EBU5403 Solutions A 2016/17
Page 1 of 11
SOLUTIONS
Module: Internet Protocols
Module Code EBU5403 Paper A
Time allowed 2hrs Filename Solutions_201617_EBU5403_A
Rubric ANSWER ALL FOUR QUESTIONS
Examiners Prof Yue Chen Dr Richard Clegg Dr Michael Chai
EBU5403 Solutions A 2016/17
Page 2 of 11
Question 1
a. Give the name (not number) of the two layers which are in the OSI network stack but not the
TCP/IP network stack?
[2 marks]
Session + Presentation
b. Give the name (not number) of the layer of the TCP/IP stack which:
[3 marks]
i. Transmits bits over a medium using the electronic, optical or mechanical properties of
that medium.
(1 mark)
ii. Is responsible for moving packets from their original source on the network to their final
destination.
(1 mark)
iii. Ensures that (if required) transmitted data arrives in order at the final destination,
without loss or corruption.
(1 mark)
i. Physical
ii. Network
iii. Transport (1/2 marks if numbers not names given).
c. In Carrier Sense Multiple Access/Collision Avoidance, give the full name and briefly explain
the role of ACK, RTS and CTS
[9 marks]
ACK = Acknowledgement is used to indicate that a message has been successfully received (1
mark expansion 2 marks definition)
RTS = Request to Send is used by a sender to ask for permission to send a message and reserve
space on the channel (1 mark expansion 2 marks definition)
CTS = Clear to Send indicates the sender is allowed to send their message. (1 mark expansion 2
marks definition)
d. Put ACK, RTS, CTS and Data transmission into the correct order that they would be sent in a
successful CSMA/CA transmission
[1 mark]
RTS, CTS, Data Transmission, ACK
e. Briefly define full-duplex, simplex and half-duplex transmission.
[4 marks]
Full duplex means a link can send in both directions at the same time.
Simplex means a link can send in only one direction.
Half duplex means a link can send in both directions but not in both directions at the same time
(2 marks for this part, one for both directions, one for not at the same time).
EBU5403 Solutions A 2016/17
Page 3 of 11
f. The figure below shows a learning bridge:
Its routing table is currently empty. Draw the routing table for the learning bridge after:
i. C sends a frame to B;
ii. Then D sends a frame to A;
iii. Then B sends a frame to A.
[6 marks]
Route Port
C 2
Route Port
C 2
D 2
Route Port
C 2
D 2
B 1
2 marks for column headings: (Headings can be Route/Destination/Host/Source and Port/Interface)
1 mark for realising you need three tables
1 mark for final table containing routes to C, D and B only (order does not matter in table)
2 marks if the correct ports are used for the correct route
EBU5403 Solutions A 2016/17
Page 4 of 11
Question 2
a) Answer the following questions with regard to Internet Protocol version 4 (IPv4) addressing and
Internet Protocol version 6 (IPv6) addressing.
[8 marks]
i) Identify the IP class of a host with IPv4 address 227.34.78.1. Calculate all the
subnetwork addresses of 227.34.78.0/26
(5 marks)
Class D (1 mark)
Network address: 227.34.78.0/26 (1 mark)
Subnetwork address: 227.34.78.64/26 (1 mark)
Subnetwork address: 227.34.78.128/26 (1 mark)
Subnetwork address: 227.34.78.192/26 (1 mark)
ii) Below is the IPv4 mapped IPv6 address.
0.0.0.0.0.0.0.0.0.0.255.255.14.0.20.80
- Identify when IPv4 mapped IPv6 address is used.
When host that supports both IPv4 and IPv6 communicates with host that supports only
IPv4 (1 mark)
- Convert it to Colon Hexadecimal Notation.
0::FFFF:0E00:1450 (2 marks)
(3 marks)
b) Answer the following questions related to routing.
[8 marks]
i) As referring to Figure 1, describe the undesirable behaviour that can arise between nodes A and
B because of a link failure such as between nodes B and C. And explain how the split-horizon
technique can be of benefit.
