1. CS 405G: Introduction to Database Systems 13b Exercise Chen Qian University of Kentucky

2. 6/28/2016Chen Qian @ University of Kentucky2 2 nd Normal Form Note about 2 nd Normal Form by definition, every nonprimary attribute is functionally dependent on every key of R In other words, R is in its 2 nd normal form if we could not find a partial dependency of a nonprimary key to a key in R.

3. 6/28/2016Chen Qian @ University of Kentucky3 Review WorkOn (EID, Ename, email, PID, hour) We say X -> Y is a partial dependency if there exist a X’  X such that X’ -> Y e.g. EID, PID -> Ename Otherwise, X -> Y is a full dependency e.g. EID, PID -> hours EIDPIDEnameemailPnameHours 123410John Smithjsmith@ac.comB2B platform10 11239Ben Liubliu@ac.comCRM40 12349John Smithjsmith@ac.comCRM30 102310Susan Sidhukssidhuk@ac.comB2B platform40

4. Third normal form 3NF requires that there are no non-trivial functional dependencies of non-key attributes on something other than a superset of a candidate key. Recall: non-trivial FD means LHS has no intersection with RHS. In summary, all non-key attributes are mutually independent. 6/28/2016Chen Qian @ University of Kentucky4

5. Boyce-Codd normal form (BCNF) BCNF requires that there are no non-trivial functional dependencies of attributes on something other than a superset of a candidate key (called a superkey). All attributes are dependent on a key, a whole key and nothing but a key (excluding trivial dependencies, like A->A). 6/28/2016Chen Qian @ University of Kentucky5

6. Quiz 4 Q1 (6pt): Consider the following relation: Explain why it is not in 1NF (3pt). Convert it to multiple tables that satisfy 1NF and 2NF. (3pt) 6/28/2016 Chen Qian @ University of Kentucky 6

7. Quiz 4 Because EmpDegrees is not atomic. 6/28/2016 Chen Qian @ University of Kentucky 7

8. Quiz 4 Question 2 (4pt) Consider the following relation and its functional dependencies: Is it in 3NF? (1pt) Is it in BCNF? (2pt) Explain why (1pt) 6/28/2016 Chen Qian @ University of Kentucky 8

9. Quiz 4 It is in 3NF but not in BCNF Student_no, course_no  instr_no Instr_no  course_no Because the determinant (LHS) of the FD Intr_no -> course_no is not a key attribute, so it is not in BCNF. However, since the RHS is a key attribute, it does not violate 3NF. 6/28/2016 Chen Qian @ University of Kentucky 9

10. Exercise: Functional dependency Suppose you are given a relation R with four attributes ABCD. For of the following sets of FDs, assuming those are the only dependencies that hold for R, do the following: (a) Identify the candidate key(s) for R. (b) Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). Candidate keys: B R is in 2NF but not 3NF. 6/28/2016Chen Qian @ University of Kentucky10

11. Exercise: Functional dependency Suppose you are given a relation R with four attributes ABCD. For of the following sets of FDs, assuming those are the only dependencies that hold for R, do the following: (a) Identify the candidate key(s) for R. (b) Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). Candidate keys: BD R is in 1NF but not 2NF. 6/28/2016Chen Qian @ University of Kentucky11

12. Exercise: Functional dependency Suppose you are given a relation R with four attributes ABCD. For of the following sets of FDs, assuming those are the only dependencies that hold for R, do the following: (a) Identify the candidate key(s) for R. (b) Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). 3. ABC → D, D → A Candidate keys: ABC, BCD R is in 3NF but not BCNF 6/28/2016Chen Qian @ University of Kentucky12

13. Exercise: Functional dependency Suppose you are given a relation R with four attributes ABCD. For of the following sets of FDs, assuming those are the only dependencies that hold for R, do the following: (a) Identify the candidate key(s) for R. (b) Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). 4. A → B, BC → D, A → C Candidate keys: A R is in 2NF but not 3NF (because of the FD: BC → D). 6/28/2016Chen Qian @ University of Kentucky13

14. Exercise: Functional dependency Suppose you are given a relation R with four attributes ABCD. For of the following sets of FDs, assuming those are the only dependencies that hold for R, do the following: (a) Identify the candidate key(s) for R. (b) Identify the best normal form that R satisfies (1NF, 2NF, 3NF, or BCNF). 5. AB → C, AB → D, C → A, D → B (a) Candidate keys: AB, BC, CD, AD R is in 3NF but not BCNF (because of the FD: C → A). 6/28/2016Chen Qian @ University of Kentucky14

15. Exercise: Concurrency control Consider the following actions taken by transaction T 1 on database objects X and Y : R(X), W(X), R(Y), W(Y) 1. Give an example of another transaction T 2 that, if run concurrently to transaction T without some form of concurrency control, could interfere with T 1. If the transaction T2 performed W(Y ) before T1 performed R(Y ), and then T2 aborted, the value read by T1 would be invalid and the abort would be cascaded to T1 (i.e. T1 would also have to abort). 6/28/2016Chen Qian @ University of Kentucky15

16. Exercise: Concurrency control Consider the following actions taken by transaction T 1 on database objects X and Y : R(X), W(X), R(Y), W(Y) 2. Explain how the use of Strict 2PL would prevent interference between the two transactions. Strict 2PL would require T2 to obtain an exclusive lock on Y before writing to it. This lock would have to be held until T2 committed or aborted; this would block T1 from reading Y until T2 was finished, thus there would be no interference. 6/28/2016Chen Qian @ University of Kentucky16

