Tuesday, August 21, 2007

Assignment Sollution MCS-035

Q1 : Consider developing a system for Tickets Reservation for Commonwealth Games and perform the followings:
(a) Suggest the most appropriate Software Engineering model for developing this project with appropriate justification.
(b) Write at least two functional and non functional requirements.
(c) Develop a test plan for the system. You can make necessary assumptions and specify them.
(d) Write the risk management plan for the system.
(e) Estimate the efforts of software projects. Make necessary assumptions.
(f) Suppose it was revealed that the poor knowledge of the tool is responsible for the problems that are being encountered for timely completion of the project. Write type of remedies do you suggest for such type of problem ? justify your answer.
(g) Suppose there exists some old systems and wants to replace it, suggest the changes with respect to the software and hardware requirements.
(h) Describe what types of testing strategies/ techniques can be followed for a web based Ticket Reservation System.
(i) Develop a design review plan for the system. Also, list the deficiencies, if any, in the SRS for the same.
Ans :- 1 a
(a) The Linear sequential model re waterfall model is appropriate for Tickets Reservation system. It suggest a systematic, sequential approach to software Develop that begins at the system level and progresses through analysis, design, Coding and testing and support. These steps can be applied in Ticket reservation system in the following manner :

Software requirement analysis : to understand the nature of the program to be Built, the software engineer must understand the information domain for the Software, as well as the required function, behavior, performance and interface.

In a Ticket reservation system, various requirements of the whole system are Viewed. It is determined what are the requirements of the system, who are the users of the system, what are various functions in the system etc. then ER – diagram, Data flow diagram etc are prepared.

Design : Software design is actually a multistep process that focuses on four distinct attributes of a program : data structure, software architecture, interface representations and procedural detail.

Code generation : The code generation the design into machine readable form. If design performed in a detailed manner, code generation can be accomplished mechanically.

Testing : Once the code has been generated, program testing begins. The testing process focuses on the logical internals of the software, ensuring that all statements have been tested and on the functional externals.

Support : Software will undoubtedly undergo change after it is delievered to the customer. Software supports reapplies each of the preceding phases to an exiting program rather than a new one.

The linear sequential model is appropriate for Tickets Reservation system as it follows all the steps necessary for the development of Ticket reservation system. In this system project planning can be made in advance and Ticket reservation system require comparatively less changes at later stage. Moreover in a Ticket reservation system most of the requirements can be stated in earlier stages. It provides templates for which method for analysis, design coding, testing and support can be placed. There is less need to make changes after the system is developed. Therefore waterfall model is appropriate for tickets reservation system.

Ans 1 (b)
A collection of elements which are assembled to fulfill some defined purpose. Elements may be hardware or software components, organizational policies and procedures and operational processes.
Systems have properties which are emergent i.e. they only come to light when the parts are put together, they have structure and mechanisms for communication and control.

Important emergent properties of a system are :-
Ø Performance
Ø Reliability
Ø Safety
Ø Security
Ø Usability
These are non functional properties they do not relate to any specific functionality of the system. Some or all of these properties are usually more important than detailed system functionality,
Physical environment
What are the characteristics of the physical environment in which the system will be installed, what other systems must coexist with it ?
User environment
Will the system be used by training or untrained users; will it be used as part of an individual or a group process ?
Ans 1c
Test plan for student Admission System (SAS) :-
We consider “Logical Module for our purpose :- Since this is a human & Computer interface the number of test associated can be estimate by :
Number of transitions contained in the state transition representation
Number of data object that move across the interface.
number of data element that input or output
As we were testing only one module so unit testing will be performed on the module.
White Box Testing fundas will be applied during modules testing like :-
Module interface test to check proper data in flaw and outflow like username and passport is properly accepted.
Local data structure which is common source of error.
Boundary conditions testing.
Independent path testing.

All error handling paths are tested

Ans 1 (d)
Risk management is the process of measuring or assessing risk and developing strategies to manage it. Strategies include transferring the risk to another party, avoiding the risk, reducing the negative effect of the risk, and accepting some or all of the consequences of a particular risk. Traditional risk management focuses on risks stemming from physical or legal causes (e.g. natural disasters or fires, accidents, death, and lawsuits). Financial risk management, on the other hand, focuses on risks that can be managed using traded financial instruments.

In ideal risk management a prioritization process is followed whereby the risks with the greatest loss and greatest probability of occurring are handled first, and risks with lower probability of occurrence and lower loss are handled later. In practice the process can be very difficult, and balancing between risks with a high p Risk management also faces difficulties allocating resources. This is the idea of opportunity cost. Resources spent on risk management could have been spent on more profitable activities. Again, ideal risk management minimizes spending while maximizing the reduction of the negative effects of risks.

Establish the context
Establishing the context involves
Panning the remainder of the process.
Mapping out the following the scope of the exercise, the identity and objectives of stakeholders, and the basis upon which risks will be evaluated.
Defining a framework for the process and an agenda for identification.
Developing an analysis of risk involved in the process.
After establishing the context, the next step in the process of managing risk is to identify potential risks. Risks are about events that, when triggered, cause problems. Hence, risk identification can start with source of problems, or with problem itself.

Source analysis Risk sources may be internal or external to the system that is the target of risk management. Examples of risk sources are : stakeholders of a project, employees of a company or the weather over an airport.
Problem analysis Risk are related to identify threats. For example : the threat of losing money, the threat abuse of privacy information or the threat of accidents and casualties. The threats may exist with various entities, most important with shareholder, customers and legislative bodies such as the government.
Decide on the combination of methods to be used for each risk. Each risk management decision should be recorded and approved by the appropriate level of management . for example, a risk control implementation and responsible persons for those actions . the risk management concept is old but is still not very effectively measured.


Follow all of the planned methods for mitigating the effect of the risks. Purchase insurance policies for the risks that have been decided to be transferred to an insurer, avoid al risks that have been decided to be transferred to an insurer, avoid all that can be avoided without sacrificing the entity’s goals, reduce others, and retain the rest.

Review and evaluation of the plan
Initial risk management plans will never be perfect. Practice, experience, and actual loss results will necessitate changes in the plan and contribute information to allow possible different decisions to be made in dealing with the risks being faced.

Risk analysis results and management plans should be updated periodically. There are two primary reasons for this :
1. to evaluate whether the previously selected security controls are still applicable and effective, and
2. to evaluate the possible risk level changes in the business environment. For example, information risks are a good example of rapidly changing business environment.

To properly adopt software Inspections practices, each participant is trained in the structured review process, defined roles of participants, system of process and product checklists, and forms and reports. The lost opportunity cost to acquire the knowledge, skills, and behaviors is twelve hours per practitioner. In addition, each manager is trained in the responsibilities for rolling out the technology and the interpretation and use of measurements taken. The management training is accomplished in four hours.

