Download Marks for each part of each question are

Department of Computer Science
University College London
Cover Sheet for Examination Paper to be sat in
May 2004
COMPC329-resit / COMPD015-resit
Communications and Networks
Time allowed 2.5 hours
Answers
Calculators are permitted
Answer THREE questions
Checked by First Examiner:
Date:
Approved by External Examiner:
COMP0D15
Date:
1
Turn Over
COMPC329 / COMPD015 Communications and Networks
Answer THREE questions
Marks for each part of each question are indicated in
square brackets
Calculators are permitted
Formulae which may be useful in the exam
Information Theory
Capacity C, bandwidth W, signal to noise ratio S/N, number of signal levels M.
Signal to Noise Ratio in decibels = 10log10(S/N)
Nyquist’s Theorem C = 2Wlog2M.
log2(1+S/N)
Shannon's Theorem C = W
Miscellaneous
log a b 
log c b
log c a
1. a) Asynchronous Transfer Mode (ATM) is a virtual-circuit (VC) based technology in which
data is transmitted in cells. Explain the terms cell and virtual-circuit. Explain how VCs
are identified within an ATM network and the constraints that apply when identifiers are
allocated. Outline some of the reasoning that has been advanced for the use of VCs in
ATM. [13 marks]
Cell = a small (53-byte) fixed-size packet.
Virtual circuit = a path set up between two hosts through the switches of a
packet network. All packets sent on the VC follow the same path.
VCs are identified by a series of numbers (VC identifiers) applied to each
link traversed. A VCI must be unique within the scope of a link. Each packet
carries a VCI in its header. Switches map VCIs appropriately before
transmission on the next link.
COMP0D15
2
Continued
In the ATM case VCIs have two components: The virtual path identifier
(VPI) identifies (usually) a path between two hosts. VPIs must be unique
within the scope of links. The virtual channel identifier (VCI) acts as a submultiplexor identifying a particular activity on the VPI. VCIs must be
unique within the scope of a VPI. In this model switches process VPIs ,
hosts process VCIs.
Cells were chosen since a) small size reduces store and forward delay, b)
fixed size makes for rapid processing.
Small size means cells cannot carry global addresses. This demands a VC
architecture in which cells need only carry VCIs which can be small due to
their limited scope.
b) i) Describe the Switched Virtual Circuit mode of the Classical IP over ATM (CIPA)
mechanism for implementing IP over ATM.
[5 marks]
There are three main problems to solve:
Encapsulation
Answers
ATM cells are generally too small to carry IP datagrams. The IP
fragmentation mechanism would be inefficient. Therefore an ATM-specific
framing mechanism (AAL5) is used to group cells into larger frames.
Address Resolution
An ATM ARP Server must be present at a well-known address. Typically
hosts establish VCCs with the AAS and register their mappings at start-up.
When an IP datagram is to be send to a host for which no mapping is
cached, the AAS is consulted.
Connection management
Host to host VCCs are set up on demand. They may be discarded after an
idle timeout.
ii) Discuss the inefficiencies that may arise when two IP subnets are implemented on a
single ATM network and outline a possible enhancement.
[4 marks]
The classical IP architecture requires a) that the two subnets use different
network prefixes and b) that inter-subnet traffic goes via a router.
This works but means that inter-subnet traffic traverses the ATM network
twice even though a more efficient path is available.
One solution is to use the Next-Hop Resolution Protocol. This requires a
Next Hop server in addition to the AAS. The NHS is informed of mappings
on both subnets. Hosts consult the NHS as well as the AAS. If the NHS
returns an address on the local ATM network then a diorect VCC is set up.
c) The diagram below shows two Ethernet LANs and an ATM network.
COMP0D15
3
Turn Over
A
ATM
B
Ethernet
Ethernet
Discuss some options for providing connectivity between hosts on these networks,
indicating some of their strengths and weaknesses.
[11 marks]
Some points that could be included … (obviously answers will vary. I am
looking for evidence of understanding of the various ways of connecting
things together)
(A) IP
Make A and B routers and deploy three IP subnets on the three networks.
Use CIP and ATM for example.
