A Company has several business units. Each business unit has multiple projects. Employees must be assigned to one business unit. One or more employees are assigned to a project, but an employee may be on vacation and not assigned to any projects. One of the assigned employees will be project manager for that project.

1. Submitted By: Group-3
• Ganesh Pattnayak (13/IT/19)
• Himanshu (13/IT/20)
• Jyotish Kumar Thakur (13/IT/21)
• Krishna Kumar Pankaj (13/IT/22)
• Kritika Raj (13/IT/23)
• Kumar Gaurav (13/IT/24)
• Madhvendra Kumar (13/IT/25)
• Md Absaar Aamir (13/IT/26)
• Md Muktasid Alam (13/IT/27)
Case Study Analysis

2. Entity
Attribute
-> Simple & Composite
-> Single & Multi-valued
Super key, Candidate key, Primary key
Composite key
Cardinality
-> One to one, One to many, Many to many
Partial Participation &Total Participation

3. A Company has several business units.
Each business unit has multiple projects.
Employees must be assigned to one business
unit.
One or more employees are assigned to a
project, but an employee may be on vacation
and not assigned to any projects.
One of the assigned employees will be
project manager for that project.

4. Company
Project
Business Unit
Employee
Comp
-BU
works
for
c-id
add1 add2 add3
b-name
b-loc
b-id
add2
add1
dob
c-name
emp-no
e-name
e-type p-id p-name
p-durphone email
no of hrs
worked

5. c-id c-name add1 add2 add3
Company
b-id b-name b-loc c-id
Business Unit
p-id p-name p-dur b-id
Project
emp-no e-name dob add1 add2 e-type b-id
Employee
R1
R2
R3
R4

6. emp-no phone
Emp-phn
emp-no email
Emp-email
R6R5
emp-no p-id no of hours worked
Emp-proj
R7

7. Attributes must be atomic in nature.
No multi-valued attribute must be present.
->
It is in 1NF
Non Prime attributes fully functionally
dependent on Primary Key

8. It is in 2NF No non prime
attribute is dependent
on another non-prime
attribute
In our Schema,
 R1 has no dependencies b/w npa(c-name,add1,add2,add3)
 R2 has no dependencies b/w npa(b-name,b_loc)
 R3 has no dependencies b/w npa(p-name,p-dur)
 R1 has no dependencies b/w npa(e-name,dob,add1,add2,e-type)

9. • B is a subset of A (trivial dependency)
• A must be the super key
For all functional dependencies such as A -> B
Atleast one of the following conditions must hold:

10. • In our Schema we have already verified 2NF
for R1 to R4,which shows that dependencies
are preserved
All the functional dependencies must be preserved
in our relational schema(R1 to R4)

11. R1 and R2 are said to be lossless join ,if one
of the following conditions hold:
• R1 intersection R2->R1
• R1 intersection R2-> R2
In our Schema,
R1 and R2 are in lossless join(c-id -> R1)
R2 and R3 are in lossless join(b-id -> R2)
R2 and R4 are in lossless join(b-id -> R2)

12. Thank you!!