Database Management Systems (COP 5725) Homework 2

Transcription

1 Database Management Systems (COP 5725) Homework 2 Instructor: Dr. Daisy Zhe Wang TAs: Yang Chen, Kun Li, Yang Peng yang, kli, ypeng@cise.uf l.edu October 8, 2013 Name: UFID: 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: Score: i

2 COP5725, Fall 2013 Homework 2 Page 1 of 6 I. [25 points] Advanced SQL. Consider the following database concerning World War II capital ships: Classes(class, type, country, numguns, bore, displacement) Ships(name, class, launched) Battles(name, date) Outcomes(ship, battle, result) Ships are built in classes from the same design, and the class is usually named for the first ship of that class. The relation Classes records the name of the class, the type ( bb for battleship of bc for battlecruiser), the country that built the ship, the number of main guns, the bore (diameter of the gun barrel, in inches) of the main guns, and the displacement (weight, in tons). Relation Ships records the name of the ship, the name of its class, and the year in which the ship was launched. Relation Battles gives the name and date of battles involving these ships, and relation Outcomes gives the result (sunk, damaged, or ok) for each ship in each battle. Write a SQL query for the following: (1) [5 points] For each class, find the year in which the first ship of that class was launched. Sort the results by launch years in decreasing order. SELECT C.class, MIN(S.launched) AS First_Launched FROM Classes C, Ships S WHERE C.class = S.class GROUP BY C.class ORDER BY First_Launched DESC; (2) [5 points] Find the countries of the classes with at least one ship sunk in a battle. SELECT DISTINCT C.country FROM Classes C, Ships S WHERE C.class = S.class AND S.name IN ( SELECT ship FROM Outcomes O WHERE O.result= sunk (3) [5 points] For each class with at least three ships, find the number of ships of that class sunk in battle. SELECT M.class, COUNT(result) FROM (SELECT class, name

3 COP5725, Fall 2013 Homework 2 Page 2 of 6 FROM Ships WHERE class IN ( SELECT class FROM Ships GROUP BY class HAVING COUNT(*) >= 3) ) M LEFT JOIN (SELECT ship, result FROM Outcomes WHERE result = sunk ) S ON M.name = S.ship GROUP BY M.class (4) [5 points] Find the names of the ships whose number of guns was the largest for those ships of the same bore. SELECT S.name FROM Ships S, Classes C WHERE S.Class = C.Class AND numguns >= ALL (SELECT numguns FROM Classes C2 WHERE C2.bore = C.bore Write the following database modifications: (5) [2 points] Modify the Classes relation so that gun bores are measured in centimeters (1 inch = 2.5 centimeters) and displacements are measured in metric tons (1 metric ton = 1.1 tons). UPDATE Classes SET bore = 2.5 * bore, displacement = displacement/1.1; (6) [3 points] Delete the classes with no ships. DELETE FROM Classes WHERE NOT EXISTS ( SELECT * FROM Ships WHERE Ships.class = Classes.class II. [15 points] Assertions and Triggers.

4 COP5725, Fall 2013 Homework 2 Page 3 of 6 (1) [5 points] Imagine you are running a massive database schema (such as EBAYs server) with thousands of tables and millions of transactions per day. Is it a good idea to use ASSERTIONS? Whether you chose YES or NO, please explain why, briefly. As databases get bigger, it becomes inefficient to use assertions and it essentially slows down the system. It is quite difficult to automatically figure out an efficient way to check only the tables that are being affected by specific queries, so DBMS usually ends up checking all the tables and this is simply not practical when you have millions of tables in your database system. (2) You own a bar and you have the following two tables to manage your database: 1.Drinkers(name, tolerance) 2.Drinks(drink name, drinker name, drink power) In table Drinkers, the tolerance is used to indicate the drinking tolerance of any drinker. In table Drinks, drink power is used to indicate the strength of the Drink. Both tolerance and drink power are integers. When a drinker orders a drink with certain power value, an entry about this action will be added to the Drinks table. i. [5 points] We need to gurantee that the tolerance of any drinker is inside the closed interval [0, 10]. Use ASSERTION to enforce this constraint. Version 1 Version 2 Version 3 CREATE ASSERTION Check_Tolerance CHECK ( NOT EXISTS ( SELECT * FROM DRINKER WHERE TOLERANCE IS NULL OR TOLERANCE < 0 OR TOLERANCE > 10 ) CREATE ASSERTION Check_Tolerance CHECK ( ALL (SELECT TOLERANCE FROM DRINKER) >= 0 AND ALL (SELECT TOLERANCE FROM DRINKER) <= 10 CREATE ASSERTION Check_Tolerance CHECK ( (SELECT MIN(TOLERANCE) FROM DRINKER) >= 0$ AND (SELECT MAX(TOLERANCE) FROM DRINKER) <= 10$