17. Exercise: disks Explain the terms seek time, rotational delay, and transfer time. 1. Seek time is the time taken to move the disk heads to the track on which a desired block is located. 2. Rotational delay is the waiting time for the desired block to rotate under the disk head; it is the time required for half a rotation on average, and is usually less than the seek time. 3. Transfer time is the time to actually read or write the data in the block once the head is positioned, i.e., the time for the disk to rotate over the block. 6/28/2016Chen Qian @ University of Kentucky17

18. Exercise: disks If you have a large file that is frequently scanned sequentially, explain how you would store the pages in the file on a disk. A: The pages in the file should be stored ‘sequentially’ on a disk. We should put two ‘logically’ adjacent pages as close as possible. In decreasing order of closeness, they could be on the same track, the same cylinder, or an adjacent cylinder. 6/28/2016Chen Qian @ University of Kentucky18

19. Exercise: disks Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 msec. 1. What is the capacity of a track in bytes? What is the capacity of each surface? What is the capacity of the disk? bytes/track = bytes/sector × sectors/track = 512 × 50 = 25K bytes/surface = bytes/track × tracks/surface = 25K × 2000 = 50, 000K bytes/disk = bytes/surface× surfaces/disk = 50, 000K × 5 × 2 = 500, 000K 6/28/2016Chen Qian @ University of Kentucky19

20. Exercise: disks Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 msec. 3. If the disk platters rotate at 5400 rpm (revolutions per minute), what is the maximum rotational delay? If the disk platters rotate at 5400rpm, the time required for one complete rotation, which is the maximum rotational delay, is The average rotational delay is half of the rotation time, 0.006 seconds. 6/28/2016Chen Qian @ University of Kentucky20

21. Exercise: disks Consider a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 msec. 4. If one track of data can be transferred per revolution, what is the transfer rate? The capacity of a track is 25K bytes. Since one track of data can be transferred per revolution, the data transfer rate is 6/28/2016Chen Qian @ University of Kentucky21

22. Exercise: disks Consider … average seek time of 10 msec. suppose that a block size of 1024 bytes is chosen. Suppose that a file containing 100,000 records of 100 bytes each is to be stored on such a disk and that no record is allowed to span two blocks. 5. What time is required to read a file containing 100,000 records of 100 bytes each sequentially? A file containing 100,000 records of 100 bytes needs 40 cylinders or 400 tracks in this disk. The transfer time of one track of data is 0.011 seconds. Then it takes 400 × 0.011 = 4.4seconds to transfer 400 tracks. This access seeks the track 40 times. The seek time is 40 × 0.01 = 0.4seconds. Therefore, total access time is 4.4+ 0.4 = 4.8seconds. 6/28/2016Chen Qian @ University of Kentucky22

23. Exercise: disks 6. What is the time required to read a file containing 100,000 records of 100 bytes each in a random order? Assume that each block request incurs the average seek time and rotational delay. For any block of data, averageaccesstime = seektime + rotationaldelay + transfertime The average access time for a block of data would be 16.44 msec. For a file containing 100,000 records of 100 bytes, the total access time would be 164.4 seconds. 6/28/2016Chen Qian @ University of Kentucky23

24. Tree structure Each intermediate node can hold up to five pointers and four key values. Each leaf can hold up to four records Name all the tree nodes to be fetched to answer the following query: “Get all records with search key greater than 38.” 6/28/2016Chen Qian @ University of Kentucky24

25. Tree structure Name all the tree nodes to be fetched to answer the following query: “Get all records with search key greater than 38.” I1, I2, and everything in the range [L2..L8]. 6/28/2016Chen Qian @ University of Kentucky25

26. Tree structure inserting a record with search key 109 6/28/2016Chen Qian @ University of Kentucky26

27. inserting a record with search key 109 6/28/2016Chen Qian @ University of Kentucky27

28. Tree structure deleting the record with search key 81 from the original tree. 6/28/2016Chen Qian @ University of Kentucky28

29. 6/28/2016Chen Qian @ University of Kentucky29

30. Tree structure Name a search key value such that inserting it into the (original) tree would cause an increase in the height of the tree. Anything to L4 or L5, e.g. 57 6/28/2016Chen Qian @ University of Kentucky30

31. Tree structure What we can infer about subtrees A, B, and C. ? 6/28/2016Chen Qian @ University of Kentucky31

32. Tree structure We can infer several things about subtrees A, B, and C. First of all, they each must have height one, since their “sibling” trees (those rooted at I2 and I3) have height one. Also, we know the ranges of these trees (assuming duplicates fit on the same leaf): subtree A holds search keys less than 10, B contains keys ≥ 10 and < 20, and C has keys ≥ 20 and < 30. In addition, each intermediate node has at least 2 key values and 3 pointers. 6/28/2016Chen Qian @ University of Kentucky32

33. Tree structure Suppose that this is an ISAM index. What is the minimum number of insertions needed to create a chain of three overflow pages? 6/28/2016Chen Qian @ University of Kentucky33

34. Tree structure If this is an ISAM tree, we would have to insert at least nine search keys in order to develop an overflow chain of length three. These keys could be any that would map to L4, L5, L7, or L8, all of which are full and thus would need overflow pages on the next insertion. The first insert to one of these pages would create the first overflow page, the fifth insert would create the second overflow page, and the ninth insert would create the third overflow page (for a total of one leaf and three overflow pages). 6/28/2016Chen Qian @ University of Kentucky34

35. Want to present on 4/21? HW4 due on 4/21 6/28/2016Chen Qian @ University of Kentucky35