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UNION OF MYANMAR MINISTRY OF SCIENCE AND

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UNION OF MYANMAR MINISTRY OF SCIENCE AND Powered By Docstoc
					                                UNION OF MYANMAR
                    MINISTRY OF SCIENCE AND TECHNOLOGY
        DEPARTMENT OF TECHNICAL AND VOCATIONAL EDUCATION
       GOVERNMENT TECHNOLOGICAL COLLEGES AND INSTUTUTES


                             IT 04104 Computer Network
                                 B-Tech Year 2 (2006)
23-10-2006                                                                  12:30 -15:30


1(a) Describe the principles that were applied to the seven layers of OSI reference model.
1(b) Define ATM cell. Explain briefly the advantages of cell switching.         (20 marks)


2(a) Differentiate between multimode and single mode fiber.
2(b) What are the applications of microwave? Describe its advantages over others.
                                                                                 (20 marks)


3(a) Discuss the advantages of digital transmission over analog transmission.
3(b) Describe the three major components of telephone system.
3(c) Draw the block diagram of the computer to computer call and explain briefly the
transmission process.                                                           (20 marks)


4(a) Sketch the Manchester encoding for the bit stream: 000 111 0101.
4(b) What is the purpose of padding field in 802.3 Ethernet frame format?       (20 marks)


5(a) What is a bridge? How does it work?
5(b) How can faster speed be achieved in Fast Ethernet LAN? Discuss three types of Fast
Ethernet Cabling.                                                            (20 marks)


6(a) What is the main function of network layer?
6(b) What are the special IP addresses? Discuss briefly.
6(c) What is main purpose to design TCP protocol?                               (20 marks)
                                          Sample Answer
                                   IT 04104 Computer Network
                                        B-Tech Year 2 (2006)
1.Solution
1(a)
         This model is based on a proposal developed by the International Standards
Organization (ISO) as a first step toward international standardization of the protocols
used in the various layers. The model is called the ISO OSI (Open System
Interconnection) Reference Model because it deals with connecting open systems- that is,
systems that are open for communication with other systems. We will usually just call it
the OSI model for short.
         The OSI model has seven layers. The principles that were applied to arrive at the
seven layers are as follows:
   1. A layer should be created where a different level of abstraction is needed.
   2. Each layer should perform a well defined function.
   3. The function of each layer should be chosen with an eye toward defining
        internationally standardized protocols.
   4. The layer boundaries should be chosen to minimize the information flow across
        the interfaces.
   5. The number of layers should be large enough that distinct functions need not be
        thrown together in the same layer out of necessity, and small enough that the
        architecture does not become unwieldy.
1.(b)
        The basic idea behind ATM is to transmit all information in small, fixed size
packets called cells. The cells are 53 bytes long, of which 5 bytes are header and 48 bytes
are payload, as shown in Figure.

                                    5                   48
                           Bytes
                                    Header              User data

                                             Figure: An ATM cell
        ATM is both a technology and potentially a service. Sometimes the service is
called cell relay, as an analogy to frame relay. The use of a cell switching technology is a
gigantic break with the 100-year old tradition of circuit switching within the telephone
system. There are a variety of reasons why cell switching was chosen, among them are
the following. First, cell switching is highly flexible and can handle both constant rate
traffic and variable rate traffic easily. Second, at the very high speeds envisioned, digital
switching of cells is easier than using traditional multiplexing techniques, especially
using fiber optics. Third, for television distribution, broadcasting is essential; cell
switching can provide this and circuit switching cannot.
       ATM networks are connection-oriented. Making a call requires first sending a
message to set up the connection. After that, subsequent cells all follow the same path to
the destination. Cell delivery is not guaranteed, but their order is. If cells 1 and 2 are sent
in that order, then if both arrive, they will arrive in that order, never first 2 then 1.
      ATM networks are organized like traditional WANs, with lines and switches
(router). The intended speeds for ATM networks are 155Mbps and 622Mbps, with the
possibility of gigabit speeds later. The 155-Mbps was chosen because this is about what
is needed to transmit high definition television. The exact choice of 155.52Mbps was
made for compatibility with AT&T's SONET transmission system. The 622Mbps speed
was chosen so four 155-Mbps channels could be sent over it. By now it should be clear
why some of the gigabit testbeds operated at 622Mbps: they used ATM.
        When ATM was proposed, virtually all the discussion was about video on demand
 to every home and replacing the telephone system, as described above. Since then, other
developments have become important. Many organizations have run out of bandwidth on
their campus or building-wide LANs and are being forced to go to some kind of switched
            system that has more bandwidth than does a single LAN. Also, in client-server
     computing, some applications need the ability to talk to certain servers at high speed.
ATM is certainly a major candidate for both these applications. Nevertheless, it is a bit of
   a letdown to go from a goal of trying to replace the entire low-speed analog telephone.
                                                                                    (20 marks)
2.Solution
2.(a)
        A light ray incident at or above the critical angle is trapped inside the fiber, as
shown in figure(b), and can propagate for many kilometers with virtually no loss.


