CLICK HERE FOR VARIOUS NEW JOBS 
CLICK HERE FOR VARIOUS EDUCATIONAL NEWS 
CLICK HERE FOR NEW SCHOLARSHIPS 
CLICK HERE FOR ADMISSION NOTICES 
Click And Follow On Google+ To Get Updates
Please Wait 10 Seconds... OR You CanSkip

ADMISSION NOTICES
Scholarships

Scholarship-300x291

BUDGET 2014-15
budget_2014-2015
New Date Sheets
VU SOLVED ASSIGNMENTS
Recent Posts

Posts Tagged ‘SPRING 2011’

VU CS605- Software EngineeringII FinalTerm solved unsolved past papers Spring 2011

VU CS605- Software EngineeringII FinalTerm Solved Unsolved Past Papers Spring 2011

 

 

CS 605 Software Engineering 2

Spring 2011 Final Term Subjective ( Solved)

1)  what is software refactoring

 

Software refactoring is the process of changing a software system such that the external behavior of

the system does not change while the internal structure of the system is improved. This is

sometimes called “Improving the design after it has been written”.

 

2)  what is software safety

 

Software Safety is a software SQA activity that focuses on identification of potential hazards that

may affect software negatively         and cause an entire system to fail. Modeling and analysis process is

conducted as part of software safety and hazards are identified and categorized by                criticality and

risk.

 

3)  Difference between DD& CD.

 

The difference between DD and                                                                                                  CD is that in Democratic Decentralized (DD) there                                                                                         is no permanent

leader and decisions are taken          with mutual consensus. While in Controlled Decentralized (CD) there

is a team leader who coordinates some specific tasks.

 

4)  uses of time boxing

 

Time-boxing is used in severe deadline pressure. It is a use incremental strategy where tasks

associated with each increment        are time-boxed in the following manner:

 

• Schedule for each task is adjusted by working backward from the delivery date.

 

• A box is put around each task

 

• When a task hits the boundary of the box, work stops and next task begins

The principle behind time-boxing is the 90-10 rule (similar to Pareto Principle) – rather than

becoming stuck on the 10% of a        task, the product proceeds towards the delivery date in 90% of the

case

 

5) Difference between check-in and check-out processes.

 

The difference between check in and check out is that when a change is needed in an object than it

is checked out on that time it is locked so that no other software engineer can lock  this object. After

the change has been done than                                                                                                    the object is again incorporated to the project this                                                                                                     process is called

checked in. at the same time the object is unlocked again.

1)  What should be included  in software tracking methods?

 

 

A schedule is meaningless if  it is not followed and tracked. Tasks and milestones defined

 

WWW.VUSTUDENTS.NING.COM
MC090409351@vu.edu.pk

 

 

 

CS 605 Software Engineering 2

Spring 2011 Final Term Subjective ( Solved)

in   a   project   schedule   must                          be   tracked   and   controlled   as   project   proceeds.   Tracking

methods include:

• Periodic project status meetings

• Evaluating the results of all reviews

• Determine whether project milestones have been accomplished by the scheduled date

• Comparing actual start date to planned start date

• Informal meetings with the practitioners

• Using earned value analysis

• Error tracking

 

 

2)  Hazards associated with the computer based cruise of an automobile

• Causes uncontrolled acceleration that cannot be stopped

• Does not respond to depression of brake pedal

• Does not engage when switch is activated

• Slowly loses or gains speed

3)  Waterfall’s major drawback is it may not fulfill client’s requirements. In   your opinion

which model handles this shortcoming?

In my point of view incremental model will fulfill its short comings. In the incremental

models, as opposed to the waterfall model, the product is partitioned into smaller pieces

which are then built and delivered to the client in increments at regular intervals. Since each

piece is much smaller than the whole, it can be built and sent to the client quickly. This

results in quick feedback from the client and any requirement related errors or changes can

be incorporated at a much lesser cost. It is therefore less traumatic as compared to the

waterfall model.

 

5) Working of check-in and check-out processes

 

 

 

 

 

 

 

WWW.VUSTUDENTS.NING.COM
MC090409351@vu.edu.pk

 

 

 

CS 605 Software Engineering 2

Spring 2011 Final Term Subjective ( Solved)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

In SCM, the processes of Check-in and Check-out take a central stage. These are two

important elements of change control and provide access and synchronization control. Access

control   manages   who   has   the   authority  to   check-out   the   object   and synchronization

control ensures that parallel changes by two different people do not overwrite one another.

Synchronization control implements a control on updates. When a copy is checked-out, other

copies can be checked out for use only but they cannot be modified. In essence, it

implements a single-writer multiple-readers protocol. This process is depicted in the

following diagram.

1)  Risks involve in migration of legacy systems.

