Computer Network Principles & Applications
Exam Paper Answers 1999/2000 and 2000/2001

These answers may not be correct so don't just take my word for it! Do the questions yourself by cheating and using your course notes and books. If they look like mine, then the chances are quite good that we are both correct. It does not matter if it takes you more than one hour for each question, the fact you have got an answer will stick in you mind and will help during the real exam.

Learn to read the question properly and only answer what is asked and not what you think they want. Don't forget to do all the parts of each question. Explain any assumptions you have made if the question is not clear. Read all the questions before you write anything down, there may be an easier one further down the paper. Tick off the ones you can answer as you read the paper, then go back and assess which will be the best question to start with.

I have looked at both of these past papers and there is a common theme of 8 questions from which you must select 5. There is no compulsary question so you can just cross out the three you don't like in the exam. All questions are worth the same 20 marks (hence 5 = 100%) although the sub questions do vary the the marks they attract. I cannot see any parts with less than 2 marks although some of the larger ones do attract 10 or more on occasion.

The questions usually group all their parts around a common subject area so there should be no rogue parts to these questions.

In common with previous exams I have assumed that you get a point for each statement, bullet point or sentence that gives a complete answer and uses all the buzz-words. If your answer does not do this only expect half a point per statement. Sometimes it is easier to draw a diagram than try to describe something in words (a TCP packet layout for instance). In this case you will get marks for every correct physical object drawn, and every correct label that explains what it is.

Keep an eye on the time, there are 5 questions to answer in 3 hours that's 15 mminutes to read the paper, 30 minutes per question and 15 minutes checking at the end. That works out at about 90 seconds per mark, so if you are still working on a 3 mark question after 5 minutes, stop, leave a few lines blank and go back to it at the end.

I have not bothered to repeate the questions before I answer them. Please print out the exam papers by downloading the zip file in the CNPA Index page. This contains two PDF files, one of each exam paper.

Exam Paper Answers 1999/2000

1.a

(7 marks)		 (to remember the ISO OSI 7 layer model use PANTSPA reading from the bottom up)
  1. Application Client program that uses the network utilities (FTP, TELNET) and tools for the application programmer.
  2. Presentation Abstraction of application data format (ASCII, EBCDIC, Integer Length, Video Stream, etc.) between peers
  3. Session Name Space to tie together descrete data streams that are part of a single application e.g. Videoconference
  4. Transport End to End (process to process) communication of messages
  5. Network Addressing and Routing of packets to their final destination
  6. dAtaLink Transmission and reception of frames over the physical medium
  7. Physical The physical medium over which the communicating bit stream is carried

1.b

(5 marks)		 (show HTTP request inside TCP packet inside Ethernet Frame - Easier to use a diagram for this one)

1.c

(3 marks)		 (The only example I can think of is the ATM LAN Emulation scenario)
Here the Datalink layer has been replaced with the three ATM layers to support LAN Emulation over an ATM network.

Just above the Physical layer is the ATM layer which deals with the delivery of CELLS carrying 48 bytes of payload.

Above this is the AAL5 layer which handles fragmentation and re-assembly to provide an Ethernet Frame lookalike interface for the layers above.

Above this is the Signaling and LAN Emulation layer which mimics the shared medium approach of a LAN and provides both Broadcast and Multicast functionality.

1.d

(5 marks)		 (Refer to the two diagrams on page 38 of the CNPA Course Book)
Diagram 1.18 shows the 4 layers with Applications at the top, then Transports, then IP and lastly Network at the bottom. Diagram 1.19 shows how the applications can make direct calls down to the Network or IP layers and ignore the TCP or UDP Transport layers. This gives a lot of flexibility that the ISO OSI model does not allow.

The why is slightly more difficult. The IP part is centric to the whole thing and as long as the interfaces to IP follow the API in both directions it matters not what transport layer is above or what network layer is below, IP will still work and keep it all together.

The four layer model fell out as a de-factor standard due to the fact that in the early 1980's the software was shipped as part of Berkley UNIX on a number of UNIX platforms (most manufacturers provided the Berkley Extensions when they shipped their own UNIX flavour).

If anyone has a more complete answer I'd like to hear it.