The cost of performing software Inspections includes the individual preparation effort of each participant before the session and the conduct effort of participants in the inspections session. Typically, 4-5 people participate and expend 1-2 hours of preparation and 1-2 hours of conduct each. This cost of 10 to 20 hours of total effort per session results in the early detection of 5- 10 defects in 250-5—lines of new development code or 1000-1500 lines of legacy.

Ans 1 (f)
The knowledge of all tools required white making of project. Review of the project analysis before starting the project. Before starting the project. Before starting the making project software engineer should the person who is going to make is well aware that the all tools which required making the project. Cost estimated should be under the control. Time limitation should be there so that software will complete on time.

Ans :- 1 (g)
Software Requirements :-
Operating System.
Ms office
The required s/w which we make in platform the s/w

Hardware Requirement :-
Inter P1v Dual core processor
Intel p 1 V Motherboard
Ram 512 MB
ATX Cabinet
Optical Mouse

Ans 1 (h)
When components are being connected to create large components they have to pass through integration testing, whose main purpose is to detect any inconsistency between the connected components. Once the integration test is completed , a component test will have to be performed on the new component, consisting of several smaller ones. Some properties, however, can be said to belong to the system as a whole, like certain quality attributes. The system, beings a huge component, will hence be tested in its entirety to verify that these requirements are met in a satisfactory way, and this is what is referred to a system testing. Starting with unit testing, then moving on to integration testing before component testing can be performed. Finally, a system test is executed.

Black Box and White Box Testing
When a tester wants to test an application or a particular part of it to detect bugs, he can look at the system from a user’s perspective and expose it to different types of input, and then check whether or not the resulting output is in accordance with the specification. This is know as functional testing or black box testing. With this type of testing, any type of error can be detected, but it would take an infinite amount of time to do so.

Distinctions and elements of Web Applications
15 types of error and refers to memory leaks as a typical example of an error that will escape functional testing.

Web server
Client Database server
Application code
White box
Black box
Application data

Structural testing or white box testing, on the other hand, implies studying the program code and testing the different parts of it. This type of testing is not capable of finding all kinds of errors, but to its advantage it is easier to determine when you have tested enough
That white box testing is capable all lines of code when given an finite amount of time, but points out that such testing might not uncover errors with respect to component integration Seem to contradict each other. In spite of this, our interpretation suggests that both of them find a combination of white box and black box testing.

Ans 1 (i)
Software Inspections are a disciplined engineering practice for detecting and correcting defects in software artifacts, and preventing their leakage into filed operations

Experienced software practitioners and managers understand that software development is a process of experimentation involving the continuous discovery of technical information associated with the function, form, and fit of the software product. Software Inspections are an integral practice in the process of experimentation.

Software inspections provide value in improving reliability, availability, and maintainability

Technical Detail
Software Inspections are strict and close examinations conducted on requirements, specifications, architectures, designs, code, test plan and procedures, and other artifacts. Leading software indicators of excellence for each artifact type provide the exit criteria for the activities of the software life cycle. For example, these indicators include completeness, correctness, style, rules of construction, and multiple views.

Completeness is based on traceability of the requirements to the code, essential for maintainability. Correctness is based on the clear specification of intended function and its faithful elaboration in code, essential for reliability and availability. Style is based on consistency of resorting, essential for maintainability. Rules of construction are based on the software application architecture and the specific protocols, templates, and conventions used to carry it out, essential for reliability and availability. Multiple views are based on the various perspectives and viewpoints required to be reflected in the software product, essential for maintainability. By detecting defects early and preventing their leakage into subsequent activities, the need for later detection and work (which is essential for reduced cycle time and lower cost ) is eliminated.

Software inspections are a reasoning activity performed by practitioners playing the defined roles of moderator, recorder, reviewer, reader, and producer. Each role carries with it the specific behaviors, skills, and knowledge needed to achieve the expert practice of Software Inspections.

The adoption of software Inspections practice is competency enhancing and meet little resistance among practitioners trained in their use. The adopting organization benefits by improved predictability in cost and schedule performance, reduced cost of development and maintenance, reduced defects, in the filed, increased customer satisfaction, and improved morale among practitioners.

In order for Software Inspections to be systematically used in statistical process control, there must be a life cycle model with defined software artifacts. In this context, software Inspections provide the exit criteria for each life cycle activity. Furthermore, the standard of excellence of leading indicators for each type of artifact must be specified and used in practice.

Q2 : (a) You are browsing a web based application but it is taking too much to open, list
Any five reasons for the same.

Ans 2(a)
To many temporary file can lead the show speed.
Bad Connection.
Pink Break
Site is running slowly for everyone.
Site has too many views so it is being bogged down.
Server Busy
Small bandwidth
RAM size small
Client System busy.
Internet speed slow.

(a) Do you anticipate any situation where usage of Clean room Software Engineering for application development is not appropriate ? Explain your answer.

Clean room software engineering is an engineering and managerial process for like development of high quality software with certified reliability. The name “Clean room” was taken from the electronics industry, where a physical clear room exists to prevent introduction of defects during hardware fabrication. Clean room software engineering reflects the same emphasis on defect prevention rather than defect removal , as well as certification of reliability for the intended environment of use.

The focus of Clean room involves moving from traditional, craft based software development practices to rigorous, engineering based practices. Clean room software engineering yields software that is correct by mathematically sound design, and software that is certified by statistically valid testing. Reduced cycle time results from an incremental development strategy and the avoidance of rework.

Clean room reduces the cost of errors during development and the incidence of failures during operation; thus the overall life cycle cost of software developed under Clean room can be expected to be far lower than industry average.
The following ideas form the foundation for clean room based development :
Incremental development under statistical quality control (SQC). Incremental development as practiced in Clean room provides a basis for statistical quality control of the development process. Each increment is a complete iteration of the process, and measures of performance in each increment are compared with reestablished standards to determine whether or not the process is in control. If quality standards are not met, testing of the increment ceases and development return to the design stage.

Software development based on mathematical principles. In clean room development a key principle is that a computer program is an expression of a mathematical function. The Box Structure method is used for specification and design and functional verification is used to confirm that the is a correct implementation of the specification.

Software testing based on statistical principal. In clean room, software testing is viewed as a statistical experiment. A representative subset of all possible uses of the software is generated, and performance of the subset is used as a basis for conclusion about general operational performance.

Usage considerations
Clean room has been documented to be very effective in new development and reengineering (whole system or major subunits) contexts. The following discussion highlights areas where Clean room affects or differs from more conventional practice :

Team based development. A Clean room project team is small typically six to eight persons, and works in a disciplined fashion to ensure the intellectual control of work in progress. Clean room teamwork involves peer review of individual work, but does not supplant individual creativity. Once the system architecture has been established and the interfaces between subunits have been defined, individuals typically work alone on a given system component. Individual designs are working drafts that then reviewed by the team. In a large project, multiple small teams may be formed, with one for the development of each subsystem, thus enabling concurrent engineering after the top level architecture has been established.