 

 

 

Legacy system migration is    not an easy task and there are a number of risks involved that

need to be mitigated. First     of all, there is rarely a complete specification of the system

available.   Therefore, there is  no straight forward   way   of   specifying   the services provided

by a legacy system. Thus, important business rules are often embedded in      the software and

may not be documented elsewhere. Business processes and the way legacy    systems   operate

are   often   intertwined.   New        software   development   may   take several years.   New software

WWW.VUSTUDENTS.NING.COM
MC090409351@vu.edu.pk

 

 

 

CS 605 Software Engineering 2

Spring 2011 Final Term Subjective ( Solved)

development is itself risky as changes to one part of the system inevitably involve further

changes to other components. We therefore need to assess a legacy system   before a decision

for migration is made.

2) What problem can be caused by the following bad smells?

  • large classes
  • shotgun surgery

Large classes try to do too much and as a result of it they reduce cohesion.

Shotgun strategy requires so many small changes in various different classes

3) what is the advantage of having consistency in work product?

4)   How can we calculate notion of availability in statistical measurement?

5)   What does mean by the term “Software Reengineering”?

Software solutions often automate the business by implementing business rules and business

processes. In many cases, the  software makes the business processes. As these rules and

processes change, the software must also change. A time comes when these changes become

very difficult to handle. So reengineering is re-implementing legacy systems  to make them

more maintainable.

6)   How can quality of design is measured quantitatively?

Statistical SQA is a technique that measures the quality in a quantitative fashion. It implies

that information about defects is collected and categorized and an attempt is   made to trace

each defect to underlying cause. It uses Pareto Principle to identify vital causes

(80% of defects can be traced to 20% of causes) and moves to correct the problems that have

caused the defects.

7)   Subtasks have interdependencies based on their sequence; Give an example  of subtask

interdependencies?

The interdependency of each compartmentalized activity or task must be determined.

Some tasks must occur in sequence while others can occur in parallel. Some   activities cannot

commence until the work product produced by another is available.

8)   What is the guideline for organizations so that they will not depend on single individuals

for there growth?

9)   What is the difference between     code and design re-structuring?

Program is restructured after the reverse engineering phase. In this case we modify source

code and data in order to make it amenable to future changes. This includes code as well as

data restructuring. Code restructuring requires redesign with same function with higher

quality than original program  and data restructuring involves restructuring the database or the

database schema. It may also involve code restructuring.

10) Give some recommendation to      make a Walkthrough to be effective?

• Avoid drift

 

WWW.VUSTUDENTS.NING.COM
MC090409351@vu.edu.pk

 

 

 

CS 605 Software Engineering 2

Spring 2011 Final Term Subjective ( Solved)

• Limit debate and rebuttal

• Enunciate problem areas but don’t try to solve all problems

• Take written notes

• Limit the number of participants and insist upon advanced preparation

• Develop a checklist for each product that is likely to be reviewed

• Allocate resources and schedule time for FTRs

• Conduct meaningful training for all reviewers

• Review your early reviews

• Determine what approach works best for you

 

11) What would this model dep ict?

Defects           Detection

Errors Passed

Errors from                Through         Percentage

previous step         Amplified Errors   Efficiency    Errors passed

1:x              For error      To next step

Newly generated    detection

errors

 

Developme nt Step

 

 

 

This model depicts that each development step inherits certain errors from the previous step.

Some of these errors are just  passed through to the next step while some are worked on and

hence are amplified with a ratio of 1:x. In addition, each step may also generate some new

errors. If each step has some mechanism for error detection, some of these errors may be

detected and removed and the rest are passed on to the next step.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

VU CS605- Software EngineeringII FinalTerm Solved Unsolved Past Papers Spring 2011

VU CS610- Computer Network FinalTerm solved unsolved past papers SPRING 2011

VU CS610- Computer Network FinalTerm Solved Unsolved Past Papers SPRING 2011

FINAL TERM EXAMINATION
SPRING 2011 (20 JULY 2011) BY IMRANGEE
CS610 – COMPUTER NETWORKS

 


Time: 120 min
Marks: 80

 

Total 53 Questions
40 x MCQs
5 x 2 Marks Questions
5 x 3 Marks Questions
3 x 5 Marks Questions

Subjective Portion is as under:-

 

 

Question No: 41 ( Marks: 2 )

What is meant by throughput in networking?

Question No: 42 ( Marks: 2 )
What is Hop Count Metric in context of Routing?

Question No: 43 ( Marks: 2 )
What will be the network bits of following network address?

130.4.102.1/22

Question No: 44 ( Marks: 2 )
How many types of Connection Services are there in TCP/IP?

Question No: 45 ( Marks: 2 )
Why NAT variants are used?

Question No: 46 ( Marks: 3 )
How ICMP can be used to trace route?

Question No: 47 ( Marks: 3 )
In client server architecture, if no signal is there then how an application knows when the communication arrives?

Question No: 48 ( Marks: 3 )
How shift operator works in CRC?

