IGNOU BCA MCA

IGNOU Solved Assignment July-December 2019

2019-10-02T06:39:00.000-07:00

Hello, Friends, I am back again to solve your problems. I can help you in solving the assignment related to IGNOU BCA and MCA. If you have any problem then you can comment on it and if you want a complete solution, then you can comment the description of the question in the comment box. I hope you will get benefit from my knowledge and I will be able to help you in a justified manner. Keep visiting on this site and guide me with your comments.
Best of luck for your future.

MCS 043 Free Assignment April 2011

2011-03-06T11:00:00.001-08:00

Question 2

Solution

(a) Select Supplier.su_id, supplier.Name.su (Supply of per quantity of supply) from supplier

On supplier.su_id inner join supply of par su_id group by su_id

(b) Begin Transaction Trans

Use Adventure work 2008 R

SELECT + Money

From Adventurework 2008 R

Human Resource.Account

Where Acount No. = 123456

Commit Transaction @ Account No

= Begin Trasaction Tran 2

Use Adventure Work 2008 R2

Insert into Adventure work 2008 R2

VALUES (10000, ACCOUNT No. )

Commit Transaction @ Account No.

(c) Create view supplier

Select * FROM Supplier

Where su_id = @ su_id

= create view system manager

Select * from supplier su_id, supplier.su_name

Supplier.su_address, supplier.su_status

Supply of part.su_id, supply of part. Pa_id, supply of part.

Date of supply, supply of part.quanitity of

Supply, supply of part.su_price from

Supplier on supplier.su_id inner join

Supply of part.su_id

(d) Transaction processing is designed to maintain a computer system (typically a database or some modern filesystems) in a known, consistent state, by ensuring that any operations carried out on the system that are interdependent are either all completed successfully or all canceled successfully.

For example, consider a typical banking transaction that involves moving $700 from a customer's savings account to a customer's checking account. This transaction is a single operation in the eyes of the bank, but it involves at least two separate operations in computer terms: debiting the savings account by $700, and crediting the checking account by $700. If the debit operation succeeds but the credit does not (or vice versa), the books of the bank will not balance at the end of the day. There must therefore be a way to ensure that either both operations succeed or both fail, so that there is never any inconsistency in the bank's database as a whole. Transaction processing is designed to provide this.

Deadlock Detection

Deadlock can occur whenever two or more processes are competing for limited resources and the processes are allowed to acquire and hold a resource (obtain a lock) thus preventing others from using the resource while the process waits for other resources.

Two common places where deadlocks may occur are with processes in an operating system (distributed or centralized) and with transactions in a database. The concepts discussed here are applicable to any system that allocates resources to processes.

Locking protocols such as the popular Two Phase Locking (see concurrency control) give rise to deadlock as follows: process A gets a lock on data item X while process B gets a lock on data item Y. Process A then tries to get a lock on Y. As Y is already locked, process A enters a blocked state. Process B now decides to get a lock on X, but is blocked. Both processes are now blocked, and, by the rules of Two Phase Locking, neither will relinquish their locks.

This is a deadlock: process A is waiting for a resource held by process B and process B is waiting for a resource held by process A. No progress will take place without outside intervention. Several processes can be involved in a deadlock when there exists a cycle of processes waiting for each other. Process A waits for B which waits for C which waits for A.

Four conditions must hold for deadlock to occur:

1. Exclusive use – when a process accesses a resource, it is granted exclusive use of that resource.

1. 2. Hold and wait – a process is allowed to hold onto some resources while it is waiting for other resources.

2. 3. No preemption – a process cannot preempt or take away the resources held by another process.

3. 4. Cyclical wait – there is a circular chain of waiting processes, each waiting for a resource held by the next process in the chain.

QUESTION-3

(a) Select from Guide Table Group by Guide_Id

(b)

<Name> Rajat Sharma </name>

<Work Experience> 3 Years </work experience>

<published materials> Data Structure </published materials>

</Teacher>

</University>

=Begin Trasaction Trans 3

Use Adsventrue Work 208 R2

Update Adventure work 2008 R2 Set

Interest = @ interest where

Account No. = @ Account No.