               Air/silica Air
                           1      2
                                                                        Total internal
                                           3                            reflection
               boundary




                  silica 1 2 3               Light source
                                 (a)                              (b)
                Figures: (a)Three examples of a light ray from inside a silica fiber impinging on the
                 air/silica boundary at different angles.(b) Light trapped by total internal reflection.



        The sketch of figure (b) shows only one trapped ray , but since any light ray
incident on the boundary above the critical angle will be reflected internally, many
different rays will be bouncing around at different angles. Each ray is said to have a
different mode so a fiber having this property is called a multimode fiber.
         However, if the fiber's diameter is to a few wavelengths of light, the fiber acts
like a wave guide, and the light can only propagate in a straight line, without bouncing,
yielding a single-mode fiber. Single mode fibers are more expensive but can be used for
longer distances. Currently available single-mode fiber can transmit at several Gbps for
30 km. Even higher data rates have been achieved in the laboratory for shorter distances.
Experiments have shown that powerful lasers can drive a fiber 100 km long without
repeaters, although at lower speeds. Research on erbium-doped fibers promises even
longer runs without repeaters.
2(b)
         Above 100 MHz, the waves travel in straight lines and can therefore be
narrowly focused. Concentrating all the energy into a small beam using a parabolic
antenna gives a much higher signal to noise ratio, but the transmitting and receiving
antennas must be accurately aligned with each other. In addition, this directionality
allows multiple transmitters lined up in a row to communicate with multiple receivers in
a row without interference. Before fiber optics, for decades these microwaves formed the
heart of the long distance telephone transmission system. In fact, the long-distance carrier
MIC's name first stood for Microwave Communication, Inc., because its entire system
was originally built on microwave towers.
            Since the microwaves travel in a straight line, if the towers are too far apart, the
earth will get the way. Consequently, repeaters are needed periodically. The higher the
towers are, the further apart they can be. The distance between the repeaters goes up very
roughly with the square root of the tower height. For 100-m high towers, repeaters can be
spaced 80 km apart.
           Unlike radio waves at lower frequencies, microwaves do not pass through
buildings well.
                                                                                   (20 marks)
3.Solution
(a)
          First is that the almost attenuation and distortion are more severe when sending
two signal level than when using modems, it is easy to calculate how the signal can
propagate and still can recognizable. Digital signal can pass through an arbitrary number
of regenerators with no loss on signal and thus travel long distances with no information
loss.
         Second, a digital transmission such as voice, data, music and image can be
interspersed with more efficient use the circuit and equipment. Another advantage is
much higher data rate using the existing lines.
         Third, a digital transmission is cheaper than analog transmission so it is not
necessary to reproduce the analog waveform after it has been passed through the
potentially hundreds of existing transcontinental call.
          Finally, maintenance of the digital signal is much easier than maintenance of
analog one. A transmitted bit is either received whether or not, it can be transmitted.
3(b) Describe the three major components of telephone system.
         The three major components of telephone system are:
      1. Local loop (twisted pair, analog transmission)
      2. Trunk (optical fiber or microwave, mostly digital)
      3. Switching office
3(c)
       When a computer wants to transmit digital data over communication channel, it
first converts to analog by modem for transmission over local loop, transform back into
digital for transmission over long-haul trunks and convert to analog over the local loop at
the receiving end and back into digital by another modem for storage into the destination
computer.
        For leased line, it is possible digital transmission from to finish, but it is very
expensive and used for building in a company, private network.
                                                            Digital
                                               Analog     (telephone              Customer premises
                                                           company        Analog      equipment
                                               (local
                                                            trunks)        (local
                                                loop)
                                                                           loop)
                                                      Codec          Codec