2.a

(5 marks)		 (This is about IPv4 Address Classes)
All IP Addresses in Version 4 are 32 bits long and are written as four groups of numbers separated by dots ie: 19.159.62.21

Each group represents an eight bit pattern written as a number between zero and 255. The actual bits in the address represent both a Network part and a Host part of the node address. The dividing line between Network and Host gives rise to four Classes as follows:

  1. Class-A 7 bit Network 24 bit Host (First Bits = 0) Network Numbers: 1 - 126
  2. Class-B 14 bit Network 16 bit Host (First Bits = 10) Network Numbers: 128 - 191
  3. Class-C 21 bit Network 8 bit host (First Bits = 110) Network Numbers: 192 - 223
  4. Class-D Multicast Addresses (First Bits = 1110) Network Numbers: 224 - 247
  5. Class-E Reserved (First Bits = 1111) Network Numbers: 248 - 255
The number of bits used for the identification of the Network or the Host determins the absolute number allowed for each part. So for Class-A with 7 bits in the Network part, it can only support 126 networks (0 and 127 are reserved) but each one can support up to 16,777,216 Hosts. For Class-C that equates to 2,097,152 networks supporting 255 Hosts each.

If the Network part of two addresses is the same (ignoring subnetting) then the Hosts are said to be on the same Network.

The following are reserved addresses which have special meaning:

  1. 127 Loopback Used for Adaptor testing
  2. 255.255.255.255 Broadcast Sent to all hosts !!!

2.b

(3 marks)		 (Remember the Network and Host parts of the IP Address)
The Router looks at the first few bits of the Destination IP Address to determin the Class of the address. Then it works out how many bits make up the Network part of the IP Address. Then it looks this up in it's Routing Table to see what Port is associated with that Network.

2.c

(6 marks)		 (SubNet - Multiple LANS withing one Site looks like One from outside the Site. SuperNet - A group of Networks treated as One Network)
SubNetting was interoduced in an attempt to improve the efficiency of address allocation. It is enabled by the use of a SubNetMask which, when logically ANDed with the Destination IP Address, reveals the Network and Host parts of the Address as follows:
		Desination IP Address:    128.016.044.012
		SubNetMask:               255.255.240.000

		IP Addr:  1000 0000 . 0001 0000 . 0010 1100 . 0000 1100 (AND)
		SubNetM:  1111 1111 . 1111 1111 . 1111 0000 . 0000 0000 =
		Network:  1000 0000 . 0001 0000 . 0010 0000 . 0000 0000 +
		Host:     0000 0000 . 0000 0000 . 0000 1100 . 0000 1100

		Destination Network:      128.016.032.000
		Destination Host:         000.000.012.012
Note: The spaces and leading zeros are for clarity only

SuperNetting was introduced to reduce the scaling problems associated with globally unique addressing schemes. As the number of Networks goes up, then so does the size of the Routing Tables held in the switches and routers. Searches through the Routing Tables to find the correct port for datagram delivery take longer as the Routing Table expands. The solution is to reduce the Routing Table to show only the shortest Address Bit Patterns which uniquely identify the outgoing Port. When Datagram nears it's final destination the Port Addresses get more specific and the Bit Pattern matching the Destination Address eventually line up on the final Port routing. See diagram below showing simplified 8 bit addresses:


Note: R = Router, N = Node

2.d

(6 marks)		 (NAT=Network Address Translation, DNS = Domain Name Service or Server)
MAC Addresses are globally unique, but IP Addresses don't have to be under some circumstances. Obviously, if you have a private network that is not connected to the Internet, then addresses used on this Private LAN can be duplicates of IP Addresses on the Internet. However as soon as you connect this private LAN to the Internet, you must make sure there is no conflict with previously used addresses. One way to do this is to use a NAT (Network Address Translator) box. As it's name implies, this has the ability to convert the Source Address field in an outgoing TCP packet from it's local (duplicate) address to a globally unique address when passed on to the Internet. The following diagram shows how this works:

Host A on NAT 1 sends a message to Host C on NAT 2. As the message arrives at NAT 1, it's sources address is changed to 199.1.2.3 by NAT 1. DNS is asked for the address of Host B and it replies with 199.5.6.7 which is used to set the Destination Address. The message is sent and when it arrives at NAT 2, the NAT box updates the Destination Address to 10.0.0.1 and delivers the message to Host C. If Host C needs to reply back to Host B, the same procedure is used in the opposite direction. The DNS always lies because it was lied to by the NAT boxes when they anounced their hosts.

A large private network can be hidden from the Internet by a NAT box with a small number of Ports. Not everyone in the company using the private LAN will be sending messages out side the LAN at the same time.

3.a

(7 marks)		 (Fragmentation of an IP packet - Page 256 of course book)
The IP header contains an Ident Field which is unique per packet over a time frame (depends on the sliding window size). When a packet is fragmented into smaller packets each of the fragments carries the same Ident so that they can be related back together at the collection point. The Offset Field in the header is used to sequence the fragmented packets back into the correct order during the re-assembly into the single large packet.