Time allocation across life cycle phases. Because one of the major objectives of clean room is to prevent errors from occurring, the amount of time spent in the design phase of a clean room development is likely to be greater than the amount of time traditionally development to design. Clear room, however, is not a more time consuming development methodology, but its greater emphasis on design and verification often yields that concern. Management understanding and acceptance of this essential point that quality will be achieved by design rather than through testing must be reflected in the development schedule. Design and verification will require the greatest portion of the schedule. Testing may being later and be allocated less time than is ordinarily the case. In large clean room projects, where historical data has enabled comparison of traditional and clean room development schedules, the clean room schedule has equaled or improved upon the usual development time.

Costs and Limitations
Using clean room to accomplish piecemeal, isolated changes to a system not development using clean room is not considered an effective use of this technology. Training is required and commercially available. Available courses range from overviews to a detailed focus on particular aspects of clean room . for some training classes, it is most productive if software manager and technical staff take the training together. Manager need a thorough understanding of clean room imperatives, and a core group of practitioners needs sufficient orientation in clean room practices to be able to adapt the process to the local environment (this includes establishing a local design language, local verification standards, etc.)

2 (c) for what types of applications is the approach of Re Engineering applied ? Explain
your answer.

“Reengineering” is the fundamental rethinking and radical redesign of business professes to achieve dramatic improvements in critical, contemporary measures of performance, such as cost, quality service, and speed.

Seven principles of reengineering
1. Organize around outcomes not tasks.
2. Identify all processes in an organization and prioritize them in order of redesign urgency.
3. integrate information processing work into the real work that produces information.
4. Tread geographically dispersed resources as though they are centralized.
5. Link parallel activities in the workflow instead of just integrating their results.
6. Put the decision point where the work is performed, and build control into the process.
7. Capture information once and at the source.

The Benefits of Reengineering
The hard task of re examining mission and how it is being delivered on a day to day basis will have fundamental impacts on an organization, especially in terms of responsiveness and accountability to customers and stakeholders. Among the many rewards, reengineering :
Empowers employees
Eliminates waste, unnecessary management overhead, and obsolvte or inefficient processes.
Produces often significant reductions in cost and cycle times.
Enables revolutionary improvement in many business processes as measured by quality and customer service
Helps top organizations stay on top and low achievers to become effective competitors.
The Benefits of Reengineering
The hard task of reexamining mission and how it is being delivered on a day to day basis will have fundamental impacts on an organization, especially in terms of responsiveness and accountability to customers and stakeholders. Among the many rewards, reengineering.

Empower employees
Eliminates water, unnecessary management overhead, and obsolete or inefficient processes.
Enables revolutionary improvements in many business processes as measured by quality and customer service.
Help top organizations stay on top and low achievers to become effective competitors.
DOD has suggested that the following six tasks be part of any functional management approach to reengineering projects:
Define the framework . define functional objectives; determine the management strategy to be followed in streamlining and standardizing processes; and establish the process, data and information systems baselines from which to being process improvement.
Analyze. Analyze business process to eliminate non value added processes; simplify and streamline processes of little value; and identify more effective and efficient alternatives to the process, data, and system baselines.
Evaluate. Conduct a preliminary, functional, economic analysis to evaluate alternatives to baseline processes and select a preferred course of action.
Plan. Develop detailed statement of requirements, baseline impacts, costs, benefits, and schedules to implement the planned course of action.
Approve. Finalize the functional economic analysis using information from the planning data, and present to senior management for approval tp proceed with proposed process improvement and any associated data or system changes.
Execute. Execute the approved process process and data changes, and provide functional management oversight of any associated information system changes.

Wednesday, August 15, 2007

Assignment Sollution ( MCS-035 )


Create the necessary Ledger and Groups to maintain the accounts as follows:
(Maintain Company as “Accounts only”)
Balance Sheet for the Year Ending 31-03-02
Test1 (User’s name)

Details of Creditors bill wise:

Yahoo India
i) Bill No: Yahoo/001/390
Rs. 25000
Dated 01-10-01
ii) Bill No: Yahoo/001/391
Rs. 45000
Dated 01-11-01

Gama India
i) Bill No: Gama/002/400
Rs. 17250
Dated 02-09-01
ii) On Account:
Rs. 12750

Square India
i) Bill No: Square/003/500
Rs. 20000
Dated 02-12-01

Details of Debtors bill wise:

BNP Traders
i) Bill No: Bnp/001/401
Rs. 25000
Dated 01-01-02
ii) Bill No: Bnp/001/402
Rs. 15000
Dated 31-03-02

Electronic India
i) Bill No: Elec/002/301
Rs. 30000
Dated 02-02-02

Smart India
i) Bill No: Smart/003/401
Rs. 10000
Dated 02-08-01
ii) Bill No: Smart/003/402
Rs. 35,000
Dated 02-10-01
iii) On Account
Rs. 15,000

Set the credit days for all debtors is 45 days.

Session 2
Open the Test1 Company and now onwards maintain the Company as “Accounts with Inventory”. Go to Company Feature (F11) > turn on the option “Allow Integrate with Accounts”. Put the details of the following products

Note: After putting the opening balances of products see the final accounts.

Session 3

Open “Test 1” Company and pass the following Transactions:
1. General expenses has been paid Rs. 2500 in cash dated 01-04-02
2. Outstanding salary paid Rs. 8000 dated on 02-04-02 from UTI, cheque no. 740521
3. Stationary paid Rs. 1200 dated on 02-04-2 by cash (Category: stationary, Centre: Paper, Pencil, Rubber; amount will be equally distributed among the centers)
4. Rs. 20,000 paid to Yahoo India against their Bill No. Alfa/001/390 by cheque issued from UTI on 01-05-02 cheque no. 740522
5. Smart India paid Rs. 50,000 for their Bill Nos, Smart/003/401, Smart/003/402 respectively and rest from on account balance on 01-05-02 cheque deposited into SBI Bank (Cheque No. 239309)
6. Machinery has been purchased Rs. 35000 from ABC Ltd. Dated on 02-05-02, Bill no. Abc/004/001
7. Cash withdrawn from SBI Rs. 10,000 dated on 02-05-02 cheque no. 150934.
8. Amount paid Rs. 28,000 to Gama Ltd. against their full dues through SBI cheque no. 150935 dated on 01-06-02
9. Cash deposited into UTI Rs. 13000 dated on 01-06-02.
10. Full amount paid to SquareLtd. with a discount @ 5% dated on 02-06-02.
11. General expenses paid Rs. 1200 from cash dated on 02-07-02 (category: Mis. Exp., Center: Postage, Tiffin divide the amount 40% and 60% respectively).
12. Furniture purchased from Raj & Raj. Rs. 15500 with sales tax @ 10% and freight Rs. 230 on 02-07-02 Bill no. Raj/005/001
13. Rs. 27500 received from Electronic India against their full dues and deposited into UTI on 02-07-02; cheque no: 450987.
14. Paid Rs. 48500 to Yahoo India against Bill no. Alfa/001/390 and Alfa/001/391 through cheques of UTI Rs. 39500 (ch. No: 740523) and SBI Rs. 9000 (ch no.: 150936), rest part of the bill will be treated as discount on 02-07-02.