                     Computer          Modem      End         Tool          End     Modem       Computer

                              Digital            office       office       office           Digital
                           (short cable)                                                 (short cable)




  Figure: The use of both analog and digital transmission for a computer to computer call.
Conversion is done by the modems and codecs.
                                                                                                           (20 marks)
4.Solution
(a)
                             0     0       0     1        1   1        0   1    0    1


              Bit stream



              Manchester
               encoding




          Manchester encoding divided each bit into two equal intervals. A binary 1 is
sent by having the voltage high during the first interval and low to the second one. A
binary 0 is just reverse, first low and then high. This scheme ensures that there is always
a transition in the middle of interval. This scheme helps the receiver to synchronize with
the sender.
4(b)
           The 802.3 frame structure is shown in figure. Each frame byte contains
preamble 7 bytes, each containing bit patterns 1010101010. Start each frame byte
containing 10101011 to denote at the start of the frame itself.
           The frame contains two addresses; one for the destination and one for the
source. The standard allows 2 byte and 6 byte addresses but the parameters defined for
the 10 Mbps broadband standards use only for 6 byte address.


             Bytes    7    1    2 or 6      2 or 6   2    0-1500 0-46     4
                               Destination Source
               Preamble                                   Data   Pad Checksum
                                address    address
                         Start of                Length of
                     frame delimiter             data field
                                Figure: The 802.3 frame format


           To make it easy to distinguish between valid frames from garbage, 802.3 states
that valid frame is must be at least 64 kbytes long from the destination address to the
checksum. If the data portion of a frame is less than 46 bytes, the pad field is used to fill
out the frame to the minimum size.
                                                                                  (20 marks)
5.Solution
(a)
          Many organizations have multiple LANs and want to connect them. This desire
requires to have a device called bridge which operates at the data link layer.
           Different departments choose different LANs, without regard to what other
departments are doing. Sooner or later there is a need for interaction. So bridges are
needed.
          Some organizations are widely spread. It is cheaper to have different LANs on
each buildings and connecting them by bridges and infrared links than to run a single
coaxial cable for the entire site.
          Normally files are kept on file sever machine and are downloaded to user
machine upon request. The enormous scale of this system require putting separate LANs.
So bridges are needed.
         A single LAN would be adequate in terms of the load but the physical distant
between the most distinct machine is for too grate. By inserting LANs and bridge the
total physical distinct cover can be increased.


5(b)
         There is matter of reliability. Bridges can be programmed to exercise discretion
about what it does forward and what it does not forward.
         Last, bridges are contributed to the organizational security. LANs contain the
promiscuous mode. By inserting LANs and bridges and being careful not to forward
sensitive traffic so that traffic cannot escape and fall into wrong band.


                          Host A                                                                    Host B




                                Pkt                                                                      Pkt


                                                                  Bridge

                                Pkt                                                                      Pkt

                                                                  Pkt

                       802.3     Pkt                                                            8024     Pkt
                                                 802.3     Pkt          8024    Pkt



                       802.3     Pkt                                                            802.4    Pkt
                                                 802.3     Pkt          8024    Pkt



                                       802.3   Pkt                                    802.4   Pkt


                               CSMA/CD LAN                                     Token bus LAN



                                   Figure: Operation of a LAN bridge from 802.3 to 802.4
6.Solution
(a)
         The main function of the network layer is routing packets from the source
machine to the destination machine. In most subnets, packets will require multiple hops
to make the journey.


