EXAMEN BGP

1. 159.334 Computer Networks Final Examination Page 1 of 10
Solution sheet – Final Exam 2009
Question 1 – Short Answer Questions (Compulsory)
1.1 The frequency of failure and network recovery time after a failure are
measures of the _______ of a network.
a) Performance
b) Reliability
c) Security
d) Feasibility [2 marks]
1.2 In cyclic redundancy checking, what is the CRC?
a) The divisor
b) The quotient
c) The dividend
d) The remainder [2 marks]
1.3 In modulo-2 arithmetic, __________ give the same results.
a) addition and multiplication
b) addition and division
c) addition and subtraction
d) none of the above [2 marks]
1.4 The ________ between two words is the number of differences
between corresponding bits.
a) Hamming code
b) Hamming distance
c) Hamming rule
d) none of the above [2 marks]
1.5 A simple parity-check code can detect __________ errors.
a) an even-number of
b) two
c) no errors
d) an odd-number of [2 marks]
1.6 The Hamming distance between 100 and 001 is ________.
a) 2
b) 0
c) 1
d) none of the above [2 marks]
1.7 A ________receives a signal and, before it becomes too weak or
corrupted, regenerates the original bit pattern. It then sends the
refreshed signal.
a) passive hub
b) repeater
c) bridge
d) router [2 marks]
1.8 Which one is not a contiguous mask?
a) 255.255.255.254
b) 255.255.224.0
c) 255.148.0.0
d) all are [2 marks]

2. 159.334 Computer Networks Final Examination Page 2 of 10
1.9 What is the first address of a block of classless addresses if one of
the addresses is 12.2.2.127/28?
a) 12.2.2.0
b) 12.2.2.96
c) 12.2.2.112
d) none of the above [2 marks]
1.10 An organization is granted a block; one address is 2.2.2.64/20. The
organization needs 10 subnets. What is the subnet prefix length?
a) /20
b) /24
c) /25
d) none of the above [2 marks]
1.11 An IPv6 address can have up to __________ hexadecimal digits.
a) 16
b) 32
c) 8
d) none of the above [2 marks]
1.12 ICMP is a _________ layer protocol.
a) data link
b) transport
c) network
d) none of the above [2 marks]
1.13 An ICMP message has _____ header and a variable-size data
section.
a) a 16-byte
b) a 32-byte
c) an 8-byte
d) none of the above [2 marks]
1.14 For purposes of routing, the Internet is divided into ___________.
a) wide area networks
b) autonomous networks
c) autonomous systems
d) none of the above [2 marks]
1.15 The ports ranging from 49,152 to 65,535 can be used as temporary or
private port numbers. They are called the ________ ports.
a) well-known
b) registered
c) dynamic
d) none of the above [2 marks]
Total [30 marks]

3. 159.334 Computer Networks Final Examination Page 3 of 10
Question 2 – OSI Model
2.1 Name, in order, the seven OSI Layers AND provide a one sentence
statement as to the function of each layer in the OSI Reference Model.
Answer:
(Answers that approximate this will be accepted. 5 marks for getting
the order and names right and 5 marks for the statements.) [10 marks]
2.2 The IP Protocol and most LAN and WAN protocols are classified as
“unreliable”. Briefly explain what this “unreliable” description refers to.
Answer:
In computer networking, a reliable protocol is one that provides
reliability properties with respect to the delivery of data to the intended
recipient(s), as opposed to an unreliable protocol, which does not
provide notifications to the sender as to the delivery of transmitted
data. [3 marks]
2.3 The so-called “hidden station” problem can arise in an 802.11x Basic
Service Set. Briefly explain how this difficulty is overcome.
Answer:
Hidden nodes in a wireless network refer to nodes that are out of range
of other nodes or a collection of nodes. In a wireless network, it is likely
that the node at the far edge of the access point's range, which is
known as A, can see the access point, but it is unlikely that the same
node can see a node on the opposite end of the access point's range,
B. These nodes are known as hidden nodes. A problem occurs when
nodes A and B start to send packets simultaneously to the access
point. Since node A and B cannot sense the carrier, Carrier Sense
Multiple Access with collision avoidance (CSMA/CA) does not work,
and collisions occur, scrambling the data. To overcome this problem,
handshaking is implemented in conjunction with the CSMA/CA
scheme. RTS/CTS can be implemented – it still might not overcome
the problem.
Other suggested methods for overcoming the problem include
(Wikipedia)
 Increase Transmitting Power From the Nodes
 Use omnidirectional antennas [5 marks]