Open the Test 1 Company and enter the following transactions:

Turn off Cost Centre allocation from Sales and Purchase account.

1. Placed a purchase order to TVS & Co. for 15 pcs. Keyboard (Group: Hardware, Category: Input/Output) on 02-07-02: Order no: Tvs/02-03/001
Note: Use “Allow 0 valued entries in Voucher” from Company Feature (F11)
2. Received 10 pcs. Keyboard from TVS & Co. against the order (Tvs/02-03/001) on 01-08-02, Challan no. Tvs/ch/001 store at Malda godown.
3. Received another challan against it order Tvs/02-03/001 for the ret goods from TVS & Co. on 02-08-02 and store it at same godown. Challan no. Tvs/ch/002.
4. Received an invoice against challan nos. Tvs/ch/001 & Tvs/ch/002, with rate @ 750 including W.B.S.T. @ 10% on 01-09-02 bill no. Tvs/006/001
5. Purchased 10 pcs Mouse (Group: Hardware Category: Input/Output) @ Rs. 2.30 from TC Ltd. and get another 2 pieces as free samples on 01-09-02, keep it within Malda godown (Bill No.- Tc/007/001)
6. Returned 5 Pcs. Keyboard to TVS Co. on 02-09-02. (Against Bill no. Tvs/006/001, Debit note no. Tvs/006-ret/001)
Note: W.B.S.T @ 10% will be adjusted with return
7. On 02-09-02 the full dues paid to TVS & Co. with 5% discount.


Do the following entries:
1. Sales order received from Milind Agency for 8 pcs. Keyboard on 02-09-02; Order no. Mil/02-03/001
2. Opened a new Bank A/c, City Bank A/c with Rs. 10000 on 01-09-02.
3. 5 Pcs Keyboard delivered from Malda godown to Milind Agency against order no. Mil/02-03/001 on 01-10-02 (challan no. Mil/ch/001).
4. The rest goods against order no Mil/02-03/001 delivered from the same godown on the next day (challan no. Mil/ch/002).
5. Earned 4% interest on IDBI investment for last six months on 01-10-02.
6. Furniture purchased for office use on 01-10-02 of Rs. 5000 from Furnishes & Co. (Bill no. Fur/008/001).
7. Returned 2 pcs Keyboard by Milind Agency from their first challan due to manufacturing defect on 01-10-02 (Ref. no. Mil/004-ret/001) (Use Rejection In). Immediately send the Keyboards to TVS & Co. for future replacement.
Note: Use Stock Journal to give the effect as Stock Outward.
8. Send a bill (Bill no. Mil/004/001) against the pending challans Mil/ch/001 & Mil/ch/002 @ 900 each with the following billing terms on 02-11-02
Trade Discount 2%
Sales Tax 10%
Delivery Charge Rs. 120
9. Paid advertisement bill Rs. 1000 from new City bank (ch. No. 920531) on 01-12-02.
10. Received from Milind Agency 50% due amount against the bill no. Mil/004/001 through cheque (ch. No. 110234) on 01-12-02 and deposited to City bank on the same day.
11. Transfer 12 pcs Mouse from Malda godown to Saithia godown with current stock rate on 01-12-02.
Prepare the following Voucher Classes only for sales voucher:
1. Name of voucher sales: Sales-WBST @ 5%
Allow 95% of sale value to Sales A/c and 5% of sale value to Sales Tax A/c.
2. Name of voucher class: Sales- WBST @ 4% Sur@ 1%
Allow 100% of Sale value to Sales A/c
Additional charges will be as follows:
Sale Tax- @4% on total sales
Surcharge- @1% as Surcharge
Freight (Sale)- Rs. 5 based on quantity.
3. Name of Voucher class: Sales-No tax
No additional charges will be added.
4. Do the following entries appropriate Voucher Class
i. Sold 3 Pcs. Floppy Drive @1200, 10 Pcs Picture Tube @ 2500 to Pinak & Co. with Sales Tax 4%, Surcharge 1% and Freight Rs. 5 per quantity from Malda godown on 01-12-02 (Bill no. Pnk/005/001, order/challan is not required).
ii. Sold 500 Ltr Pepsi (300 ml) @22 including Sales Tax 5% to Cola House on 01-12-02 from Saithia godown (Bill no. Cola/006/001).
iii. Sold 800 Ltr Cock (1.5 ltr) @23 to Pine & Co. on 01-12-02 from Malda godown. (Bill no.Pine/007/001)
iv. 300 Ltr Cock (1.5 ltr) distributed as free among school students to get the market on 01-12-02 from Malda godown.
v. BNP Traders paid Rs. 39000 for full settlement of their dues through cheque (Ch. No. 230897) on 01-12-02 and the cheque deposited into SBI on the next day.
vi. Paid Rs. 35000 to ABC Ltd. against Bill no. ABC/004/001 from SBI (ch no. 150937) on 01-12-02.
vii. Raj & Raj charged 5% interest p.a. on their due amount on 01-12-02.
viii. Cola house returned 100 Ltd. Pepsi (300 ml) at Saithia godown on 02-12-02 (Bill no. Cola/006-ret/001), sales tax will be adjusted accordingly.


1. Purchased the following items from Gama Ltd. with WBST @ 5% on 01-01-03 and stored into Saithia godown (Bill no. Gama/002/401):

1. Produce 8 Pcs Radio and store to Malda godown Aditional cost: Labor charges Rs. 100 per Radio, Power Rs. 210, Carriage Rs. 120.
Note: Make all payment entries for Additional cost from cash.
2. Salary paid to Mr. Aniket (employee) Rs. 6000 for the month of December 2002, through SBI (ch. No. 150938) dated on 01-01-03.
3. Received Rs. 20000 from Pinak & Co. on full settlement through cheque (ch. No. 450258) on 02-01-03 and deposit the amount to SBI.
4. Paid Rs. 43000 to Gama Ltd. through cheque issued from SBI (ch no. 150939) to clear all dues on 02-02-03.
5. Received 2 Pcs. Keyboard at Malda godown from TVS & Co. as replacement on 01-02-03.
6. Rs. 8800 received from Cola House on 01-02-03.
7. Sold 5 Pcs. Radio to Sony & Co. at 5% profit on the manufacturing price including WBST @5% (Bill no: Sony/008/001).
8. Sony & Co. paid a cheque (ch no. 458023) on full settlement and discount allowed by Rs. 500 duly deposited into UTI on 02-02-03.

