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College Management System Project



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College Management System Project

  2. 2. 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.
  3. 3. 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. 5. 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
  6. 6. Data Flow Diagrams (DFD’s) AMS Admin User Report Get info Manage Dat a Gener at e Repor t Det ails Level 0 DFD
  7. 7. 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
  8. 8. 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
  9. 9. 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.
  10. 10. 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.
  11. 11. 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,
  12. 12. 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
  13. 13. 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
  14. 14. • }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(); • } • } • } 8 9 10 11 12 13 14 15
  15. 15. 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
  16. 16. THANK YOU