4. 159.334 Computer Networks Final Examination Page 4 of 10
 Remove obstacles
 Moving the node
 Use protocol enhancement software
 Use antenna diversity
2.4 Briefly explain the difference between a port address, a logical address
and a physical address.
Answer:
The physical address (link address) is the address of a node as
defined by its LAN or WAN, it is the lowest level address and is
assigned to a physical device.
The logical address is an address used in the network layer to identify
a host that is connected to the Internet.
The port address is the address of a session running on a host as
used by the transport layer. [6 marks]
2.5 Explain the difference between the two packet data transfer techniques
referred to as “connectionless” and “connection-orientated”.
Answer:
In “connection-orientated” transfer, resources need to be reserved
along a path before packets can traverse the path. In “connectionless”
transfer, packets pass from point to point in a store and forward
movement using buffers at each intermediate router before passing to
the next router. Since packets can be routed independently in the
connectionless mode they may arrive out of sequence, whereas for
connection-oriented transfer packets should remain in the sequence
that they have been sent. [6 marks]
Total [30 marks]
Question 3 – Network Layer
3.1 a) In a few brief sentences, explain what is meant by “Equal Cost
MultiPath” (ECMP) as implemented in the OSPF routing
protocol.
Answer:
It means that more than one path can be found with the same
total metric cost. If the ECMP facility is switched on for OSPF,
[6+ 3 = 9
marks]

5. 159.334 Computer Networks Final Examination Page 5 of 10
then traffic can be divided between all paths that have this
equal cost. A problem that occurs here is the possibility of
packet streams splitting across multiple paths leading to out of
order packets appearing at the destination. Hash tables based
on the IP addresses could be used to ensure that packets from
the same session use the same path.
b) How can ECMP be used to reduce the possibility of network
congestion in a network using the OSPF routing protocol?
Answer:
An approach that can be used is to modify link metrics in such a
way that equal cost paths can be created which, in turn, results
in traffic being split over multiple paths and steering traffic away
from local congestion hotspots.
3.2 An organisation is granted the block 125.238.0.0/16. The administrator
wants to create 512 subnets:
a. Find the subnet mask required
b. Find the number of addresses in each subnet
c. Find the first and last allocatable addresses in subnet 1
d. Find the first and last allocatable addresses in subnet 14
Answer:
a. This is a class A address range so that the default mask is /8.
We need sufficient bits to produce 512 subnets so 2x
= 512
where x is the number of bits required. Hence x = 9, thus we
require the mask to be /8+9 = /17. OR 255.255.128.0 is also an
acceptable answer.
b. There will be 15 bits remaining so the number of hosts per
subnet is 32768 – 2 = 32766 since 2 addresses are not able to
be used for hosts (they are reserved addresses).
c. The first and last allocatable addresses in subnet 1 are:
The first address for the block is found by ANDing the address
125.238.0.0 with the subnet mask /16 giving
01111101 11101110 00000000 00000000 (125.238.0.0)
But this is not an allocatable address, so the next one is
01111101 11101110 00000000 00000001 (125.238.0.1)
Finally, we note that the broadcast address is the last one of
the 32768 addresses so the last allocatable one is the one
before that broadcast address so it must be
01111101 11101110 00000000 00000000 (125.0.127.254)
d. The first and last allocatable addresses in subnet 14 are:
To find the 14th
subnet we have to add 13 to the subnet number
of the first subnet. We can add this amount to the first
allocatable address of the first subnet as we know that the first
address in every block is not allocatable as it is a special
address.
The format is: 0nnnnnnn.ssssssss.shhhhhhh.hhhhhhhh [6 marks]

6. 159.334 Computer Networks Final Examination Page 6 of 10
1310 = 11112 added to bit positions 9 -17 inclusive
10000010 00111000 00000000 00000001
+
00000000 00000111 10000000 00000000
10000010 00111111 10000000 00000001 (125.6.128.1)
As there are 32768 addresses in the range with the first and
last being reserved, the last allocatable one must be 61
addresses after the above first allocatable address. Now
6110 = 1111012 is added to the first allocatable address to get:
10000010 00111000 00000111 11000001 (130.56.7.193)
+
00000000 00000000 00000000 00111101 (0.0.0.61)
10000010 00111000 00000111 11111110 (130.56.7.254)
Thus the final allocatable address must be 130.56.7.254
3.3 There are four basic message types for BGP:
a) OPEN
b) KEEPALIVE
c) UPDATE
d) NOTIFICATION
Provide a brief explanation of the function of each message type.
a) After establishing the connection, an OPEN message is sent by
one or other party.
b) A KEEPALIVE message is periodically sent between BGP
routers using TCP protocol often enough to ensure that timers
do not expire in the routers.
c) UPDATE messages are used to transfer routing information
between BGP peers. The information in the UPDATE packet
can be used to construct a graph describing the relationships of
the various autonomous systems. By applying appropriate rules
routing information loops and some other anomalies may be
detected and removed from inter-AS routing.
d) A NOTIFICATION message is sent when there is an error
condition. BGP usually closes the connection after receipt of
such a message. [4 marks]
3.4 BGP is regarded as a Path Vector Protocol, how does it differ from Link
State and Distance Vector Protocols?
Answer:
The primary function of a BGP system is to exchange network
reachability information with other BGP systems. BGP is rule based
rather than metric based as distinct from the case of the other two
protocols. Link state protocols pass around topology and network
status information using a flooding protocol. [5 marks]
3.5 List three reasons why most network specialists prefer Link State
Routing instead of Distance Vector Routing. [6 marks]