End of April Month send Bank Statement of UTI


Do the following using Optional voucher mode and check all possible effects using Scenario:

Company wants to check the written down value of the existing assets right now without affecting the account. [charge depreciation @ 40% p.a.]
If a sale occurs for 5 CD of Tally 6.3 with a 10% profit margin, then show what will be the effect on Gross Profit? But don’t give the effect on the Books of Accounts.
Show the Scenario result at the end of March 2003 before providing the salary of Rs. 25000 from UTI for the month of March 2003.
Make two Price Levels for (i) Retailer and (ii) Wholesaler. [Turn on “Set/Modify other Company Features?” from Company Features (F1), then go for the option “Use Multiple Price Levels for Invoicing”]
Set the following ranges of pepsi (300 ml) for Retailer:

Sold of 60 Ltr Pepsi (200 ml) to Pepsi House (Wholesaler) based on Price Levels on 02-02-03 (Bill no. Pep/009/001).

Monday, August 13, 2007

Assignment Sollution ( MCS-033 )

Advanced Discret Mathematics


(i) True, because an depends on all its previous term, therefore order is not defined and since, an is not like a polynomial, so the degree is also not defined.
(ii) False. The generating function for given R, R is 1/(1-x)
(iii) False
(iv) False, because it a tree has n verities then it has n-1 edges.
(v) False.

Answer: 2
(a) (i) The general form of the solution is
k10+(k20+k21 n) (-2)n + (k10+k31n+k32n2)(2)n ________________
(ii) k10+(k20+k21 n) (-2)n + (k30+k31n+k32n2)(2)n + (A0 + A1n)n32n+A2n2)(-2)n

(b) (i) an = 2an-1 + 10000n, n≥1
a0 = 40,000

(ii) an = 60,000 (2)n – 10000n – 20000)
= 10000 (6.2n-n-2)

(iii) Salary for fifth year a1 = 900000 (9 lakh)

Answer: 3

(a) The number of solution to the linear equation x1+x2+x3 where
1≤x1≤3, -1≤x2≤1 and x1≥3, is given by the coefficient of xn in the expansion of (x+x2+x3) (x4+1+x) (x3+x4+…+xn)
(b) We have
10an-1 + 3an-2 – 6an-1 + an = 3(-1)n +5(1/2)n, n≥0 ……….(1)

The characteristic equation is
ahn = C1(-1)n + C2(1/2)n + C3(1/5)n

For apn, Since –1& ½ are roots of characteristic equation with multiplicity one. Thus the particular solution is
apn = A0 n(-1)n + A1 n (1/2)n

From (1) ,we get
An [-18] (-1)n + A1 (9/4)(1/2)n = 3(-1)n+5(1/2)n

On comparing we have
A0 = -1/6, A1 = 20/9
apn = -1/6 n (-1)n –(20/9)n(1/2)n

The general solution is
an = an h + an p = C1 (-1)n + C2 (1/2)n + C3 (1/5)n -1n/6 (-1)n – 20n/9 (1/2)n

First note that by the bionomial theorem C (2n,n) is the coefficient of xn in (1+x)2n
However, we also have (1+x)2n = [(1+x)n]2 = [C(n,0) + C(n, 1)x + … + C(n,n)xn]2

The coefficient of xn in this expression is C(n,0) C(n,n) + C (n,1) C (n, n-1)+ C(n,2) C (n,n-2) + …. + C (n,n) C (n,0)
= S [C(n,k)]2


(a) we have
bn = bn-1+n2+n(n+1)/2

è bn-bn-1 = 3n2/2 + 1n/2

Using generating function
x x x x
Sbnzn - Sbn-1zn = 3/2Sn2zn + 1/2Snzn
n=1 n=1 n=1 n=1

G(z)-bn-ZG (Z) = 3/2 Z (1+z)/(1-z) + 1z/2(1-z)2

G(z) – b0/(1-z) + 3/2 Z (1+z)/(1-z)4 + 1z/2(1-z)3

è bn = b0 + 3/2n3 + ¼ n(n+1)

For this question we obtain a sequence like
1, 2, 3, 5, 13 ….

For this
an = an-2 + an-1, where a1 = 2, a0=1

(i) a1 = 2, a2 = 3, a3 = 6, a4 = 8
(ii) an = an-2 + an-1, a3 = 2, a0 =1

The characteristics equation is

r2 = r-1 = 0 è r1= 1+Ö5 /2, r2 = 1-Ö5 /2

an = C1[1+Ö5 /2]n + C2 [1-Ö5/2]n

To find C1 & C2, we use a0 = 1, a1=2

a0=1 è C1+C2 = 1

\a1=2 = C1 [1+Ö5 /2] + C2 [1-Ö5 /2 = 2]

è C1 = 1/Ö5, C2 = -1/Ö5

\ an = 1/Ö5 [1+Ö5 /2]n + 1/Ö5 [1+Ö5 /2]n


(i) {2, 2, 3, 3, 3, 3, 3, 3, 6}
(ii) v5 v4 v2 v1 v9 v8 v7 v6 v5
(iii) k3(iv) D(G) = Largest vertex degree of G = 6
d(G) = min. degree of vertex = 2

(b) (i) complement of the graph is shown below:

(ii) (a) A«a, g«C, d«H, b«B, f«D, E«C, G«e.
(b) e«A, g«B, C«D, d«C
(c) f«A, b«B, C«d, g«E, e«C

(c) (i) {V1, V3,V5,V7} & {V2, V4, V6, V8}

Complete matching: if every vertex in V1 is matched against some vertex in V2.

(ii) Not bipartite

(a) a b c a g c d f a d e a
(b) Not Hamiltonian

Since, S = {b, d} Î V (G) & S is a propersubset of V(G), by them we should have C (G-S) ≤ S, if G is Hamiltonian.
But S = 2 and C (G-S)= 3 which is a contradiction.
Therefore, G not Hamiltonian

(c) {v1, v2, v3, v4, v5, v1} weigh = 22 + 18 + 17 + 20 + 24 = 101
{a, b, d, c e, a} weight = 22 +15 + 17 + 17 + 24 = 95
{a, d, c, b, c, a} weight = 11 + 17 + 18 + 12+ 24 = 82

(i) Since D(G) = 6. Since there is triangle subgraph of G.
3≤x (a)≤6
(ii) X (G) = 4 colouring in Fig.
(iii) {(a, c, g)(b, d, f)(e, h), (i)}
(iv) No

(b) (i) Non-planar
(ii) Yes.

10an-3 + 3an-2 + 6an-1 + an = 3(-1)n + 5/2n, n≥ 0.