Advantages. Full IP connectivity between hosts on all networks. Simple,
well understood. Routers may have useful firewall features
Disadvantages. Excludes non-IP traffic. Routers comparatively expensive.
(B) LAN
Use LANE on the ATM network, A and B are bridges.
Advantages. Supports IP and non-IP. Only need one IP subnet
Disadvantages. May make for a very large subnet. LANE complex, not
especially efficient
(B) A and B half bridges or routers joined by ATM VCC
Advantages. Very simple use of ATM network.
Disadvantages. Excludes ATM hosts
[Total 33 marks]
COMP0D15
4
Continued
2. a) A wireless channel is subject to a signal to noise ratio of –20dB. The channel is required
to provide a capacity of 1 Mbps.
i) Show that the required capacity may be achieved provided the channel bandwidth is
greater than about 70 MHz.
[4 marks]
SNR = -20dB => S/N = 0.01
From Shannon, C = Blog2(1.01) = 0.0144B
For C = 106 we must have B = 106 / 0.0144 = 69.7 MHz (3.s.f)
ii) Outline briefly why a conventional modulation scheme such as Quadrature Amplitude
Modulation (QAM) would not be effective on this channel.
[4 marks]
16-QAM, for example would need to transmit at 106 / 16 = 62500 baud.
Modulation would produce a signal of the order of 262500 = 0.125 MHz –
a long way short of 70Mhz. In practice the noise would make the scheme
impractical
Answers
iii) A possible transmission technique for this channel is Code Division Multiple Access
(CDMA). A CDMA scheme uses the following chip sequences for stations A, B and
C:
A 00111001
B 01001101
C 10011100
By reference to this scheme, explain how CDMA works and why it is effective on a
noisy channel.
[10 marks]
–
Operation is best understood if a zero chip is represented as 1.
The chip sequences are orthogonal – i.e. A.B = 0 for all pairs. Also, A.A = 1
for all codes.
To transmit a 1 bit, a station transmits its chip sequence. To transmit a 0 bit
it transmits the complement.
A receiving station must compensate for differential power levels and lack
of synchrony between stations. If this is done, the combined, received signal
may be regarded as the sum of the transmissions. Thus, if A, B and C
transmit 0, 1, 1 respectively the combined signal will be:
1 1 -1 -1 -1
-1 1 -1 -1 1
1 -1 -1 1 1
1 1 -3 -1 1
1 1 –1
1 -1 1
1 -1 –1
3 -1 -1
To extract the signal for (say) B we form the normalised scalar product with
B’s code which gives (-A+B+C).B = -A.B + B.B + C.B = B.B = 1, i.e.:
COMP0D15
5
Turn Over
-1 + 1 + 3 + 1 + 1 + 3 + 1 + -1 = 8 => 8/8 = 1
In practice the noise would mean that some chips would be received in error
so that the result is not quite 1(0). However, most of the time sufficient will
be received correctly so that we can correctly recover the bit.
Clearly this scheme is using 8 times the capacity, hence the bandwidth, of a
conventional modulation scheme. By varying the chipping ratio we can
spread the signal over an arbitrary bandwidth.
b) i) The sequence 1010 1110 0011 0111 1111 1001 0011 0011 1001 11
is a message followed by a Cyclic Redundancy Check (CRC) generated using the
polynomial x5+x3+1. The left-most bit is transmitted first. Write down the message
and the CRC. [N.B. the bits have been grouped in fours to aid reading – there is no
other significance.]
[2 Marks]
message 1010 1110 0011 0111 1111 1001 0011 0011 1, CRC
00111
ii) Show how the sequence in b) i) above would be encoded in the data link layer if the
system used transparent bit-oriented framing (such as that used in HDLC for
example).
[3 marks]
The data link layer will perform bit stuffing and will delimit with flags and
the resulting string will be:
0111 1110 1010 1110 0011 0111 1101 1100 1001 1001 1100 1110 1111 110
c) The diagram shows two network switches S1 and S2 on the path between hosts A and B.
The diagram also shows the link capacities. The propagation time on each link is p sec.
A
a bps
S1
b bps
S2
c bps
B
i) “S1 and S2 are store-and-forward packet switches”. Explain what this statement means
and derive a formula for the delay when sending a packet of s bits from A to B.
