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Employee: (EID, Name, City, Address) Department: (DID, Department_Name, City, budget) Works IN: (EID, DID, Annual_Salary, Start_Date) Manages: (ManagerID, DepartmentID, rank) DID and EID in Works_IN relation are foreign keys referencing the primary keys of the Department and Employee relations respectively. Manager_ID and Department_ID in the Manages relation are foreign keys referencing the primary keys of the Employee and Department respectively. From the relational schemas illustrated above, use the different relational algebra operators and operations (p, o, 7, n, U, -, A) discussed in class to build the following queries: 1. Display employees (by name and address) from “Chicago" and with an annual salary between 60000 (included) and 90000 (excluded) then rename the output as "ChicagoEmployees" 2. For each employee, who works in the “Design" department since January 2009, display the employee's name and the employee's annual salary 3. Find the name and rank of the manager who manages the “Research" department, also display the budget of that department. 4. Find the name of employees who earn more than $100,000 per year, along with the names of the departments they work in 5. Assume employees can work in more than one department, find employees (by name and salary) who work in the "Design" department but not the "Research" department.

Employee: (EID, Name, City, Address) Department: (DID, Department_Name, City, budget) Works IN: (EID, DID, Annual_Salary, Start_Date) Manages: (ManagerID, DepartmentID, rank) DID and EID in Works_IN relation are foreign keys referencing the primary keys of the Department and Employee relations respectively. Manager_ID and Department_ID in the Manages relation are foreign keys referencing the primary keys of the Employee and Department respectively. From the relational schemas illustrated above, use the different relational algebra operators and operations (p, o, 7, n, U, -, A) discussed in class to build the following queries: 1. Display employees (by name and address) from “Chicago" and with an annual salary between 60000 (included) and 90000 (excluded) then rename the output as "ChicagoEmployees" 2. For each employee, who works in the “Design" department since January 2009, display the employee's name and the employee's annual salary 3. Find the name and rank of the manager who manages the “Research" department, also display the budget of that department. 4. Find the name of employees who earn more than $100,000 per year, along with the names of the departments they work in 5. Assume employees can work in more than one department, find employees (by name and salary) who work in the "Design" department but not the "Research" department.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Employee: (EID, Name, City, Address) Department: (DID, Department_Name, City, budget) Works_IN: (EID, DID, Annual_Salary, Start_Date) Manages: (ManagerID, DepartmentID, rank) DID and EID in Works_IN relation are foreign keys referencing the primary keys of the Department and Employee relations respectively. Manager_ID and Department_ID in the Manages relation are foreign keys referencing the primary keys of the Employee and Department respectively. From the relational schemas illustrated above, use the different relational algebra operators and operations (p, 6, n, n, U, -, A) discussed in class to build the following queries: Display employees (by name and address) from "Chicago" and with an annual salary between 60000 (included) and 90000 (excluded) then rename the output as "ChicagoEmployees" 1. 2. For each employee, who works in the "Design" department since January 2009, display the employee's name and the employee's annual salary 3. Find the name and rank of the manager who manages the "Research" department, also display the budget of that department. Find the name of employees who earn more than $100,000 per year, along with the names of the departments they work in 4. 5. Assume employees can work in more than one department, find employees (by name and salary) who work in the "Design" department but not the "Research" department.
Transcribed Image Text:Employee: (EID, Name, City, Address) Department: (DID, Department_Name, City, budget) Works_IN: (EID, DID, Annual_Salary, Start_Date) Manages: (ManagerID, DepartmentID, rank) DID and EID in Works_IN relation are foreign keys referencing the primary keys of the Department and Employee relations respectively. Manager_ID and Department_ID in the Manages relation are foreign keys referencing the primary keys of the Employee and Department respectively. From the relational schemas illustrated above, use the different relational algebra operators and operations (p, 6, n, n, U, -, A) discussed in class to build the following queries: Display employees (by name and address) from "Chicago" and with an annual salary between 60000 (included) and 90000 (excluded) then rename the output as "ChicagoEmployees" 1. 2. For each employee, who works in the "Design" department since January 2009, display the employee's name and the employee's annual salary 3. Find the name and rank of the manager who manages the "Research" department, also display the budget of that department. Find the name of employees who earn more than $100,000 per year, along with the names of the departments they work in 4. 5. Assume employees can work in more than one department, find employees (by name and salary) who work in the "Design" department but not the "Research" department.
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Step 1

All the above details that are mentioned above in the question we will take to solve the question:

so after taking all the input from the question here is the solution:-

//person.java:

public class person {

private String name;

private int age;

person(String name, int age){

this.name = name;

this.age = age;

}

public String getName() {

return name;

}

public void setName(String n) {

name = n;

}

public int getAge() {

return age;

}

public void setAge(int a) {

age = a;

}

@Override

public String toString() {

return "Name: " + name + "\nAge: " + age;

}
}

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