Saturday, August 11, 2007

Assignment Sollution ( MCS - 032 )

Object Oriented Analysis and Design

Q.1 Explain basic characteristics of the object oriented modeling with examples.
Ans. Object-Oriented Modeling, or OOM, is a modeling paradigm mainly used in computer programming. Prior to the rise of OOM, the dominant paradigm was functional programming, which emphasized the use of discreet reusable code blocks that could stand on their own, take variables, perform a function of them, and return values.

The Object-Oriented paradigm assist the programmer to address the complexity of a problem domain by considering the problem not as a set of functions hat can be performed but primarily as a set of related, interacting Objects. The modeling task then is specifying, for a specific context, those Objects (or the Class the Objects belong to), their respective set of Properties and Methods, shared by all objects members of the Class.

As an example, in a model of a Payroll System, a Company is an Object. An Employee is another Object. Employment is a Relationship or Association. An Employee Class (or Object for simplicity) ahs Attributes like Name, Birthdate, etc. The Association itself may be considered as an Object, having Attributes, or qualifiers like Position, etc. An employee Method may be Promote, Raise, etc.

The Model description or Schema may grow in complexity to require a Notation. Many notations has been proposed, based on different paradigms, diverged, and converged in a more popular one known as UML.

Characteristics of Object Oriented Technology

The term Object Oriented means that we organize the software as a collection of discrete objects. An object is a software package that contains the related data and the procedures. Although objects can be used for any purpose, they are most frequently used to represent real-world objects such as products, customers and sales orders. The basic idea is to define software objects that can interact with each other just as their real world counterparts do, modeling the way a system works and providing a natural foundation of building system to manage that business.

In principle, packaging data and procedures together makes perfect sense. In practice, it raises an awkward problem. Suppose we have many objects of the same general type- for example a thousand product objects, each of which could report its current price. Any data these objects contained could easily be unique for each object. Stock number, price, storage dimensions, stock on hand, reorder quantity, and any other values would differ from one product to the next. But the methods for dealing with the data might well be the same. Do we have to copy these methods and duplicate them in every object? Capturing these commodities in a single place. That place is called a class. The class acts as a kind of template for objects of similar nature.

Polymorphism is used to express the fact that the same message can be sent to many different objects and interpreted in different ways by each object. For example, we could send the message “move” to many different kinds of objects. They would all respond to the same messages, but they might do so in very different ways. The move operation will behave differently for a window and differently for a class piece.

Polymoprhism allows client programs to be written based only on the abstract interfaces of the objects which will be manipulated (interface inheritance). This means that future extension in the form of new types of objects is easy, if the new objects conform to the original interface. In particular, with object-oriented polymorphism, the original client program does not event need to be recompiled (only relinked) on order to make sue of new types exhibiting new (but interface-conformant) behavior.

Inheritance is the sharing of attributes and operations among classes on a hierarchical relationship. A class can be defined as a generalized form and then it specialized in a subclass. Each subclass inherits all the properties of its superclass and adds its own properties in it. For example, a car and a bicycle are subclasses of a class road vehicle, as they both inherits all the qualities of a road vehicle and add their own properties to it. Consider types of bank accounts. Figure below shows how both checkingAccount and SavingsAccount classes inherit from the BankAccoun class.

Q.2 Explain major advantages of object oriented modeling..

Ans. A major factor in the invention of Object-Oriented approach is to remove some of the flaws encountered with the procedural approach.

Some benefits for OO approach are:
Faster Development: OOD has long been touted as leading to faster development. Many of the claims of potentially reduced development time are correct in principle, if a bit overstated.
Reuse of Previous work: This is the benefit cited most commonly in literature, particularly in business periodicals. OOD produces software modules that can be plugged into one another, which allows creation of new programs. However, such reuse does not come easily. It takes planning and investment.
Increased Quality: Increases in quality are largely a by-product of this program reuse. If 90% a new application consists of proven, existing components, then only the remaining 10% of the code has to be tested scratch. That observation implies an order-of-magnitude reduction in defects.
Modular Architecture: Object-oriented systems have a natural structure for modular design: objects, subsystems, framework, and so on. Thus, OOD systems are easier to modify. OOD systems can be altered in fundamental ways without ever breaking up since changes are neatly encapsulated. However, nothing in OOD guarantees or requires that the code produced will web modular. The same level of care in design and implementation is required to produce a modular structure in OOD, as it is for any form of software development.
Client/Server Applications: By their very nature, client/server applications involve transmission of messages back and forth over a network, and the object-message paradigm of OOD meshes well the physical and conceptual architecture of client/server applications.
Better Mapping to the Problem Domain: This is a clear winner for OOD, particularly when the project maps to the real world. Whether objects represent customers, machinery, banks, sensors of pieces of paper, they can provide a clean, self-contained implication which fits naturally into human thought processes.

Q.3 Explain the concept of encapsulation with an example. Also, list the benefits it during implementation.

Ans: Encapsulation is the object model concept of including processing or behavior with the object instances defined by the class. Encapsulation allows code and data to be packaged together. The definition of methods for a class is an integral part of encapsulation. A method is programming code that performs the behavior an object instance can exhibit. Calculating the age of a person would be an example of such behavior. The figure shows a way of looking a encapsulating the age method with an instance object. The code for the age method is “attached” to or encapsulated with the object rather than part of the application.

The notation for object model encapsulation is shown below. Methods are shown at the bottom of the class notation.

Code and data were not always packaged together. At one time, for example, it was necessary to define an age calculation in each application or have a library that contained the age calculation routine. Having an age calculation or any routine, replicated in many applications may make it difficult to ensure that a change in the routine was made everywhere that it is used. Using a library improves this situation if use of the library is enforced. Nevertheless, with a library, you can never be sure which routine is supposed to be used with which data. It is entirely possible to execute the right code on the wrong data or the wrong code on the right data.

Object system recognize which methods belong to which data. You cannot execute the right method on the wrong data as you could with library systems. The correct execution of methods is called dispatching, and it is handled by the object system.

Importance and benefits
Encapsulation is critical to building large complex software which can be maintained and extended. Many studies have shown that the greatest cost in software is not the initial development, but the thousands of hours spent in maintaining the software Well encapsulated components are far easier to maintain.
Once software is in place, another great expense is extending its functionally. As you add new features, there is risk that you’ll break existing parts of the application. Again, encapsulation helps to minimize this risk.
In a well designed program, each object should have a single area of responsibility. That object presents an interface which defines the services the object provides.
Your object promises to provide certain services and to provide certain information. You, as author of the object, publish an interface and say, in essence, “Here is how you interact with my object. If you make these function calls, I promise to provide valid information in this format or to take these actions.”
It is common for an object to offer interfaces based on the privileges of the client class. For example, one client might be allowed only a subset of the possible operations on your new object. In addition, an object might offer a second interface to accommodate updates and changes to the object without breaking old client code. That is, older clients use the old interface, but newer clients can use the newer and updated interface.

