**Introduction to Computer Networking - Exam
2016/2017 ACADEMIC YEAR (SECOND SEMESTER)
THE UNIVERSITY OF DODOMA
COLLEGE OF INFORMATICS AND VIRTUAL EDUCATION
SCHOOL OF INFORMATICS
UNDERGRADUATE UNIVERSITY EXAMINATIONS
SECOND SEMESTER 2016/2017
TN 122: INTRODUCTION TO NETWORKING
Date: 7 July, 2017
Time Allocated: 3 Hours
Instructions:
- This examination consists of six questions.
- Answer only FOUR questions; Section A is compulsory and any other two (2) questions from section B.
- Do not use this examination paper for rough work. All rough work must be done in the answer book last page
- All University of Dodoma examination regulations apply.
SECTION A (40 Marks)
Question One
A: Multiple Choice questions (1 Mark Each)
i. FDDI uses which type of physical topology?
- A) Bus
- B) Ring
- C) Star
- D) Mesh
Answer (Click to show)
B) Ring
ii. Loss in signal power as light travels down the fiber is called?
- A) Propagation
- B) Attenuation
- C) Scattering
- D) Interruption
Answer (Click to show)
B) Attenuation
iii. Which color coding of cable is used to connect similar devices?
- A) Straight cable
- B) Crossover cable
- C) Serial cable
- D) All mentioned
Answer (Click to show)
B) Crossover cable
iv. In the layer hierarchy as the data packet moves from the lower to the upper layers, headers are
- A) Added
- B) Removed
- C) Modified
- D) Rearranged
Answer (Click to show)
B) Removed
v. In OSI model dialogue control and token management are responsibilities of what layer?
- A) Physical Layer
- B) Presentation Layer
- C) Application Layer
- D) Session Layer
Answer (Click to show)
D) Session Layer
vi. MAC address is of
- A) 24 bits
- B) 36 bits
- C) 42 bits
- D) 48 bits
Answer (Click to show)
D) 48 bits
vii. Bluetooth transceiver devices operate in ______ band.
- A) 2.5GHz
- B) 2.6GHz
- C) 5GHz
- D) 2.4GHz
Answer (Click to show)
D) 2.4GHz
viii. Message ______ means that the data must arrive at the receiver exactly as sent.
- A) Confidentiality
- B) Availability
- C) Authentication
- D) Integrity
Answer (Click to show)
D) Integrity
ix. Switch is a device which operates at what layer?
- A) Physical
- B) Data Link
- C) Network
- D) Transport
Answer (Click to show)
B) Data Link
x. In IPV 4, the IP address (200,345/200,200) belongs to
- A) Class A
- B) Class B
- C) Class C
- D) None mentioned
Answer (Click to show)
D) None mentioned
B: True and False Questions: (Write T for True and F for False) (1 mark Each)
i. Because all systems on the bus network connect to a single backbone, a break in the cable prevents all systems from accessing the network.
Answer (Click to show)
True
ii. Coaxial cable supports TV signals only.
Answer (Click to show)
False
iii. UDP is a connection-oriented protocol that creates and verifies a connection with a remote host before sending information.
Answer (Click to show)
False
iv. An IPv4 packet sent to all workstations on the network is called a Multicast.
Answer (Click to show)
False (It is a Broadcast)
v. In Circuit switching type of network; data is broken up into packets before it’s sent over the network?
Answer (Click to show)
False
vi. Like a virus, a worm needs the user to perform an action such as starting a program or opening an e-mail attachment to start the infection.
Answer (Click to show)
False
vii. Three security goals are Confidentiality, Integrity and Authentication.
Answer (Click to show)
True
viii. The network interface card converts serial data from the computer into parallel data for transmission over the network cable.
Answer (Click to show)
False (It converts parallel data from the computer to serial for transmission)
ix. A LAN becomes a WAN when you expand the network configuration beyond your own premises and must lease data communication lines from a public carrier.
Answer (Click to show)
True
x. The most fundamental differences between SM and MM Optic Fibers are the size of the fiber’s core and the associated attenuation or loss and bandwidth of the fiber
Answer (Click to show)
True
C: Short Answers Questions (1 Mark Each)
i. Which layer among the OSI layers is responsible for data compression and encryption?
Answer (Click to show)
Presentation Layer
ii. What is the frequency range used by 802.11a?
Answer (Click to show)
5 GHz
iii. Given (11111011\ 01101111\ 11111011\ 01001111); Give out its dotted-decimal notation.
