UGC NET SOLVED-COMPUTER SCIENCE

JANUARY 2017 - PAPER II

2017-09-14T03:56:00.003-07:00

JANUARY 2017 – PAPER II

OPERATING SYSTEMS

37. In a paging system, it takes 30 ns to search translation Look-a-side Buffer (TLB) and 90 ns to access the main memory. If the TLB hit ratio is 70%, the effective memory access time is

(1) 48ns

(2) 147ns

(3) 120ns

(4) 84ns

Ans :- 2

Explanation:-

This is a paragraph taken from “Operating systems” book by Silberschatz.

In order to calculate the effective memory access time, we are going to apply the explanation given in the above mentioned book.

A 70 percent ratio means that we find the desired page number in the TLB 70 percent of the time. If it takes 30 ns to search the TLB and 90 ns to access memory, then a mapped-memory access takes 30+90 ns when the page number is in the TLB. If we fail to find the page number in the TLB(30), then we must first access memory for the page table and frame number(90ns) and then access the desired byte in the memory(90ns), for a total of 210ns. To find the effective memory access time, we weight each case by its probability,

Effective access time = 0.70 X 120 + 0.30 X 210

= 84 + 63 = 147ns

So, the correct answer is 147ns. The option is 2.

DIFFERENCE BETWEEN C,C++ AND JAVA

2017-09-14T03:10:00.002-07:00

DIFFERENCES BETWEEN C,C++ AND JAVA IN A NUTSHELL

S.NO	C	C++	JAVA
1.	Structured programming language	Object-oriented programming language	Object-oriented programming language
2.	Compiled	Compiled	Compiled and interpreted
3.	32 keywords Eg:- case, char	60 keywords Eg:- class, protected	50 keywords Eg:- interface, extends
4.	Rules for naming identifier 1. Should begin with alphabet 2. Can have numbers,underscore 3. Keywords cannot used as identifiers	Rules for naming identifier 1. Should begin with alphabet 2. Can have numbers,underscore 3. Keywords cannot used as identifiers	Rules for naming identifier 1. Can begin with alphabet, underscore or dollar symbol. But cannot begin with a number 2. Keywords cannot used as identifiers
5.	Comments:- /…./ (multi line comment), //(single line comment)	Comments:- /…./ (multi line comment), //(single line comment)	Comments:- /…./ (multi line comment), //(single line comment), /** …. */(documentation comment)
6.	Data types :- int,char,float,double,void	Data types :- int,char,float,double,void	Data types:- byte,short,int,long, float,double, char,boolean
7.	Bitwise shift operators are : >>(right shift),<<(left shift)	Bitwise shift operators are : >>(right shift),<<(left shift)	Bitwise shift operators are : >>(right shift),<<(left shift),>>>(triple right shift operator)
8.	Char is 8 bits wide	Char is 8 bits wide	Java uses Unicode to represent characters. It is 16 bits wide.
9.	Preprocessor directives allowed Eg:- #include,#define	Preprocessor directives allowed Eg:- #include,#define	No preprocessor directives
10.	Variables need not be initialized before using it	Variables need not be initialized before using it	Variables must be initialized before using it
11.	Array should be declared mentioning its size Eg:- int num[10];	Array should be declared mentioning its size Eg:- int num[10];	Arrays are declared without mentioning its size. int num[]; num=new int[10] (or) int num[]=new int[10];
12.	Explicit string data type not available	Explicit string data type not available	String data type available in java. It is a class
13.	Supports pointers	Supports pointers	Does not support pointers
14.	Structures and unions available	Structures and unions available	Structures and unions not available
15.	-	Supports operator overloading	Does not support operator overloading
16.	-	Programs can be written in c++ without using a class	All the code in java program needs to be written in one or more classes
17.	Global variables and functions are allowed	Global variables and functions are allowed	No global variables or functions
18.	-	Allows default arguments in functions	Not allowed in java
19.	-	Supports multiple inheritance	Multiple inheritance not supported
20.	-	Supports constructors and destructors	Supports constructors, no destructors
21.	Supports typedef	Supports typedef	Does not support typedef
22.	Unsigned integers can be declared	Unsigned integers can be declared	No unsigned integers
23.	Allows goto	Allows goto	Does not allow goto
24.	-	delete operator available	No delete operator
25.	-	Objects may be passed by value or by reference	Objects are passed by reference only in Java

JANUARY 2017 - OPERATING SYSTEM QUESTION - PAPER III

2017-09-14T03:08:00.000-07:00

JANUARY 2017 – PAPER III

OPERATING SYSTEM QUESTIONS

49. A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively :

(1) 14 and 15

(2) 14 and 29

(3) 15 and 14

(4) 16 and 32

Ans:- 3

Explanation:

Always page size = frame size for minimizing the internal fragmentation.