The two main advantages of encapsulation are:
The data structure that represents the state can be re-implemented without affecting the implementation of the other objects. For example if the state of a point is represented internally with Cartesian coordinates then this might be changed to Polar coordinates without changing the interface of the object, i.e., without changing which messages with which arguments are understood by the object and what the type of the result of these messages will be. The state of the object can be guarded by the methods. The data structure that represents the state of the object may allow certain values that are not considered meaningful, e.g., a percentage ma be represented by an integer that allows numbers larger than 100. The methods that update this integer can then ensure that it never rises above 100. Many languages allow the programmer to bypass encapsulation, or to specify varying degrees of encapsulation for specific object types. In Java, for example, a class might be defined like this:

Class Cow extends Mammal {
public Tail m_tail;

private Horns m_horns;

public void eat(Food f) {


chewcud ();


private void chewcud ()


With the Cow type as defined above, some piece of code external to Cow’s implementation would be allowed to access m_tail directly, without going through any of the methods that Cow has chosen to expose as its interface. Such code would not, however, be able to access m_horns. Likewise, the method eat() is exposed as part of the Cow’s interface, so any other object would be able to cause the cow object to eat by calling that method, passing it the offered food item as an argument to the method. External code would not be able to directly cause the cow to call its chewcud ()method (nor would it even know such a method existed; only the cow itself knows or cares that eating involves chewing its cud).

Q.4 What is object modeling? Explain how objects and classes are identified with an example.
Ans: The object model describes the structure of the objects in the system – their identity, their relationship to other objects, their attributes, and their operations. The object model depicts the primary view of how the real world in which the system interacts is divided and the overall decomposition of the system. The object model provides the framework into which the other models are placed.

The object model is represented graphically with object class diagrams containing the object classes and their relationships. Each application-domain concept from the real world that is important to the application should be modeled as an object class. Classes are arranged into hierarchies sharing common structure and behavior and are associated with other classes. Classes define the attributes carried by each object instance and the operations that each object performs or undergoes.

The object class diagram should adhere to OMT’s notation and exploit the capabilities of OMT, such as links and vasodilatations, inheritance (generalization and specialization), and aggregation. The attributes and operations need not be included in this diagram, as they will be described below in the object class specification.

Elements of an Object Model
Object classes with attributes, relations and connections.
Classes with constraints no their environment
Behavior associated with an object model

The activity of establishing an object model consists of going through all the elements of an object model.

The purpose of object modeling is to describe objects. For example, Joe Smith, Simplex company, and the top window are objects. An object is simply something that makes sense in an application context. Objects serves two purposes: they promote understanding of the real world and provide a practical basis for computer implementation. Decomposition fo a problem into objects depends on judgments and the nature of the problem. There is no one correct representation.

All objects have identity and are distinguishable. Two apples with the same color, shape and texture are still individual apples; a person can eat one and then eat the other.

An object class describes group with similar properties, common behavior, common relationships to other objects, and common semantics, company, animal, process, and window are all object classes. Ach person has an age, IQ, and may work at a job. Ac process has an owner, priority, and list of required resources. Objects and object classes often appear as nouns in problem description.

The object in class share a common semantic purpose, above and beyond the requirement of common attributes and behavior. Each object knows its class.

A simple class diagram is shown below:

Identifying Objects (and Class)
Identify the nouns( objects) in the problem description.
Identify more objects in your developing problem solution.
From these objects, determine what new types (classes) you need.

Example: Black Jack Program
A deck
A dealer
Sample Design Questions:
Are the dealer and player instances of the same class? i.e., objects of the same type?
Should there be a separate Hand class?

Q.5 What is state diagram? Differentiate between a simple state diagram and a composite state diagram with an example.

Ans. State diagrams (also called State Chart diagram) are used to help the developer better understand any complex/unusual functionalities or business flows of specialized areas of the system. In short, State diagrams depict the dynamic behavior of the entire system, or a sub-system, or even a single object in a system. This is done with help of Behavioral elements.

Elements of a State diagram
Initial state: This shows the starting point or first activity of the flow. Denoted by a solid circle. This is also called as a “pseudo state,” where the state has no variables describing it further and no activities.
State: Represents the state of object at an instant of time. In a state diagram, there will be multiple of such symbols, one for each state of the Object we are discussing. Denoted by a rectangle with rounded corners and compartments (such as a class with rounded corners to denote an Object).

Transition: An arrow indicating the Object to transition from one state to the other. The actual trigger event and action causing the transition are written beside the arrow, separated by a slash. Transition that occur because the state completed and activity are called “triggerless” transitions. If an event has to occur after the completion of some event or action, the event or action is called the guard condition. The transition takes place after the guard condition. The transition takes place after the guard condition occurs. This guard condition/event/action is depicted by square brackets around the description of the event/action.

History States: A flow may require that the object go to occur is called as an event or action. Every transition need not occur due to the occurrence of an event or action directly related to the state that transitioned from one state to another.

Final State: the end of the state diagram is shown by a bull’s eye symbol also called a final state. A final state is another example of a pseudo state because it does not have any variable or action described.

Simple Set
A simple state is a state that does not have substates, i.e. it has no regions and it has no submachine state machine.

Q.6 (a) What is Unidirectional implementations? Explain how it is different from Bi-directional implementation.
Ans. Associations represent relationships between instances of types (a person works for a company, a company has a number of offices…). The interpretation of them varies with the perspective. Conceptually they represent conceptual relationships between the types involved. In specification these are responsibilities for knowing, and will be made explicit by access and update operations. This may mean that a pointer exists between order and customer, but that is hidden by encapsulation. A more implementation interpretation implies the presence of a pointer. Thus it is essential to know what perspective is used to build a model in order to interpret it correctly.

Associations may be bi-directional (Can be navigated in either direction) or unidirectional (can be navigated in one direction only). Conceptually all associations can be thought of a bi-directional, but unidirectional associations are important for specification and implementation models. For coupling. IN implementation models a bi-directional association implies coupled sets of pointers, which many designers find difficult to deal with.

Undirectional one-to-one associations

The simplest type of association to realize is unidirectional one-to-one association between two classes.
With a unidirectional association, you can navigate from one class to another (As indicated by the direction of the arrow) but not vice-versa.
It is conveniently implemented using a reference.


Bi-directional one-to-one associations
The following diagram indicates a bi-directional association no arrows on any end of the association


It indicates that we must be able to navigate from an account to the corresponding customer and vice versa. There are three approaches to bi-directional implementation:
Implement as an attribute in one direction only and perform a search when a backward traversal is required.
Implement as attributes in both directions.
Implement as a distinct association object, independent of either class.

(b) Explain the concept of collaboration with example.