Question No: 49 ( Marks: 3 )
If a network Engineer wants to send a packet to group of nodes then which IP address class will be used?

Question No: 50 ( Marks: 3 )
When a packet is lost then what is the procedure adopted in TCP/IP?

Question No: 51 ( Marks: 5 )
What are the characteristics of Border Gateway Protocol?

Question No: 52 ( Marks: 5 )
In UDP context describe end point identification with protocol port numbers?

Question No: 53 ( Marks: 5 )
A question was related to write a Routing Table, in which Subnet Mask and Next Hop was required to filled.

 

VU CS610- Computer Network FinalTerm Solved Unsolved Past Papers SPRING 2011

VU CS601 – DATA COMMUNICATION FINALTERM solved unsolved past papers SPRING 2011

VU CS601 – DATA COMMUNICATION FINALTERM Solved Unsolved Past Papers SPRING 2011

 

 

 

CS601-Data Communication

 

Latest Solved subjective from Final term Papers

 

 

FINAL TERM EXAMINATION

SPRING 2011

 

What is the difference between guided and unguided media?

Answer:             Click here for detail

Guided Media are those media that provide a conduit from one device to another. Guided

Transmission Media uses a “cabling” system that guides the data signals along a specific path while
Unguided Transmission Media consists of a means for the data signals to travel but nothing to guide
them along a specific path. It passes through a vacuum; it is independent of a physical pathway.

 

Write Commercial advantage and characteristics of token bus

Answer: (Page 232)

Ø   Other LANs are not suitable for this purpose

Ø   Token Bus has no commercial application in data communications Ø   Token Ring allows each station to send one frame per turn
Ø   Access method: Token passing

What is the difference between FDM and TDM

Answer:    Click here for detail

1) FDM-Frequency division multiplexing where as TDM mean Time division Multiplexing.

2) In FDM spectrum is divided into frequency whereas in TDM divided into time slot.

3) FDM is used in 1st generation analog system whereas TDM is used in 2nd generation analog system.

 

Write the types transmission noise

Answer: (Page 143)

Thermal Noise: Due to random originally sent by TX

Induced Noise: Comes from sources like Motors and Appliances Crosstalk:  Effect of one wire on another

Impulse Noise: Spike (A signal with high energy in a very short period of timepower lines, lightening
etc.

 

What is power bandwidth

Answer:  Click here for detail

The power bandwidth of an amplifier is sometimes taken as the frequency range (or, rarely, the

upper frequency limit) for which the rated power output of an amplifier can be maintained (without excessive distortion) to at least half of the full rated power.

OR

 

 

 

 

Power Bandwidth

Answer: (Page 67)

Frequency band in which 99% of the total power resides.

 

What does the CRC generator append to data unit? [2]

Answer: (Page 175)

Appending it to the end of the data must make the resulting bit sequence exactly divisible by the divisor

How much bandwidth for modem is required in case of FSK? [2]

Answer:

BW required for FSK is equal to the Baud rate of the signal plus the frequency shift.

Because of the limitations of voice-grade telephone lines, these modems are restricted to a bandwidth of about 3 kHz

 

What is even parity generator in VRC error detection mechanism? [2]

Answer: (Page 172)

Even parity generator counts the 1‟s and appends the parity bit (1) to the end.

What is the difference between angle of incident and angle of reflection? [2]

Answer: (Page 126)

The difference between them is that Angle of refraction passes from less dense to denser medium whereas angle of incidence passes from more dense to less dense medium.

What is daisy chaining in 1Base 5 star Lan? [2]

Answer: (Page 229)

Slower speed in star lan can be increase by the use of DAISY CHAINING.

 

What is the responsibility of Application layer? [3]

Answer: Page 53

Enables the user either human or software to access the network

It provides user interface and support for the services such as Electronic mail, Remote File access and Transfer, Shared Database Management and other services

 

What is critical angle?

Answer: (Page 127)

We have a beam of light moving from a more dense to a less dense medium. We gradually

increase the angle of incidence measured from vertical axis.  As angle of incidence increases, so does the angle of refraction.

The angle at which refracted line lies on the horizontal axis is called Critical Angle

 

Hamming code-Redundancy bit (5 marks Q)

Answer: (Page 181)

Redundancy Bits (r)

Ø   r must be able to indicate at least m+r+1 states Ø   m+r+1 states must be discoverable by r bits Ø   Therefore, 2r ≥  m+r+1

 

 

 

 

Ø   If m=7, r=4 as 24 ≥  7+4+1

 

Tree topology advantages (3 or 5 marks)

Answer: (Page 31)

Ø   Because of Secondary Hub, More devices can be attached to a Central Hub and
therefore increase the distance a signal can travel

Ø   Enables    Differentiated    Services:    Allows    to    prioritize    communication,    e.g. computers
attached to one secondary hub can be given priority over others