Commit Transaction @ Acocount, @ Interest

= Begin Trasaction Trans 4

Use adventure Work 208 R2

Select * from Adventure work 2008 Rs

Commit Transaction

Go.

MCS 041 Operating System Solution

2011-02-24T02:31:00.000-08:00

Dear Friends here we post the available solution to the assignments if you have any doubt about any question please comment it.

Question 1:

(a) Assume you have the following jobs to execute with one processor

Process Burst Time Arrival Time

P1 75 0

P2 40 10

P3 25 10

P4 20 80

P5 45 85

Support a system uses RR scheduling with a quantum of 15.

a) Create a Gantt chart illustrating the execution of these processes.

b) What is turn around time for process P3.

c) What is the average wait time for the processes?

(9 Marks)

Solution:

Gantt chart

0 15 30 45 60 75 90 105 120 125 140 155 165 180 195 210

Process	Processing Time	Turnaround Time= Process completed-process Submitted	Waiting Time=Turnaround Time- Processing Time
P1	75	210-0=210	210-75=135
P2	40	165-0=165	165-40=125
P3	25	120-0=120	120-25=95
P4	20	125-0=125	125-20=105
P5	45	180-0=180	180-45=135

(b) The around tine for the process p3 is =120

Q1. b)) Consider a swapping system in which memory consists of the following hole size in memory order: 10K, 4K, 20K, 18K, 7K, 9K, 12K and 15K.which hole is taken for successive segment requests of

i) 12K ii) 10K iii) 9K

for first fit, best fit and worst fit.

Solution:(i) First Fit – 20K

Best fit – 12k

Warst fit – 20k

(ii) First fit – 10 k

Best fit – 10 k

Worst fit – 20k

(iii) First fit – 9 k

Best fit – 9 k

Worst fit – 20k

Question 2:

Write a semaphore based solution to Dining Philosopher problem and explain all the assumptions and the algorithm.

Ans.

Solution written in Java

import java.util.concurrent.Semaphore;

import java.util.Random;

import java.util.Vector;

public class Philosopher extends Thread

    private static final Random rand = new Random();

    private static int event=0;

    private static String binary="";

    private int id;

    private Semaphore sem;

    private static Vector<Object[]> vector = new Vector<Object[]>();

    public Philosopher(int i, Semaphore s)

        id = i;

        sem = s;

        binary = binary + "0";

    private void busy()

try

            sleep(rand.nextInt(1000));

        } catch (InterruptedException e){}

    private void thinking()

        String str = "Philosopher " + id + " is thinking";

        vector.add( this.addToObject(System.currentTimeMillis(), event , str) );

        event++;

        busy();

    private void eating()

        String str ="Philosopher " + id + " is hungry and is trying to pick up his chopsticks";

        vector.add( this.addToObject(System.currentTimeMillis(), event , str) );

        event++;

        busy();

        str = "Philosopher " + id + " is eating";

        this.oneToBinary(id);

        vector.add( this.addToObject(System.currentTimeMillis(), event , str) );

        event++;

        busy();

        str = "Philosopher " + id + " finished eating, and puts away his chopsticks";

        this.zeroToBinary(id);

        vector.add( this.addToObject(System.currentTimeMillis(), event , str) );

        event++;

    private Object[] addToObject(long t, int i,String s ){

        Object[] o = new Object[4];

        o[3] = s;

        o[2] = i;

        o[1] = binary;

        o[0] = t;

        return o;

    private void oneToBinary(int i){

        binary = binary.substring(0,i) + "1" + binary.substring(i+1);

    private void zeroToBinary(int i){

        binary = binary.substring(0,i) + "0" + binary.substring(i+1);

    @Override

    public void run()

        for (int i = 0; i < 10; ++i)

            thinking();

try

                sem.acquire();

            } catch (InterruptedException e){}

            eating();

            sem.release();

}

Question 3. a)

memory segmentation is one of the most common ways to achieve memory protection; another common one is paging. In a computer system using segmentation, an instruction operand that refers to a memory location includes a value that identifies a segment and an offset within that segment. A segment has a set of permissions, and a length, associated with it. If the currently running process is allowed by the permissions to make the type of reference to memory that it is attempting to make, and the offset within the segment is within the range specified by the length of the segment, the reference is permitted; otherwise, a hardware exception is raised.

