Absract Description of the College Management System

2. AUTOMATION
MANAGEMENT
SYSTEM

3. Introduction
• Today all the work at the time of admission of the students is done
manually by ink and paper, which is very, slow and consuming much
efforts and time. It is required to Design of Computerized Automated
Management System, to speed up and make it easy to use system. It
reduces the manpower needed to perform the entire admission and
administration task by reducing the paper works needed. The main goal
of the system is to automate the process carried out in the organization
with improved performance and realize the vision of paperless work.

4. Process Model
• The Process Model used in our projects “College Management System” is
waterfall model.
• The Waterfall Model:
• The waterfall model is a sequential design process, used in software
development processes, in which progress is seen as flowing steadily downwards
(like a waterfall) through the phases of Conception, Initiation, Analysis, Design,
Construction, Testing, Production/Implementation and Maintenance.

5. WATERFALL MODEL

6. Software Requirements Specification (SRS)
• It is required to Design of Computerized Automated Management
System, to speed up and make it easy to use system.
Specific Requirements:
• User class and characteristics:
a) Administrator
b) User
Software Interface:
• Operating system: Window XP,Vista,7,8,8.1 and higher
• Platform: .NET
• Database: SQL server
• Language: Visual Studio 2013 (ASP.NET & C#)
Hardware Interface:
• Intel Pentium 4 or higher processor
• 1.5 GHz
• 512MB of RAM or More

7. Data Flow Diagrams (DFD’s)
AMS
Admin User
Report
Get info
Manage Dat a
Gener at e Repor t
Det ails
Level 0 DFD

8. Admin
A uthent i c ati on Login Inf o
St udent File Facult y File
View Info
Generate Repor t
Repor t
User
User ID
And
Passw or d
Delet ing Ent ery
Remove
Editing Entr y
Modif yi ng
Checking ID
Removing Enr ty Updat ing Ent ry
Removing Ent ry
Updat ing Ent ry
Viewing Det ails
Get ting Repor t
Det ails
Repor t Generat ed
Verif ied
St udent info Facult y inf o
Viewing st udent info Viewing Facult y info
Level 1DFD

9. View
st udent
inf o
View
Faculty
Info
Check Per for mance
St udent
Per f or mance
Check Payment Deat ils
Fee Payment
Det ails
View per sonal inf oSt udent Pr of ile
View at t endance
Att endanc e
Wor king Days St atus
Facult y Profile
View Per sonal info
Wor king Days
St udent info
View Info
Facult y inf o
Checking inf o
Level 2 DFD
St udent File
Viewing Att endanc e
Viewing Pr of ile Viewing Fee Det ails
Viewing inf o
Facult y File
Viewing inf o
Det ails
Det ails Det ails
ViewingProfile
Details

10. Use cases
Use case 1: Update an entry of the student.
Primary Actor: Admin
Precondition: Admin has logged in.
Main Success Scenario:
1. Admin checks all the previously filled data.
2. Admin retrieve the student data which is meant to update.
3. Admin updated the selected student data from the database.
4. System confirm the modification.
Exception Scenario:
-2a) There is no such student data, which the searched for.
System shows error message.

11. Use case 2: View Attendance.
Primary Actor: User (Student).
Precondition: User should be student of that college.
Main Success Scenario:
1. Student is asked to fill his roll no. by the software.
2. Now the student’s record displayed on the screen.
3. Student is asked to choose various options (Name, Address, Attendance
etc.).
4. Student choose his attendance.
5. Attendance displayed on the screen.
Exception Scenario:
-1a) Student data is missing.
System shows error message.
-5a) The attendance is not up to date.
No error message from software.

12. FUNCTION POINT
CALCULATION
Weighting Factor
Information
Domain Values
Count Simple Average Complex
External Inputs (EIs) 2 x 3 4 6 6
External Outputs (Eos) 1 x 4 5 7 4
External Inquiries (EQs) 6 x 3 4 6 18
Internal Logical Files (ILFs) 2 x 7 10 15 14
External interface Files (EIFs) 2 x 5 7 10 10
Count Total: 52
Since complexity is simple so,
FP = count total*[0.65 + (0.01*∑ (Fi))]
And project FP is 57.2
By calculating the value adjustment factor ∑ (Fi) = 45,

13. Effort Estimation
• Work effort is the labor required to complete an activity. Work effort is typically the
amount of focused an uninterrupted labor time required to compute an activity.
• FP-based Estimation:
• Decomposition for FP-based estimation focuses on information domain values
rather than software functions.
• FP estimated =57.2
• To derive an estimate of effort on computed FP value, “productivity rate” must be
derived.
• The organizational average productivity rate for system of this type is 6.5 FP/pm.
• An estimate of the project effort is computed using:
• Estimated Effort = FP/PROD
= 57.2/6.5
= 8.8

14. Basis Path Testing
• Basis path testing, or structured testing, is a white box method for designing test cases. The method analyzes
the control flow graph of a program to find a set of linearly independent paths of execution. The method
normally uses cyclomatic complexity to determine the number of linearly independent paths and then
generates test cases for each path thus obtained. Basis path testing guarantees complete branch coverage
(all CFG edges), but achieves that without covering all possible CFG paths—the latter is usually too costly.
Basis path testing has been widely used and studied.
• To measure the logical complexity of our software we consider the following procedure:
• void view_info(){
• cout<<"Select option: n";
• cout<<"1.Student info.n";
• cout<<"2.Faculty info. n";
• char ch;
• cin>>ch;
• if(ch==1){
• cout<<"Select option: n";
• cout<<"1.Student Profile.n";
• cout<<"2.Student Performance.n";
• cout<<"3.Attendance.n";
• cout<<"4.Fee details.n";
• char ch;
• cin>>ch;
• if(ch==1){
• cout<<"Student Profile: n";
• obj.profile();
• }else if(ch==2){
• cout<<"Student Performance: ";
• obj.perfrm();
1
2
3
4
5
6
7

15. • }else if(ch==3){
• cout<<"Attendance: ";
• obj.attendance();
• }else{
• cout<<"Fee Details: ";
• obj.pay();
• }
• }else{
• cout<<"Select option: n";
• cout<<"1.Profilen";
• cout<<"2.Working Daysn";
• char ch;
• cin>>ch;
• if(ch==1){
• cout<<"Profile";
• obj.profile();
• }else{
• cout<<"Working Days";
• obj.wday();
• }
• }
• }
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10
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14
15

16. 1
2
3
4
6
5
7
8
11
12
1314
15
9
10
Cyclomatic complexity: Cyclomatic complexity V (G)
for a flow graph G is defined as
V (G) =E-N+2 = 19-15+2 = 6
So that no. of the independent path is 6.
Path 1: 1-2-3-4-5-15
Path 2: 1-2-3-4-6-7-15
Path 3: 1-2-3-4-6-8-9-15
Path 4: 1-2-3-4-6-8-10-15
Path 5: 1-2-11-12-13-15
Path 6: 1-2-11-14-15

17. THANK YOU