Ø   Therefore, TIME SENSITIVE data will not have to wait for access to the network Ø   Rest of the advantages are almost the same as STAR

 

Q #41:  Whether VRC error detection method is used for single bit error or burst error. (2)

Answer: (Page 173)

VRC can detect all single bit errors

Can also detect Burst errors as long as the total number of bits changed is ODD

 

Q # 42: Which modem was first developed commercially in 1970? (2)

Answer: (Page 114)

Bell modems

-First commercial modems by Bell Telephone Co. -Developed in early 1970s

Q # 45: Consider a major telecom company using RZ encoding for its signals
conversion. What will be  the major problem faced by using such type of

encoding? (2)

Answer: (Page 75)

Any time, data contains long strings of 1’s or 0’s, Rx can loose its timing.

The only problem with RZ encoding is that it requires two signal changes to encode one bit and therefore occupies more BANDWIDTH

 

Q # 47: Geosynchronous Satellite? (3)

Answer: (Page 139)

Ø   Line of sight propagation requires the sending and receiving antennas must be locked into
each other

Ø   To ensure continuous communication, satellites must move with the same speed as earth. So
that they seem fixes w.r.t earth

Ø   These satellites are called Geosynchronous Satellites

 

 

 

 

 

FINAL TERM EXAMINATION 2011

 

Question#1

What are the Asynchronous protocols in data communication layer? ………….Marks (10)

Answer:   (Page 206)

Asynchronous protocols Treat each character in a Bit stream independently. Employed mainly in Modems.

Transmission does not require timing coordination; Timing is done by using extra bits

Different Asynchronous Protocols

XMODEM

o  The first field is a One Byte start of header (SOH) field o  The second field is a two-byte Header.

-The first header byte , the Sequence number carries the Frame number

-The second header byte is used to check the validity of the sequence number

 

o  The fixed data field holds 128 bytes of data

o  The last field CRC checks for errors in the data field only

YMODEM

YMODEM is similar to X-MODEM with only the following major differences:

o 1024-Byte data unit

o  Two CANs to abort Transmission

o  ITU-T CRC-16 for Error Checking

o  Multiple files can be sent simultaneously

 

ZMODEM :

Newer Protocol

Combines features of XMODEM    and YMODEM.

 

BLAST

o  Blocked Asynchronous Transmission
o  More powerful than XMODEM
o  Full Duplex

o  Sliding Window Flow Control

o  Allows transfer of Data and Binary Files

KERMIT

o  Designed at Columbia University

o  Most Widely used Asynchronous Protocol

o  File Transfer protocol is similar in operation to XMODEM, with sender waiting for an
NAK before it starts TX

o  Kermit allows the transmission of control characters as Text

 

 

 

 

 

Question#2

What is Frequency division multiplexing ?…… .Marks (5)

Answer:      (Page 149)

Frequency division multiplexing (FDM)

o  An analog technique that can be applied when BW of the link is greater than the combined
BW of the signals to be TX

o  Signals generated by each sending device modulate difference carrier frequencies o  These modulated signals are then combined into a single Composite signal that
can be transported by the link

o  Carrier frequencies are separated by enough BW to accommodate the modulated signal o  These BW ranges are the channels through which the various signals travel

 

 

Question#4

What is stop and wait ARQ in error control ?….Marks (3)

Answer: Page 197

Stop-and-Wait is an extended form of flow control to include retransmission of data in case of Lost or Damaged frames.

There are four main features added in it.

1.  Sending device keeps a copy of the last frame transmitted until it receives the ACK for that
frame.

2.  Both data and ACK frames are numbered 0 and 1 alternately.

3.  A data 0 frame is acknowledged by a ACK 1 frame indicating that the receiver has received data
0 and is now expecting data 1 .

4.  For retransmission to work, 4 features are added to the basic flow control mechanism.

 

Question#5

What is Interleaving ?…    ..Marks (3)

Answer: Page 153

Synchronous TDM is considered as a very fast rotating switch. When this switch opens in front of a
device, the device has the opportunity to send a specific amount of data on to the path.
The switch moves from device to device at a constant rate and in a fixed order. This process is called
INTERLEAVING. Interleaving can be done by BITS, BYTES or by any other DATA UNIT

Question#6

What is DSU in terms of digital services?…    …Marks (3)

Answer: Page 163

DSU (Digital service unit) changes the rate of digital data created by the subscriber’s device to 56 Kbps and encodes it in the format used by service provider. It used in dialing process and is more expensive than MODEM. But it has better speed, better quality and less susceptibility to noise.

 

Question#7

Which architecture of Ethernet developed by ITU_T and

ANSI?…….. .. Marks (2)

Answer: 236

FDDI (Fiber Distributed Data Interface)architecture of Ethernet developed by ITU_T and ANSI.

