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VU CS610- Computer Network FinalTerm solved unsolved past papers Fall 2010

VU CS610- Computer Network FinalTerm Solved Unsolved Past Papers Fall 2010

Computer Networks

Final Term Paper (Fall 2010)

February 2011

Total Marks: 75

30 questions of multiple choices

2 mark questions:

1: Limitations of parity checking?

2: how can we prove that we have 2.147.483.648 addresses in class A.

3: what is meant by the client server paradigm?

4: why is internet multicast routing difficult?

5: where should an ICMP message be sent?

6: what is the basic concept of twice NAT?

3 Marks questions:

7: what are the some of the metrics used by routing protocols?

8: How can switch virtual network be established?

9: Could IP be redesigned to use hardware addresses instead of the 32-bit addresses it currently uses. Why or why not?

10: Three features of dynamic message method in ARP.

11: in internet routing how does a host join or leave a group?

12: why TCP is called end to end protocol?

5 Marks questions:

13: If IPV4 works so well. Why change it?

14: Following is the UDP data gram format:

You look UDP datagram format from handout <ok>

Answers these questions:

  • What does “UDP source Port” field contain?
  • What does “UDP Destination port” field contain?
  • What does “UDP Message Length” field contain?
  • What is the size of the entire UDP datagram (in bits)?

15: Main advantages and disadvantages of Routing Information Protocol.

The End

Rashad

 

 

VU CS610- Computer Network FinalTerm Solved Unsolved Past Papers Fall 2010

VU CS610- Computer Network FinalTerm solved unsolved past papers Fall 2010

VU CS610- Computer Network FinalTerm Solved Unsolved Past Papers Fall 2010

CS610 –Computer Networks

Fall 2010 FINAL Paper 12 FEB,  2011

BY

*~*SHINING STAR*~*($$)

 

 

WHAT IS ATM?

Asynchronous Transfer Mode (ATM) is a technology that has the potential of revolutionizing data communications and telecommunications. Based on the emerging standards for Broadband Integrated Services Digital Networks (B-ISDN), ATM offers the economically sound “bandwidth on demand” features of packet-switching technology at the high speeds required for today’s LAN and WAN networks  and tomorrow’s.

WHAT IS THE FUNCTION OF IANA?

IANA (Internet Assigned Numbers Authority) is an operating unit of the Internet Corporation for Assigned Names and Numbers (ICANN) that maintains the top-level domain, IP address and protocol number databases.

WHAT IS THE FACILITY OF IP MUTICAST?

IP Multicast is a protocol for transmitting IP datagrams from one source to many destinations in a local or wide-area network of hosts which run the TCP/IP suite of protocols. The basic facility provided by the IP protocol is a unicast transmission service: That is, the current standard for IP provides only unreliable transmission of datagrams from a single source host to a single destination host. . A packet which is destined for N recipients can be sent as just a single packet!

HOW MANY LARGE GROUPS OF ROUTERS DO U THINK?

We will draw any routers we may be using to separate major portions of our network first. Smaller private networks do not require routers, but may still use them for administrative reasons. Routers are only required if a.)Dividing our network into multiple smaller networks, b.) Allowing indirect internet access using NAT. Next, add any switches and hubs. For small networks, only one switch or hub may be necessary.

WHAT ARE COLON HEXADECIMAL NOTATION?

Colon Hexadecimal Notation is the notation used to represent IPv6 addresses. The 128 bits address is divided in 8 blocks of 16 bits separated by colons. One example is

805B:2D9D:DC28:0000:0000:FC57:D4C8:1FFF

 

WAT ARE THE IMPLEMENTATIONS OF NAT?

 

For each outgoing IP packet, the source address is checked by the NAT configuration rules. If a rule matches the source address, the address is translated to a global address from the address pool.

 

For each incoming packet, the destination address is checked if it is used by NAT.

When this is true, the address is translated to the original internal address

 

HOW ICMP TRACE A ROUTE?

 

ICMP messages can be used to check whether hosts are reachable. The ICMP command can also be used to trace routes to Internet hosts by sending UDP datagrams to unused UDP ports and interpreting the ICMP responses (Van Jacobsen and Steve Deering algorithm).

 

The ICMP trace command works similar to the ICMP TTL command but it always returns the destination address even when the responding host chooses another interface with a different IP address to send the reply. This can be used to trace a route to a host since the command returns the host that discards the packet if it does not reach the destination.

WAT IS MEANT BY BRIDGINING OF TWO BUILDINGS?

Bridge mode uses WDS (Wireless Distribution System) which is generally used for extending wireless range or bridging between two network segments wirelessly, for example, connecting two office buildings. In technical terms we say a bridge type connects a wireless network to a wired network transparently.

  • Device will act as a transparent bridge and will operate in Layer 2 – Mac layer
  • Bridge mode association is based on channel and MAC authorization only. SSID will not be used for the setting up a link.
  • If the remote is configured in a bridge mode and channel needs to be changed then channel needs to be changed at both the radios.

HOW A COMPUTER DIFFERS IP DATAGRAM AND ARP?

IP datagram receives a datagram, has exact destination address, and find  next hope address and it uses IP address across physical network to send, forward or deliver.

Where as in ARP Computer caches ARP responses in local ARP cache/table. Using ARP for each IP packet adds two packets of overhead for each IP packet.