LOGICAL ADDRESS CALCULATION

In the above question, number of pages = 64, and 512 bytes page size.

64 = 2⁶ . So, number of bits to represent pages = 6.

512=2⁹. So, number of bits to represent bytes per page size = 9.

So, the number of bits for logical address = 6 + 9 = 15

PHYSICAL ADDRESS CALCULATION

In the above question, number of page frames = 32, and frame size is the same as page size which is 512 bytes.

32 = 2⁵. So, number of bits to represent frames = 5.

512=2⁹. So, number of bits to represent bytes per frame size = 9.

So, the number of bits for physical address = 5 + 9 = 14.

Therefore, the number of bits required in logical and physical address are 15 and 14 respectively.

So, the correct answer is 3.

OPERATING SYSTEM QUESTIONS

2017-09-13T09:17:00.000-07:00

OPERATING SYSTEMS

DECEMBER 2004

36. Semaphores are used to :

(A) Synchronise critical resources to prevent deadlock

(B) Synchronise critical resources to prevent contention

(D) Facilitate memory management

Ans :- A

Explanation :-

In order to understand semaphores, we should first understand the problem which it is tryig to solve. It is called “The critical-section problem”. What is “Critical-Section Problem”?. Consider a system consisting of n processes. Each process has a segment of code, called a critical section, in which the process may be changing common variables, updating a table, writing a file etc. The important thing is that, when one process is executing in its critical section, no other process is to be allowed to execute in its critical section. Thus the execution of critical sections by the processes is mutually exclusive in time.

The critical section problem is to design a protocol that the processes can use to cooperate. To overcome this difficulty, we can use a synchronization tool called a semaphore.

37. In which of the following storage replacement strategies, is a program placed in the largest available hole in the memory?

(A) Best fit

(B) First fit

(D) Buddy

First-fit, best-fit and worst-fit are the most common strategies used to select a free hole from the set of available holes.

First-fit : allocate the first hole that is big enough.

Best-fit : Allocate the smallest hole that is big enough.

Worst-fit : Allocate the largest hole.

Ans :- C

38. Remote computing systems involves the use of timesharing systems and :

(A) Real time processing

(B) Batch processing

(D) All of the above

Ans:- B

39. Non modifiable procedures are called

(A) Serially usable procedures

(B) Concurrent procedures

(D) Topdown procedures

Ans :- C

Explanation : -

We would have guessed that it comes under the “Memory management” topic. Non modifiable procedures are referred to when a discussion on advantages of paging is done. One important advantage of paging is the possibility of sharing common code. This is particularly important in a time-sharing environment. Let us understand this with an example.

Consider a system that supports 40 users, each of whom executes a text editor. If the text editor consists of 150K of code and 50K of data space, we would need 8000K to support the 40 users. If the code is reentrant, however, it can be shared. However, each process will have its own data page.

Reentrant code also called pure code is non-self-modifying code. If the code is reentrant, then it never changes during execution. So, the correct answer for the above question is non modifiable procedures are called reentrant procedures.

40. Match the following:

(a) Disk scheduling	(1) Round robin
(b) Batch processing	(2) Scan
(C)Time sharing	(3) LIFO
(d) Interrupt processing	(4) FIFO

(A) a-3,b-4,c-2,d-1

(B) a-4,b-2,c-2,d-1

(D) a-3,b-4,c-1,d-2

Ans:- C

Explanation :-

This is a easy question. You can first choose the obvious ones. For example, Time sharing is associated with round robin. So, C-1. But there are two options where C is matched with 1. So, let us look at other alternatives. Scan is one of the disk scheduling algorithm and so, we can match a with 2. So, batch processing matches to 4 and the last one matches d with 3. So, the correct answer is C.

JUNE 2005

36.Moving Process from main memory to disk is called :

(A)	Caching	(B)	Termination
(C)	Swapping	(D)	Interruption

Ans :- C

Explanation:-

Moving process from main memory to disk and vice versa is called swapping. Swapping is one of the memory management techniques used by the operating system. Since the size of the RAM is limited and finite, all the processes or programs to be executed cannot be made to fit in it. So the disk is also treated as an extension of the memory and is referred to as virtual memory. Moving a process from main memory to disk is called swapping.