 

 

 

 

 

Question#8

What is a spike in noise term?…    …Marks (2)

Answer:143

Spike is a signal with high energy in a very short period of time that comes from power lines, lightening etc,

 

Question#9

Answer: Page 172

What is even parity generator in VRC error detection mechanism?…       ..Marks (2)

Even parity generator counts the 1’s and appends the parity bit (1) to the end.

 

Question#10

Compare line discipline methods ENQ/ACK and Poll/ Select? Answer:  Page 188-189

=>ENQ/ACK coordinates which device may start a transmission and whether or not the intended recipient is ready and enabled.

=> Using ENQ/ACK, a session can be initiated by either station on a link as long as both are of equal
rank.

=> In both half-duplex and full-duplex transmission, the initiating device establishes the session.

=> In half duplex, the initiator then sends its data while the responder waits. The responder may take over the link when the initiator is finished or has requested a response.

=> In full duplex, both devices can transmit simultaneously once the session has been established.

 

POLL/SELECT:

 

=> The poll/select method of line discipline works with topologies where one device is designated as a primary station and the other devices are secondary stations.

=> Multipoint systems must coordinate several nodes, not just two.

=> The primary device controls the link and the secondary device follow sits instruction

It is up to the primary to determine which device is allowed to use the channel data given time

 

 

Why addressing is required in Poll / Select method and not required in ENQ/ACK method?3 Answer:   (Page 190)

Addressing is required in Poll / Select method as it is a not point-to-point configuration, For the primary device in a multipoint topology to be able to identify and communicate with a specific secondary
device, there must be some addressing, while ENQ/ACK method  is a point-to-point  method and for point-to-point configuration, there is no need for addressing.

 

 

What do you know about ITU-T Modems?3

Answer:      (Page 114)

ITU-T modem :

V-series: Today’s most popular modem standards
Bell modem compatible:

V.21/22/23/26/27/29

 

 

 

 

 

Following abbreviations stands for what?3

Answer: (Page 224)

ARP…….…. (Address Resolution Protocol)

RARP……… (Reverse Address Resolution Protocol)
ICMP ……… (Internet Control Message Protocol)

 

Write names of Link Access Protocols developed by ITU-T?3

Answer:   (Page 211)

LAPs: LAPB, LAPD, LAPM, LAPZ etc. all based on HDLC

 

Write the names of different types of noise in the medium?3

Answer:     (Page 144)

Thermal Noise
Induced Noise
Crosstalk

Impulse Noise

Write down some disadvantages of star topology.3

Answer: (Page 30)

Although Cabling required is far less than Mesh

Still each node must be connected to a Hub , so Cabling is still much more

What are the two major classes of synchronous protocols at data link layer?2

Answer: (Page 206)

Character – Oriented Protocols
Bit – Oriented Protocols

Whether Hamming code is the technique used for error detection or error correction?2

Answer: (Page 181)

Hamming code is the technique used for error correction

 

Define Multiplexing? What is its advantage?2

Answer: (Page 147)

Set of techniques that allows the simultaneous transmission of multiple signals across a single data link

It allows multiple users to share total capacity of a Transmission Medium.

What is the purpose of dual ring?2

Answer: (Page 34)

Unidirectional traffic movement is overcome by dual ring technology.

 

Which modem was first  developed commercially in 1970?2

Answer: (Page 114)

Bell modems

 

Write any two functions of physical layer?2

Answer: (Page 45)

 

 

 

 

It defines characteristics of Interface between device and transmission Medium It also defines the type of transmission medium

Physical Layer is also concerned with Line Configuration

 

 

FINAL TERM EXAMINATION

2011

 

Which one has more overhead, a repeater or a bridge? Explain your answer. [3]

Answer:

A bridge has more overhead than a repeater. A bridge processes the packet at two layers; a repeater processes a frame at only one layer. A bridge needs to search a table and find the forwarding port as well as to regenerate the signal; a repeater only regenerates the signal. In other words, a bridge is also a repeater (and more); a repeater is not a bridge

Define high frequency [HF] and super high frequency [SHF], which devices uses these frequencies [3]

Answer: Page 135 and 136

High frequency.

HF uses ionospheric propagation. These frequencies move into the ionosphere where the density difference reflects them back on earth.

It is used for Citizen’s Band Radio, International Broadcasting, Military Communication, Telephone, Telegraph and Fax

Super high frequency.

SHF waves are TX using mostly line-of-sight and some Space propagation.

It is used for Terrestrial and Satellite Microwave and Radar Communication devices.

