Prepared by

Manish Sharma Manish Kumar Kushwaha

AUTOMATION

MANAGEMENT SYSTEM

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.

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.

WATERFALL MODEL

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

Data Flow Diagrams (DFD’s)

AMS

Admin User

Report

Get infoManage Data

Generate Report

Details

Level 0 DFD

Admin

Authentication Login Info

Student File Faculty File

View Info

Generate Report

Report

User

User ID And

Password

Deleting Entery

Remove

Editing Entry

Modifying

Checking ID

Removing Enrty Updating EntryRemoving Entry

Updating Entry

Viewing Details

Getting Report

Details

Report Generated

Verified

Student info Faculty info

Viewing student info Viewing Faculty info

Level 1 DFD

View student

info

View Faculty

Info

Check Performance

Student Performance

Check Payment Deatils

Fee Payment Details

View personal info Student Profile

View attendance

Attendance

Working Days Status

Faculty Profile

View Personal info

Working Days

Student info

View Info

Faculty info

Checking info

Level 2 DFD

Student File

Viewing Attendance

Viewing Profile Viewing Fee Details

Viewing info

Faculty File

Viewing info

Details

Details Details

Vie

win

g P

rofile

Details

Use casesUse 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.

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.

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,

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

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

45

67

• }else if(ch==3){

• cout<<"Attendance: ";

• obj.attendance();

• }else{

• cout<<"Fee Details: ";

• obj.pay();

• }

• }else{

• cout<<"Select option: \n";

• cout<<"1.Profile\n";

• cout<<"2.Working Days\n";

• char ch;

• cin>>ch;

• if(ch==1){

• cout<<"Profile";

• obj.profile();

• }else{

• cout<<"Working Days";

• obj.wday();

• }

• }

• }

8

9

10

11

12

13

14

15

1

2

3

4

6

5

78

11

12

1314

15

910

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

THANK YOU