37.The principle of Locality of reference justifies the use of :

(A)	Virtual memory	(B)	Interrupts
(C)	Cache memory	(D)	Secondary memory

Ans :- C

In computer science, locality of reference, also called the principle of locality, is the term applied to situations where the same value or related storage locations are frequently accessed. There are three basic types of locality of reference: temporal, spatial and sequential:

Temporal locality
Here a resource that is referenced at one point in time is referenced again soon afterwards.
Spatial locality
Here the likelihood of referencing a storage location is greater if a storage location near it has been recently referenced.
Sequential locality
Here storage is accessed sequentially, in descending or ascending order.

The reason locality occurs is often because of the manner in which computer programs are created. Generally, data that are related are stored in consecutive locations in storage. One common pattern in computing is that processing is performed on a single item and then the next. This means that if a lot of processing is done, the single item will be accessed more than once, thus leading to temporal locality of reference. Furthermore, moving to the next item implies that the next item will be read, hence spatial locality of reference, since memory locations are typically read in batches.

Locality often occurs because code contains loops that tend to reference arrays or other data structures by indices.

Increasing and exploiting locality of reference are common techniques for optimization. This can happen on several levels of the memory hierarchy. Paging obviously benefits from spatial locality. A cache is a simple example of exploiting temporal locality, because it is a specially designed faster but smaller memory area, generally used to keep recently referenced data and data near recently referenced data, which can lead to potential performance increases. Data in cache does not necessarily correspond to data that is spatially close in main memory; however, data elements are brought into cache one cache line at a time. This means that spatial locality is again important: if one element is referenced, a few neighbouring elements will also be brought into cache. Finally, temporal locality plays a role on the lowest level, since results that are referenced very closely together can be kept in the machine registers. Programming languages such as C allow the programmer to suggest that certain variables are kept in registers.

38.Banker’s algorithm is for.

(A)	Dead lock Prevention	(B)	Dead lock Avoidance
(C)	Dead lock Detection	(D)	Dead lock creation

Ans :- B

Explanation :- Banker’s algorithm is used for dead lock avoidance.

39.Which is the correct definition of a valid process transition in an operating system ?

(A)Wake up : Ready Running

(B)Dispatch : Ready Running

(C)Block : Ready Running

(D)Timer run out : Ready Blocked

Ans :- C

NOVEMBER 2017 - JRF AND NET EXAM

2017-06-13T22:26:00.002-07:00

Hi all,

CBSE has published the notification for the conduct of the JRF and NET EXAM exam in the month of November, 2017.

Click on the following link for more details....

http://cbsenet.nic.in/CMS/Handler/FileHandler.ashx?i=File&ii=49&iii=Y

DECEMBER 2015 - PAPER III - Q NO 39

2017-04-17T00:23:00.000-07:00

DECEMBER 2015 – PAPER III – Q.NO 39

This question is based on the concept of deadlock avoidance. In order to understand the theory behind this concept, refer to my earlier post on the same topic.

http://ugcnetsolved-computerscience.blogspot.in/2017/04/deadlock-avoidance.html

Let us try to understand the question and solve it based on the theory given above.

Consider a system with twelve magnetic tape drives and three processes p1,p2 and p3. Process p1 requires maximum ten tape drives, process p2 may need as many as four tape drives and p3 may need upto nine tape drives. Suppose that at time t1, process p1 is holding five tape drives, process p2 is holding two tape drives and process p3 is holding three tape drives. At time t1, system is in :
(1) Safe state
(2) Unsafe state
(3) Deadlocked state
(4) Starvation state

Explanation:-

Below is a table given depicting the maximum needs and current needs of each process.

	MAXIMUM NEEDS	CURRENT NEEDS
p0	10	5
p1	4	2
p2	9	3

Remember, there is no specific formula to solve this problem. It is more of understanding, logic and common sense for solving this.

Total number of magnetic tapes = 12

Already allocated = 10

Number of free magnetic tapes = 12 – 10 =2

From the table, we can understand that process p2 requires 2 more tapes for completion and since two tapes are free, they are allocated to process p2.

Process p2 completes its execution, and returns all the four tapes to the sytem.

So, number of free tape drives = 4.

Since process p1 is allocated 5 tape drives, but has a maximum of 10, it may then request 5 more tape drives. Since they are unavailable, process p1 must wait. Similarly, process p3 may request an additional 6 tape drives and have to wait, resulting in a deadlock. So the system is in deadlocked state.