The initial packet is shown at the top, beneath are three fragments showing the use of Ident and Offset fields and the M bit of the Flags which is set to 1 when the fragment is part of a collection. The M bit is set to 0 when it is the last fragment in the set or when it is a stand alone packet.

3.b

(6 marks)		 (Remember AAL5)
The fragmentation of IP packets over ATM is more efficiently done using the ATM mechanisms of AAL5 Segmentation and Reassembly. This AAL protocol sits above ATM and segments the data into units that fit into the ATM 48 byte Cells.

Firstly the payload data is padded to the next 48 byte boundary to ensure the payload can be split into a whole number of Cells. Then the payload is segmented and encapsulated into ATM Cells before transmission over the ATM network.

Bit 3 in the ATM header Type Field is now used to indicate the last Segment in the group or the only one when set to 1. When set to 0 then the ATM Cell is part of a group (i.e. there is more to follow).

3.c

(7 marks)		 (ATM uses VPI, VCI not addresses - Confused! )
Give us a clue someone!

4.a

(6 marks)		 (Characteristics )
.

4.b

(8 marks)		 (CRC and Sliding Window )
.

4.c

(6 marks)		 (Remember congestion control)
.

5.a

(8 marks)		 (Remember )
.

5.b

(7 marks)		 (Remember )
.

5.c

(5 marks)		 (Remember )
.

6.a

(6 marks)		 (Remember Client and Server STUBS )
.

6.b

(6 marks)		 (Pass )
.

6.c

(2 marks)		 (Pass )
The

6.d

(6 marks)		 (Remember HTTP Protocol from Last Year)
The Web Client receives a data stream back from the Server which includes a series of response headers describing the returned content. The Entity Header details the content type as "text/html" or "video/mpeg" etc. depending of file type. The Entity Header also details the data length of the returned data stream.

In the case of textual data, one of the Client Request Headers allows the client to indicate it's support for character sets (but is this only for Chinese, Russian, etc. char sets or does it include ASCII and EBCDIC as well?).

7.a

(6 marks)		 (Remember Topologies)
A network can be connected in several ways as follows:
Type Description Advantage Disadvantage
Star All machines connected at one point Dedicated resources Single point of failure
Bus All machines connected to a common Bus Effiecient use of resources Collisions happen, exponential backoff reduces throughput
Ring All machines connected to a Ring No collisions Must wait for Token before sending
Mesh All machines connected to each other No collisions Connections per machine grows exponentially with size of network

7.b

(2 marks)		 (Remember Spanning Tree)
The Spanning Tree of a Network is a subgraph of the network that contains no cycles. For the given network, the spanning tree would be as follows:

7.c

(5 marks)		 (Remember the alorithm)
The idea behind the Spanning Tree Algorithm is for bridges to select the ports over which they will forward frames. The algorithm first Elects the bridge with the smallest ID as the Root of the Spanning Tree. The Root always forwards frames over all of it's ports. Next each bridge computes the shortest path to the Root and notes which of it's ports is on this path. This port is selected as the bridges preferred path to the Root. Finally each bridge elects a designated bridge that will be responsible for forwarding frames to the Root. This is based on which one is closest - usually determined from the hop-count. The node ID is used to break ties with the smallest ID winning.

The bridges use configuation messages to form the Spanning Tree by using these three pieces of information:

  1. ID of Sending Bridge
  2. ID of Root according to Sending Bridge
  3. Hop-Count from Sending Bridge to Root

7.d

(5 marks)		 (Remember the difference between Router and Bridge)
Well I can't - give me a clue someone....!

7.e

(2 marks)		 (Remember it's dynamic)
The Spanning Trees used by different bridges on an extended LAN will probably be different but will tend to migrate towards consistency over time. This dynamic behaviour is required if the algorithm is ever to "self repare" when some part of the LAN becomes disconnected due to a communications failure. The Root node sends out periodic claims to be Root to stop the other bridges from stepping in. If the messages stop, then other bridges will start another round of elections until the network stablises in the new configuration.

8.a

(5 marks)		 (Remember basic mechanics: Distance = Speed * Time, and Speed = Wavelength * Frequency)
Therefore Speed = 200,000km/s and Frequency = 100Mb/s so Wavelength = Speed/Frequency = 2 * 10^8m/s / 1 * 10^8b/s or 2m/bit. Length of 500 bytes = 500 * 8 bits * 2m = 8000m. Number of 8km frames in 200km of cable = 25 frames.