(4 marks)
EBU5403 Solutions A 2016/17
Page 5 of 11 A B C
Figure 1:- Split Horizon Illustration
The undesirable behaviour is illustrated below:
Routing Table Extract at A
Dest Next Cost Time
C B 2 xx
Dest Next Cost Time
C B 2 xx
Routing Table Extract at B
Dest Next Cost Time
C C 16 xx
Dest Next Cost Time
C C 16 xx
Dest Next Cost Time
C B 2 xx
Dest Next Cost Time
C B 2 xx
Dest Next Cost Time
C A 3 xx
Dest Next Cost Time
C A 3 xx
Dest Next Cost Time
C B 4 xx
Dest Next Cost Time
C B 4 xx
Dest Next Cost Time
C A 5 xx
Dest Next Cost Time
C A 5 xx
1 Failure arises on link between B and C
B announces unreachable link to A
but A has a lower cost link (via B!!)
2
A tells B it can see C 2 hops away and
B thinks A has an alternative path
3
A’s low cost path times out and A now
accepts the path announced via B that’s
3 hops away (via A!!)
4
The process repeats until the cost
progressively escalates to 16 and the route
is finally considered truly unreachable
Etc...
2 marks for correct diagram or correct description of each node.
The Split Horizon update mechanism means that a router does not broadcast routing information
over the same interface over which it was received. This leads to a more rapid discovery of the
unreachable path and thus improves convergence time. (2 marks)
ii) Briefly explain how the two following enhancements to Routing Information Protocol (RIP) that
can be used to facilitate rapid re-convergence.
Hold down timer
Triggered updates
(4 marks)
Hold-Down Timer - Router does not change information about a route following a message
indicating that the destination is unreachable (for 60 seconds) (2 marks)
Triggered Updates - Route changes require a router to issue a broadcast message rather than
waiting for the normal broadcast interval (2 marks)
c) With reference to the topology shown in Figure 1, use Dijkstra’s algorithm to generate the
Shortest Path Tree originating from Node E and use it to construct the corresponding routing table.
It is important to clearly show how you arrived at your answer.
EBU5403 Solutions A 2016/17
Page 6 of 11
[9 marks]
Candidate Cost to Root Tree
D,D,0
D,C,3
D,E,3
3
3
D,D,0
D,E,3
C,E,2
C,F,1
C,B,6
C,A,4
3
5
4
9
7
D,D,0
D,C,3
C,F,1
C,B,6
C,A,4
E,F,1
4
9
7
4
D,D,0
D,C,3
D,E,3
C,B,6
C,A,4
9
7
D,D,0
D,C,3
D,E,3
C,F,1
C,B,6
A,B,1
9
8
D,D,0
D,C,3
D,E,3
C,F,1
C,A,4
D,D,0
D,C,3
D,E,3
C,F,1
C,A,4
A,B,1
EBU5403 Solutions A 2016/17
Page 7 of 11
(2 marks for correct choosing candidate to the Tree; 2 marks for correct identification
of Cost to Root; 1 mark for appropriate comments) Total of 5 marks awarded for an
appropriate Dijkstra calculation;
2 marks for correct SPT
2 marks for correct routing table
Routing table
Destination Next hop Cost
D - -
C C 3
E E 3
F C 4
A C 7
B C 8
Question 3
a) Answer the following questions with regard to Internet Control Message Protocol (ICMP).
[12 marks]
i) What is the purpose of ICMP? Explain how ICMP Type-3 messages are used for Path
Maximum Transfer Unit (MTU) Discovery.
(4 marks)
ICMP enables a router or destination host to communicate with a source host, for example,
to report an error in datagram processing. (2 marks)
Path MTU discovery uses the Type-3 Destination Unreachable ICMP messages as follows:
TCP negotiates initial MTU size – usually
TCP/IP sends datagram with Don’t Fragment flag set
If datagram is too large an ICMP Destination Unreachable is received
Reduce Maximum Transfer Unit (MTU) and try again
EBU5403 Solutions A 2016/17
Page 8 of 11
(2 marks)
ii) Explain the main function of the ICMP Type-4 Source Quench message and explain why its
use is not always advisable?
Type-4 Source Quench ICMP messages used for congestion control for overloaded routers.
(2 marks)
ICMP Source Quench is sent to source for each discarded packet. Reception lowers packet
rate in theory. However congestion can give rise to a storm of Type-4 messages towards the
originating source – possibly compounding the problem. (2 marks)
(4 marks)
iii) Describe how ICMP’s traceroute works for debugging.