Answer (Click to show)
251.111.251.79
iv. Mention the topology in which each computer on the network connects to every other, creating a point-to-point connection between every device on the network.
Answer (Click to show)
Mesh Topology
v. Mention the TCP/IP architecture layer which makes use of ports?
Answer (Click to show)
Transport Layer
vi. Name the authority that oversees global IP address allocation.
Answer (Click to show)
IANA (Internet Assigned Numbers Authority)
vii. How long is an IPv6 address as far as bits are concerned?
Answer (Click to show)
128 bits
viii. Mention three “versions” of Ethernet available for commercial use.
Answer (Click to show)
10Base-T, 100Base-TX, 1000Base-T
ix. Name the OSI Layer that covers all mechanical, electrical, functional and procedural aspects.
Answer (Click to show)
Physical Layer
x. What is the subnetwork address if the destination address is 200.45.34.56 and the subnet mask is 255.255.240.0?
Answer (Click to show)
200.45.32.0
xi. As far as Transmission rate is concerned; which is much faster between 802.11a and 802.11b?
Answer (Click to show)
802.11a
xii. When it comes to Port Address, which port number does HTTP use?
Answer (Click to show)
Port 80
xiii. Given a class C IP address, you need to subnet a network that has 12 subnets, each with at least 10 hosts. Which subnet mask would you use?
Answer (Click to show)
255.255.255.240
xiv. If 10 nodes are communicating using a Hub on a 10-Mbps network, then what will each node get as far as bandwidth is concerned?
Answer (Click to show)
1 Mbps
xv. With Ethernet there is either straight through or crossover connections; what connection between the two is used when connecting two switches using their normal ports?
Answer (Click to show)
Crossover Cable
Question Two
a) What was a major motive towards IPv6? (2 Marks)
Answer (Click to show)
The major motive was the exhaustion of IPv4 addresses. IPv6 provides a much larger address space to accommodate the growing number of internet-connected devices.
b) Convert listed IP address into binary notation; in your conversion include the class where each address belongs. (2 Marks Each)
i. 149.12.164.42
Answer (Click to show)
Class B Binary: 10010101.00001100.10100100.00101010
ii. 45.13.4.209
Answer (Click to show)
Class A Binary: 00101101.00001101.00000100.11010001
iii. 251.126.63.223
Answer (Click to show)
Class E Binary: 11111011.01111110.00111111.11011111
iv. 122.17.167.222
Answer (Click to show)
Class A Binary: 01111010.00010001.10100111.11011110
c) UDOM is granted the site address 207.168.45.0; the university has 6 colleges of which each one has its own network. Design the subnets. In your design you must show the following (15 Marks)
i. The subnet mask ii. The number of addresses in each subnet iii. In each subnet determine the start address, range of usable IP address and the broadcast address.
Answer (Click to show)
Given: Network Address: 207.168.45.0 (Class C) Required Subnets: 6. We need at least 3 subnet bits (2³=8 subnets).
i. Subnet Mask: Default Class C mask: 255.255.255.0 With 3 borrowed bits: 255.255.255.224 (or /27)
ii. Number of addresses in each subnet: Host bits = 5. 2⁵ = 32 addresses per subnet. Usable hosts = 32 - 2 (Network & Broadcast) = 30.
iii. Subnet Details:
Subnet Network Address Usable Range Broadcast Address 0 207.168.45.0 207.168.45.1 - 207.168.45.30 207.168.45.31 1 207.168.45.32 207.168.45.33 - 207.168.45.62 207.168.45.63 2 207.168.45.64 207.168.45.65 - 207.168.45.94 207.168.45.95 3 207.168.45.96 207.168.45.97 - 207.168.45.126 207.168.45.127 4 207.168.45.128 207.168.45.129 - 207.168.45.158 207.168.45.159 5 207.168.45.160 207.168.45.161 - 207.168.45.190 207.168.45.191 6 (Unused) 207.168.45.192 207.168.45.193 - 207.168.45.222 207.168.45.223 7 (Unused) 207.168.45.224 207.168.45.225 - 207.168.45.254 207.168.45.255
SECTION B (40 Marks)
Question Three
a) Define Computer Networking; provide two arguments in favor of the use of a computer network in an organization and one argument against. (4 Marks)
Answer (Click to show)
Computer Networking is the practice of interfacing two or more computing devices with each other for the purpose of sharing data. Arguments in Favor:
- Resource Sharing: Networks allow for the sharing of hardware (like printers) and software, reducing costs.