[Ignore all other possible sources of delay].
[4 marks]
Store-and-forward switches must receive the complete packet on the input
link before they begin transmission on the output link.
Assume transmission starts at t = 0 sec
Last bit of packet leaves A at t = s/a sec
Last bit of packet arrives at S1 at t = p + s/a sec
Last bit of packet arrives at S2 at t = 2p + s/a + s/b sec
COMP0D15
6
Continued
Last bit of packet arrives at B at t = 3p + s/a + s/b + s/c sec
ii) A single error occurs as a packet traverses the inter-switch link. The error is detected
and recovered as a result of a timeout of T sec and successful retransmission. In one
scheme, detection and recovery take place in the Data-link Layer; in another scheme,
detection takes place at the destination host (B) and recovery is in the Transport
Layer. Find, in each case, the effect this error has on the delay calculated in part c) i).
Briefly discuss the strengths and weaknesses of the two recovery schemes.
[6 marks]
In both cases the last bit of the packet arrives at B at t = T +3p+ s/a + s/b
+ s/c sec,. However, the timeout in the Data-Link case can be shorter since
it depends on the RTT on a single link. This will be both shorter than the
end-to-end RTT and will likely have lower variance.
The disadvantage of this is that it forces the application to accept error
recovery and the consequential variable delay whether it wants it or not.
[Total 33 marks]
Answers
3. a) A web server runs on a computer with domain name haig.cs.ucl.ac.uk. A browser
accesses a page with URL http://haig.cs.ucl.ac.uk/foo.html. Both the browser host, the
web server host and the DNS host are connected to the same Ethernet LAN. Give an
account of the steps that take place as the browser requests the page. [Assume no useful
cached information is available at the start of the exchange. You do not need to account
for every packet sent!]
[8 marks]
The browser software extracts the domain name from the URL and
constructs a look-up request to send to the DNS server. The IP address of
the DNS server must be known.
IP routing will determine that the packet can be sent direct. The IP address
will need to be resolved using ARP. Assuming the look-up is successful, the
browser now has the web server’s IP address. This is used to set up a TCP
connection to the web server. If this is successful, the HTTP request for the
page is sent over the TCP connection. (Again, ARP will be used to resolve
the server’s IP address)
b) Explain what is meant by a user-level session.
[2 marks]
User-level session – an association between (typically) two applications.
There is a synchronisation phase at the start, a message exchange phase,
then a termination phase.
COMP0D15
7
Turn Over
i) Explain one method whereby a user-level session spanning several HyperText
Transfer Protocol (HTTP) requests may be implemented.
[3 marks]
For example, the cookie mechanism …
Cookie – a piece of named information understood at the server
Server might include Set-Cookie: SessionId="1234"; in an
HTTP response. In this case the cookie identifies a session at the server. The
client must include the cookie in all subsequent requests to the server.
Typically the client stores the cookie on a local disc. Cookies generally have
timeouts associated with them.
ii) How is the concept of a user-level session realised in the Internet File Transfer
Protocol (FTP)? Outline the principle events that occur at the transport layer and
above as a file is transferred from the server.
[4 marks]
Client establishes a control channel with the server. This is achieved by
opening a TCP connection to the server and logging in. The user-level
session is 1-1 with the transport connection.
Having established a session, client may request a file transfer via the
control connection. The server validates the request and opens a new TCP
connection back to the client which is used to transfer the file. Each file
transfer requires the set-up of a new TCP connection. TCP guarantees the
reliable delivery of the file.
c) Discuss the role played by cacheing and proxies in the world-wide web. What impact, if
any, do proxies have on user-level session maintenance?
[9 marks]
Caches of frequently accessed pages speed up operation across slow
sections of the Internet. They also help the network by minimising the
amount of data transferred.
Not all data can sensibly be cached. There is no point cacheing dynamically
generated content. Some files may be updated frequently and so should not
be cached. A server may indicate in the HTTP header whether or not a page
should be cached.
Most browsers keep a cache of pages on the local disc. The local cache will
be searched before a request is made to the server. Sometimes the currency
of the cached version is checked with the server.