5 COP5725, Fall 2013 Homework 2 Page 4 of 6 ii. [5 points] When someone wants to order a drink, we need to check after this entry inserted into Drinks table, whether the sum of Drinks.drink power for that drinker stay below the drinker s tolerance value. If the tolerance is less than the the sum of drink power, we should cancel this transaction. Use a TRIGGER to enforce this constraint. (Hint: Use ROLLBACK to cancel the transaction) CREATE TRIGGER Cab_Trigger AFTER INSERT ON DRINKS REFERENCING NEW ROW AS NEWDRINK FOR EACH ROW WHEN ( (SELECT TOLERANCE FROM DRINKERS WHERE NAME=NEWDRINK.DRINKER_NAME) < (SELECT SUM(POWER) FROM DRINKS WHERE DIRNKER_NAME=NEWDRINK.DRINKER_NAME) ) ROLLBACK; III. [20 points] Disk Management. 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. Suppose a block size of 1024 bytes is chosen. A file containing 100k records of 100 bytes each is to be stored on such a disk and that no record is allowed to span two blocks. (1) [5 points] If the disk platters rotate 5400 rpm, 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 1 60 = 0.011s The average rotational delay is half of the rotation time, 0.006s. Note: The definition of maximum rotational delay is the time it takes to do a full rotation excluding any spin-up time. If you use half rotation time according to the slides, we will not deduct points. (2) [5 points] 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 25K = 2, 250KB/s

6 COP5725, Fall 2013 Homework 2 Page 5 of 6 (3) [5 points] What is the time required to read the file sequentially? What if the disk were capable of reading/writing from all heads in parallel? 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 seconds. Then it takes =4.4 seconds to transfer 400 tracks. This access seeks the track 40 times. The seek time is =0.4 seconds. Therefore, total access time is =4.8 seconds. If the disk were capable of reading/writing from all heads in parallel, the disk can read 10 tracks at a time. The transfer time is 10 times less, which is 0.44 seconds. Thus total access time is =0.84 seconds. (4) [5 points] What is the time required to read each block in the file in a random order? (assuming that each block request incurs the average seek time and rotational delay) For any block of data, average access time = seek time+rotational delay+transfer time. seek time = 10ms rotational delay = 6ms 1K transfer time = 2, 250K/s = 0.44ms The average access time for a block of data would be 16.44ms. For a file containing 100k records of 100 bytes, the total access time would be seconds. IV. [15 points] Buffer Management. Suppose you have a Buffer Pool that can hold 3 pages. There are 26 blocks on your disk; for simplicity we will refer to each block with a letter of the English alphabet between A and Z. An access pattern is a string of letters that log the requests to the Buffer Pool for pages. For each access pattern below, figure out the number of I/Os that would occur starting with an empty buffer pool each time, for different replacement policies. (Spaces in the access patterns are there just for legibility.) (1) [5 points] For access pattern ZYXW ZYXW, what are the numbers of I/Os that would occure for FIFO, LRU and MRU. Explain how you get the resutls. FIFO 8, LRU 8, MRU 5 (2) [5 points] For access pattern ABCD DCBA ABCD, what are the numbers of I/Os that would occure for FIFO and LRU. Explain how you get the resutls. FIFO 8, LRU 6 (3) [5 points] For access pattern ACEG BDFH ACEG BDFH, what are the numbers of I/Os that would occure for LRU and MRU. Explain how you get the resutls.

7 COP5725, Fall 2013 Homework 2 Page 6 of 6 LRU 16, MRU 13