 

Write all steps of checksum method. [3]

Answer: (Page 179)

o The sender subdivides data units into equal segments of ‘n’ bits(16 bits) o These segments are added together using one’s complement

o The total (sum) is then complemented and appended to the end of the original data unit as redundancy bits called CHECKSUM

o The extended data unit is transmitted across the network

o The receiver subdivides data unit as above and adds all segments together and complement the result o If the intended data unit is intact, total value found by adding the data segments and the checksum field should be zero o If the result is not zero, the packet contains an error & the receiver rejects it

 

Differentiate internet and the internet? [3]

Answer:   (Page 240)

INTERNET

o  An internet is a generic term used to mean an interconnection of individual networks o  To create an internet, we need networking devices called routers and gateways
o  An internet is different from the Internet

o  Internet is the name of a specific worldwide network

 

 

 

 

 

What is the differences in between bit oriented and character oriented protocols [5]

Answer:   (Page 206)

Character – Oriented Protocols

o  Also called Byte- Oriented Protocol

o  These protocols interpret a transmission frame or packet as a succession of characters,
each usually composed of one byte

o  All control information is in the form of an existing character encoding system

Bit – Oriented Protocols

o  Character -Oriented Protocols are not as efficient as bit – oriented protocols and are seldom used o  They are easy to comprehend and employ the same logic as bit-oriented protocols
o  Their study will provide the basis for studying the other data link layer protocols
o  IBN’s BSC is the best known character oriented protocol

 

FINAL TERM EXAMINATION

2011

Question#1

What is the formula to calculate the number of redundancy bits required to correct a bit error in a given number of data bits? [2]

Answer:              Click here for detail

Messages(frames) consist of m data (message) bits, yielding an n=(m+r)-bit codeword.

 

Question#2

What is R G rating of coaxial cable?

Answer:-      (Page 126)

Different coaxial cable designs are categorized by their Radio government ( RG ) ratings

Each cable defined by RG rating is adapted for a specialized function: RG-8

   Used in Thick Ethernet
RG-9

   Used in Thick Ethernet
RG-11

   Used in Thick Ethernet
RG-58

   Used in Thin Ethernet
RG-59

   Used for TV

 

Question#3

What are the advantages of thin Ethernet?

Answer:      (Page 228)

The advantages of thin Ethernet are:

1.  reduced cost

2.  ease of installation

Because the cable is lighter weight and more flexible than that used in Thick net

 

 

 

 

 

Question#4

What is the difference between a unicast, multicast, and broadcast address? [3]

Answer:          Click here for detail

Broadcast: transmitting a packet that will be received by every device on the network Unicast: the sending of information packets to a single destination
Multicast: delivery of information to a group of destinations.
Question#5

T lines are designed for Digital data how they can be used for Analog Transmission?

Answer:     (Page 166)

o DS-1 requires 8 Kbps of overhead

o To understand this overhead, let’s examine format of a 24-voice channel frame

o Frame used on T-1 line is usually 193 bits divided into 24 slots of 8 bits each + 1 bit for synchronization (24*8+1=193)

o 24 segments are interleaved in one frame

o If a T-1 carries 8000 frames, the data rate is 1.544 Mbps (193 * 8000=1.544 Mbps) which is capacity of the line

Question#6

What are the three types of Guided Media?

Answer:        (Page 120)

1.  Coaxial cable

2.  Twisted-pair cable

3.  Fiber optic cable.

 

Question#7

Why do we need Inverse Multiplexing? [5]

Answer:        (Page 159)

   An organization wants to send data, voice and video each of which requires a different data rate    To send voice it needs 64Kbps,

   To send data, it needs 128 Kbps link

   To send video it may need 1.544 Mbps link

   It can lease a 1.544 Mbps line from a common carrier and only use it fully for sometime    Or it can lease several separate channels of lower data rates

   Voice can be sent over any of these channels

   Data & Video can be broken into smaller portions using Inverse Multiplexing and TX

 

Question#8

Describe method of checksum briefly?

Answer:         (Page 180)

o The sender subdivides data units into equal segments of ‘n’ bits(16 bits) o These segments are added together using one’s complement

o The total (sum) is then complemented and appended to the end of the original data unit as redundancy bits called CHECKSUM

o The extended data unit is transmitted across the network

o The receiver subdivides data unit as above and adds all segments together and complement the result o If the intended data unit is intact, total value found by adding the data segments and the checksum field should be zero o If the result is not zero, the packet contains an error & the receiver rejects it

 

 

 

 

 

Question#9

Explain Asynchronous Time Division Multiplexing in detail? Also discuss its advantages over synchronous TDM?

Answer:

Asynchronous time-division multiplexing (ATDM) is a method of sending information that resembles
normal TDM, except that time slots are allocated as needed dynamically rather than pre-assigned to
specific transmitters. ATDM is more intelligent and has better bandwidth efficiency than TDM.
asynchronous time-division multiplexing comprising receive circuits (CRl/i) supplying cells received
via input links, transmit circuits (CTl/j) transmitting retransmitted cells on output links, a buffer
memory (MT) storing the received cells and delivering the cells to be retransmitted and a buffer
memory addressing device (SMT) including a write address source (SAE) and a read address source
(fsl/j).

