Database Management Systems - 1 CS561 Midterm Exam

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Database Management Systems - 1 CS561 Midterm Exam Name : Sai Vignesh Chiruvella CWID : 20007437 1. a) (3 points) Construct an instance of the skill table that contains a violation of the table's key constraint - or explain why it is impossible to construct such an instance.

Answer: INSERT INTO skill VALUES ('0002934654','SAI','12229475'); Error report : ORA-00001 : Unique Constraint Violation (ADMIN.SKILL_PK) Violation of the table constraint because we are trying to insert a row with the same SSN,PROD_ID, MANUF_CO as one row already exists in the table b) (3 points) Construct an instance of the product table that contains a violation of the table's key constraint - or explain why it is impossible to construct such an instance Answer: INSERT INTO product VALUES('SAI','12229475','DIV_007','IOT','NO_612'); Error report : ORA-00001 : Unique Constraint Violation (ADMIN.SKILL_PK) Violation of the table constraint because we are trying to insert a row with the same PROD_ID as one row already exists in the table c) (4 points) Construct instances of both the skill and product tables that contain violations of all the relevant foreign key reference constraints -- or explain why it is impossible to construct such instances Answer: INSERT INTO skill VALUES ('0002934758','RAM','12229842'); Error report: ORA-02291: Integrity Constraint Violated-parent key not found(ADMIN.SKILL_PK) Violation of relevant foreign key reference constraints of table because we are trying to insert a row in SKILL that don’t exists in PRODUCT which is the parent of the foreign key column.

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In all cases where a violation occurs, specify which row the violation occurs in and why it is a violation.

2. (10 points) Write the following query in the pseudo-code used in the lectures: Find every division (of any company) that has at least two subdivisions Answer: for i=1 to no_rows(company) do for j = 1 to no_rows(company) do if(company[i][govt_id] = division[j][co_id] and division[i][co_id] <> division[j][co_id]) then output(division[j][co_id]) SQL Query : SELECT u.name, v.ssn, w.co_id, count(distinct v.div_name) as no_of_div, count(distinct w.loc) as no_of_loc FROM person u, works_for v, site w WHERE u.ssn=v.ssn AND w.co_id=v.co_id GROUP BY u.name, v.ssn, w.co_id HAVING no_of_div >=2 and no_of_loc >=2 3. (10 points) Write the following query in unextended relational algebra: Find every person who works for at least two different divisions of the same company Answer: name(  ( u1(DIVISION) X u2 (DIVISION)) X WORKS_FOR) u1.ssn = u2.ssn AND u1.co_id != u2.co_id AND COMPANY.govt_id = u1.co_id AND COMPANY.govt_id = u2.co_id ).

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4. (10 points) Write the following query in extended relational algebra: Find the number of products manufactured by the company with govt_id 123-456789 Answer:

∏COUNT(*)(∏Prod_id(σ(PRODUCTxDIVISION) (σCOMPANY.govt_id=DIVISION.co_id)ANDσCOMPANY.govt_id=”123-45-6789”)) 5. (10 points) Write the following query in SQL without any embedded subqueries: Find every person who works for a company that has at least two divisions each of which has at least two sites.

Answer: select u.name, v.ssn, w.co_id, count (distinct v.div_name) as div_count, count (distinct w.loc) as loc_count from person u, works_for v, site w where u.ssn=v.ssn and w.co_id=v.co_id group by u.name, v.ssn, w.co_id having count(distinct w.div_name)>=2 and count(distinct w.loc)>=2; 6. Consider the following query: Find every company that employs at least one person who has no skills at any of the products that the company manufactures. a) (10 points) Write the query in SQL using an embedded query, but without using any of the embedding operators (IN, NOT IN, EXISTS, NOT EXISTS).

Answer: SELECT co_name FROM company WHERE (SELECT COUNT(*) FROM SKILL, WORKS_FOR

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WHERE (SKILL.ssn = PERSON.ssn) AND (WORKS_FOR.ssn = PERSON.ssn) AND (SKILL.manuf_co = WORKS_FOR.co_id) = 0 ) b) (10 points) Write the query in SQL with an embedded query that uses either IN or NOT IN, whichever is appropriate. (You may not use EXISTS or NOT EXISTS).

Answer: SELECT co_name FROM company AS U WHERE govt_id IN (SELECT co_id FROM works_for WHERE ssn NOT IN (SELECT ssn FROM skill WHERE manuf_co = U.govt_id)) c) (10 points) Write the query in SQL using an embedded query that uses either EXISTS or NOT EXISTS, whichever is appropriate. (You may not use IN or NOT IN).

Answer: SELECT co_name FROM company AS U WHERE EXISTS (SELECT * FROM works_for AS V WHERE co_id = U.govt_id AND NOT EXISTS (SELECT * FROM skill AS W WHERE W.ssn=V.ssn AND W.manuf_co=U.govt_id))

7. (10 points) Write the following query in SQL using both a GROUP BY and a HAVING clause. (You may not use any of the following operators: IN, NOT IN, EXISTS, NOT EXISTS, <rel op> ALL, <rel op> ANY, <rel op SOME>) Answer: select division.div_name, avg(works_for.salary) from division, person, works_for where division.div_name = works_for.div_name and person.ssn = works_for.ssn group by division.div_name

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having count(*) >= 200;

8. (10 points) Write the query of Question 7 in SQL using a GROUP BY but no HAVING clause. (You may not use any of the following operators: IN, NOT IN, EXISTS, NOT EXISTS, <rel op> ALL, <rel op> ANY, <rel op SOME>) Answer: select u.div_name, v.avg_sal from (select div_name, avg(salary) as avg_sal from works_for group by div_name) u, (select div_name, count(ssn) as count from works_for group by div_name) v where u.div_name=v.div_name and count >=200;

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