Moreover, as well as its set of permissions and length, a segment also has associated with it information indicating where the segment is located in memory. It may also have a flag indicating whether the segment is present in main memory or not; if the segment is not present in main memory, an exception is raised, and the operating system will read the segment into memory from secondary storage. The information indicating where the segment is located in memory might be the address of the first location in the segment, or might be the address of a page table for the segment, if the segmentation is implemented with paging. In the first case, if a reference to a location within a segment is made, the offset within the segment will be added to address of the first location in the segment to give the address in memory of the referred-to item; in the second case, the offset of the segment is translated to a memory address using the page table.

In most systems in which a segment doesn't have a page table associated with it, the address of the first location in the segment is an address in main memory; in those systems, no paging is done. In the Intel 80386 and later, that address can either be an address in main memory, if paging is not enabled, or an address in a paged "linear" address space, if paging is enabled.

paging is one of the memory-management schemes by which a computer can store and retrieve data from secondary storage for use in main memory. In the paging memory-management scheme, the operating system retrieves data from secondary storage in same-size blocks called pages. The main advantage of paging is that it allows the physical address space of a process to be noncontiguous. Before paging, systems had to fit whole programs into storage contiguously, which caused various storage and fragmentation problems.

Paging is an important part of virtual memory implementation in most contemporary general-purpose operating systems, allowing them to use disk storage for data that does not fit into physical Random-access memory (RAM). Paging is usually implemented as architecture-specific code built into the kernel of the operating system.

Combined segmentation and paging In a combined paging/segmentation system a user address space is broken up into a number of segments at the discretion of the programmer

. Each segment is in turn broken up into a number of fixed-size pages which are equal in length to a main memory frame. If a segment is less than a page in length, the segment occupies just one page. From the programmer`s point of view, a logical address still consists of a segment number and a segment offset. From the system`s point of view, the segment offset is viewed as a page number and page offset for a page within the specified segment.

Question 3. b). Compare direct file with indexed sequential file organization.

Solution: Indexed sequential file organization. In indexed sequential file organization, the records are stored in sequence according to a primary key and an index is created to allow random access of the file. This type of organization also allows the file to be accessed sequentially. Indexed sequential is the most commonly used type of file organization. l Direct file organization. In direct file organization, the records are stored and retrieved using a relative record number, which gives the position of the record in the file. This type of organization also allows the file to be accessed sequentially

Question 4. What are the necessary and sufficient conditions for a deadlock to occur? Explain Banker’s algorithm for deadlock detection with and example. Also suggest 2 mechanisms of recovery from a deadlock.

Ans. There are four necessary and sufficient conditions for a Coffman deadlock to occur, known as the Coffman conditions from their first description in a 1971 article by E. G. Coffman.

Mutual exclusion condition: a resource that cannot be used by more than one process at a time
Hold and wait condition: processes already holding resources may request new resources
No preemption condition: No resource can be forcibly removed from a process holding it, resources can be released only by the explicit action of the process
Circular wait condition: two or more processes form a circular chain where each process waits for a resource that the next process in the chain holds

Banker’s Algo

The Banker's algorithm is run by the operating system whenever a process requests resources.^[2] The algorithm prevents deadlock by denying or postponing the request if it determines that accepting the request could put the system in an unsafe state (one where deadlock could occur). When a new process enters a system, it must declare the maximum number of instances of each resource type that may not exceed the total number of resources in the system. Also, when a process gets all its requested resources it must return them in a finite amount of time.

function bankers_algorithm(set of processes P, currently available resources A) {

    while (P not empty) {

        boolean found = false

        for each (process p in P) {

            C_p = current_resource_allocation_for_process(p)

            M_p = maximum_resource_requirement_for_process(p)

            if (M_p − C_p ≤ A) {

                // p can obtain all it needs.

                // Assume it does so, terminates, and releases what it already has.

