CSC317/MCS9317. Database Performance Tuning. Class test

Transcription

1 CSC317/MCS9317 Database Performance Tuning Class test 7 October 2015 Please read all instructions (including these) carefully. The test time is approximately 120 minutes. The test is close book and close notes, no other written material. There are 5 questions, with a total of 20 marks for the test. Please read through the entire question set before getting started, in order to plan your strategy accordingly. Please write your solutions in the spaces provided below the questions. You can also use the blank areas and back of the test pages for scratch work. Please do not bring any scratch paper with you. Please write neatly and clearly. You may get as few as 0 points if we cannot understand your solution. Student Number: Do not forget to sign the pledge below: I declare that I have never given nor received assistance on this test. SIGNATURE:

2 Introduction The questions 1, 2, 3, 4, and 5 of the midsession test are related to the following simplified version of TPC-H benchmark database used in the laboratory classes. CUSTOMER( C_CUSTKEY NUMBER(12) NOT NULL, C_NAME VARCHAR(25) NOT NULL, C_ADDRESS VARCHAR(40) NOT NULL, C_NATIONKEY NUMBER(12) NOT NULL, CONSTRAINT CUSTOMER_PKEY PRIMARY KEY(C_CUSTKEY) ); PART( P_PARTKEY NUMBER(12) NOT NULL, P_NAME VARCHAR(55) NOT NULL, P_BRAND CHAR(10) NOT NULL, P_AMOUNT NUMBER(12) NOT NULL, P_RETAILPRICE NUMBER(12,2) NOT NULL, CONSTRAINT PART_PKEY PRIMARY KEY (P_PARTKEY) ); PARTSUPP( PS_PARTKEY NUMBER(12) NOT NULL, PS_SUPPNAME VARCHAR(55) NOT NULL, PS_AVAILQTY NUMBER(12) NOT NULL, CONSTRAINT PARTSUPP_PKEY PRIMARY KEY (PS_PARTKEY,PS_SUPPNAME) ); ORDERS( O_ORDERKEY NUMBER(12) NOT NULL, O_CUSTKEY NUMBER(12) NOT NULL, O_TOTALPRICE NUMBER(12,2) NOT NULL, O_DATE DATE NOT NULL, CONSTRAINT ORDERS_PKEY PRIMARY KEY (O_ORDERKEY), CONSTRAINT ORDERS_FKEY1 FOREIGN KEY (O_CUSTKEY) REFERENCES CUSTOMER(C_CUSTKEY) ); LINEITEM( L_ORDERKEY NUMBER(12) NOT NULL, L_PARTKEY NUMBER(12) NOT NULL, L_LINENUMBER NUMBER(12) NOT NULL, L_QUANTITY NUMBER(12,2) NOT NULL, L_SHIPDATE DATE NOT NULL, CONSTRAINT LINEITEM_PKEY PRIMARY KEY (L_ORDERKEY, L_LINENUMBER), CONSTRAINT LINEITEM_FKEY1 FOREIGN KEY (L_ORDERKEY) REFERENCES ORDERS(O_ORDERKEY), CONSTRAINT LINEITEM_FKEY2 FOREIGN KEY (L_PARTKEY) REFERENCES PART(P_PARTKEY) ); Assume that, the relational tables listed above occupy the following amounts of disk storage: CUSTOMER 100 Mbytes PART 400 Mbytes PARTSUPP 600 Mbytes ORDERS 700 Mbytes LINEITEM 900 Mbytes

3 Question 1 (5 marks) Consider a fragment of simple JDBC application listed below. The application connects to a sample database described on Introduction page of the test paper. Rewrite the application such that the new version of executes faster than the original one. Try to speed up the application as much as you can. Assume that the order in which the output lines are printed can be different from an order produced by the original application. Write the comprehensive explanations why the improved application is more efficient than the original one. Statement stmt1 = conn.createstatement (); Statement stmt2 = conn.createstatement(); long total = 0; long part_key1; ResultSet rset1 = stmt1.executequery( " SELECT P_PARTKEY " + " FROM PART " ); while ( rset1.next() ) { part_key1 = rset1.getint(1); ResultSet rset2 = stmt2.executequery( " SELECT L_PARTKEY " + " FROM LINEITEM " + " WHERE L_ORDERKEY < 1000 " ); long part_key2; while ( rset2.next() ) { part_key2 = rset2.getint(1); if (part_key1 == part_key2) total++; } } System.out.println( total );