On an Idle FDDI Network, a Node wishing to transmit may have to wait a complete Token Rotation Time (TRT). As a rule this will depend of the length of the transmission medium and the signal velocity over that medium. But it can also be delayed by Token Holding Time (THT) at each node - so networks with many nodes will have longer TRT's. Assuming the above network consists of only two nodes situated diametrically opposed along the 200km ring, the TRT would be 1ms (200km / 200,000km/s).

Sounds pretty good to me! Any comments from anyone else?

8.b

(3 marks)		 (Remember segment length = 500m max)
The main problem with using Ethernet in a WAN is the limit of 500 meters per segment. As we can see from above, this is not an issue for FDDI.

8.c

(5 marks)		 (Remember )
Does anyone have clue what algorithm he is talking about here? The only one I can find is the THT, TRT, TTHT, details starting at the foot of page 133 in the course book. How does this relate to Broadcast and Multicast?

8.d

(7 marks)		 (Remember the mobile phone and base station examples)

The diagram above shows a Base Station (B) and two Mobile Stations (A and C). The circles represent their broadcast limitations. It is clear that both A and C fould transmit to B at the same time and frames would collide. However, neither of these broadcasters would be aware of the collision. This is the Hidden Terminal problem. How can A and C transmit to B if they cannot know that the other terminal is already engaged in communication with B.

Similarly, if B transmits to A, then this is seen by C although C can safely transmit to D even though frame collisions are evident at both B and C.

The MACAW (Multiple Access Collision Avoidance for Wireless) algorithm is used to mitigate both of these problems as follows:

  • The Sender transmits an RTS (Request to Send) Frame
  • The Receiver replies with a CTS (Clear to Send) Frame
  • The Sender sends the Data Frame
  • The Receiver sends an ACK after successful Frame recept
There are some other rules which allow this simple protocol to operate in a distributed arena:
  • The RTS and CTS contain a Field Length which implies a time that the transmission cannot exceed.
  • Any node that sees a CTS must stay quiet for this time in order not to disturb the transmission. Or until it sees the Receiver's ACK.
  • If a node sees the RTS but not the CTS then it is not close enough to the transmitter to interfere with the transmission and so is clear to also transmit.
  • If two or more Senders cause RTS collision, they follow the rule of exponential backoff as used on the Ethernet before retrying.

Exam Paper Answers 2000/2001

1.a

(1 mark)		 (Remember: increasing diversity of PDA's)
Because growth of minaturisation in the area of electronic technology has allowed intelligent devices (PDA, Laptop PC) to become smaller and lighter, this has lead to more use of these devices out of the office away from the corporate network infrastructure. To stay in touch requires that there be a wireless network to allow communicatiion to continue while on the move. Mobile phone technology and WAP has erged this process forward.

1.b

(5 marks)		 (Remember: Read Up on 802.11)
.

1.c

(8 marks)		 (Remember: Read up on MACA)
.

1.d

(4 marks)		 (Remember: Read up on Mobile Nodes and Access Points)
.

1.e

(2 marks)		 (Remember: Introduce Wirless to internet)
.

2.a

(2 marks)		 (Remember: Token Ring)
Advantage of Token Ring: No collisions
Disadvantage: Have to wait for Token before transmitting

2.b

(3 marks)		 (Remember: IBM Token Ring)
If a node on the ring dies then the Token cannot be sent on. In IBM Token Ring (with MSAU), a relay closes and the token can continue along the ring to the next node. MSAU = Multi Station Access Unit.

2.c

(7 marks)		 (Remember: Token Ring Process)
Station A waits while the Token is passed around the ring. When the token arrives at A it is taken off and the frame is sent to station B. As soon as the frame is sent (early token release) or after the rotation of the frame around the ring (late token release), the token is placed back on the ring. Station A sees the frame to B circulate around the ring. If B has seen the frame it sets the A bit (Ack) in the Frame Status field. If station B has accepted the frame it sets the C bit in the Frame Status field. The token monitor will remove frames which have been round the ring once without acknowledgement by inspecting the monitor bit.

2.d

(6 marks)		 (Remember: Token Monitor Duties)
The monior is elected mased on the highest (lowest) address. The monitor must maintian the integrity of the ring by removing corrupt frames or tokens, re-inserting a new token, and checking existance of nodes on the ring. If a node does not respond then the monitor alerts the other nodes of it's death. If the monitor dies, the remaining nodes elect a new monitor.

2.e

(2 marks)		 (Remember: Toten Rotatio Time [TRT])
The delays involved in signal propagation between ground based and orbital based stations is very large. Having a token ring where one station was in orbit would extend the token rotation time into the realms of unacceptability.