Ans. An object-oriented system is composed of objects. The behavior of the system is achieved through collaboration between these objects, and the state of the objects in it. Collaboration between objects involves then sending messages to each other. The exact semantics of message sending between objects varies depending on what kind of system is being modeled.
Collaboration diagrams (interaction diagrams) illustrate the relationship and interaction between software objects. They require use cases, system operation contracts, and domain model to already (instances). A diagram is created for each system operation that relates t the current development cycle (iteration).

Object: Objects are instances of classes, and are one of the entity types that can be involved in communications. An Object is drawn as a rectangular box, with the class name inside prefixed with the object name (optional) and a semi-colon.
Object 1

Actor: Actors can also communicate with Objects, so they too can be listed on Collaboration diagrams. An Actor is depicted by a stick figure.


Message: Message, modeled as arrows between objects, and labeled with an ordering number, indicate the communications between objects.

Here is an example of an Administrator using a Web Application to manage a user account. Notice how you can follow the process from object to object, according to the outline below:

Find User
1.1 LookUp User
Update User
2.1 Validate user
2.2 Update user
Web App:
User Interface

2.2 Update User
User DB
User Validator
2.1 Validator
1. Find User
2. Update User
1.1 Look UpUser

Thursday, August 9, 2007

Term-End Examination, Dec -2005 ( CS-13 )

Term-End Examination, December, 2005

CS13 : Operating Systems

Time: 3 hoursMaximum Marks: 75

Note : Question No. 1 is compulsory. Answer any three questions from the rest.

1. (a) Write an algorithm to explain the producer/consumer problem with an unbounded buffer in concurrent programming. (7)

(b) Expiain the functional specifications for partition allocation of memory in a system with static partitioning. List the advantages and disadvantages of it. Also describe the necessary hardware support for protection. (9)

(c) When does a page fault occur ? Describe the action taken by the operating system when a page fault occurs. (6)

(d) What is process migration in a distributed system ? List its advantages. Write the step-by-step procedure to migrate a process to the destination node. (8)

2. (a) A file system, resident on the disk should have the following characteristic features:(i) The file records should be read sequentially as well as randomly.(ii) There should be a practical limit of the file size.(iii) The file space allocated, should neither be under-utilized nor over-utilized.Explain the most suitable file space allocation scheme keeping in mind the features above. Justify your answer. (4)

(b) Explain tbe Flynn's classification of parallel computer architecture. Also, give the classification of shared memory multiprocessors on the basis of memory architecture and access delays. (7)

(c) What is the data structure that records all the information about a particular process in an O/S ? List a the fields for recording various aspects of process execution and resource usage. (4)

3. (a) What is a process ? How is it different from a program ? What are the different states of a process ? How does a process change from one state to another ? (5)

(b) Write, and explain, the deadlock detection algorithm. (6)

(c) List at least four common responsibilities of the file management system in an O/S. (4)

4. (a) What is the primary goal of Authentication ? How can you achieve this goal through (i) the password mechanism, and (ii) artifact-based mechanism ? (7)

(b) Wlth the help of a block diagram explain the client/server division of labour of file system in a workstation based model of distributed computing. (8)

5. (a) Write an algorithm for implementing Dining philosophers problem using semaphores. (7)

(b) Explain the principles of operation of virtual memory. Write any two distinct replacement algorithms. (8)

Wednesday, August 8, 2007

Term-End Examination JUNE, 2007 (CS-06)

Term-End Examination JUNE, 2007


Time : 3 hours...............Maximum Marks : 75

Note : question number 1 is compulsory. Answer any three questions from the rest.

1.(a) A Bank maintains information about customers and their accounts. Each customer has a name, address (ouse number, area, city and state code) and telephone number. Account has number, type and balance. We need to record customers who own an account. Account can be operated individually or jointly.
Design and draw an ER diagram, clearly indicating the attributes, keys, the cardinality ratio and participation constraints. [ 7 ]

(b). What is meant by data abstraction? Explain the differences between physical level, conceptual level and view level of data abstraction. [ 6 ]

(c). Explain at least five important characteristics of OORDBMS. [ 5 ]

(d). For the relation R and S given below, compute natural join and outer join, [ 6 ]

A.... B.... C
1.... 2.... 3
4.... 5.... 6
7.... 8.... 9


B.... C.... D
2.... 3.... 10
2.... 3.... 11
6.... 7.... 12

(e). Given the following relations: [ 6 ]
Vehicle (reg_no, colour, type)
Person (eno, name, address)
Owner (eno,reg_no)
Write expressions in the relational algebra to answer the following queries:
(i). List the reg_no of vehicles owned by “Sandeep”.
(ii). List the names of persons who own “Indica” cars.
(iii). List details of the black coloured vehicles.

2.(a) Describe DROP TABLE command of SQL with both the options CASCADE and RESTRICT. [ 5 ]

(b). Explain the recovery process after system failure, using checkpoint. [ 5 ]

(c). How is the knowledge representation done through semantic network? Give an example of semantic network.[ 5 ]

3.(a) How does data fragmentation support the distributed a\databses? Explain horizontal and vertical Project (Project_name, Emp_no, Emp_name, Job_class, Emp_location, Salary) [ 6 ]
Note: Assume the suitable data.

(b). What is NULL? Give an example to illustrate testing for NULL in SQL. [ 4 ]

(c). Define and differentiate between ordered indexing and hashing. [ 5 ]

4.(a) Construct a B+ tree for the following set of key values where the number of key values that fit in a node is 3. [ 8 ]
Key Values : (12,2,15,4,123,45,6,7,9,1,3)
Show the steps involved in the deletion of key values 7 and then 6.

(b). Why is BCNF a more desirable normal form than any of the lower order normal forms? Give an example of a relational schema that is in 3NF but not in BCNF. [ 7 ]

5.(a) Define the following with respect to SQL. Also give an example of each. [ 6 ]
(i). UNIQUE function
(ii). ORDER-BY clause
(iii). LIKE predicate

(b). Explain different methods of implementing the following data models: [ 6 ]
(i). Hierarchical Data Model
(ii). Network Data Model
(c). What is the difference between a data base schema and a database state? [ 3 ]

Monday, August 6, 2007

Examination Fees Applicable from DEC2007

Important Notice! IGNOU finally has declare examination fees from TERM END DEC2007 for all its courses. And also change the schedule of submission of form. Now its Rs. 50/- per paper. So if you want to give 10 papers then you have to submit Rs. 50*10= Rs. 500 as examination fees. In their website give it details. They also have change the format of examination form, so now this time when you are filling the examination form be sure that it is new one.
FORMS FOR DEC TEE---------------------------LATE FEE

1ST OCTOBER TO 20TH OCTOBER-------------------RS.100/-
21ST OCTOBER TO 15TH NOVEMBER-----------------RS.500/-
16TH NOVEMBER TO 28TH NOVERBER----------------RS.1000/-

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