The old and new values of IP source field and destination field are shown with their directions.NAT device stores state information in table. The value is entered in the table when NAT box receives outgoing datagram from new

 

AS OSPF ALLOW AN AUTONOMOUS SYSTEM INTO MULTIPLE AREASOSPF PRODUCED COMPLEX OR POWERFUL NETWORKS?

I think it reduces overhead (less broadcast traffic). Because it allows a manager to partition the routers and networks in an autonomous system into multiple areas, OSPFcan scale to handle a larger number of routers than other IGPs.

Transmission Control Protocol is a transmission protocol which ensures reliable and sequential data delivery. It establishes so called virtual connections and provides tools for error correction and data stream control. It is used by most of applications protocols which require reliable transmission of all data, such as HTTP, FTP, SMTP, IMAP, etc.

DIFFEENTITATE DITANCE VECTOR AND LINK STATE?

“Distance Vector” and “Link State” are terms used to describe routing protocols which are used by routers to forward packets between networks. The purpose of any routing protocol is to dynamically communicate information about all network paths used to reach a destination and to select the from those paths, the best path to reach a destination network.

Distance vector protocols use a distance calculation plus an outgoing network interface (a vector) to choose the best path to a destination network. The network protocol (IPX, SPX, IP, Appletalk, DECnet etc.) will forward data using the best paths selected.Common distance vector routing protocols include Appletalk RTMP,IPX RIP,IP RIPand IGRP . It’s well Supported. Protocols such as RIP have been around a long time and most, if not all devices that perform routing will understand RIP.

 

Link State protocols track the status and connection type of each link and produces a calculated metric based on these and other factors, including some set by the network administrator. Link state protocols know whether a link is up or down and how fast it is and calculates a cost to ‘get there’  Link State protocols will take a path which has more hops, but that uses a faster medium over a path using a slower medium with fewer hops.

WHAT ARE THE DISADVANTAGES OF RIP?

The disadvantages of the Routing Information Protocol (RIP) are as follows:

  • RIP generates additional protocol traffic, as it propagates routing information by periodically transmitting the entire routing table to neighbor routers.
  • It is not suitable for large networks, because RIP packet size increases as the number of networks increases.
  • It requires a lengthy convergence time.
  • In RIP, the count-to-infinity problem persists when a link between two routers is broken down and the remaining routers recalculate the hop count to infinity

 

WRITE A NOTE ON IPV6 BASE HEADER FORMAT?.

The network header of the currently deployed IPv4 protocol is 20 bytes (plus options).

IPv6 omits the group of fields in the second 32-byte word of the IPv4 header. A decision was made that IPv6 routers would not support fragmentation within the network, following poor performance for router-based IPv4 fragmentation. IPv6 still supports host-fragmentation and transparent link/tunnel fragmentation (where the packet is reassembled at the next-hop).

The IPv6 header also omits the network checksum. This was removed on the basis that routers were reliable, and that checksum processing incurred an unnecessary overhead in high-speed routers. Instead, IPv6 relies upon the presence of the pseudo-header in the transport checksum to validate that a packet has been delivered to the intended recipient.

The resulting IPv6 base header is 40 bytes. This increase in header size was not accompanied by an increase in complexity; the IPv6 base header is much simpler, comprising just 8 fields. The main reason for the larger size is to accommodate a pair of larger network addresses, increasing the size from 32 to 128 bits

 

 

VU CS610- Computer Network FinalTerm Solved Unsolved Past Papers 2011

VU CS201 – Introduction to Programmming FinalTerm solved unsolved past papers Fall 2010

VU CS201 – Introduction To Programmming FinalTerm Solved Unsolved Past Papers Fall 2010

CS201 Fall  2010 Paper

Mc090400760

MCS

 

 

 

 

 

 

Question:

What is self assignment mean, why should it be avoided?                        [3 marks]

 

Question:

Is it possible to overload operators for primitive data type? Justify           [3 marks]

 

Question:

What are the limitations of the friendship relations between classes?        [3 marks]

 

Question:

C and C++ are free format languages. What does mean by free format?   [2 marks]

 

Question:

What are the two types of conversions for user defined data type?            [1 marks]

 

Question:

Describe the three important constructs to solve a given problem.            [5 marks]

 

Question:

How can we increase the size of dynamic allocated memory in C?           [5 marks]

 

Question:

Does the following statement create new variable? If not, why?               [1 marks]

int &ref = val

 

 

Question:    [10 marks]                                                                                          

Write a program which contains a class Account. The class should contains three data members Name, AcNo. and Balance. The class shouls further contains two constructors i.e. default constructor and parameterized constructor. Overload the stream insertion operator (<<) for this class.

 

Question:    [10 marks]

Write a program that reads a number that says how many integer numbers to be stored in an array. Create an array to for the exact size of data and read in that many numbers into the array.

 

 

There were 30 MCQs among which following three:

1)  C/C++ string constant is enclosed in

Small braces

Curly braces

Double quote

Single quote

 

 

2) Unary operator takes orgument

Zero

One

Two

Three

 

3) When define array of objects

Constructor will call

Destructor will call

 

 

 

 

 

 

 

 

   

VU CS201 – Introduction To Programmming FinalTerm Solved Unsolved Past Papers Fall 2010

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