3.a

(3 marks)		 (Remember: What is a VC)
Connection oriented, packet switched network. Each descrete step in the total path is defined as a separate virtual circuit. Requies a setup, transfer, and teardown for each use of the VC. Each segment of the VC will have its own VCI.

3.b

(9 marks)		 (Remember: Switch Routing Tables)
Switch sees a message on an incomming port and notes it in the table. It assigns an outgoing port for the message to use on to the next switch. At the next switch it sees the incomming message on a different input port and assigns an output port for delivery to the host. The reply from the host comes back to the second switch and arrives on an input port. An entry in the switches table is made for the new port and an associated VCI is set up to reach switch 1. Lastly, switch 1 sees the incomming message from switch 2 and sets up an outgoing port to host A and assigns a VCI to that. At the end of the round-trip communication for the VCI setup, both switch tables are populated with the 4 VCI's that associate input and output ports in each switch. The VCI's in each switch are usually always different because each swich cannot know in advance what circuits are free on another switch.

3.c

(6 marks)		 (Remember: Adv and Disadv of VC)
Disadvantages of VC's are:
  • the time taken for set-up and teardown during which no useful data can flow,
  • if a link is dropped the whole set-up has to be repeated,
  • dedicated resource for the duration of the connection buffers is costly.
Advantages of VC's are:
  • Smaller address (VCI) = low overhead,
  • no collisions,
  • Quality of Service

3.d

(2 marks)		 (Remember: Main advantage of VC)
Quality of service on a DG network could be improved by intellegent flow control and caching.

4.a

(3 marks)		 (Remember: IP)
  • Has it's own globally unique addressing scheme
  • Datagram - makes no assuptions about underlying network
  • Addresses have a structure - hierarchy
  • Allows fragmentation and reassembly

4.b

(8 marks)		 (Remember: Address mapping strategies)
  • Manually map IP addresses to MAC addresses
  • Network Administrator maintains a table to do the mapping
  • Use ARP (Address Resolution Protocol)

4.c

(9 marks)		 (Remember: Draw and describe fragmentation stages including M-bit in the header. Mention MTU discovery.)
.

5.a

(8 marks)		 (Remember: Forwarding Algorithm)
Firstly the router will look at the first few bits of the address to determin the network class (A, B, C, etc), then it will look at the network part of the classified address and check if any outgoing port sits on the same network. If is does, it is delivered to that port. If not the router checks the routing table for the closest match of the destination address and sends the packet out on the associated port for that address. The process is repeated on every router on the path until the packet is delivered to the destination network.

5.b

(2 marks)		 (Remember: Link State, Distance Vector)
.

5.c

(5 marks)		 (Remember: Note - NODE-C is root - Draw graph and create table)
.

5.d

(2 marks)		 (Remember: )
If a node dies, the remaining nodes and re-evaluate the spanning tree.

5.e

(3 marks)		 (Remember: Read up on Link State Topology and how the nodes find out)
.

6.a

(2 marks)		 (Remember: )
Flow Control is the handshaking between the two communnication devices (Sliding Window). Congestion Control is limiting network or router saturation (DEC bit, RED).

6.b

(3 marks)		 (Remember: Sliding Window)
This is done by the receiver sending back an ack packet registering acceptance of the last sequence and containing an advertised window size of zero.

6.c

(2 marks)		 (Remember: )
Because TCP uses implicit acknowledgements - it assumes all packets up to and including the acknowledged packet have been received. Therefore it cannot ack out of sequence packets.

6.d

(4 marks)		 (Remember: Draw the receive buffer and mark the segments received)
.

6.e

(4 marks)		 (Remember: )
.

6.f

(3 marks)		 (Remember: )
.

6.g

(2 marks)		 (Remember: )
.

7.a

(5 marks)		 (Remember: )
.

7.b

(6 marks)		 (Remember: )
.

7.c

(2 marks)		 (Remember: )
.

7.d

(2 marks)		 (Remember: )
.

7.e

(5 marks)		 (Remember: )
.

8.a

(5 marks)		 (Remember: )
.

8.b

(6 marks)		 (Remember: )
.

8.c

(7 marks)		 (Remember: )
.

8.d

(2 marks)		 (Remember: )
.

Exam Paper Answers 2001/2002

Only Joking!

Information Systems Techniques

Program Design

Systems & Networks 1

Artificial Intelligence

Systems Design

Systems & Networks 2

Computer Network Principles & Applications

Database Systems