 

Advantages asynchronous TDM:

In asynchronous TDM, the timeslots are not fixed. They are assigned dynamically as needed.

The objective would be to switch from one user to another user whenever the one user is idle, and to
asynchronously time multiplex the data. With such an arrangement, each user would be granted access
to the channel only when he has a message to transmit. This is known as an Asynchronous Time
Division Multiplexing System (ATDM). A segment of a typical ATDM data stream is shown in Figure

2. The crucial attributes of such a multiplexing technique are:

1. An address is required for each transmitted message, and

2. Buffering is required to handle the random message arrivals.

 

 

FINAL TERM EXAMINATION

2011

 

Question No: 31      ( Marks: 2 )

What is hybrid topology?

Answer:

Hybrid topology is a kind of topology, In which Several topologies combined in a larger topology

 

Question No: 32      ( Marks: 2 )

What is combined station of HDLC?

Answer: (Page 211)

A combined station is one of a set of connected peer devices programmed to behave either as a primary or as a secondary depending on the nature and the direction of the transmission.

Question No: 33      ( Marks: 2 )

What kind of error is undetectable by the checksum?

Answer: (Page 180)

Error is invisible if a bit inversion is balanced by an opposite bit inversion in the corresponding digit of another segment

 

Question No: 34      ( Marks: 2 )

What’s the name of the telephone service in which there is no need of dialing?

Answer: (Page 161)

 

 

 

 

In Analog Leased Service there is no need of dialing

 

Question No: 35      ( Marks: 3 )

How Bit Rate & Baud rate are related?

Answer: (Page 85)

Bit rate equals the baud rate times the no. of bits represented by each signal unit.

The baud rate equals the bit rate divided by the no. of bits represented by each signal shift. Bit rate is always greater than or equal to Baud rate

 

Question No: 36      ( Marks: 3 )

Following abbreviations stands for what?

Answer: (Page244)

1.  ARP: Address Resolution Protocol

2.  RARP: Reverse Address Resolution Protocol

3.  ICMP: Internet Control Message Protocol

 

Question No: 37      ( Marks: 3 )

Differentiate between Polling and Selecting.

Answer:   (Page 189)

If the primary wants to receive data, it asks the second-arise if they have anything to send, This is called Polling

If the primary wants to send data, it tells the target secondary to get ready to receive, This function is called Selecting.

 

Question No: 38      ( Marks: 3 )

T lines are designed for Digital data how they can be used for Analog Transmission?

Answer: repeat

 

Question No: 39      ( Marks: 5 )

What is the difference between character oriented and bit oriented protocols?

Answer: repeat

 

Question No: 40      ( Marks: 5 )

Write down the function of Primary-Secondary communication in line discipline.

Answer: (Page 189)

1.  Poll method works with topologies where one device is designed as a Primary station
and the other devices are Secondary stations

2.  The primary device controls the link and the secondary device follow sits instruction

3.  It is up to the primary to determine which device is allowed to use the channel at a
given time.

4.  The primary therefore is always the initiator of the a session

5.  Whenever a multipoint link consists of a primary device and multiple secondary devices
using a single TX line , all exchanges must be made through the primary device even
when the ultimate destination is a secondary device

 

 

 

 

 

FINAL TERM EXAMINATION

2010

 

Question No: 31      ( Marks: 2 )

What kind of error is undetectable by the checksum? [2]

Answer: Page 180 (repeated)

 

Question No: 32      ( Marks: 2 )

What are properties of signals?

Answer: (Page 17)

Capable of being propagated over TX. Medium ,Interpretable as data at the Receiver

 

 

Question No: 33      ( Marks: 2 )

Whether in Asynchronous or Synchronous TDM, addressing is used?

Answer: (Page 158)

Addressing is used only in Asynchronous TDM.

 

Question No: 34      ( Marks: 2 )

What is the basic purpose of Router?

Answer:           Click here for detail

Basic purpose of Router:

A router is a device that extracts the destination of a packet it receives, selects the best path to that destination, and forwards data packets to the next device along this path. They connect networks together; a LAN to a WAN for example, to access the Internet.

Question No: 35      ( Marks: 3 )

Why we need a Null Modem?

Answer:   (Page 106)

A null modem provide DTA -DTE interface w/o DCEs

Question No: 36      ( Marks: 3 )

Count LRC for the following bits?

10011010   10100101  1101 0110

Question No: 37      ( Marks: 3 )

What are the categories of multiplexing?

Answer: Page 148

There are three catogaries of  multiplxing
FDM

TDM

Have a two other catagories
Synchronous and asyrounce
WDM

Question No: 38      ( Marks: 3 )

What are the three purposes of control frames at data link layer?

Answer:   (Page 209)

Control frames serve 3 purposes:

 

 

 

 

Establishing Connections

Maintaining Flow and Error Control during Data Transmission Terminating Connection

 

Question No: 39      ( Marks: 5 )

Compare line discipline methods ENQ/ACK and Poll/ Select?