(4 marks)
traceroute sends a series of packets for which it sets the IP “Time To Live” (TTL) hop-count
field to different values. Setting the TTL to N will result in the Nth router generating a
“Time Exceeded” ICMP message which it sends back to the originating host. (2 marks)
Because ICMP messages are sent using IP packets, traceroute can extract from their source
address the IP address associated with the router at the Nth hop. (Strictly speaking it’s not
the IP address associated with the router, since routers have one IP address for each of their
interfaces.) (2 marks)
b) Provide brief definition of the following terminologies related to Transport layer.
[7 marks]
i) Flow control
(3
marks)
Flow control protocol is required to prevent data items being discarded at consumer
when over pushing by the producer. It involves preventing senders from overrunning the
capacity of the receivers (2 marks).
Receiver controls the Send window size, and adjust its own Receive window, usually
opening or closing to control the data flow (1 mark).
ii) Congestion control
(2 marks)
Congestion control prevents too much data from being injected into the network, thereby
causing switches or links to become overloaded (2 marks).
iii) Error control
EBU5403 Solutions A 2016/17
Page 9 of 11
(2 marks)
Error control detects and discards corrupted packets; Keeps track of lost and discarded
packets and resends them; Recognises duplicate packets and discards them; Buffers outof-order
packets until the missing packets arrive (2 marks).
c) Explain how flow control can be used to prevent receive buffer overflow using a
Sliding Window approach.
[6 marks]
(1 mark for each point)
Rx tells Tx how much data to send - window (sent in Acks).
Rx can vary window size (increase or decrease)
Reduce window to 1: stop-and-wait or zero to halt flow
Other window sizes allow for progressive flow control that can easily adapt to the rate of
application layer pull.
ACK from Rx contains SN for data to be acknowledged
Tx can send (W - u) packets: where W is window size for Tx and u is the number of
packets outstanding
EBU5403 Solutions A 2016/17
Page 10 of 11
Question 4
a) Explain the purpose of the following TCP segment header fields:
[8 marks]
i) Source / Destination Port Numbers
(2 marks)
Used to identify source and destination application instances. (2 marks)
ii) SYN Flag
(2 marks)
Used to signal the opening of a new TCP connection – 3-way handshake – causes Snum to
be set at the receiver if connection accepted. (2 marks)
iii) Data Offset
Header is variable size so data offset is the number of 32bit chunks comprising a particular
header. (2 marks)
(2 marks)
iv) Acknowledgement Number
Index value of the next byte expected in the ordered sequence of data to be received. By
definition, all preceeding bytes must have been received to form a contigious stream up until
this point. (2 marks)
(2 marks)
b) With the aid of diagrams, explain how TCP maintains the transmission as a connection-oriented
transport layer protocol.
[6 marks]
TCP use three-way handshaking scheme to establish the bi-direction connection before
transmitting data. While the termination process is carried out separately for two directions (2
marks).
(2 marks)
EBU5403 Solutions A 2016/17
Page 11 of 11 www.3daixie.com
(2 marks)
c) Answer the following questions with regard to Transmission Control Protocol (TCP).
[11 marks]
i) What can be deduced from the reception of duplicate acknowledgements? How is this
information used in TCP Reno to maintain channel utilisation?
(5 marks)
The communication channel is unlikely to be congested as ACKs are still getting through (2
marks). Missequenced data will cause dup-ACKs so to prevent false triggers three dupACKs
used (1 mark - extra).
Reno uses fast retransmit and fast recovery (2 marks). The aim is to prevent premature
congestion window collapse (1 mark).
ii) Describe the mechanism of TCP’s Path MTU Discovery.
(3 marks)www.daixie3.com
After connection, TCP uses minimum of outgoing I/F MTU or receiver advertised MSS (1
mark). DF bit set in segments (1 mark);
If ICMP “can’t fragment” returned (1 mark), TCP decreases segment size and retransmits (1
mark).
iii) Why is the TCP transport service unsuitable for real-time data? Use a real-time voice flow
as an example within your answer.
(3 marks)
The ability to resend missing segments is inappropriate for a service where low delay and
jitter are required. For RT voice, the reception of “late” samples would not add to the quality.
Delaying during this intervening time is worse than playing silence or repeating last sample (1
mark for each point)
因为专业，所以值得信赖。如有需要，请加QQ：99515681 或邮箱：[email protected]
微信：codinghelp