                A = A + C_p

                remove_element_from_set(p, P)

                found = true

        if (not found) {

            return UNSAFE

    return SAFE

Recover from deadlock

Once it is determined that a deadlock exitx, the system needs to recover from the deadlock. For this, one or more transactions are rolled back to break the deadlock. While performing the roll back operation, the following issues need to be addressed :

a. Selection of a victim : in the situation of a deadlock, you first need to determine the transaction (or transaction) that should be rolled back to break the deadlock. Such a transaction is called the victim transaction. The transaction that will lead to minimum loss, in terms of cost, should be chosen for rollback. The following factors determine the cost of a rollback :
How much execution has the transaction completed and for how much more time the transaction will execute to complete its task?§
How many data items were used by the transaction?§
How many more data items does the transaction need to complete?§
How many transaction will be included in the rollback?§

b. Rollback : after determining the transaction to be rolled back, you need to determine how far the transaction is to be rolled back. The easlest answer to this problem is to do a total rollback, which means that the transaction will be aborted and restarted. However, it is better to roll back the transaction only till the point where the deadlock can be broken. This method requires the DBMS to maintain information about all current transaction.

c. Starvation : when the selection of a victim is based on cost factors, it might happen ihat the same transaction is selected as a victim every time a deadlock occurs. Due to this, the transaction might not be able to complete its task. Such a situation is called starvation. To avoid starvation, you need to ensure that a transaction is picked as a victim for only a fixed number of times. To ensure this, you can select a victim based on the number of rollbacks along with the cost factor.

Question 5. a)

Explain take-grant model for operation system security with an example. Also explain the mechanisms of security in WIN 2000 operating system.

Solution :Modeling and analysis of information system vulnerabilities helps us to predict possible attacks to networks using the network configuration and vulnerabilities information. As a fact, exploiting most of vulnerabilities result in access rights alteration. In this paper, we propose a new vulnerability analysis method based on the Take-Grant protection model. We extend the initial Take-Grant model to address the notion of vulnerabilities and introduce the vulnerabilities rewriting rules to specify how the protection state of the system can be changed by exploiting vulnerabilities. Our analysis is based on a bounded polynomial algorithm, which generates the closure of the Take-Grant graph regarding vulnerabilities. The closure helps to verify whether any subject can obtain an access right over an object. The application of our results have been examined in a case study which reveals how an attacker can gain an unauthorized access right by exploiting chain of vulnerabilities.

Question 5. b) Explain Bell and La-Padula Model for security and protection. Why is security a critical issue in a distributed OS environment?

Solution: The Bell-LaPadula Model (abbreviated BLP) is a state machine model used for enforcing access control in government and military applications. It was developed by David Elliott Bell and Leonard J. LaPadula, subsequent to strong guidance from Roger R. Schell to formalize the U.S. Department of Defense (DoD) multilevel security (MLS) policy. The model is a formal state transition model of computer security policy that describes a set of access control rules which use security labels on objects and clearances for subjects. Security labels range from the most sensitive (e.g."Top Secret"), down to the least sensitive (e.g., "Unclassified" or "Public").

The Bell-LaPadula model focuses on data confidentiality and controlled access to classified information, in contrast to the Biba Integrity Model which describes rules for the protection of data integrity. In this formal model, the entities in an information system are divided into subjects and objects. The notion of a "secure state" is defined, and it is proven that each state transition preserves security by moving from secure state to secure state, thereby inductively proving that the system satisfies the security objectives of the model. The Bell-LaPadula model is built on the concept of a state machine with a set of allowable states in a computer network system. The transition from one state to another state is defined by transition functions.

Question 6). Write and explain an algorithm used for ordering of events in a distributed environment. Implements\ the algorithm with an example and explain?

Solution:

Vector clocks is an algorithm for generating a partial ordering of events in a distributed system and detecting causality violations. Just as in Lamport timestamps, interprocess messages contain the state of the sending process's logical clock. A vector clock of a system of N processes is an array/vector of N logical clocks, one clock per process; a local "smallest possible values" copy of the global clock-array is kept in each process, with the following rules for clock updates:

Example of a system of vector clocks

Initially all clocks are zero.
Each time a process experiences an internal event, it increments its own logical clock in the vector by one.
Each time a process prepares to send a message, it increments its own logical clock in the vector by one and then sends its entire vector along with the message being sent.
Each time a process receives a message, it increments its own logical clock in the vector by one and updates each element in its vector by taking the maximum of the value in its own vector clock and the value in the vector in the received message (for every element).