Answer:   (Page 188-189)

=>ENQ/ACK coordinates which device may start a transmission and whether or not the intended recipient is ready and enabled.

=> Using ENQ/ACK, a session can be initiated by either station on a link as long as both are of equal
rank.

=> In both half-duplex and full-duplex transmission, the initiating device establishes the session.

=> In half duplex, the initiator then sends its data while the responder waits. The responder may take over the link when the initiator is finished or has requested a response.

=> In full duplex, both devices can transmit simultaneously once the session has been established.

 

POLL/SELECT:

=> The poll/select method of line discipline works with topologies where one device is designated as a primary station and the other devices are secondary stations.

=> Multipoint systems must coordinate several nodes, not just two.

=> The primary device controls the link and the secondary device follow sits instruction

It is up to the primary to determine which device is allowed to use the channel data given time

 

 

Question No: 40      ( Marks: 5 )

What is the difference between character oriented and bit oriented protocols?

Answer: repeat

 

 

FINAL TERM EXAMINATION

2010

 

Question No: 31      ( Marks: 2 )

What are the advantages of a multipoint connection over a point-to-point connection?

Answer:

Point-to-point connection is limited to two devices, where else more than two devices share a single link in multipoint connection. Multipoint connection can be used for fail-over and reliability.

Question No: 32      ( Marks: 2 )

What’s the name of the telephone service in which there is no need of dialing.

Answer:    (Page 164)

DSS (digital data service)    is the telephone service in which there is no need of dialing.

 

Question No: 33      ( Marks: 2 )

Which type of frames is present in BSC frames?

Answer:   (Page 206)

 

 

 

 

There are two types of frames that are present in BSC.

1.  Control Frames and

2.  Data Frames

 

Question No: 34      ( Marks: 2 )

What methods of line discipline are used for peer to peer and primary secondary communication?

Answer:   (Page 187)

Line discipline is done in two ways:

1.  ENQ/ACK       (Enquiry Acknowledgement)

 

This is used for peer to peer communication.

2.  Poll/ Select

This method is used for primary secondary communication.

 

 

Question No: 35      ( Marks: 3 )

How does the checksum checker know that the received data unit is undamaged? [3]

Answer: (Page 179)

Checksum Checker or generator:

The sender subdivides data units into equal segments of ‘n’ bits(16 bits)

1.  These segments are added together using one’s complement.

2.  The total (sum) is then complemented and appended to the end of the original data unit
as redundancy bits called CHECKSUM.

3.  The extended data unit is transmitted across the network.

4.  The receiver subdivides data unit and adds all segments together and complement the
result.

5.  If the intended data unit is intact, total value found by adding the data segments and the
checksum field should be zero.

6.  If the result is not zero, the packet contains an error & the receiver rejects it.

 

Question No: 36      ( Marks: 3 )

Which one has more overhead, a repeater or a bridge? Explain your answer. [3]

Answer:

A bridge has more overhead than a repeater. A bridge processes the packet at two layers; a repeater processes a frame at only one layer. A bridge needs to search a table and find the forwarding port as well as to regenerate the signal; a repeater only regenerates the signal. In other words, a bridge is also a repeater (and more); a repeater is not a bridge.

 

 

Question No: 37      ( Marks: 3 )

Write down disadvantages of Ring Topology.

Answer:   (Page 33)

Disadvantages of Ring Topology

¾   Unidirectional Traffic

9    A break in a ring I.e. a disabled station can disable the entire network

 

 

 

 

¾   Can be solved by using:
9    Dual Ring or

9    A switch capable of closing off the Break

Question No: 38      ( Marks: 3 )

How parity bits are counted in VRC error detection method technique in case of odd parity generator?

 

Question No: 39      ( Marks: 5 )

How are lost acknowledgment and a lost frame handled at the sender site? [5]

Answer:

A lost or damaged frame is handled in the same way by the receiver; when the receiver receives a damaged frame, it discards it, which essentially means the frame is lost. The receiver remains silent about a lost frame and keeps its value of R.

Question No: 40      ( Marks: 5 )

Explain Protocol Data Unit (PDU)?

Answer:          (Page 221)

Protocol Data Unit (PDU)

The data unit in the LLC level is called the Protocol Data unit (PDU) The PDU contains 4 fields familiar from HDLC:

-A destination service access point (DSAP)
-A source service access point (SSAP)
-A control field

-An Information field

 

VU CS601 – DATA COMMUNICATION FINALTERM Solved Unsolved Past Papers SPRING 2011

ALL NEW RESULTS
Educational News

Updated Educational News

Categories
POSTS BY DATE
December 2016
M T W T F S S
« Sep    
 1234
567891011
12131415161718
19202122232425
262728293031