Some sites deploy proxies which maintain site-wide caches. Browsers send
requests initially to the proxy which returns the cached version if present.
The strategy is effective where many users at a site access the same pages.
COMP0D15
8
Continued
Cookies are an end-to-end mechanism. Proxies should not cache cookies
nor should they reply to cookie requests on behalf of users.
d) Tag bytes in the ISO Basic Encoding Rules (BER) are constructed as follows:
Type
8 7 6 5 4 3 2 1
Class
Bit
Tag Value
Where:
Class: 00 = UNIVERSAL, 01 = APPLICATION
Type: 0 = Primitive, 1 = Constructed
Tag Value: 00010 = INTEGER
10000 = SEQUENCE
00100 = OCTET STRING
Given the following ASN.1 syntax definition:
IpAddress ::= [APPLICATION 0] IMPLICIT OCTET STRING (SIZE (4))
SocketAddress ::= SEQUENCE {
ipAddr IpAddress,
port INTEGER (0..65535)
}
Answers
Illustrate how an instance of SocketAddress would be encoded for transmission.
[7 marks]
There should be a correct interpretation of IMPLICIT and the INTEGER
SIZE. Possible answer (in hex)
30 0A
SEQUENCE
40 04 80 10 08 44
128.16.8.68
02 02 12 34
IMPLICIT IpAddress
INTEGER 0x1234
This would be transmitted as 30 0A 40 04 80 10 08 …
4. a) The diagram shows two hosts A and B communicating via a router R.
A
320 Kbps
R
64 Kbps
B
A is transmitting data packets to B of length 200 bytes (including 40 byte headers). B is
replying with acknowledgement packets of length 40 bytes. The time-sequence diagram
below shows one exchange.
[Assume throughout part a) that no errors occur and that all parties may overlap
transmissions and receptions. Note also that processing and propagation times are
assumed to be negligible].
COMP0D15
9
Turn Over
A
R
B
0
5
10
15
20
25
30
35
40
Time (ms)
i) Assuming that A transmits continuously (i.e. no flow control), describe the evolution
of the system over the next few packets and calculate their round-trip times as
observed by A.
A queue will build up at the router.
RTT for the first packet is 36 ms
Transmission of the second data packet on the A-B link will begin at 5ms
Transmission of the second data packet on the R-B link will begin at 30ms.
Its ack will reach A 25 + 5 + 1 = 31 ms later, i.e. at t = 61 ms. Thus RTT is
61 –5 = 56ms
Clearly behaviour is governed by the data transmission time R-B (25 ms).
Thus acks will reach A at 25 ms intervals.
Since packet transmissions at A begin at 5ms intervals, RTTs will increase
by 20ms per packet – i.e. 36, 56, 76, 96 etc.
[6 marks]
ii) Assume now that A’s transmissions are constrained by an idle-RQ protocol. What
effect does this have on the round-trip times? What effect does it have on the data
throughput of the system?
[4 marks]
RTTs are now constant since no queue builds at the router.
Throughput was constrained by capacity of R-B link that was continually
busy. This implies 64 kbps or 64 * 160/200 = 51.2 Kbps of data
With stop and wait, one packet is transmitted every 36ms, so throughput is
160*8/36 = 35.5 Kbps
iii) Assume now that A’s transmissions are constrained by a continuous request protocol
with window size w packets where w is big enough to allow continuous transmission
on the R->B link. Assuming that protocol operation has reached a steady state, at what
rate will acknowledgement packets arrive at A and what will be the state of A’s
transmit window each time an acknowledgement packet arrives?
[4 marks]
Continuous transmission on the R-B link implies one data packet arrives
and one ack is generated each 25 ms. Therefore acks arrive at A at 25 ms
intervals.
COMP0D15
10
Continued
Clearly A can transmit data packets more frequently than this and will do so
until its window runs out. Therefore, each ack changes the transmit window
from 0 to 1 and allows the transmission of one data PDU.
b) i) Briefly describe the flow control mechanism used in the Internet Transmission
Control Protocol (TCP).
[5 marks]
Transmitter has a window measured in bytes.
Transmitter may have bytes outstanding up to the limit imposed by the
window. When this limit is reached transmitter must pause.
