Identify a query plan that a decent query optimizer would choose. π D.dname, F.budget π F.did, F.budget π E.did π D.did, D.dname F σ E.sal 59000, E.hobby="yodelling" σ D.floor=1 (3) Suppose that the following additional information is available: E Unclustered B+ tree indexes exist on Emp.did, Emp.sal, Dept.did, and Finance.did (each leaf page contains up to 200 entries). The systems statistics indicate that employee salaries range from 10,000 to 60,000, employees enjoy 200 different hobbies. The company owns two floors in the building. There are a total of 50,000 employees and 5,000 departments (each with corresponding financial information) in the database. The DBMS used by the company has just one join method available, namely index nested loops. D

COP5725, Fall 2013 Homework 3 Page 6 of 7 i. [3 points] For each of the query s base relations, estimate the number of tuples that would be initially selected from that relation if all of the non-join predicates on that relation were applied to it before any join processing begins. Emp: 50000 1000 50000 1 200 = 5. Dept: 5000 1 2 = 2500. Finance: 5000. ii. [8 points] Under the System R approach, determine a join order that has the least estimated cost. Compute the cost of your plan (in terms of the number of disk I/O s). ((D E) F ). First, we use the fact that there is a B-tree index on salary to retrieve the tuples from E such that E.salary >= 59000. We estimate that (50000/50) = 1000 such tuples selected out, with a cost of 1 tree traversal (say 3 I/O s to get to the leaf) + the cost of scanning the leaf pages (1000/200 + 1-1 = 5) + the cost of retrieving the 1000 tuples (since the index is unclustered each tuple is potentially 1 disk I/O) = 3 + 5 + 1000 = 1008. Of these 1000 retrieved tuples, do an on-the fly select out only those that have hobby = "yodelling", we estimate there will be (1000/200) = 5 such tuples. Pipeline these 5 tuples from E one at a time to D. By using the B+ tree index on D.did and the fact the D.did is a key, we can find the matching tuples for the join by searching the D.did B+ tree and retrieving at most 1 matching tuple per tuple from E. The cost of E D is hence total cost of index nested loop. 5 (tree traversal of D.did Btree + record retrieval) = 5 (3 + 1) = 20. Now select out the 5/2 = 3 tuples that have D.floor = 1 on the fly and pipeline it to the next level F. (This is done after E D is done). Use the B+ tree index on F.did and the fact that F.did is a key to retrieve at most 1 tuple for each of the 3 pipelined tuples. This cost is at most 3 (3 + 1) = 12. Ignoring the cost of writing out the final result, we get a total cost of 1008 + 20 + 12 = 1040. IV. [25 points] Transactions and Concurrency Control. (1) For each of the following schedules: a) r 1 (A); r 2 (B); w 1 (B); w 2 (C); r 3 (C); w 3 (A); b) r 1 (A); r 2 (A); r 1 (B); r 2 (B); r 3 (A); r 4 (B); w 1 (A); w 2 (B); Answer the following questions: i. [4 points] What is the precedence graph for the schedule? ii. [4 points] Is the schedule conflict-serializable? If so, what is an equivalent serial schedules? a) i. T 2 T 1, T 2 T 3, T 1 T 3. ii. Yes, equivalent schedules: T 2 T 1 T 3. b) i. T 2 T 1, T 3 T 1, T 1 T 2, T 4 T 2. ii. No, there are cycles in the precedence graph (T 2 T 1, T 1 T 2 ). (2) [17 points] Consider the following two transactions: T 1 : w 1 (C); r 1 (A); w 1 (A); r 1 (B); w 1 (B); T 2 : r 2 (B); w 2 (B); r 2 (A); w 2 (A);

COP5725, Fall 2013 Homework 3 Page 7 of 7 Say our scheduler performs exclusive locking only (i.e., no shared locks). For each of the following three instances of transactions T 1 and T 2 annotated with lock and unlock actions, say whether the annotated transactions: 1. obey two-phase locking, 2. will necessarily result in a conflict serializable schedule (if no deadlock occurs), 3. will necessarily result in a strict schedule (if no deadlock occurs), 4. will necessarily result in a serial schedule (if no deadlock occurs), and 5. may result in a deadlock. a) T 1 : l 1 (B); l 1 (C); w 1 (C); l 1 (A); r 1 (A); w 1 (A); r 1 (B); w 1 (B); Commit; u 1 (A); u 1 (C); u 1 (B); T 2 : l 2 (B); r 2 (B); w 2 (B); l 2 (A); r 2 (A); w 2 (A); Commit; u 2 (A); u 2 (B); b) T 1 : l 1 (C); l 1 (A); r 1 (A); w 1 (C); w 1 (A); l 1 (B); r 1 (B); w 1 (B); u 1 (A); u 1 (C); u 1 (B); Commit; T 2 : l 2 (B); r 2 (B); w 2 (B); l 2 (A); r 2 (A); w 2 (A); Commit; u 2 (A); u 2 (B); c) T 1 : l 1 (C); w 1 (C); l 1 (A); r 1 (A); w 1 (A); l 1 (B); r 1 (B); w 1 (B); Commit; u 1 (A); u 1 (C); u 1 (B); T 2 : l 2 (B); r 2 (B); w 2 (B); l 2 (A); r 2 (A); w 2 (A); Commit; u 2 (A); u 2 (B); Format your answer in a table with Yes(Y)/No(N) entries. 2PL Necessarily conflict Serializable Necessarily strict schedule Necessarily Serial schedule May result in deadlock a) Y Y Y Y N b) Y Y N Y Y c) Y Y Y N Y