EKT332 COMPUTER NETWORK
TEST I
DATE : 29 MARCH 2012
TIME : 12.00 PM – 01.00 PM
NAME :
MATRIC NO :
Instructions:
1. This is a close book test. No books and notes are allowed to be referred to during the test.
2. There are 5 (five) questions. Answer ALL questions. Please write your answer in the column provided.
3. Time given : 1.0 Hour.
4. Number of pages (including this page) : 4
TOTAL MARKS / 501. a) By comparing the performance of circuit-switched network and packet-switched network, what advantage does a circuit-switched network have over a packet-switched network?
A circuit-switched network can guarantee a certain amount of end-to-end bandwidth for the duration of a call. Most packet-switched networks today (including the Internet) cannot make any end-to-end guarantees for bandwidth.
b) Suppose Jane and John are sending packets to each other over a computer network. Suppose Trudy positions herself in the network so that she can capture all the packets sent by Jane and send whatever she wants to John. She can also capture all the packets sent by John and send whatever she wants to Jane. List some of the malicious things Trudy can do from this position. [2 marks]
Trudy can pretend to be John to Jane (and vice-versa) and partially or completely modify the message(s) being sent from John to Jane. For example, she can easily change the phrase “Jane, I owe you $1000” to “Jane, I owe you $10,000”. Furthermore, Trudy can even drop the packets that are being sent by John to Jane(and vise-versa), even if the packets from John to Jane are encrypted.
2. a) Draw and compare the two dominant standards architectures; the ISO OSI Reference Model and TCP/IP architecture. [6 marks]
7. Application / 4. Application6. Presentation / 3. Transport
5. Session / 2. Internet
4. Transport / 1. Network Access
3. Network
2. Data Link
1. Physical
ISO OSI TCP/IP
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 OSI layer is responsible for the following? [3 marks]
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
c) Differentiate between a physical address, a network address, and a domain name.
[6 marks]
The physical address is the unique hardware address that identifies an interface of a machine on a physical network such as a LAN. Physical addresses are used in the data linklayer.
A network address is a machine's logical address on a network. The network address is used in the network layer. The network address used on the Internet is the IP address.
Domainnamesare used as an aid to identify hosts and networks in the Internet, since names are easier to remember than numbers. The DNS system is used to translate between domain names and IP addresses. The domain name for the network address 128.100.132.30 is toronto.edu.
* Since IP addresses are rather difficult to remember (and are not particularly descriptive), the Internet also allows you to specify a computer by anamerather than a number string. For example, the machine at MIT with the IP address 18.72.0.3 can also be referred to as:bitsy.mit.edu.
This whole string is known as the computer'shost name. In this string, the first part ("bitsy") is the name of the machine itself, while everything else ("mit.edu") is thedomain name.
The domain name is the name of a network associated with an organization. For sites in the United States, domain names typically take the form:org-name.org-type
d) List some of the physical media that Ethernet can run over? [2 marks]
Twisted-pair copper wire, coaxial cable, fibers optic links
3. 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:
1 2 ms 1 ms 1 ms 10.208.129.27
2 3 ms 3 ms 3 ms vl162-61 [202.179.118.122]
3 4 ms 3 ms 3 ms 218.100.45.127
4 8 ms 4 ms 3 ms 10.55.64.39
5 9 ms 7 ms 6 ms 210.187.75.23
6 7 ms 6 ms 5 ms mcl-getaway.gw.unimap.edu.my [58.27.57.1]
7 7 ms 9 ms 9 ms 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) Examine the output from traceroute in Figure 1, try to identify how many different networks and service providers 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.
[2 marks]
By sending ICMP Echo messages to specified host requesting a reply. Sender sends a datagram with a type 8 Echo message and a sequence number to detect a lost, reordered, or duplicated message. The receiver changes the type to Echo Reply and returns the datagram.
4 a) What does the application layer on the source host do as soon as it creates a
message for its on another machine?
They immediately pass the message down to the next-lower layer.
Note: They do not do encapsulation; that is done at the next-lower layer.
b) Briefly explain the encapsulation process.
Encapsulation is the placing of a message in the data field of another message.
c) When the Layer N process passes the message down to the Layer N-1 process, which layer performs encapsulation?
Layer N-1 performs the encapsulation.
d) Suppose an application layer entity wants to send an L-byte message to its peer process, using an existing TCP connection. The TCP segment consists of the message plus 20 bytes of header. The segment is encapsulated into an IP packet that has an additional 20 bytes of header. The IP packet in turn goes inside an Ethernet frame that has 18 bytes of header and trailer. Calculate the percentage of the transmitted bits in the physical layer as corresponds to message information if L is 100 bytes, 500 bytes and 1000 bytes. [6 marks]
The message overhead includes:
· TCP: 20 bytes of header
· IP: 20 bytes of header
· Ethernet: total 18 bytes of header and trailer.
Therefore
L = 100 bytes, 100/158 = 63%
L = 500 bytes, 500/558 = 90%
L = 1000 bytes, 1000/1058 = 95%
4. a) For the client-server application over TCP in socket programming, why must the server program be executed before the client program? [2 marks]
Server needs to establish a connection first with the client.
b) For the client-server application over UDP, why may the client program be executed before the server program? [2 marks]
No connection establishment is needed prior to communicating. Client may send request even when the server is not ready.
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