Receiver sends acknowledgements which state which bytes have been
received (acks are cumulative). Acks also include a new window size.
If receiver allocates a large enough window then transmitter never has to
pause.
If receiver allocates a small (esp. zero) window, then transmitter will have
to pause or halt.
Answers
Receiver may now send a larger window to allow transmit rate to increase.
ii) A TCP transmitter is in slow-start, using a segment size of 1460 bytes and has just
successfully sent 16 segments. It will now carry on transmitting. Assuming an RTT of
100ms and that its next windows worth of transmission are all successfully
transmitted and acknowledged, evaluate the transmission rate in bits/second.
[3 marks]
Next window will be 32 segments (slow start), which will be delivered in a
single round-trip time, that is 32  1460bytes  8bits/byte / 0.1s = 3 737 600
b/s  3.7Mb/s.
[Question 4 continued on next page]
COMP0D15
11
Turn Over
[Question 4 continued]
c) i) Describe the operation of a token bucket scheme that may be used to shape traffic
from a source.
[5 marks]
A token bucket specification has two parameters; the token replenishment
rate r, and the bucket size b. Logical tokens drip into the bucket at a
constant rate r bytes/s. A packet of length s bytes logically consumes s bytes
of tokens when it leaves. If there are not s bytes in the bucket, then the
source is "not conforming" and transmissions must wait. If there is a break
in packet generation then the bucket fills - but only until it contains b bytes
of tokens.
ii) Host A is transmitting data to host B across a link with a capacity of p bps. Host A
must conform to a token bucket of size b bits and a replenishment rate r bps. Derive a
formula for the maximum time at which A may transmit at the full link speed.
[4 marks]
In the best case A has a full bucket of tokens.
Suppose the maximum burst time is t sec. In this time pt bits are transmitted
and rt token bits arrive – so a total of b+rt token bits are consumed.
We must have pt <= b+rt, so t <= b/(p-r)
iii) Suppose that in c) ii) p = 10Mbps and that the transmission process on host A
transmits in a regular pattern of 50 ms transmission at the peak rate followed by 50 ms
of silence. What are the smallest values of r and b that will accommodate this pattern?
[2 marks]
Average data rate is 107 x 0.05 = 5 x 105 bits in 100 ms => r = 5 x 106 bps
During a 50 ms transmission we consume 5 x 105 bits and r x 0.05 = 2.5 x
105 token bits arrive. Therefore the bucket size must be at least 2.5 x 105
bits.
[Total 33 marks]
5. The diagram shows computers in a private home connected to the public Internet via ADSL.
X is a “home gateway” machine implementing several functions including; ADSL modem, IP
router and LAN bridge. Y is a router at the ISP.
COMP0D15
12
Continued
F
Internet
Y
ADSL
E
X
A
B
D
C
Addresses are allocated as follows:
A
B
C
D
E
F
PC - Ethernet
PC - Ethernet
Laptop – Wireless LAN
Home gateway Ethernet
Home gateway - ADSL
ISP router - ADSL
192.168.0.1/24
192.168.0.2/24
192.168.0.3/24
192.168.0.50/24
217.32.65.90/24
217.32.65.200/24
Answers
192.168.0.0/24 is a private IP network, 217.32.65.0/24 is public.
a) Give a detailed account of the routing and address resolution steps that occur when an IP
datagram is sent from A to D for the first time. Assume all relevant caches are initially
empty and state what information is cached during the operations.
[10 marks]
A and D are on the same IP network so A will assume it can contact D
directly. This means that X must act as a bridge between the Ethernet and
wireless LAN.
(1) A does not know D’s MAC address so it broadcasts an IP ARP request
on the Ethernet giving 192.168.0.3 as target
(2) The bridge function at X receives the broadcast frame and forwards it
on the wireless LAN. It notes that A is accessible via its Ethernet
interface
(3) D receives the ARP request, recognises itself as target and constructs
an ARP reply. It also caches A’s AR mapping
(4) D transmits the ARP reply to A’s MAC address
(5) The bridge function at X recognises A’s MAC address and forwards
the frame on the Ethernet. It notes that D is accessible via the wireless
interface.
(6) A receives the reply and adds D to its AR cache.