7. 159.334 Computer Networks Final Examination Page 7 of 10
Answer:
1) Fast, loopless convergence.
Two phases are used: (a) Rapid transmission via flooding and,
(b) Local computation of the routes. Cf Distance vector
protocols.
2) Support of precise metrics and multiple metrics - if desired.
Avoids the problem of counting to infinity, finer grain metrics are
now possible.
3) Support of multiple paths to a destination.
If there are some almost equivalent routes in terms of cost,
these may be used in link state protocol systems.
4) Separate representation of external routes.
Total [30 marks]
Question 4 – IP and Routing
4.1 Briefly describe the factors influencing the need to adopt IPv6 and replace IPv4.
Answer:
IPv4 was never intended for the Internet that we have today, either in terms of the
number of hosts, types of applications, or security concerns.
IPV6 was developed to address each of these items; viz:
 Number of hosts - IPV4 address space inefficiently used and is going to
exhaust in a relatively short period of time.
 Types of Applications - greater emphasis on Real time services, Voice,
video, music etc. Need a quality of service capability
 Security - IPV4 had no security so methods needed to be deployed in the
new system to address these concerns.
[6
ma
rks]
4.2 Briefly describe TWO possible methods for enabling a smooth transition from IPv4
to IPv6. Briefly comment on their suitability.
Answer:
Several approaches have been proposed, viz:
a) Dual-stack methods
b) Tunnelling
c) Header translation
1) Dual stack means running IPv4 and IPv6 simultaneously until everyone is
ready for IPv6. (To figure out which version of IP to use when sending a
packet to the destination, the source must send a query to the DNS server.
If the DNS server returns an IPv4 address then the source sends an IPv4
packet, else it sends an IPv6 packet.)
[6
ma
rks]

8. 159.334 Computer Networks
Figure
2) Tunnelling involves encapsulating an IPv4 packet inside an IPv6 packet or
vice versa and setting the protocol indicator for the packet as “IP”
Figure
3) Header translation used if the receiver doesn’t understand IPv4and you are
using IPv6. Translation is required at the router that connects to the IPv4
destination network.
Figure 3: Demonstrating header translation
4.3 The following diagram shows the IPv4 header
Final Examination
Figure 1: Demonstrating Dual Stack
Tunnelling involves encapsulating an IPv4 packet inside an IPv6 packet or
vice versa and setting the protocol indicator for the packet as “IP”
Figure 2: Demonstrating tunnelling process
tion used if the receiver doesn’t understand IPv4and you are
using IPv6. Translation is required at the router that connects to the IPv4
destination network.
: Demonstrating header translation
diagram shows the IPv4 header
Page 8 of 10
Tunnelling involves encapsulating an IPv4 packet inside an IPv6 packet or
vice versa and setting the protocol indicator for the packet as “IP”
tion used if the receiver doesn’t understand IPv4and you are
using IPv6. Translation is required at the router that connects to the IPv4
[12
ma
rks]

9. 159.334 Computer Networks Final Examination Page 9 of 10
In a few brief sentences explain the function of each field in the header.
Answer:
a) Version- IP version number (Version 4)
b) IHL – Internet Header Length
c) Type of Service – Type of service
d) Total Length – Total length of the packet
e) Identification -
f) Flags – Indications as to whether a packet has been segmented
g) Fragment offset – Offset if a packet must be segmented
h) Time to live – number of hops before packet is dicarded
i) Protocol – identifier for the program to be run at the destination (TCP, UDP
OSPF, etc etc)
j) Header Checksum – A checksum CRC-32 for the header only
k) Source address – a 32 bit address for the source address
l) Destination address – a 32 bit address for the destination address
m) Options – Extra options required to be passed to the destination
n) Padding – padding to complete 32 bit fields if options don’t use all 32 bits of
a word.
4.4 a) Briefly describe what is NAT (Network Address Translation) and how it
operates.
b) How does it assist with the problem of IPv4 address exhaustion?
[3 + 3
= 6
ma
rks]
Total [30
ma
rks]
Question 5 – Multimedia
5.1 What is meant by the term “jitter”? How does jitter impact real time
audio/video? [10 marks]
5.2 Discuss in some detail how SIP (Session Initiated Protocol) is used in
the transmission of multimedia. Include in your discussion the various
types of SIP message, addressing and a simple session description. [12 marks]
5.3 The JPEG process consists of three phases as shown in the figure
below. Briefly outline the process involved in transforming an image
into a compressed image using these three phases as a guide to your
descriptions:
[8 marks]
Total [30 marks]

10. 159.334 Computer Networks Final Examination Page 10 of 10
Question 6 – TCP, UDP and Security
6.1 Describe in detail the interpretation of the following 4 fields that form
the header of a UDP packet:
[8 marks]
6.2 Briefly explain the principal differences between UDP and TCP and the
reasons for choosing between them for a given internet application. [8 marks]
6.3 Explain the congestion avoidance scheme used in TCP. [4marks]
6.4 In order to answer the question "Where in the protocol stack does
security belong?", choose any 3 OSI layers and describe a network
security measure that can be implemented in each of these selected
layers. [6 marks]
6.5 Briefly compare the relative merits of Secret Key and Public Key
cryptography as used to provide network security. [4marks]
Total [30 marks]