6(b) What are the special IP addresses? Discuss briefly.
         Every host and router on the Internet has an IP address, which encodes its
network number and host number. The combination is unique: no two machines have the
same IP address. All IP addresses are 32 bits long and are used in the Source address and
Destination address fields of IP packets. The formats used for IP address are shown in
figure (a). Those machines connected to multiple networks have a different IP address on
each network.


                                                32 Bits
                                                                                  Range of host
                         Class
                                                                                  addresses
                           A     0 Network                  Host                 1.0.0.0 to
                                                                                 127.255.255.255
                           B     10      Network                   Host          128.0.0.0 to
                                                                                 191.255.255.255

                           C     110            Network                   Host   192.0.0.0 to
                                                                                 223.255.255.255

                           D     1110              Multicast address             224.0.0.0 to
                                                                                 239.255.255.255
                                                                                 240.0.0.0 to
                           E     11110          Reserved for future use
                                                                                 247.255.255.255
                                         Figure(a): IP address formats



          The class A, B, C, and D formats allow for up to 126 networks with 16 million
hosts each, 16,382 networks with up to 64K hosts, 2 million networks, with up to 254
hosts each, and multicast, in which a datagram is directed to multiple hosts. Addresses
beginning with 11110 are reversed for future use. Tens of thousands of networks are now
connected to the Internet, and the number of doubles every year. Network numbers are
assigned by the NIC (Network Information Center) to avoid conflicts.
          Network addresses, which are 32-bit numbers, are usually written in dotted
decimal notation. In this format, each of the 4 bytes is written in decimal, from 0 to 255.
For example, the hexadecimal address C0290614 is written as 192.41.6.20. The lowest IP
address is 0.0.0.0 and the highest is 255.255.255.255.
          The values 0 and 1 have special meanings, as shown in figure (b). The value 0
means this network or this host. The value of -1 is used as a broadcast address to mean all
host on the indicated network.


                                  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 This host
                                  0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0



                                  00   ….     ...    000
                                                       0                  Host

                                                                                                    Broadcast on the
                                  1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
                                  11111111111111111111111111111111                                  local network

                                                                                                    Broadcast on a
                                         Network
                                        Network            1111               ...        1111       distant network


                                        127
                                       127                          (Anything)                      Loopback


                                               Figure: Special IP addresses




          The IP address 0.0.0.0 is used by hosts when they are being booted but is not
used afterward. IP addresses with 0 as network number refer to the current network.
These addresses allow machines to refer to their own network without knowing its
number. The address consisting of all 1s allows broadcasting on the local network ,
typically a LAN. The addresses with a proper network number and all 1s in the host field
allow machines to send broadcast packets to distant LANs anywhere in the Internet.
Finally, all addresses of the form 127.xx.yy.zz are reversed for loopback testing. Packets
sent to that address are not put out onto the wire; they are processed locally and treated as
incoming packets. This allows packets to be sent to the local network without the sender
knowing its number. This feature is also used for debugging network software.
6(c)
        Every byte on a TCP connection has own 32-bit sequence number. For a host
blasting away at full speed on a 10-Mbps LAN, theoretically the sequence numbers could
wrap around in an hour, but in practice it takes much longer. The sequence numbers are
used both for acknowledgements and for the window mechanism , which use separate 32-
bit header fields.
        The sending and receiving TCP entities exchange data in the form of segments. A
segment consists of a fixed 20-byte header followed by zero or more data bytes. The TCP
software decides how big segments should be. It can accumulate data from several writes
into one segment or split data from one write over multiple segments. Two limits restrict
the segment size. First, each segment, including the TCP header, must fit in the 65,535
byte IP payload. Second, each network has a maximum transfer unit or MTU, and each
segment must fit in the MTU. In practice, the MTU is generally a few thousand bytes and
thus defines the upper bound on segment size. If a segment passes through a sequence of
networks without being fragmented and then hits one whose MTU is smaller than the
segment, the router at the boundary fragments the segment into two or more smaller
segments.
                                                                             (20 marks)

				
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