(7) A constructs a frame containing the IP datagram and having D’s
MAC address. It transmits this on the Ethernet
(8) Bridge function at X recognises D’s MAC address and forwards the
frame on the wireless LAN.
COMP0D15
13
Turn Over
(9) D receives the datagram
[Note that B learns A’s AR mapping at step (1)]
b) Assume now that all relevant address resolution mappings are known. Give a detailed
account of the steps that occur as an IP datagram is sent from A to a remote Internet host
and a reply is received.
[6 marks]
(1) A recognises the destination IP address as on a different subnet and
knows it must send the datagram to X which will perform the router
role.
(2) A transmits to X
(3) X recognises that the destination is not local and so will forward the
datagram via its ADSL interface. However, because the local network
is private, X must perform NAT.
(4) X replaces the source IP address with 217.32.65.90 and the port
with a globally unique value. It stores the mapping between the old
and new address/port pairing
(5) X transmits the IP datagram to Y, which forwards it across the
Internet.
(6) The reply will have destination address/port as allocated in step (4). X
recognises these and replaces them with the saved values.
(7) X recognises the new destination address as local and transmits the
datagram to A across the Ethernet.
[Question 5 continued on next page]
COMP0D15
14
Continued
[Question 5 continued]
c) The host with interface A belongs to a Virtual Private Network (VPN) managed by a
server with address 128.16.10.34. Accordingly, it has a second (virtual) interface which
has the address 128.16.114.4.
i) State two benefits that may result from employing such a VPN scheme.
[2 marks]
For example:
Can ensure all traffic on public Internet is encrypted
Can have strong authentication of remote hosts so that local privileges may
be extended to them
Can ensure only VPN traffic is allowed – no general Internet traffic
Cheaper to build a VPN this way rather than leasing private lines
ii) Some lines from the VPN host’s routing table are presented below:
Network Destination
128.16.0.0
128.16.114.4
192.168.0.0
192.168.0.1
128.16.10.34
192.168.0.50
Default
192.168.0.50
Netmask
255.255.0.0
Interface
128.16.114.4
Gateway
Answers
255.255.255.0
192.168.0.1
255.255.255.255
192.168.0.1
192.168.0.1
Give an account of the processing that will occur on this host and on the VPN server
when a datagram is sent to a destination with address 128.16.8.78.
[6 marks]
The original datagram has destination address 128.16.8.78
The destination address matches line 1 so the datagram is routed via the
virtual interface 128.16.114.4. The source address is therefore
128.16.114.4.
Packets routed via the virtual interface are processed by software that
encapsulates the datagram in a new one with destination address
128.16.10.34. Encryption may occur at this point. The new datagram is
passed back to the routing module.
The new datagram matches line 3 and so is routed via the Ethernet interface
and the home gateway. Its source address will be 192.168.0.1 (this will be
modified by NAT).
At the server the encapsulation will be stripped and the original datagram
will be recovered. This datagram can now be forwarded to its final
destination.
d) Suppose now that the home-owner wants to set-up a home-office for his/her home
business. S/he acquires the public network 214.17.52.192/29. Show, using a clearly
COMP0D15
15
Turn Over
labelled diagram with supporting explanations, how the home owner may now use these
addresses to set up a network at home. The home network:
i) will have a publicly accessible WWW server
ii) will have a publicly accessible FTP server
iii) should allow at least 12 other machines to have access to the Internet, protected by a
firewall
You diagram should indicate clearly, with explanations, any specific, routing and
addressing features, and any special application-level servers that are used.
[9 marks]
WWW server
Internet
firewall +
router
FTP server
NAT+ firewall +
router
other network
segments
(main site)
site/office network
A diagram similar to that above is expected, with clear indication of where
public and private addressing is used.
Explanations should be given on the use of:
a) public IP addressing
b) private addressing (on “main site”) – private addressing will have to be
used as there are too few public addresses to use for all the machines
c) use of the NAT
COMP0D15
16
Continued
d) use of firewall(s) (a single firewall at the Internet boundary is acceptable)
e) how the NAT and any application level gateways (ALGs) are used
[Total 33 marks]
Answers
COMP0D15
17
Turn Over