- Enhanced Communication: Networks enable fast and efficient communication tools like email and instant messaging. Argument Against:
- Security Risks: Networks introduce vulnerabilities, making shared data accessible to unauthorized users if not properly secured (e.g., hackers, viruses).
b) Using their definitions, give the difference between LAN, MAN & WAN (include example of each case). (6 Marks)
Answer (Click to show)
- LAN (Local Area Network): A network connecting devices in a small geographical area like a single building or campus. It is privately owned and offers high-speed data transfer.
- Example: A network in a university computer lab.
- MAN (Metropolitan Area Network): A network that spans a city or a large campus. It connects multiple LANs and is typically operated by a single entity (e.g., a city council or a large company).
- Example: A network connecting all branches of a bank within a city.
- WAN (Wide Area Network): A network that covers a broad geographical area, such as a country or continent. It often uses leased telecommunication lines and is not limited to a single organization.
- Example: The Internet.
c) List only 4 design factors to be looked at when choosing a networking media. (4 Marks)
Answer (Click to show)
- Bandwidth
- Cost
- Distance
- Security
d) Write short notes about the listed topologies (include diagrams in your notes) (2 Marks Each)
i. Star Topology
Answer (Click to show)
Description: All nodes are connected to a central device, like a switch or hub. Advantage: Easy to install and manage; failure of one node doesn’t affect others. Disadvantage: Failure of the central device brings down the entire network. Diagram:
[PC1] [PC2] \ / \ / [Switch] / \ / \ [PC3] [PC4]
ii. Mesh Topology
Answer (Click to show)
Description: Every node is connected directly to every other node. Advantage: Highly fault-tolerant; multiple paths for data. Disadvantage: Expensive and complex to install and manage due to the large number of connections. Diagram:
[PC1] ----- [PC2] | \ / | | \ / | | [PC3] | | / \ | | / \ | [PC4] ----- [PC5]
iii. Ring Topology
Answer (Click to show)
Description: Each node is connected to exactly two other nodes, forming a circular pathway for data. Advantage: orderly network where every device has access to the token and can transmit. Disadvantage: A single node failure can break the loop and disable the entire network. Diagram:
[PC1] ---> [PC2] ^ | | v [PC4] <--- [PC3]
Question Four
a) Why protocols in computer networking? (2 Marks)
Answer (Click to show)
Protocols are sets of rules that govern all aspects of data communication between devices on a network. They are essential for ensuring that communication is successful, reliable, and that devices from different vendors can interoperate seamlessly.
b) Why Layered Model in ISO’s OSI Model; Provide 4 reasons. (4 Marks)
Answer (Click to show)
- Modularity and Ease of Maintenance: Changes in one layer do not affect other layers.
- Interoperability: Standards defined for each layer allow for multi-vendor compatibility.
- Easier to Learn: The complex process of communication is broken down into smaller, more manageable parts.
- Simplified Protocol Design: Functions are separated, making it easier to design and debug protocols for specific tasks.
c) List the seven layers of ISO’s OSI model starting from layer number 1, when listing briefly explain the function of each layer (the order of arrangement should be adhered) (8 Marks)
Answer (Click to show)
- Physical Layer: Responsible for the physical connection between devices. It transmits raw bit streams over a physical medium (cables, radio waves).
- Data Link Layer: Responsible for node-to-node delivery. It frames packets, handles physical addressing (MAC addresses), and error control.
- Network Layer: Responsible for the source-to-destination delivery of a packet across multiple networks. It handles logical addressing (IP addresses) and routing.
- Transport Layer: Responsible for end-to-end delivery of the entire message. It ensures complete data transfer, error recovery, and flow control.
- Session Layer: Responsible for establishing, managing, and terminating sessions (connections) between applications.
- Presentation Layer: Responsible for translation, encryption, and compression of data to ensure it is in a usable format for the application layer.
- Application Layer: Provides an interface for the user to access network services (e.g., web browsers, email clients).
d) Briefly explain how the collision is avoided in the below listed LAN technologies i. Ethernet (3 Marks)
Answer (Click to show)
Ethernet uses CSMA/CD (Carrier Sense Multiple Access with Collision Detection).
- A device listens to the network (Carrier Sense) before transmitting.
- If the network is free, it transmits.
- While transmitting, it listens for a collision (Collision Detection).
- If a collision is detected, it stops transmission, sends a jam signal, and waits a random backoff time before retrying.
ii. Token Ring (3 Marks)
Answer (Click to show)
Token Ring uses a token-passing mechanism.
