Retrieve the names of all employees in department 5 who work more than 10 hours per week on the ‘ProductX’ project. p10ssn ← (Π essn (σ hours > 10 (works-on |X| pno = pnumber(Π pnumber (σ pname = ‘ProductX’ (project)))) Π lname, minit, fname (σ dno = 5 (employee |X| ssn = essn (p10ssn))) select lname, minit, fname from employee, works-on, project where ssn = essn AND pno = pnumber AND dno = 5 AND hours > 10 AND pname = ‘Product X’;

select lname, minit, fname from employee, dependent List the names of all employees who have a dependent with the same first name as themselves. Π lname, minit, fname (employee |X| fname = dependent-name AND ssn = essn (dependent)) select lname, minit, fname from employee, dependent where fname = dependent-name AND ssn = essn;  

Find the names of all employees who are directly supervised by ‘Franklin Wong’. Π lname, minit, fname (employee |X| superssn = ssn (Π ssn (σ fname = ‘Franklin’ AND lname = ‘Wong’ (employee)))) select employee.lname, employee.minit, employee.fname from employee emp where emp.lname = ‘Wong’ AND emp.fname = ‘Franklin’ AND employee.superssn = emp.ssn;  

For each project, list the project name and the total hours per week (by all employees) spent on that project. pname F sum hours (project |X| pnumber = pno (works-on)) select pname, sum(hours) from project, works-on where pno = pnumber group by pname;  

select dname, average(salary) from department, employee For each department, retrieve the department name and the average salary of all employees working in that department. dname F average salary (department |X| dnumber = dno (Π dno, salary, ssn (employee))) select dname, average(salary) from department, employee where dnumber = dno group by dname;  

Retrieve the average salary of all female employees. F average salary (σ sex = ‘F’ (employee)) select average (salary) from employee where sex = ‘f’;