The vector clocks algorithm was independently developed by Colin Fidge and Friedemann Mattern in 1988

Free Assignments of BCA MCA all semester on Request.

2011-02-22T22:02:00.000-08:00

Dear Friends we are glad to yours response and always dedicated to work for you. If you are a student of IGNOU and want solution of the assignments of your semester please join us and leave your demanding semester so that we can upload the answer of the requested semester. Please note that as we work for free of cost so keep visiting the site as we are always looking for the appropriate solution to the question and leave the comment in case of any doubt.
Have a Nice Day :)

Free solution Of Ignou Assignments

2011-02-14T19:33:00.000-08:00

If you have any problem related to your assignment or you want to get solution of particular course we are there to help you for absolutely free of cost. Please keep visiting and make your valuable feedback regarding the blog so that we can improve our services. We only provide all the solution for which our reader demands, so keep visiting and ask for solution.

BCA MCA ALL SEMESTER ANSWERS ON DEMAND FOR FREE

2011-01-28T05:10:00.000-08:00

If you have any problem in solving your problems during assignments or you want to find the solution to the answer please write the particular question of your assignment and we will respond to you with correct answer. Have a nice day and best of luck to your future.

2011-01-11T05:16:00.001-08:00

Join IGNOU ASSIGNMENT

@
www.SirfDosti.com

MAKE MONEY FROM HOME

2011-01-08T00:58:00.001-08:00

Earn upto Rs. 9,000 pm with PaisaLive.com!
Hi ,

I have something interesting for you - you can easily earn regular income online via PaisaLive.com!

It’s really amazing! You get paid to open & read the contents of PaisaLive mails. You also receive special discount coupons, promotions and free passes to various events in your city.

Join now and get Rs. 99 instantly, just for joining. What more, as a special bonus you get paid for inviting your friends also!

Create your PaisaLive Account by clicking below and spent just 15 minutes and earn money.

http://www.PaisaLive.com/register.asp?1149451-5925202

PaisaLive - Get Paid to read emails

{Insert your Name Here}

mcs042

2011-01-05T00:13:00.000-08:00

What is count to infinity problem? Explain through an example.

Count-To-Infinity Problem

Followed illustration shows an imagined network and denotes the distances from router A to every other router. Until now every thing works fine.

The illustration shows that link (A, B) is broken. Router B observed it, but in his routing table he sees, that router C has a route to A with 2 hops.
The problem is, that router B doesn't know that C has router B as successor in his routing table on the route to A.
That occurs followed count-to-infinity problem. B actualizes his routing table and takes the route to A over router C.
In the next picture, we can see the new distances to A. In C's routing the route to A contains router B as next hop router, so if B has increase his costs to A, C is forced to do so. Router C increases his cost to A about B + 1 = 4.
Now we see the consequence of the distributed Bellman-Ford protocol: Because router B takes the path over C to A, he reactualizes his routing table and so on!
At the end this problem is going to immobilize the whole network.

MCA 1st semester

2011-01-03T22:13:00.000-08:00

Answer of Assignment of current April 2011 session will be posted soon please keep visiting.

BCA 2nd Semester

2011-01-03T22:11:00.000-08:00

Solution of assignment of june 2011 will be posted soon. Know about particular topic please go through comments.

BCA 1st Semester

2011-01-03T22:08:00.000-08:00

Solution of BCA 1st Semester of April 2011 session will be posted soon. Please keep visiting.

IGNOU MCA 4th Semester

2011-01-03T00:21:00.000-08:00

MCS-042

Explain the operation of leaky and token bucket algorithms?

The Token Bucket(Leaky Bucket) Model

This is introduction and basic explanation of the model, it is not completed . First part is similar to the document which submitted by Dr. Chimento. The other parts briefly explain the idea.