The correct option is 3.

DEADLOCK AVOIDANCE

2017-04-16T23:19:00.001-07:00

Deadlock avoidance is a very important topic in operating system. So, first let us understand the basics related to deadlock.

What is a deadlock?

In a multiprogramming environment, there are several processes running concurrently. Each of these processes may compete for a limited or finite number of resources. If a process requests resource, and it is not available, then it has to wait for the resource, and so it enters into wait state. But the waiting process may never change state, because the resources they have requested are held by other waiting processes. Such a situation is called deadlock.

Safe state

A system is in a safe state if it can allocate resources to each process in some order and still avoids a deadlock.

ILLUSTRATION

Let us consider a system with 12 magnetic tape drives and 3 processes. p0,p1 and p2. Process p0 requires 10 tape drives, process p1 requires 4 tape drives and process p2 may need up to 9 tape drives. Suppose, at time t0, process p0 is holding 5 tape drives, process p1 is holding 2 tape drives and process p2 is holding 2 tape drives.

	MAXIMUM NEEDS	CURRENT NEEDS
p0	10	5
p1	4	2
p2	9	2

Now, the systems is in safe state. The question is, can the sequence,<p1,p0,p2> satisfy the safety condition.

EXPLANATION

TOTAL NUMBER OF TAPE DRIVES = 12

NUMBER OF FREE TAPE DRIVES = 12-9=3

p1 requires 2, allotted 2, so number of free tape drives = 1

Since p1 got its maximum need, it finishes its work and returns all the 4 tape drives.

NUMBER OF FREE TAPE DRIVES NOW = 1+4 = 5

p0 requires 5, allotted 5, so number of free tape drives = 0

Since p0 got its maximum need, it finishes its work and returns all the 10 tape drives.

NUMBER OF FREE TAPE DRIVES NOW = 10

p2 requires 7, allotted 7, and p2 also got its maximum need and it finishes its work and returns all the tape drives.
The system is in safe state if the sequence is <p1,p0,p2>.

UNSAFE STATE

Let us consider, that at time t1, process p2 requests 1 tape drive and is allotted 1 more. So, the maximum and current needs are depicted as below.

	MAXIMUM NEEDS	CURRENT NEEDS
p0	10	5
p1	4	2
p2	9	3

TOTAL NUMBER OF TAPE DRIVES = 12

NUMBER OF FREE TAPE DRIVES = 12 - 10=2

p1 requires 2, allotted 2, so number of free tape drives = 0

Since p1 got its maximum need, it finishes its work and returns all the 4 tape drives.

Now, neither p0 nor p2 can be allotted the rest for completion. So, the system is in unsafe state.

Let us try to solve some problems on deadlock avoidance.

JUNE 2013-PAPERIII QUESTION NO 61

An operating system using banker’s algorithm for deadlock avoidance has ten dedicated devices (of same type) and has three processes P1, P2 and P3 with maximum resource requirements of 4, 5 and 8 respectively.
There are two states of allocation of devices as follows : State 1 Processes P1 P2 P3
Devices allocated 2 3 4
State 2 Processes P1 P2 P3
Devices allocated 0 2 4
Which of the following is correct ?
(A) State 1 is unsafe and state 2 is safe.
(B) State 1 is safe and state 2 is unsafe.
(C) Both, state 1 and state 2 are safe.
(D) Both, state 1 and state 2 are unsafe.

Ans:- A

EXPLANATION:-

	MAXIMUM NEEDS	CURRENT NEEDS
p1	4	2
p2	5	3
p3	8	4

NUMBER OF FREE DEVICE = 10
NUMBER OF FREE DEVICE = 10-9=1
Since the number of free device is just 1, it can be allocated to any of the processes p1,p2 or p3.
p1 requires 2, as it is allocated only 2.
p2 requires 2, because it is already allocated 3 and its maximum need is 5.
p3 requires 4, because it is already allocated 4 and its maximum need is 4.
Since, none of the processes requirement can be satisified, the state 1 is in unsafe state.

	MAXIMUM NEEDS	CURRENT NEEDS
p1	4	0
p2	5	2
p3	8	4

Let us see whether state 2 is in safe or unsafe state.
TOTAL NUMBER OF FREE DEVICES = 10
NUMBER OF FREE DEVICE=10-6=4

Process p1 requires a maximum of 4 devices and they can be allocated, because the number of free device right now is 4. Since p1 gets its maximum need, it completes the process and returns all of the 4 devices.
Process p2 requires a maximum of 5, which can be allocated right now. p2 completes its process and returns all the 5 devices.
Process p3 requires a maximum of 8, which also can be allocated right now. So, p3 also can finish its execution successfully.
So, state 2 is in safe state So, the correct answer for the question is State 1 is in unsafe state and state 2 is in safe state which is option A.

