CS 561 Database - Stevens Institute Of Technology _Midterm Exam

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CS561 Midterm Exam Exams must be submitted as single Word or PPT documents; other formats will not be accepted.

All questions on this exam refer to the following relational database schema company(co_name, govt_id, ceo_ssn, hq_loc) division(co_id, div_name, subdiv_of, dir_ssn, div_hq) site(co_id, div_name, loc) product(prod_id, manuf_co, manuf_div, loc, prod_descr) person(ssn, name, address) works_for(ssn, co_id, div_name, salary, emp_id, sup_ssn) skill(ssn, prod_id, manuf_co) The foreign key references can be found in the lectures.

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 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 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 In all cases in which there is a violation, specify in which row the violation is to be found and why it’s a violation. ***************************************************************************** Note: The following questions are intended to measure your learning of SQL operators that are discussed in the course’s lectures. Do not use operators, such as JOIN and different versions of JOIN that aren’t discussed in the lectures. ***************************************************************************** 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 Select div_name From divisions Where count(sub_div) >= (Select *, Count(subdiv_of) as count1 From divisions Group by div_ name) 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

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πssn , Count(div_name) Count (σ WORKS_FOR Group by 2) πname( σ PERSON Where ssn = x.ssn and cnt>=2 ) 4. (10 points) Write the following query in extended relational algebra: Find the number of products manufactured by the company with govt_id 123-45-6789

π

COUNT(prod_id)

(

σ (WORKS_FOR) manuf_co=”123-45-6789”

5. (10 points) Write the following query in SQL without any embedded sub-queries: Find every person who works for a company that has at least two divisions each of which has at least two sites.

Select p.name, w.ssn, s.co_id, Count (Distinct w.div_name) As div_count, Count (Distinct s.loc) As loc_count From PERSON p, works_for w, SITE s Where p.ssn=w.ssn And s.co_id=w.co_id Group By p.name, w.ssn, s.co_id Having Count(Distinct s.div_name)>=2 and count(Distinct s.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).

Select skill.manuf_co From SKILL Where SKILL.manuf.co In (SELECT COUNT(*) From WORKS_FOR, PRODUCT Where (WORKS_FOR.ssn = SKILL.ssn)AND (SKILL.prod_id = PRODUCT.prod_id AND (PRODUCT.manfu_co = WORKS_FOR.co_id)) = 0 ) ssb) (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).

Select co_name From COMPANY As C Where gov_id In (Select co_id From WORKS_FOR Where ssn Not In (Select ssn From SKILL Where manuf_co = C.gov_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).

Select co_name From COMPANY As C Where Exists (Selec * From WORKS_FOR AS W Where co_id = C.gov_id And Not Exists (Select * From SKILL As S Where C.ssn=W.ssn and S.manf_co=C.gov_id))

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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>) For each division (of any company) that employs at least 200 people, find the average salary of all the people the division employs.

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 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>) Select A.div_name, A.avg_sal From (Select div_name, avg(salary) As avg_sal From WORKS_FOR group by div_name) A, (Select div_name, Count(ssn) As Count From WORKS_FOR Group by div_name) B Where A.div_name=A.div_name And Count >=200;

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