Question 5Given the relation schema R(A, B, C) and S(C, D, E). please match each of the following relational algebra expressions to the corresponding SQL (Structured Query Language) statement.(Assume R and S are union compatible)1. RUSA. SELECT FROM RMINUSSELECT * FROM S:B. SELECT FROM S2. ROSSELECTFROM R:3. R-SC. SELECT FROM R CROSS JOIN S;4.S RD. SELECT * FROM R NATURAL JOIN S;5. R* SE. SELECT6. RX SF. SELECT7. RRB S.BS G. SELECT FROM R12SELECT FROM S:8. RRC S.CS H. SELECT FROM RINTERSECT456MINUS78UNIONFROM R JOIN S ON R.C=S.C;FROM R JOIN S ON R.B = S.B;SELECT FROM S:GHABDсFE

Given the relation schema R(A, B, C) and S(C, D, E), please match each of the following relational algebra expressions to the corresponding SQL (Structured Query Language) statement. (Assume R and S are union compatible) 1. RUS A. SELECT FROM R MINUS SELECT FROM S: B. SELECT FROM S MINUS SELECT C. SELECT D. SELECT E. SELECT F. SELECT 7. RRB S.BS G. SELECT 2. ROS 3. R-S 4.5 R 5. R* S 6. RX S 1 SELECT 8. RRC S.CS H. SELECT INTERSECT 2 3 4 5 6 7 UNION 8 FROM R; FROM R CROSS JOIN S; FROM R NATURAL JOIN S: FROM R JOIN S ON R.C=S.C; FROM R JOIN S ON R.B= S.B; FROM R FROM S: FROM R SELECT FROM S: G H A B D с F E

Given the relation schema R(A, B, C) and S(C, D, E), please match each of the following relational algebra expressions to the corresponding SQL (Structured Query Language) statement. (Assume R and S are union compatible) 1. RUS A. SELECT FROM R MINUS SELECT FROM S: B. SELECT FROM S MINUS SELECT C. SELECT D. SELECT E. SELECT F. SELECT 7. RRB S.BS G. SELECT 2. ROS 3. R-S 4.5 R 5. R* S 6. RX S 1 SELECT 8. RRC S.CS H. SELECT INTERSECT 2 3 4 5 6 7 UNION 8 FROM R; FROM R CROSS JOIN S; FROM R NATURAL JOIN S: FROM R JOIN S ON R.C=S.C; FROM R JOIN S ON R.B= S.B; FROM R FROM S: FROM R SELECT FROM S: G H A B D с F E

Database Systems: Design, Implementation, & Management

12th Edition

ISBN:9781305627482

Author:Carlos Coronel, Steven Morris

Publisher:Carlos Coronel, Steven Morris

Chapter8: Advanced Sql

Section: Chapter Questions

Problem 18RQ

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Please answer the remaining questions 4,5!! Please answer this DBMS ques asap! Suppose that we have three relations R1={A, B, C}, R2={A, B, D} and R3={B, C, D} under the relational data model. The relation R1 has 5 tuples, the relation R2 has 6 tuples and the relation R3 has 7 tuples. There are no NULL values in these relations. For each of the following relational algebra expressions, list the minimum and maximum numbers of tuples in the result that the relational algebra expression may have. (4) R1 ⋈ R2(5) ρX (R2) ⋈X.B≠Y.B ρY (R2) You can use the following format to organise your answers. (4) max=____ , min=____ ;(5) max=____ , min=____ ;
Consider two relations Order and Product with the schema Order (Oid, Odate, Pid, qty) and Product (Pid, manufacturer_id, unit_price). i) Can the Oid attribute in the relation Order be a candidate key of the relation? Justify your answer. ii) Can the Pid attribute in the relation Product be a candidate key of the relation? Justify your answer. iii) Find the candidate key for the relation Product. iv) Using the primary keys identified for the above relations write a SQL query that retrieves all tuples of the form (Oid, Odate, Pid, unit_price).
Consider the following relational data model. Employee (empid, ename, address, title, headid) Works (Empid,Deptid,joindate) Department (Deptid, dname, block) Payment (title, salary) A) Write relational algebra for Display the employee name and department name of each employee. Display the employee who works as senior lecturer and belongs to BEI or BCT department. Display the employee name with corresponding department head. B) Write sql query for Creating the above relations; including appropriate versions of all primary and foreign key constraints Find the employee name and address whose address name ends with substring ‘pur’. Update the salary of senior lecturer by 25%.(Solve A - 1,2,3 only)
1.Consider the relation schema given in below fig. Design and draw an ER diagram that capture the information of this schema.(Analyse)(Apr/May 2017)Employee(empno,name,office,age)Books(isbn,title,authors,publisher)Loan(empno,isbn,date)Write the following queries in relational algebra and SQL.Find the names of employees who have borrowed a book published by McGraw-Hill.Find the names of employees who have borrowed all books published by McGraw-Hill.
produce a list of relations indicating the primary key, columns and foreign keys. Consider the following attributes for tables in a relational model designed to keep track of information for a moving company that moves residential customers, usually from one home or apartment to another:customerID, customerName, customerCurrentAddress, customerCurrentPhone, customerNewAddress, customerNewPhone, pickupLocation, dropOffLocation, dateOfMove, startingTime, estimatedWeight, estimatedCost, truck#Assigned, driverName, driverLicNo, actualCost, amountOfDamages, truckCapacity, costOfTolls, tax, finalAmount, invoiceNumber, amountPaid, datePaid, checkNumber, amountDueAssume:- Although in most cases the pickupLocation is the customer’s old address and the dropOffLocation is the new address, there are exceptions, such as when furniture is moved to or from storage.-An estimate is provided before the move using a pre-printed invoice containing a unique invoice number. The actual cost is recorded on…
Please answer this DBMS ques asap! Suppose that we have three relations R1={A, B, C}, R2={A, B, D} and R3={B, C, D} under the relational data model. The relation R1 has 5 tuples, the relation R2 has 6 tuples and the relation R3 has 7 tuples. There are no NULL values in these relations. For each of the following relational algebra expressions, list the minimum and maximum numbers of tuples in the result that the relational algebra expression may have. (1) πA,B(R1)(2) R2 × πB(R3)(3) πB(R2) ∪ πB(R3)(4) R1 ⋈ R2(5) ρX (R2) ⋈X.B≠Y.B ρY (R2) You can use the following format to organise your answers.(1) max=____ , min=____ ;(2) max=____ , min=____ ;(3) max=____ , min=____ ;(4) max=____ , min=____ ;(5) max=____ , min=____ ;
Question- Translate the following ER diagram into the appropriate relational schema(s):
In SQL, foreign key constraints can reference only the primary key attributes ofthe referenced relation or other attributes declared to be a superkey using theunique constraint. As a result, total participation constraints on a many-to-manyrelationship set (or on the “one” side of a one-to-many relationship set) cannotbe enforced on the relations created from the relationship set, using primarykey, foreign key, and not null constraints on the relations. Explain how to enforce total participation constraints using complexcheck constraints or assertions (see Section 4.4.8). (Unfortunately, thesefeatures are not supported on any widely used database currently.)
In SQL, foreign key constraints can reference only the primary key attributes ofthe referenced relation or other attributes declared to be a superkey using theunique constraint. As a result, total participation constraints on a many-to-manyrelationship set (or on the “one” side of a one-to-many relationship set) cannotbe enforced on the relations created from the relationship set, using primarykey, foreign key, and not null constraints on the relations. Explain why

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