DECEMBER 2011 - PAPER II

2015-11-12T05:49:00.002-08:00

36. X – = Y + 1 means

(A) X=X– Y +1
(B) X = –X – Y – 1
(C) X=–X+Y+1
(D) =X– Y – 1

Ans:-A

37. Handoff is the mechanism that

(A) transfer an ongoing call from
one base station to another
(B) initiating a new call
(C) dropping an ongoing call
(D) none of above

Ans:-A

38. Which one of the following statement is false ?

(A) Context-free languages are closed under union.
(B) Context-free languages are closed under concatenation.
(C) Context-free languages are closed under intersection.
(D) Context-free languages are closed under Kleene closure.

Ans:-C

Explanation:- CFL's are closed under union, concatenation, and Kleene closure. But not under intersection or difference. So the correct answer is option C.

39. All of the following are examples of real security and privacy risks except

(A) Hackers
(B) Spam
(C) Viruses
(D) Identify theft

Ans:-B

40. Identify the incorrect statement :

(A) The ATM adoption layer is not service dependent.
(B) Logical connections in ATM are referred to as virtual channel connections.
(C) ATM is streamlined protocol with minimal error and flow control capabilities
(D) ATM is also known as cell delays.

Ans:-A

41. Software risk estimation involves following two tasks :

(A) Risk magnitude and risk impact
(B) Risk probability and risk impact
(C) Risk maintenance and risk impact
(D) Risk development and risk impact

Ans:-B

42. The number of bits required for an IPV6 address is

(A) 16
(B) 32
(C) 64
(D) 128

Ans:-D

43. The proposition ~ q–vp is equivalent to

(A)p→q
(B) q→p
(C) p↔q
(D) p∨q

Ans:-B

44. Enterprise Resource Planning (ERP)

(A) has existed for over a decade.
(B) does not integrate well with the functional areas other than operations.
(C) is inexpensive to implement.
(D) automate and integrates the majority of business processes.

Ans:-D

45. Which of the following is false concerning Enterprise Resource Planning (ERP) ?

(A) It attempts to automate and integrate the majority of business processes.
(B) It shares common data and practices across the enterprise.
(C) It is inexpensive to implement.
(D) It provides and access information in a real-time environment.

Ans:-C

46. To compare, overlay or cross analyze to maps in GIS

(A) both maps must be in digital form
(B) both maps must be at the same equivalent scale.
(C) both maps must be on the same coordinate system
(D) All of the above

Ans:-D

47. Web Mining is not used in which of the following areas ?

(A) Information filtering
(B) Crime fighting on the internet
(C) Online transaction processing
(D) Click stream analysis.

Ans:-B

48. A telephone conference call is an example of which type of communications ?

(A) same time / same place
(B) same time / different place
(C) different time / different place
(D) different time / same place

Ans:-B

49. What is the probability of choosing correctly an unknown integer between 0 and 9 with 3 chances ?

(A) 963/1000
(B) 973/1000
(C) 983/1000
(D) 953/1000

Ans:-A

The number of nodes in a complete binary tree of height h (with roots at level 0) is equal to (A) 2⁰ + 2¹ + ..... 2^h
(B) 2⁰ + 2¹ + ..... 2^{h – 1}
(C) 2⁰ + 2¹ + ..... 2^{h + 1}
(D) 2¹ + ..... 2^{h + 1}

Ans:-A

JUNE 2012 - PAPER III

2015-11-12T05:48:00.000-08:00

4. What is Granularity ?

(A) The size of database
(B) The size of data item
(C) The size of record
(D) The size of file

Ans:-B

Explanation:- Granularity refers to the size of any data item. It could be a tuple, relation, database,anything. So the option is B.

5. Suppose that a given application is run on a 64-processor machine and that 70 percent of the application can be parallelized. Then the expected performance improvement using Amdahl’ s law is

(A) 4.22
(B) 3.22
(C) 3.32
(D) 3.52

Ans:-B

Explanation:-
According to Amdahl's law, in case of parallelization,if P is the proportion of the program that can be made parallel, then (1-P) is the proportion that cannot be parallelized. Then the maximum speedup that can be achieved by using N processor is, S(N)=1/(1-P)+P/N where N refers to the no of processors, and P refers to the proportion that can be parallelized.
In the question above, no of processor, N=64
The proportion of the program that can be made parallel, P = 70%=0.7
So, substituting in the formula, we get, 1/(1-0.7)+0.7/64=1/0.3+0.0109375=1.0.310975
=3.215
=3.22
Therefore, the option is B.

6. If two fuzzy sets A and B are given with membership functions μA(x) = {0.2, 0.4, 0.8, 0.5, 0.1} μB(x) = {0.1, 0.3, 0.6, 0.3, 0.2} Then the value of μ ––– willbe A∩B

(A) {0.9, 0.7, 0.4, 0.8, 0.9}
(B) {0.2, 0.4, 0.8, 0.5, 0.2}
(C) {0.1, 0.3, 0.6, 0.3, 0.1}
(D) {0.7, 0.3, 0.4, 0.2, 0.7}

Ans:-A

Explanation:-
The fuzzy intersection of two fuzzy sets A and B on universe of discourse X: μA∩B(x) = min [μA(x), μB(x)] , where x∈X
But here in the question, they are asking for complement of AintersectionB and so the answer would be 1-min[A(x),B(x)]. The minimum of 0.2 and 0.1 will be 0.1, and 1-0.1 will be 0.9. The second value is min(0.4,0.3)=0.3 and 1-0.3=0.7
The third value is min(0.8,0.6)=0.6 and 1-0.6=0.4
The fourth value is min(0.5,0.3)=0.3 and 1-0.3=0.7
The last value is min(0.1,0.2)=0.1 and 1-0.1=0.9
The only option which has got the values 0.9,0.7,0.4,0.7 and 0.9, although the fourth value is given as 0.8 instead of 0.7 is option A. So the answer is option A.

OPEN SYSTEMS INTERCONNECTION MODEL

2015-11-12T05:45:00.000-08:00

This model was created by the International Organization for Standardisation(ISO). This provides a framework for communication between any two networks. There are 7 layers in this model. The lowest is the physical layer and the highest one is the application layer.The purpose and design issues of each of the 7 layers are different. Each layer performs a particular activity concerned with the task of communication between devices. Although I will be giving you a concise explanation for the design issues of each layer, go through the following diagram well. It explains a lot about the different layers of OSI.

Following image taken from google images.

Following are the different layers of the OSI model and their purpose.

7.APPLICATION LAYER
This layer provides network services to user applications. It consists of a set of protocols that define specific user-oriented applications such as e-mails, file transfers and virtual terminals. Some of the protocols used in this layer are TELNET(Virtual terminal),FTP(File transfer),E-mail(SMTP and POP), and also HTTP.

6.PRESENTATION LAYER
This layer is responsible for how the data looks or is formatted. It is concerned with the syntax and semantics of data getting transmitted. Data compression, encryption and decryption happen at this layer. In practice, this layer is usually incorporated with the application layer.

5.SESSION LAYER
Provides for how two devices establish, maintain and manage a connection. Also responsible for enforcing the rules of dialog,synchronising the flow of data and reestablishing a connection in the event a failure occurs.

4.TRANSPORT LAYER
Ensures error-free delivery of data. Accepts data from the session layer, partitions the data into smaller packets, passes them to the network layer and ensures that the data arrives correctly at the destination.

3. NETWORK LAYER
Responsible for end-to-end routing or switching of data using intermediate routers.

2. DATA LINK LAYER
This layer groups the bits that are received(or to be transmitted)into what are called as frames. This layer provides for error detection, framing and flow control. Also handles lost, damaged or duplicate frames.

1. PHYSICAL LAYER
This is the lowest layer of the OSI reference model. This layer is responsible for the actual transmission of bits from one end to another.

The message to be transmitted is referred by different terms in different layers. Let us go through them.
7. Application layer - Message
6. Presentation layer - Message
5. Session layer - Message
4. Transport layer - Message
3. Network layer - packets
2. Data link layer - Frames
1. Physical layer - Bits

PREVIOUS EXAM QUESTIONS AND ANSWERS

1. Match the following:
(a) Physical layer (i) Allow resources to network access
(b) Data link layer (ii) Move packets from one destination to another
(c) Network layer (iii) Process to process message delivery
(d) Transport layer (iv) Transmission of bit stream
(e) Application layer (v) Formation of frames

Codes:	a	b	c	d	e
(A)	iv	v	ii	iii	i
(B)	v	iv	i	ii	iii
(C)	i	iii	ii	v	iv
(D)	i	ii	iv	iii	v

June 2014 - Paper II Q.No 5
Ans:- A
Explanation:- If you find many options confusing choose one which you are very confident and about and then compare the rest of the options. Here, Transmission of bit streams is the work of the physical layer is something which one can be very sure about. So, a is matched to iv. There is only one option which does that and it is A. So just compare the rest of the codes and confirm that they are right. Data link layer is formation of frames. Network layer move packets from one destination to another. Transport layer is process to process message delivery and finally Application layer is allow resources to network access. So the correct answer is A.

2. Which layers of the OSI reference model are host-to-host layers?
(A) Transport, session, presentation, application
(B) Network, transport, session, presentation
(C) Data link, network, transport, session
(D) Physical, data link, network, transport
June 2014-Paper III
Ans:- A
Explanation:- The host-to-host layers are the four from the upper layers. Application, presentation, session, transport. The last three are referred to as media layers.

3. Encryption and decryption is the responsibility of _____layer.
(A) Physical
(B) Network
(C) Application
(D) Data link
December 2013 - Paper II
Ans:- C
Explanation:- Since the presentation layer is not available as an option, the right answer is Application. In the explanation above for Presentation layer, there is a mention that in practice it is combined with Application. So the right answer is C.

4. Which layer of OSI reference model is responsible for decomposition of messages and generation of sequence numbers to ensure correct re-composition from end to end of the network?
(A) Physical
(B) Data link
(C) Transport
(D) Application
December 2013 - Paper III
Ans:-C
Explanation:- Actually physical, data link and application are completely ruled out for the above question and so the right answer is C.

There are many more questions from the same topic which will be covered later.

RSA Algorithm

2015-11-12T05:43:00.000-08:00

Using RSA algorithm, what is the value of cipher text C, if the plain text M = 5 and p = 3, q = 11 & d = 7 ?
(A) 33
(B) 5
(C ) 25
(D )26

Ans:- D

Solution:-

Given p=3,q=11. n=(p X q) = (3 X 11)=33
   m=(p-1)X(q-1) = (2 X 10)=20
Find a small odd integer e, that is relatively prime to m.
If e=3, then GCD(3,20)=1. e should be small and prime and so we let e=3.
d is given, d=7.
Public key = (e,n). (Values of e and n are known).
To encrypt a message, we apply the public key to the function
E(s) = se(mod n),
where s is the given message and e and n represent the public key integer pair. In the above question, the plain text M = 5. Plain text needs to be encrypted using the above formula.
E(s) = s^e(mod n)
   =5³(mod 33)
   = 125 (mod 33)
   = 26
As a result, the encrypted message E(s) = 26. This is what gets transmitted.

Coupling and cohesion - Software Engineering

2015-10-14T18:19:00.001-07:00

Columnar Transposition Cipher

2015-07-21T09:19:00.001-07:00

Fractional Knapsack Problem

2015-07-21T07:44:00.002-07:00

Click on the following link to understand fractional knapsack problem, with an example.

Fractional Knapsack

KSET 2015 NOTIFICATION

2015-06-29T08:16:00.001-07:00

KSET 2015 NOTIFICATION

Applications are invited from eligible candidates for Karnataka state eligibility test for Lecturer/Assistant professor. The examination is to be held on 6th December 2015. Click on the following link for more information and application details .

KSET 2015 NOTIFICATION

Visit this blog for computer science papers preparation and also tips on cracking paper - I.

DECEMBER 2010 - QUESTION NO 25

2015-06-19T08:35:00.000-07:00

This particular question is not clear for few readers. So, I am taking it up once again.

25. Which of the following can be the sequence of nodes examined in binary search tree while searching for key 88 ?

(A) 90, 40, 65, 50, 88
(B) 90, 110, 80, 85, 88
(C) 190, 60, 90, 85, 88
(D) 65, 140, 80, 70, 88

Ans:- B

Explanation:-
I am taking the liberty of retaining an explanation given for a similar question from yahoo answers.

Best Answer: For a binary search tree, the entire tree is sorted such that for any node, every node to the left is less than the current node value and every node to the right is more than the current node value.

When walking a BST tree, you "zero in" on the value by following the correct nodes down the tree. Ideally you work closer and closer to your answer, kind of like the "guess the number" game where you give "nope, more!" and "nope, less!" hints.

a) 2, 399, 387, 219, 266, 382, 381, 278, 363
This sequence is possible.

b) 935, 278, 347, 621, 299, 392, 358, 363
This sequence is not possible. Since 621 is to the right of 347, every other node under 621 must ALSO be to the right of 347. 299 is not -- so this is an impossible walk. Basically, once you pass 347, 299 is swinging too far in the opposite direction.

c) 924, 220, 911, 244, 898, 258, 362, 363
This sequence is possible.

d) 925, 202, 911, 240, 912, 245, 363
Not possible -- 240 is to the left of 911, so every other node must also be less than 911 (but still may be to the right of 240). 912 is not to the left of 911.

e) 2, 252, 401, 398, 330, 344, 397, 363 This sequence is possible. Just barely though -- 397 is to the left of 398 but just barely!

I am taking the example b which explains why a particular sequence is not possible. Since 621 is to the right of 347, every other node under 621 must ALSO be to the right of 347.

Let us apply the same rule to all the sequences given above for us in the question.
The first sequence is 90,40,65,50,88. Do not consider any particular element as the root. But start analysing the numbers from first. 90 is the first number. 40 is the second one. So, 40 will be to the left of 90, since it is less than 90. Since 40 is to the left of 90, all numbers following 40 also should be to the left of 90, which is true in this sequence. 65,50, and 88 will be to the left of 90. So this sequence is possible.

Let us consider the second sequence which is 90,110,80,85,88. Again 90 is the first number. 110 will be to its right since it is greater than 90. So all the numbers following 110 also should be to the right of 90,but 80,85 and 88 fall to its left. So, this sequence is not possible.

Let us consider the third sequence which is 190,60,90,85,88. So, 190 is the first number. 60 is to its left. So all the other numbers following 60 should be to the left of 190, which is holding good here. 90,85 and 88 will be to the left of 190. So, this sequence is possible.

Let us consider the fourth sequence which is 65,140,80,70,88. 65 is the first number. 140 will be to its right. All the numbers following 140 should be to the right of 65 which is true. 80,70 and 88 will be to the right of 65 as well.

So, the correct answer is option B and not C.
Thank you all, for bringing this question for discussion as we have zeroed in on the right answer.

Calculating cyclomatic complexity

2015-06-15T08:37:00.001-07:00

Determining GNF and CNF

2015-06-11T03:48:00.001-07:00

Eliminate ∈ - productions from the grammar

2015-06-11T02:18:00.001-07:00

B-Trees

2015-06-10T02:00:00.001-07:00

Huffman tree

2015-06-09T02:38:00.003-07:00

Huffman tree

Please find the explanation for the Huffman coding in the below link...

Huffman tree

Given inorder and preorder tree traversal, finding its postorder

2015-06-09T02:33:00.001-07:00

Evaluating a postfix expression

2015-06-08T04:53:00.001-07:00

Conversion of infix expression to postfix expression

2015-06-08T00:13:00.001-07:00

DISCRETE STRUCTURES

2015-05-20T03:52:00.001-07:00

DISCRETE STRUCTURES

Q. A relation R in {1,2,3,4,5,6} is given by {(1,2),(2,3),(3,4),(4,4),(4,5)}. This relation is :

(A) reflexive

(B) symmetric

(D) not reflexive, not symmetric, and not transitive

Ans:- D

Explanation:-

i) Reflexive : A relation R on a set A is reflexive if aRa for every a ϵ A, that is, if (a,a) ϵ R for every a ϵ A.

In the above relation, since the set contains the six elements 1,2,3,4,5 and 6. A relation R on A is reflexive if it contains the six pairs (1,1),(2,2),(3,3),(4,4),(5,5) and (6,6). Since the relation only contains (4,4) and not the rest, it is not reflexive.

ii) Symmetric : A relation R on a set A is symmetric if whenever aRb then bRa, that is if whenever (a,b) ϵ R then (b,a) ϵ R. R is not symmetric if there exists a,b ϵ A such that (a,b) ϵ R but (b,a) does not belong to R.

The above relation is not symmetric because (1,2) ϵ R but (2,1) does not belong to R. The same argument holds good for (2,3),(3,4) and (4,5) as well. So, the relation is not symmetric.

iii) Transitive: A relation R on a set A is transitive if whenever aRb and bRc then aRc, that is, whenever (a,b),(b,c) ϵ R then (a,c) ϵ R. The above relation is not transitive because (1,2), (2,3) ϵ R but (1,3) does not belong to R.

S, the above relation is not reflexive, not symmetric, and not transitive.