This document tries to explain basic token bucket model(some document use leaky bucket), its parameters and its usage in diffserv environment. Token bucket represents the Policing function of Traffic Conditioning Block of diffserv. A token bucket flow is defined by (r,b), r denotes the rate at which tokens(credits) are accumulated and b is the depth of the token pool(in bytes).



b



Token
wait

Packets



As it shown above the new token are adding to the bucket at rate of r tokens/sec, the maximum token can be accumulated is b bytes. If the bucket is full, the incoming tokens will be thrown away. The Token Bucket(TB) profile contains three parameters: an average rate, a peak rate, and burst size. Lets look at the meaning of these parameters:

Average rate: The network may wish to limit the long-term average rate at which a flow’s packets can be sent into the network. The important point here is the interval of time over which the average rate will be policed. For example; a flow which has 10000 packets/sec is more flexible than a flow which has 10 packets/msec. Even though both have the same average rate, their behaviors are different. The flow which has 10000 packets/sec can send 100 packets in a one msec long-term, but the flow which has 10 packets/msec cannot sent 100 packets in a one msec long-term. This is one of the important feature we should consider.

Peak rate: Defines the maximum rate at which packets can be sent in short interval time. In above example; even though the flow’s long term average rate is 10000 packets/sec peak rate can be limited to 200 packets/msec.

Burst size: Burst size is the maximum number of packets which can be transmitted extremely in a short interval time. In other word, when time interval approaches zero, the burst size limits the number of packets which can be transmitted to the network. when we don’t consider time interval as zero, burst size depends on bucket depth b, and maximum speed of output link . We will clarify in the next section in more detail.

In any interval t [S,S+t], let say in time S, and the number of tokens in the bucket are a(a<=b), and during time interval [S,S+t] the counter accumulate rt tokens more. The total number of tokens available for source to transmit packets is a+rt, (a+rt<=b+rt). When the bucket size is b and during interval time [S,S+t], b+rt tokens accumulated. In this case bucket gets empty by the maximum rate of output link. If we call maximum speed of output B. The time which buckets gets empty b+rt=Bt è t=b/(B-r). Choosing all these parameters can be a bit tricky. One of the potential problem with token bucket is it allows burst again. If another burst come during this time interval, it cannot be handled, or in the case of bursts come very often the output speed will be maximum for the period of burst which can cause congestion. This problem can be reduced by carefully choosing parameters.
Essentially, the network has to simulate the algorithm and make sure that no more packets or bytes are being sent than are permitted. Depends on the agreements the excess packets can be dropped, remarked etc.

When we concern about delay and reliability, we could think some other aspects of the parameters. If we consider single flow, delay can be expressed as
Delay=b/R+C/R +D R is the maximum transmit rate of output link, D is non-rate-dependent delay, C is the delay which depends on flow transmission rate. C and D are local parameters. The delay in node must be less than the above value[6]. The end-to-end delay for this flow is the sum of all the delay through the route.
The leaky bucket is an algorithm used in packet switched computer networks and telecommunications networks to check that data transmissions conform to defined limits on bandwidth and burstiness (a measure of the unevenness or variations in the traffic flow). The leaky bucket algorithm is also used in leaky bucket counters, e.g. to detect when the average or peak rate of random or stochastic events orstochastic processes exceed defined limits.
The Leaky Bucket Algorithm is based on an analogy of a bucket (figure 1) that has a hole in the bottom through which any water it contains will leak away at a constant rate, until or unless it is empty. Water can be added intermittently, i.e. in bursts, but if too much is added at once, or it is added at too high an average rate, the water will exceed the capacity of the bucket, which will overflow.
There are actually two different methods of applying this analogy described in the literature . These give what appear to be two different algorithms, both of which are referred to as the leaky bucket algorithm. This has resulted in confusion about what the leaky bucket algorithm is and what its properties are.
In one version , the analogue of the bucket is a counter or variable, separate from the flow of traffic, and is used only to check that traffic conforms to the limits, i.e. the analogue of the water is brought to the bucket by the traffic and added to it so that the level of water in the bucket indicates conformance to the rate and burstiness limits. This version is referred to here as the leaky bucket as a meter. In the second version ^[2], the traffic passes through a queue that is the analogue of the bucket, i.e. the traffic is the analogue of the water passing through the bucket. This version is referred to here as the leaky bucket as a queue. The leaky bucket as a meter is equivalent to (a mirror image of) the token bucket algorithm, and given the same parameters will see the same traffic as conforming or nonconforming. The leaky bucket as a queue can be seen as a special case of the leaky bucket as a meter .