Download test 1 - answer

EKT335 PRINCIPLES OF COMPUTER NETWORK
TEST I - ANSWER
DATE
: 17 OCT 2012
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1. a)
Draw and compare the 7-layer defined in the ISO OSI Reference Model and 5-layer
TCP/IP layer architecture.
[3 marks]
Application
Presentation
Session
Transport
Network
Data Link
Physical
7-Layer ISO OSI Reference Model
Application
Transport
Network
Link
Physical
5-Layer TCP/IP Model
The key difference is that the 5-layer Internet model does not have the presentation and
session layers found in the OSI model. Otherwise the models are essentially the same .
b)
Which TCP/IP layer is responsible for the following?
i)
Determining the best path to route packet.
Network Layer
ii)
Providing end-to-end communications with reliable service.
Transport Layer
iii)
Providing node-to-node communications with reliable service.
Data Link Layer
[3 marks]
c) Briefly explain the process of encapsulation happening in layered architecture. [4 marks]
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The Application layer sends message to the Transport layer.
The Transport layer converts the Application layer message to segments by
appending transport-layer header information to the Application layer message and
sends it down to the Network layer.
The Network layer converts the segments to packets by adding the Network layer
header information to the segments and sends them to the Data Link layer.
The Data Link layer converts the packets to frames by adding the Data Link layer
header and sends them to the Physical layer.
The Physical layer converts the frames to 1's and 0's (electrical signals) and sends
them across the network.
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2. The following question is based on Figure 1 below:
Tracing route to webserver.unimap.edu.my [58.27.57.229]
over a maximum of 30 hops:
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2
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3
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7
ms
ms
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ms
ms
ms
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ms
ms
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1
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ms
10.208.129.27
vl162-61 [202.179.118.122]
218.100.45.127
10.55.64.39
210.187.75.23
mcl-getaway.gw.unimap.edu.my [58.27.57.1]
www.unimap.edu.my [58.27.57.229]
Trace complete.
Figure 1
‘traceroute’ command is used to determine the path from your home PC to your university’s
main web page as shown in Figure 1.
a) Using the output from traceroute in Figure 1, identify how many different routers are
traversed.
[2 marks]
6 different networks and service providers. The university getaway router is reached at hop
6.
b) Briefly explain how this command can be used as a debugging tool by network managers.
[3 marks]
The traceroute command discovers the routes packets follow when traveling to their
destinations. The traceroute command works by using the error message generated by switch
routers when a datagram exceeds its time-to-live (TTL) value. First, probe datagrams are sent
with a TTL value of one. This causes the first switch router to discard the probe datagrams and
send back "time exceeded" error messages. The traceroute command then sends several
probes, and displays the round-trip time for each. After every third probe, the TTL increases by
one.
Each outgoing packet can result in one of two error messages. A "time exceeded" error message
indicates that an intermediate switch router has seen and discarded the probe. A "port
unreachable" error message indicates that the destination node has received the probe and
discarded it because it could not deliver the packet to an application. If the timer goes off before
a response comes in, the traceroute command displays an asterisk (*).
3. Consider a packet of length L which begins at end system A and travels over 3 links to a
destination end system B. These 3 links are connected by 2 packet switches. Let di , si , and Ri
denote the length, propagation speed, and the transmission rate of link i, for i = 1, 2, 3. The
packet switch also delays each packet by dproc.
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a) Assuming no queuing delays, express the total end-to-end delay for the packet in terms of
di , si , Ri , (i = 1, 2, 3), and L.
[5 marks]
dend-end = L/R1 + L/R2 + L/R3 + d1/s1 + d2/s2 + d3/s3+ dproc+ dproc
b) Suppose now the packet is 1,500 bytes, the propagation speed on all 3 links is 2.5 x 108 m/s,
the transmission rates for all 3 links are 2 Mbps, the packet switch processing delay is 3
msec, the length of the first link is 5000 km, the length of the second link is 4000 km, and
the length of the last link is 1000 km. Calculate the end-to-end delay for this packet to be
transmitted from A to B.
[5 marks]
dend-end = 6 + 6 + 6 + 20+16 + 4 + 3 + 3 = 64 msec
4. Suppose two hosts, A and B, are separated by 20,000 kilometers and are connected by a direct
link of R = 2 Mbps. Suppose the propagation speed over the link is 2.5 x 108 m/s.
a) Calculate the propagation delay, dprop.
[2 marks]
dprop = d/s = 20000 x 103/(2.5 x 108) = 0.08 s
b) Calculate the maximum number of bits that can be in the link (i.e the bandwidth-delay
product, R x dprop. ).
[2 marks]
BDP = R x dprop = (2 x 106) x 0.08 = 160, 000 bits
c) Consider sending a file of 800,000 bits from Host A to Host B. Suppose the file is sent
continuously as one large message. Calculate the total time taken to send the file.
[4 marks]
ttrans = L/R = 800,000 / (2 x 106) = 0.4 s
Time = tprop + ttrans = 0.08 + 0.4 = 0.48 s
5. a) Define the usage of a web cache (i.e., what benefits are derived from using a web cache)?
Where can a web cache be located or placed?
[3 marks]
A web cache may be used to reduce response time as experienced by a user, reduce load on
a link, and/or reduce load on a web server. A web cache may be placed in the user client
(e.g., within the browser application), at “my” network edge (we call this a proxy server),
and/or at the edge of the network containing the server (we call this a transparent cache).
b) Define what is DNS and what is it used for? Explain what would happen to the Internet if all
DNS servers “crashed” (taken offline).
[3 marks]
DNS is Domain Name Service and is used to associate host names (as in say,
www.yahoo.com or [email protected]) with IP addresses. If all DNS servers were to be
crashed one would be unable to use hostnames and could only use IP addresses when using
Internet services.
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c) Consider what happens when a browser (that is, an HTTP client), running on some user’s
host, requests the URL www.somesite.com/index.html. In order for the user’s host to be
able to send an HTTP request message to the Web server www.somesite.com, the user’s
host must first obtain the IP address of www.somesites.com. Explain the steps through
which the IP address for such a hostname is obtained by the client.
[4 marks]
Steps:
a) The browser extract hostname, www.somesites.com, from the URL and passes the
hostname to the client side of the DNS application.
b) The DNS client sends a query containing the hostname to a DNS server.
c) The DNS client receives a reply, which includes the IP address for the hostname.
d) Once the browser receives the IP address from DNS, it can initiate a TCP connection
to the HTTP server process located at port 80 at the IP address.
d) For the client-server application over TCP, why must the server program be executed before
the client program? For the client-server application over UDP, why may the client program
be executed before the server program?
[2 marks]
TCP:
Client needs to establish a TCP connection first with the server. If you run client program
first, then the client will attempt to make a TCP connection with a non-existent server
process. A TCP connection will not be made.
UDP:
No connection establishment is needed prior to communicating. Client may send request
even when the server is not ready. UDP client doesn't establish a TCP connection with the
server. Thus, everything should work fine if you run UDP client program first.
-oooOooo-
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