**Introduction to Computer Networking - Exam
THE UNIVERSITY OF DODOMA
College of Informatics and Virtual Education
Department of Computer Science and Engineering
End of Semester II University Examinations
For the 2019/2020 Academic Year
- Course Name: Introduction to Computer Networks
- Paper Code Number: CN 121
- Date of Examination: 21st August, 2020
- Time: 08:00AM - 11:00AM
- Duration: 3 Hours
- Venue: Auditorium and LRB 105
- Sitting programme(s): Bsc. BIS1, CE1, CNSE1, CS1, CSDFE1, HIS1, ICT_IDIT1, IS1, MTA1, SE1 and TE1
SECTION A (40 Marks)
Answer ALL questions from this section.
QUESTION ONE
Choose the most correct answer.
(1 Mark Each)
i. Which of the following is not a guided media?
- A. Fiber optic cable
- B. Atmosphere
- C. Coaxial cable
- D. Twisted pair cable
Answer (Click to show)
B. Atmosphere
ii. What is the major factor that makes coaxial cable less susceptible to noise than twisted-pair cable?
- A. Thickness of inner-conductor
- B. Cable diameter
- C. Outer conductor
- D. Insulating material
Answer (Click to show)
C. Outer conductor
iii. What is the dotted-decimal notation of 10000001 00001011 00001011 11101111?
- A. 193.131.27.255
- B. 192.168.10.9
- C. 129.11.11.239
- D. 172.16.11.3
Answer (Click to show)
C. 129.11.11.239
iv. Which multiple access technique is used by IEEE 802.11 standard for wireless LAN?
- A. CDMA
- B. CSMA/CD
- C. CSMA/CA
- D. ALOHA
Answer (Click to show)
C. CSMA/CA
v. What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?
- A. 14
- B. 32
- C. 62
- D. 30
Answer (Click to show)
D. 30
vi. A sender must not be able to deny sending a message that he or she, in fact, did send, is known as:
- A. Message Integrity
- B. Message Confidentiality
- C. Message Authenticity
- D. Message Nonrepudiation
Answer (Click to show)
D. Message Nonrepudiation
vii. The address to ping in order to establish if there is an IP connection to the same machine or computer being used by the end-user
- A. 127.0.0.0
- B. 1.0.0.127
- C. 127.0.0.1
- D. 0.0.0.0
Answer (Click to show)
C. 127.0.0.1
viii. Gigabit Ethernet has a data rate of ______ Mbps.
- A. 10
- B. 100
- C. 1000
- D. 10000
Answer (Click to show)
C. 1000
ix. The PDU in Data-Link layer is:
- A. Message
- B. Frame
- C. Packets
- D. None of the above
Answer (Click to show)
B. Frame
x. Bluetooth transceiver devices operate in ______ band.
- A. 2.5GHz
- B. 2.6GHz
- C. 5GHz
- D. 2.4GHz
Answer (Click to show)
D. 2.4GHz
QUESTION TWO
Write T for a TRUE statement and F for a FALSE statement.
(1.5 Marks Each)
i. In CSMA/CD the chance of collision can be reduced if a station senses the medium before trying to use it.
Answer (Click to show)
T
ii. The size of IP address in IPv6 is 128 bits.
Answer (Click to show)
T
iii. Message confidentiality means the data must arrive at the receiver exactly as sent.
Answer (Click to show)
F
iv. An important function of the physical layer of the OSI model is to encode frames into electrical, optical, or radio wave signals.
Answer (Click to show)
T
v. Straight through connection in UTP is used to connect dissimilar types of Ethernet equipment.
Answer (Click to show)
T
vi. Router is a layer 2 networking device.
Answer (Click to show)
F
vii. Once the network security policy is in place within the organization, the people it affects do not necessarily need to know it well.
Answer (Click to show)
F
viii. The frequency range (Bandwidth) of the IEEE 802.11a standard is 5GHz.
Answer (Click to show)
T
ix. /28 in 152.16.168.196/28 in logical addressing means that there are 28 bits in the host portion of the IP address.
Answer (Click to show)
F
x. On Ethernet LAN when collision occurs, all stations run a random back off algorithm. When the back off delay period has expired, all stations have equal priority to transmit data.
Answer (Click to show)
T
QUESTION THREE
Write down the services associated with each of the following port number.
(2 Marks Each)
i. 21
Answer (Click to show)
FTP (File Transfer Protocol)
ii. 443
Answer (Click to show)
HTTPS (HTTP Secure)
iii. 25
Answer (Click to show)
SMTP (Simple Mail Transfer Protocol)
iv. 23
Answer (Click to show)
Telnet
v. 22
Answer (Click to show)
SSH (Secure Shell)
SECTION B (60 Marks)
Answer TWO (2) questions from this section.
QUESTION FOUR
(30 Marks)
a) Explain how the collision is avoided in the below listed LAN technologies:
(3 Marks Each)
i. Ethernet
Answer (Click to show)
Uses CSMA/CD (Carrier Sense Multiple Access with Collision Detection). Stations listen to the medium before transmitting. If the medium is busy, they wait. If two stations transmit simultaneously and detect a collision, they stop transmission, wait for a random backoff time, and then retry.
ii. Token Ring
Answer (Click to show)
Uses a token-passing mechanism. A special frame called a ‘token’ circulates around the ring. Only the station possessing the token is permitted to transmit data, ensuring that only one station can transmit at any given time, thus completely avoiding collisions.
b) List down the four security objectives.
(4 Marks)
Answer (Click to show)
1. Confidentiality
2. Integrity
3. Availability
4. Authenticity
c) With the help of a diagram, explain the four types of security attacks.
(6 Marks)
Answer (Click to show)
1. Interruption (Attack on Availability): The source (S) is unable to communicate with the destination (D). The connection is broken. (Diagram: Line between S and D is cut).
2. Interception (Attack on Confidentiality): An unauthorized party (X) gains access to the message. (Diagram: X on the side, tapping the line between S and D).
3. Modification (Attack on Integrity): An unauthorized party (X) not only accesses but also tampers with the message. (Diagram: X intercepting and altering the message between S and D).
4. Fabrication (Attack on Authenticity): An unauthorized party (X) inserts a fake message into the network, pretending to be a legitimate source (S). (Diagram: X creating a new, fake message to D).
d) With the aid of diagram, briefly explain the function of each field of Data link layer frame format.
(8 Marks)
Answer (Click to show)
**A generic data link frame has the following fields:
[ Flag | Address | Control | Data | FCS | Flag ]
- Flag (1 byte): Marks the beginning and end of the frame (e.g., 01111110).
- Address (variable): Contains the physical (MAC) address of the source and/or destination.
- Control (variable): Used for flow and error control (e.g., sequence numbers, acknowledgements).
- Data (variable): The payload received from the network layer.
- Frame Check Sequence (FCS) / Trailer (variable): Used for error detection (e.g., CRC code).**
e) Briefly explain the TCP/IP Layers each with its functionalities.
(6 Marks)
Answer (Click to show)
**The TCP/IP model has four layers:
- Application Layer: Provides services to the user, such as HTTP, FTP, SMTP. It defines the format and meaning of data.
- