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Tuesday, July 8, 2014

Technical Exam preparation Qs

1. What is the decimal equivalent to the hexadecimal number E6D3?
A. 59019
B. 59091
C. 59136
D. 59093
Answer::-B.

2. Convert the decimal number 2989 to hex.
A. FDD1
B. BAD
C. TED
D. CAD
Answer::- B.

3. MAC address are _________ bits in length.
A. 12
B. 24
C. 48
D. 64
Answer::- C.

4. Where does the MAC address reside?
A. Transceiver
B. Computer BIOS
C. NIC
D. CMOS
Answer::- C.

5. Which of the following statements best describes communication between two devices on a LAN?
A. The source device encapsulates data in a frame with the MAC address of the destination device, and then transmits it. Everyone on the LAN sees it, but the devices with non-matching addresses otherwise ignore the frame.
B. The source encapsulates the data and places a destination MAC address in the frame. It puts the frame on the LAN, where only the device with the matching address can check the address field.
C. The destination device encapsulates data in a frame with the MAC address of the source device, puts it on the LAN, and the device with the matching address removes the frame.
D. Each device on the LAN receives the frame and passes it up to the computer, where software decides whether to keep or to discard the frame.
Answer::-A.

6. Which are the functions associated with framing?
A. Identifies which computers are communicating with one another
B. Signals when communication between individual computers begins and when it ends
C. Flags corrupted frames
D. All of the above
Answer::-D.

7. How does a computer on a LAN detect an error in a frame?
A. It sends a copy of the frame back to the sender for verification.
B. It checks the destination address to verify that the frame was really intended for them.
C. It compares an FCS in the frame to one that the computer calculates from the contents of the frame.
D. It calculates a checksum from the data in the frame, and then sends it back to the source for verification.
Answer::- C.

8. Which best describes a CSMA/CD network?
A. One node's transmission traverses the entire network and is received and examined by every node.
B. Signals are sent directly to the destination if the source knows both the MAC and IP addresses.
C. One node's transmission goes to the nearest router, which sends it directly to the destination.
D. Signals are always sent in broadcast mode.
Answer::- A.

9. Which best describes broadcasting?
A. Sending a single frame to many stations at the same time
B. Sending a single frame to all routers to simultaneously update their routing tables
C. Sending a single frame to all routers at the same time
D. Sending a single frame to all hubs and bridges at the same time
Answer::- A.

10. A half-duplex circuit means
A. Only one side can talk at a time
B. The signal strength is cut in half
C. The signal strength is doubled
D. Two hosts can talk simultaneously
Answer::-A.

11. Attenuation means
A. Travel
B. Delay
C. A signal losing strength over distance
D. Loss of signal due to EMI
Answer::-C.

12. Which of the following is an external source of electrical impulses that can attack the quality of electrical signals on a cable?
A. EMI caused by electrical motors
B. RFI caused by electrical motors
C. Impedance caused by radio systems
D. EMI caused by lightning
Answer::- A.

13. What is the primary cause of crosstalk?
A. Cable wires that are too large in diameter
B. Too much noise in a cable's data signal
C. Electrical motors and lighting
D. Electrical signals from other wires in a cable
Answer::-D.

14. Which of the following describes cancellation?
A. Wires in the same circuit cancel each other's electrical current flow.
B. Cancellation is a commonly used technique to protect the wire from undesirable interference.
C. The magnetic fields from one cable run cancel magnetic fields of another cable run.
D. External magnetic fields cancel the fields inside network cabling.
Answer::- B.

15. What is it called when two bits from two different communicating computers are on a shared medium at the same time?
A. Latency
B. Dispersion
C. Collision
D. Obstruction
Answer::-C.

16. Reflection does not occur with what kind of signals?
A. Electrical
B. Radio waves
C. Microwaves
D. None of the above
Answer::- D.

17. __________ means to convert binary data into a form that can travel on a physical communications link.
A. Encoding
B. Decoding
C. Encrypting
D. Decrypting
Answer::- A.

18. Which of the following is the correct order for the network layers?
A. 1: Physical 2: Data Link 3: Transport 4. Network 5: Presentation 6: Session 7: Application
B. 1: Physical 2: Data Link 3: Network 4: Transport 5: Session 6: Presentation 7: Application
C. 1: Physical 2: Data Link 3: Network 4: Session 5: Transport 6: Application 7: Presentation
D. 1: Physical 2: Network 3: Session 4: Data Link 5: Transport 6: Application 7: Presentation
Ans::- B.

19. Which layer of the OSI model handles error detection, network topology, and medium access?
A. The physical layer
B. The data link layer
C. The transport layer
D. The network layer
Ans::- B.

20. Which layer of the OSI model establishes, maintains, and manages sessions between applications?
A. The transport layer
B. The session layer
C. The presentation layer
D. The application layer
Answer::-B.

21. Which best describes the function of the presentation layer?
A. It provides data representation and code formatting.
B. It handles error notification, network topology, and flow control.
C. It provides network services to user applications.
D. It provides electrical, mechanical, procedural, and functional means for activating and maintaining the link between systems.
Answer::-A.

22. Which layer of the OSI model provides network services to user applications?
A. The transport layer
B. The session layer
C. The presentation layer
D. The application layer
Answer::-D.

23. Which layer offers provisions for data expedition, class of service, and exception reporting?
A. Session
B. Presentation
C. Network
D. Data link
Answer::-A.

24. Which of the following statements regarding encapsulation is untrue?
A. Encapsulation allows computers to communicate data.
B. If one computer wants to send data to another computer, the data first must be packaged by a process called encapsulation.
C. Encapsulation occurs at one layer.
D. Encapsulation wraps data with the necessary protocol information before network transit.
Answer::- C.

25. Which of the following correctly describes the five conversion steps of data encapsulation when one computer sends an e-mail message to another computer?
A. Data, segments, packets, frames, bits
B. Bits, frames, packers, segments, data
C. Packets, segments, data, bits, frames
D. Segments, packets, frames, bits, data
Answer::- A.

26. An e-mail message is sent from Host A to Host B on a LAN. To send this message, the data must be encapsulated. Which of the following best describes the first step of data encapsulation?
A. Alphanumeric characters are converted into data.
B. The message is segmented into easily transportable chunks.
C. A network header is added to the message (source and destination addresses).
D. The message is converted into binary format.
Answer::- A.

27. An e-mail message is sent from Host A to Host B on a LAN. Before you can send this message, the data must be encapsulated. Which of the following best describes what happens after a packet is constructed?
A. The packet is transmitted along the medium.
B. The packet is put into a frame.
C. The packet is segmented into frames.
D. The packet is converted to binary format.
Answer::- B.

28. An e-mail message is sent from Host A to Host B on a LAN. Before you can send this message, the data must be encapsulated. Which of the following best describes what happens after the e-mail message's alphanumeric characters are converted into data?
A. The data is converted into binary format.
B. A network header is added to the data.
C. The data is segmented into smaller chunks.
D. The data is put into a frame.
Answer::- C.

29. Which best describes a datagram?
A. A message sent to the source to confirm the receipt of uncorrupted data
B. A binary representation of routing information
C. A data packet less than 100 bytes in size
D. A network layer packet
Answer::- D.

30. Which of the following layers of the OSI model is not present in the TCP/IP protocol stack?
A. Transport
B. Network
C. Internet
D. Data link
Answer::- D.

More Questions with ans to solve

1. A router is an OSI Layer __________device.
A. 1
B. 5
C. 3
D. 4

Ans:-C

2. What do routers connect?
A. Bridges and repeaters
B. Bridges and hubs
C. Two or more networks
D. Hubs and nodes

Ans:- C.

3. What does a router route?
A. Layer 1 bits
B. Layer 2 frames
C. Layer 3 packets
D. Layer 4 segments

Ans:- C.

4. To construct a simple LAN of four computers, you connect them with a _________ ?
A. Cross-connect cable
B. Serial line
C. Hub
D. Router

Ans. C.

5. What is the maximum cable length for STP (without a repeater, and so on)?
A. 100 ft
B. 100 m
C. 150 ft
D. 1000 in

Ans: - B.

6. How many pairs of wires make up a UTP cable?
A. 2
B. 4
C. 6
D. 8

Ans::- B.

7. Which connector does UTP use?
A. STP
B. BNC
C. RJ-45
D. RJ-69

Ans::- C.

8. What is an advantage that coaxial cable has over STP or UTP?
A. It is capable of achieving 10-100 Mbps.
B. it is inexpensive.
C. It can run longer distances unboosted.
D. None of the above.

Ans::- C.

9. A more compact version of coaxial cable is known as
A. Thinnet
B. BNC
C. STP
D. UTP

Ans::- A.

10. A _________ fiber optic cable allows multiple streams of LED-generated light.
A. Multimode
B. Multi-channel
C. Multiphase
D. None of the above

Ans::- A.

11. What is the importance of the EIA/TIA standards?
A. They provide a framework for the implementation of the OST reference model.
B. They provide guidelines for NIC manufacturers to follow to ensure compatibility.
C. They provide the minimum media requirements for multi-product and multi-vendor environments.
D. None of the above.

Ans::- C.

12. For the horizontal cabling component, TIA/EIA-568A requires a minimum of how many connectors at each work area?
A. 1
B. 2
C. 4
D. 6

Ans::- B.

13. What does the twisting of the wires do in a CAT-5 cable?
A. It makes it thinner.
B. It makes it less expensive.
C. It reduces noise problems.
D. It allows six pairs to fit in the space of four pairs.

Ans::- C.

14. The standard 10BaseT cable is Category ___________
A. 3
B. 4
C. 5
D. 6

Ans::- C.

15. The network area within which data packets originate and collide is called a ____________?
A. Collision domain
B. Network domain
C. Collision segment
D. Network segment

Ans::- A.

16. Using repeaters __________ the collision domain.
A. Reduces
B. Has no effect on
C. Extends
D. None of the above

Ans::- C.

17. The process of installing complex networking devices that break up the domains by using bridges, switches, and routers is known as:
A. Sectioning
B. Segmentation
C. Collision Domain Reduction
D. None of the above

Ans::- B.

18. What physical topology has all its nodes connected directly to one link, and has no other connections between nodes?
A. Linear bus
B. Star
C. Ring
D. None of the above

Ans::- A.

19. What type of numbering system is characterized by 0’s and 1’s?
A. Base 4
B. Base 10
C. Binary
D. Hexadecimal

Ans::- C.

20. What is the decimal number 151 in binary?
A. 10100111
B. 10010111
C. 10101011
D. 10010011

Ans::- B.

21. What is the binary number 11011010 in decimal?
A. 186
B. 202
C. 222
D. 218

Ans::- D.

22. What best describes how much information can flow from one place to another in a given amount of time?
A. Mbps
B. Transfer-rate
C. Reliability
D. Bandwidth

Ans::- D.

23. Bandwidth is described in _____________.
A. Bytes per second
B. Bits per second
C. Megabits per millisecond
D. Centimeters

Ans::- B.

24. What term is used to describe the rated throughput capacity of a given network medium or protocol?
A. TCP/IP
B. Ethernet
C. Bandwidth
D. Routing protocol

Ans::- C.

25. At its most basic level, computer data consist of _________.
A. Bits
B. Bytes
C. Packets
D. None of the above

Ans::- A.

26. Which of the following is not a form of cable media?
A. Coaxial cables
B. Optical fibers
C. Category 5 UTP
D. None of the above

Ans::- D.

27. A protocol is not __________.
A. A set of rules
B. An agreement
C. A connection layer
D. A definitive outline

Ans::- C.

28. Which of the following best describes topology?
A. A connection of computers, printers, and other devices for the purpose of communications
B. The physical arrangement of network nodes and media within an enterprise networking structure
C. A network type that prevents collisions of data packets
D. A method for filtering network traffic to reduce the chance of bottlenecks and slowdowns

Ans::- B.

29. Which of the following best describes a star topology?
A. A LAN topology in which a central hub is connected by vertical cabling to other hubs that are dependent on it
B. A LAN topology in which transmissions from network stations propagate the length of the medium and are received by all other stations
C. A LAN topology in which endpoints on a network are connected to a common central point
D. A LAN topology in which central points on a network are connected to a common central switch by linear links

Ans::- C.

30. Which of the following best describes a node?
A. An endpoint of a network connection or a junction common to two or more lines in a network that serve as control points
B. An application that establishes, manages, and terminates sessions between devices and manages data exchange between presentation layer entities
C. An application that synchronizes cooperating devices and establishes agreement on procedures for error recovery and control of data integrity
D. All of the above

Ans::- A.

File Organization

FILE ORGANIZATION:

It refers to the way of arranging the records in a file which can be accessed in a faster way.
The criteria considered in choosing a file organization are:
1) Fast access to single record or collection of related records.
2) Easy record adding/ update/ removal without disrupting.
3) Storage efficiency
4) Redundancy as a warranty against data corruption.

The files are generally viewed as
• Logical File organization: The groups of fields which are combined to form logical
record, which can perform different operations on the file.
• Physical file organization: The data which is stored in the form of records can be
placed in any storage devices, main memory, secondary memory etc.

Types of File Organization:
I) Sequential Access File Organization
II) Direct Access File Organization
III) Index Sequential Access File Organization

I) Sequential Access File Organization:

1) All records are stored in a sequential order.
2) That is, the records are arranged in the ascending or descending order of a key field.
3) Eg: a) In a student information system, the file would contain roll number, name, division, marks
obtained in the examination.
b) In a payroll application, the records are stored with employee number as a
key field.
4) To locate a particular record in such file organization, we have to start searching
from the beginning of the file until it is found in the file.
5) It is time consuming process.
6) Normally created and maintained on magnetic tapes. Eg: Audio Cassettes.
7) There is no need for any storage space identification

Advantages:
1) Simple to understand
2) Easier to organize, maintain
3) Economical
4) Error in files remain localized

Disadvantages:
1) Entire file has to be processed
2) Transactions must be sorted in a particular sequence before processing
3) Time consuming searching
4) High data redundancy
5) Random enquiries are not possible to handle

II) Direct Access File Organization (Random Access or relative file organization).:

1) Also known as Random Access or relative file organization.
2) Records are stored in Direct Access Storage Device(DASD). Such as magnetic disk(Hard disks).
3) For direct access, the file is viewed as numbered sequence of blocks or records.
4) These blocks or records are taken as key for accessing the desired information randomly.
5) It allows arbitrary (random) blocks to be read or written.
6) It is useful for immediate access to large amount of information. They are often used in accessing
large databases.
7) This technique is called as hashing

Advantages:
1) Immediate Access of the desired records.
2) No sorting of the records is required.
3) Faster updating of several files.
4) Helps in online transaction processing system like online reservation systems.

Disadvantages:
1) Data may be accidentally erased or over-written unless special precautions are taken
2) Backup facility is needed
3) Expensive- hard disks are needed to store the records, it is expensive
4) Less efficient as compared to sequential file organization in the use of storage space
5) Only one key is used
6) Cannot be accessed sequentially

III) Index Sequential access file organization(ISAM)

1) This file organization is a synthesis of the above two file organizations.
2) It combines the positive features of both the sequential and direct access file organizations.
3) Here records are stored randomly on a direct access device such as magnetic disk by a primary key.
Hence, we can access data either sequentially or randomly using the index. The index is stored in a
file and read into memory when the file is opened.
4) It may have multiple keys. These keys can be alphanumeric
5) The key upon which the data records are ordered is called the primary key.
6) Other keys are called alternate keys

Advantages:
1) Multilpe keys – are also alphanumeric in nature
2) Both sequential and random access is possible
3) Accessing of records is fast, if the index table is properly organized

Disadvantages:

1) More storage space is needed because of the presence of Index
2) Less efficient in the use of storage space as compared to other file organizations
3) It requires special software, i.e expensive.

Call by value and Call by refrence

Explain call by value and call by reference with example.

Call by value-
When a function call is given, a relationship is established between actual and formal parameters.
A temporary storage is created for formal parameters to store the data(actual parameters) received
as input from calling function. This process of data transfer from actual parameter to formal
parameters is known as call by value.

Ex: void main()
{ int a=20,b=30,x;
int f1(int,int);
-----
x=f1(a,b);
-----
getch( );
}
int f1(int p, int q)
{
p=p+p;
q=q+q;
return(p+q);
}

Call by reference-
When a function is called, the address of variables are passed on to the
called function. The parameters receiving the address should be pointer variables. This
concept of sending the address of data from calling function to pointer variables in the
called function is known as call by reference.

Ex: viod main()
{
int a=10,b=15,c=20;
------
X=f1(&a,&b,&c);
------
getch();
}
int f1(int *p1, int *p2, int *p3)
{
*p1=*p1+10;
*p2=*p2+5;
}

More on Networking

Network services
• File Services-All the data files of an organisation are stored in one place and can be
shared by many users. Files can be organized and updated systematically.
• Print Services- When Printers available on the Network , multiple users can print to
the same printer. The network printers are usually faster and more capable. They
may have accessories such as envelope readers or multiple paper trays
• Database Services-Multiple users can have access to the database at the same time.
The Database software ensures integrity of data and provides multiple access.
• Message Services-The services include Electronic mail, voice mail, remote
access,document transfers across the internet.
• Application services- several software applications that are developed using
network architecture. The applications can be made available to multiple users. Any
up-gradation made to these applications is also instantly available to the users.

Advantages of Computer Network
•Speed:- The files can be transferred from one Computer to another instantly.
•Cost:- Application software(Network version) can be installed on one Computer and
accessed by other Computers on the Network. This makes maintenance and
upgrading of software easy.
•Security:- The information on the computer can bemade available only through
password.Further some information can be made available only for reading and
not copying.
•Resource sharing:- Expensive peripheral devices can be shared by many users in a
network.
•Easy access:- Users can access information from any computer that is available on
the network. They need go to a specific Computer which may be located in a
different building or different location to retrieve information from their account.

Sunday, July 6, 2014

Status codes

Staus codes:-

Informational
Code :-100
Phrase :- Continue
Description:-The initial part of the request has been received , and the client may continue with the request.

Code :- 101
Phrase :- Switching
Description :- The server is complying with a client request to switch protocols defined in the upgrade header.

Success
Code :- 200
Phrase :- OK
Description :- The request is successful

Code :-201
Phrase :- Created
Description :- A new URL is created .

Code :- 202
Phrase :- Accepted
Description :- The request is accepted , but it is not immediately acted upon

Code :- 204
Phrase :- No content
Description :-There is no content in the body

Redirection

Code :- 301
Phrase :- Moved permanently
Description :- The requested URL is no longer used by the server.

Code :- 302
Phrase :- Move temporarily
Description :- The requested URL has moved temporarily.

Code :- 304
Phrase :- Not modified
Description :-The document has not been modified.

Client Error

Code :- 400
Phrase :- Bad request
Description :- There is a syntax error in the request.

Code :- 401
Phrase :- Unauthorized
Description :- The request lacks proper authorization.

Code :- 403
Phrase :- Forbidden
Description :- Service is denied .

Code :- 404
Phrase :- Not found
Description :- The document is not found.

Code :- 405
Phrase :- Method not allowed
Description :- The method is not supported in this URL.

Code :- 406
Phrase :- Not acceptable
Description :- The format requested is not acceptable.

Server Error

Code :-500
Phrase :- Internal Server error
Description :- There is error , such as a crash , at the server site.

Code :- 501
Phrase :- Not implemented
Description :- The action requested cannot be performed.

Code :- 503
Phrase :- Service unavailable
Description :-The service is temporarily unavailable ,but may be requested in the future.

DNS in the Internet

DNS is a protocol that can be used in different platforms.

In the Internet the Domain Name Space is divided into three different sections :

1) Generic domains

2) Country domains and

3) The Inverse domains

1) Generic domains:-
The generic domains define registered hosts to their generic behavior.

Generic domain labels are as follow:-

aero :- Airlines and aerospace companies

biz    :- Businesses or firms

com :- Commercial organizations

coop :- Cooperative business organizations

edu    :- Educational institutions

gov      :- Government institutions

info     :-   Information service provides

int       :- International organizations

mil    :- Military groups

museum :- Museums and other non profit organizations

name    :- Personal names (individuals)

net      :- Network support centers

org :- Non profit organizations

pro     :- Professional individual organizations

2) Country domains:-
The country domains section uses two - character country abbreviations.
Example :-
us :- United States

in :-   India

uk :- United Kingdom

Second labels can be organizational ,or they can be more specific , national designations.
The United States ,for example , uses state abbreviations as a subdivisions of us example ca.us

3) Inverse domains:-
The inverse domain is used to map an address to a name .the convention of reading domain labels from bottom to top ,an IP address such as 132.34.45.121 is read as 121.45.34.132.