Information Systems for Engineers. Exercise 10. ETH Zurich, Fall Semester Hand-out Due

Transcription

1 Information Systems for Engineers Exercise 10 ETH Zurich, Fall Semester 2017 Hand-out Due (Exercise in [1]) Movies(title, year, length, genre, studioname, producercertnumber) MovieStar(name, address, gender, birthdate) StarsIn(movieTitle, movieyear, starname) MovieExec(name, address, certnumber, networth) Studio(name, address, prescertnumber) From the base tables above, give SQL code to construct each of the following views: (a) A view RichExec giving the name, address, certificate number and net worth of all executives with a net worth of at least $10,000,000. CREATE VIEW RichExec AS SELECT * FROM MovieExec WHERE networth >= ; (b) A view StudioPres giving the name, address, and certificate number of all executives who are studio presidents. CREATE VIEW StudioPres AS SELECT MovieExec.name, MovieExec.address, MovieExec.certNumber FROM (MovieExec INNER JOIN Studio ON MovieExec.certNumber = Studio.presCertNumber)); (c) A view ExecuteStar giving the name, address, gender, birth date, certificate number, and net worth of all individuals who are both executives and stars. CREATE VIEW ExecuteStar AS SELECT * FROM (MovieExec INNER JOIN MovieStar); 1

2 2. (Exercise in [1]) Which of the views above are updatable? (a) (b) (c) Updatable. Not updatable. Not all attributes in the SELECT clause (although this is tolerated if networth is nullable or has a default value). More than one table in the FROM clause. Not updatable. More than one table in the FROM clause. 3. (Exercise in [1]) Declare each of the following indices using SQL: (a) studioname CREATE INDEX studioname_index ON Studio (name); (b) address of MovieExec CREATE INDEX address_index ON MovieExec (address); (c) genre and length CREATE INDEX genre_index ON Movies (genre, length); 4. Suppose that there are three database operations that we sometimes perform on the StarsIn relation: Q 1 : We look for the title and year of movies in which a given star appeared. That is, we execute a query of the form: SELECT movietitle, movieyear FROM StarsIn WHERE starname = s; Q 2 : We look for the stars that have appeared in a given movie. That is, we execute a query of the form: SELECT starname FROM StarsIn WHERE movietitle = t AND movieyear = y; I: We insert a new tuple into StarsIn. That is, we execute an insertion of the form: Page 2

3 INSERT INTO StarsIn VALUES(t, y, s); for constants t, y, and s. We assume the following costs of operation: Examining all rows in the table has a cost of 10. On average, a star appears in 3 movies and a movie has 3 stars. If we have an index on a field, the cost of reading a value is 1. The index must be read before it can be used, and this has a cost of 1. If an index is modified, another cost of 1 is added to update the stored value. Updating a row has a cost of 2. We can use this to write the following cost table: Action No Index Star Index Movie Index Both Indices Q Q I Average 2 + 8p 1 + 8p p p 1 6 2p 1 2p 2 (a) Suppose that the cost of scanning StarsIn has an unknown cost of N instead of 10, but all other assumptions of that example continued to hold. Fill in the resulting costs in the table below: Action No Index Star Index Movie Index Both Indices Q 1 N 4 N 4 Q 2 N N 4 4 I Average 2 + (N 2)(p 1 + p 2 ) 4 + (N 4)p (N 4)p 1 6 2p 1 2p 2 (b) Assuming p = p 1 = p 2, at what of value of p does it become profitable to use both indices instead of no indices, as a function of N? 2 + (N 2)(p + p) = 6 2p 2p 2 + 2p(N 2) = 6 4p p(4 + 2(N 2)) = 4 4 p = 4 + 2(N 2) p = 2 N (c) If the cost of scanning the table is 20, at what value of p does it become profitable to uses indices on both columns instead of no indices? p = 2 N = 0.1 For more than 10% of queries to inserts, indices are profitable. Page 3

4 (d) If the fraction p of queries to total operations is 20%, at what cost N of scanning the table does it become profitable to use indices? N = = 10 For a cost of more than 10, adding an index is profitable with a fraction of queries of more than 20%. 5. Consider the table Ships(name, category, launched). We will use the same assumptions as stated in the previous question, but set the cost of scanning the table to 50. We expect four types of queries: Q 1 : SELECT * FROM Ships WHERE name = n; with probability p 1. Q 2 : SELECT * FROM Ships WHERE category = y; with probability p 2. Q 3 : SELECT * FROM Ships WHERE launched = l; with probability p 3. I: Insertion of a new tuple with probability 1 p 1 p 2 p 3. Further assume the following: name is the key of the relation. The cost of accessing all ships in a category once the location of one is known is 1. There are 25 ships are launched each year. We can create any combination of indices on name, category, and launched. For each of the following values of p 1, p 2 and p 3, what is the best choice of indices? Action No Index name category launched Q Q Q I Average 2+48(p 1 +p 2 + p 3 ) 4 2p (p 2 + p 3 ) 4 2p (p 1 + p 3 ) p (p 1 + p 2 ) Action name, name, category, All Indices category launched launched Q Q Q I Average 6 4(p 1 +p 2 )+ 44p 3 6 4p 1 +44p p p 1 4p p 3 8 6p 1 6p p 3 (a) p 1 = p 2 = p 3 = 0 Page 4

5 No indices (average cost 2.0). (b) p 1 = 0.3, p 2 = 0.05, p 3 = 0.05 name, category or all indices (both have an average cost of 6.8) (c) p 1 = 0.8, p 2 = 0.1, p 3 = 0.05 All indices (average cost 3.5). References [1] Garcia-Molina, Hector and Ullman, Jeffrey D. and Widom, Jennifer, Database Systems: The Complete Book, Second Edition. Page 5