Fundamentals of Database Systems

Transcription

1 Fundamentals of Database Systems Assignment: 4 September 21, 2015 Instructions 1. This question paper contains 10 questions in 5 pages. Q1: Calculate branching factor in case for B- tree index structure, given: page size: 16 kb, key-size: 16 bytes, pointer size = 8 bytes. A. 255 B. 256 C. 510 D. 511 Order of B-tree is defined as, where, C = Disk page size in bytes. η = Size of pointer γ = Size of key θ B tree = C η 2(γ+2η) Branching factor = maximum number of children an internal node in a tree can have. Here, braching f actor = 2θ B tree + 1 Q2: Calculate branching factor in case for B+ tree index structure, given: page size: 32 kb, key-size: 8 bytes, pointer size = 16 bytes. A. 682 B. 683 C D

2 Order of B+-tree is defined as, where, C = Disk page size in bytes. η = Size of pointer γ = Size of key θ B+ tree = C η 2(γ+η) Branching factor = maximum number of children an internal node in a tree can have. Here, braching f actor = 2θ B+ tree + 1 Q3: Which of the following is/are TRUE? a. When a query is parsed, equivalent action plans are generated. b. Time taken to process query depends only on time taken for block transfers. c. Binary search is fast and is the chosen strategy for all selection predicates. d. Disk access is the biggest factor contributing to query cost, and hence most of the query optimizers optimize on disk accesses. A. a only B. a and d C. b and d D. a and b a. When a query is parsed, equivalent action plans are generated and one of the action plan is chosen based on the optimizer. b. Time taken to process query depends on time taken for seek and block transfers. c. Binary search is the chosen strategy for selection predicates if the relation is sorted on the attribute being compared. d. Disk access is the biggest factor contributing to query cost, and hence most of the query optimizers optimize on disk accesses, and ignore other factors such as CPU time, network communication lag, etc. Q4: Given the memory can hold 5 blocks at a time. Also, it is known that size of data is 64MB and size of each block is 1 MB. How many merge passes are required to sort the whole data in external merge-sort, after the initial sort pass, when sorted runs have already been create? A. 2 B. 3 C. 5 D. 7 Page 2 of 5

3 M = number of blocks memory can hold = 5 Size of relation = 64MB Size of 1 block = 1MB Number of blocks to be sorted = b = 64 Number of merge passes = log M 1 b M Intuitively, The data is first sorted. M blocks are fetched, sorted in memory and then these sorted runs are written back to disk. This is the Sort Pass. Now, we have 13 sorted runs. Next, these sorted runs are merged. The first Merge Pass, results in 4 merged runs( sorted runs), since 4 blocks are available for each run, and the 5th serves as buffer. Now, we merge the 4 merged runs in second pass. This results in all the runs being merged together. Read the statement below to solve questions 5 and 6. Relation r has 40,000 tuples and relation s has 60,000 tuples. The block size is 150 tuples/block. Q5: What is the optimal number of block transfers required in case of Nested Loop Join (NLJ)? A. 15,960,400 B. 16,000,266 C. 16,000,267 D. 16,020,400 Number of tuples in r = n r = Number of tuples in s = n s = Block size = 150 tuples/block Number of blocks in r = b r = 40000/150 = 267 Number of blocks in s = b s = 60000/150 = 400 Optimality is achieved when the outer relation is smaller. Therefore, r is the outer relation. Number of block transfers in NLJ = b outer + n outer b inner = = Q6: What is the optimal number of block transfers required in case of Block Nested Loop Join (BNLJ)? A. 107,067 B. 106,800 C. 106,666 D. 107,200 Page 3 of 5

4 Number of tuples in rr = n r = Number of tuples in s = n s = Block size = 150 tuples/block Number of blocks in r = b r = 40000/150 = 267 Number of blocks in s = b s = 60000/150 = 400 Optimality is achieved when the outer relation is smaller. Therefore, r is the outer relation. Number of block transfers in BNLJ = b outer + b outer b inner = = Q7: SELECT * FROM R INNER JOIN S WHERE R.key1 >= S.key2; To solve the above SQL query, if merge join was used, what would be the number of block transfers and number of seeks? (r = 20,000 tuples, s = 30,000 tuples, 100 tuples/block) A. 500 B C. 50 D. not applicable Merge join is applicable only when the join condition is equality. Q8: Connect the statements in Column 1 with those in Column 2. Column 1: I - In hash join, smaller relation should be build relation. II - If relation r has more number of tuples than relation s, then it is preferable to treat s as outer table while performing block nested loop join. Column 2: p - relation needs to fit in main memory. q - cost of join operation is reduced. r - number of iterations is reduced. s - smaller relation is outer table. A. I - p, r; II - q, s B. I - q, s; II - p, r, s C. I - r, s; II - p, r D. I - p, q; II - q, r, s Page 4 of 5

5 I - In hash join, smaller relation should be build relation, since each partition needs to fit in main memory. II - If relation r has more number of tuples than relation s, then it is preferable to treat s as outer table while performing block nested loop join, since this reduces the number of times r, the larger relation, is read in memory, i.e., number of iterations is reduced, and hence the cost of join operation is lowered. And the smaller relation should be outer relation for the same reason. Q9: Relation r: rs, r1, r2, r3, r4, r5, r6 Relation s: rs, s1, s2, s3, s4, s5, s6 (Attribute rs is common to both the relations) Based on the above schema, optimize the following query: Π r1,s1 (σ r2>s3 (Π r1,r2,r3,r4,s1,s2,s3,s4 (σ s2>r4 (r s)))) A. Π r1,s1 (r (s2>r4) (r2>s3) s) B. σ (s2>r4) (r2>s3) (Π r1,s1 (r s)) C. Π r1,s1 (σ (s2>r4) (r2>s3) (r s)) D. Π r1,s1 (r (r.rs=s.rs) (s2>r4) (r2>s3) s) A. Doesn t preserve equality condition present in natural join. B. Syntactically incorrect, cannot select on r2, r4, s2 and s3 when projection is only on r1 and s1. C. Lacks the equality condition of natural join. Q10: SELECT A FROM T WHERE M > 1000 AND M < 5000; Which of the following is best to solve the above query? A. Linear search B. Binary search on M C. B+-tree with index on M D. B+-tree with index on A B+-tree wins since it is indexed on the attribute that is being compared. It is better than binary search in terms of disk operations. Page 5 of 5