- A special frame called a “token” circulates around the ring.
- A device can only transmit data when it possesses the token.
- This deterministic access method prevents collisions because only one device can transmit at any given time.
Question Five
a) Define the following i. Security Attack (1 Mark)
Answer (Click to show)
Any action that compromises the security of information.
ii. Security mechanism (1 Mark)
Answer (Click to show)
A mechanism that is designed to detect, prevent, or recover from a security attack.
iii. Security service (1 Mark)
Answer (Click to show)
A service that enhances the security of data processing systems and information transfers.
b) Is worm the same as virus? Provide briefly explanation (1 Mark)
Answer (Click to show)
No. A virus requires a host program and user action to spread, while a worm is a standalone program that can self-replicate and spread independently across a network without user intervention.
c) Provide brief explanation on why do we need Computer Network Security. (4 Marks)
Answer (Click to show)
We need computer network security to:
- Protect Confidentiality: Ensure that data is only accessible to authorized users.
- Maintain Integrity: Ensure that data is accurate and has not been altered by unauthorized parties.
- Ensure Availability: Ensure that network resources and data are accessible to authorized users when needed.
- Prevent Financial Loss and Protect Reputation: To avoid the costs and loss of trust associated with data breaches and service disruptions.
d) With diagrams, list the four types of security attacks. (6 Marks)
Answer (Click to show)
- Interruption (Attack on Availability): Destruction or disruption of system resources. Diagram: Sender ---X-⇒ Receiver (The line of communication is broken).
- Interception (Attack on Confidentiality): Unauthorized access to data. Diagram: Sender —⇒ [Attacker] —⇒ Receiver (Attacker eavesdrops).
- Modification (Attack on Integrity): Unauthorized alteration of data. Diagram: Sender —⇒ [Attacker alters data] —⇒ Receiver (Data is changed in transit).
- Fabrication (Attack on Authenticity): Creation of fake data or objects. Diagram: [Attacker forges data] —⇒ Receiver (Attacker impersonates a legitimate sender).
e) Briefly explain 3 ways one can implement security defense mechanism. (2 Marks Each)
Answer (Click to show)
- Access Control: Using mechanisms like passwords, biometrics, and firewalls to restrict access to network resources to authorized users only.
- Encryption: Converting data into a coded format (ciphertext) to prevent unauthorized access during transmission or storage.
- Intrusion Detection/Prevention Systems (IDS/IPS): Using software or hardware to monitor network traffic for suspicious activity and take action to block it.
Question Six
a) Provide reason on why copper cable used as data transfer media are always twisted? (2 Marks)
Answer (Click to show)
The wires are twisted together to reduce crosstalk and electromagnetic interference (EMI) from external sources. The twisting cancels out the noise that is picked up by the pairs of wires.
b) List 2 advantages and 2 disadvantages of Fiber optic cable. (4 Marks)
Answer (Click to show)
Advantages:
- Very high bandwidth and data rates.
- Immune to electromagnetic interference (EMI) and crosstalk. Disadvantages:
- More expensive to install and terminate than copper cables.
- Fragile and requires more careful handling.
c) Mention two advantages and 2 disadvantages of twisted pair. (4 Marks)
Answer (Click to show)
Advantages:
- Inexpensive and cost-effective.
- Easy to install and work with. Disadvantages:
- Susceptible to crosstalk and electromagnetic interference (EMI).
- Suffers from higher attenuation, limiting segment lengths.
d) Outline the two limitations of star topology. (2 Marks)
Answer (Click to show)
- Single Point of Failure: If the central hub or switch fails, the entire network becomes unusable.
- Cost and Cable Usage: Requires more cable than linear topologies (e.g., Bus) and the cost of the central device can be high.
e) What are Hubs? Briefly explain its types. (3 Marks)
Answer (Click to show)
A hub is a basic networking device that connects multiple Ethernet devices, making them act as a single network segment. Types:
- Passive Hub: Simply connects cables without any signal amplification.
- Active Hub: Regenerates and amplifies the incoming electrical signals before broadcasting them, effectively acting as a repeater.
- Intelligent Hub: Includes management features, allowing an administrator to monitor traffic and configure ports.
f) Mention 5 design factors one needs to look upon when choosing transmission media (5 Marks)
Answer (Click to show)
- Bandwidth
- Transmission Impairment (Attenuation, Distortion)
- Interference
- Number of Receivers
- Cost
END OF EXAMINATION