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5 Question 2 (3 marks) Consider SELECT statement listed below. The relational tables used in the statement are listed on Introduction page of the test paper. SELECT C_NAME, COUNT(*) FROM CUSTOMER JOIN ORDERS ON C_CUSTKEY = O_CUSTKEY WHERE TO_CHAR(O_DATE, 'YYYY') = '2015' GROUP BY C_CUSTKEY, C_NAME HAVING COUNT(*) = (SELECT MAX(COUNT(*)) FROM CUSTOMER WHERE C_CUSTKEY IN (SELECT O_CUSTKEY FROM ORDERS WHERE TO_CHAR(O_DATE, 'YYYY') = '2015') GROUP BY C_CUSTKEY, C_NAME); Use WITH clause to rewrite the statement into a more compact form that can be optimized by a query processor in a better way than the original statement. Explain why your statement with WITH clause can be optimized by a query processor.

6 Question 3 (4 marks) Transform SELECT statement given below into a syntax tree of an equivalent relational algebra expression and optimize the syntax tree by "pushing" selections and projections down the syntax tree. The relational tables used in the statement are listed on Introduction page of the test paper. Use the following operations of relational algebra: selection, projection, join, and antijoin. You can use the symbols of operations explained during the lecture classes. There is no need to write a relational algebra expression and there is no need to draw the intermediate stages of a syntax tree. The final optimized syntax tree is all right. There is no need to consider different implementations of join and antijoin operations. SELECT C_CUSTKEY, C_NAME FROM CUSTOMER WHERE EXISTS (SELECT O_ORDERKEY FROM ORDERS WHERE C_CUSTKEY = O_CUSTKEY AND O_ORDERKEY NOT IN (SELECT L_ORDERKEY FROM LINEITEM WHERE L_PARTKEY = 'P007') AND O_DATE > '01-JAN-2000');

7 Question 4 (4 marks) Consider a fragment of simple JDBC application listed below. The application connects to a sample database described on Introduction page of the test paper. Rewrite the application such that the new version executes faster than the original one. Try to speed up the application as much as you can. Assume, that the order in which output lines are printed can be different from an order produced by the original application. Write the comprehensive explanations why the improved application is more efficient than the original one. Statement stmt1 = conn.createstatement(); ResultSet rset1 = stmt1.executequery( "SELECT * FROM PART" ); String p_name = ""; String p_brand = ""; double p_retailprice = 0.0; while ( rset1.next() ) { p_name = rset1.getstring(2); p_brand = rset1.getstring(3); p_retailprice = rset1.getdouble(5); if (p_amount > 100) System.out.println( p_name ); else if (p_retailprice <= 901.0) System.out.println( p_name ); } System.out.println( "Done." );

8 Question 5 (4 marks) SELECT statements listed below are two implementations of same query. Parts are supplied by different suppliers (PARTSUPP table). Each supplier may have a different quantity of parts (PS_AVAILQTY) available. The query finds for each part supplied its key (PS_PARTKEY) and the largest available quantity. (1) SELECT PS_PARTKEY, MAX(PS_AVAILQTY) FROM PARTSUPP GROUP BY PS_PARTKEY; (2) SELECT PS_PARTKEY, PS_AVAILQTY FROM PARTSUPP WHERE PS_AVAILQTY >= ALL(SELECT PS_AVAILQTY FROM PARTSUPP PS WHERE PS_PARTKEY= PS.PS_PARTKEY); Explain how each one of SELECT statement will be processed. In your explanation you can propose the simple processing plans for each statement. Compare the processing plans for both statements and decide which plan is more efficient, i.e. implementation of better plan takes less read block operations and less time. Justify your decision. END OF TEST PAPER