Question:Consider the following relations. Employee (eid, ename, salary, age): 10,000 tuples Work_On (eid, pid): 200,000 tuples Project (pid, pname,...

Question:Consider the following relations.

Employee (eid, ename, salary, age): 10,000 tuples

Work_On (eid, pid): 200,000 tuples

Project (pid, pname, location, mgr_id): 1,000 tuples 

We assume that the size of each column is 50 bytes. The size of disk block is 4KB.

 Each tuple in Employee matches with 20 tuples in Work_On on average. 

 Each tuple in Project matches with 200 tuples in Work_On on average.

 The column Employee.eid has the secondary Index with index level = 2

 The column Work_On.eid has the secondary index with index level = 4.

 The column Work_On.pid has the secondary index with index level = 4.

 The column Project.pid has the secondary index with index level = 1. 

Assume that all the operators are pipelined.

Given the Relational Algebra (Employee ⋈ Work_On) ⋈ Project, compute I/O costs for the following 

join methods.

1) (2 points) Block nested loop join (52 buffer pages) 

2) (2 points) Index based nested loop join

3) (2 points) Index based nested loop join (with the assumption that the Work_On table is clustered 

based on eid)

Given the Relational Algebra (Project ⋈ Work_On) ⋈ Employee, compute I/O costs for the following 

join methods.

4) (2 points) Block nested loop join (52 buffer pages) 

5) (2 points) Index based nested loop join (with the assumption that the Work_On table is clustered

based on pid)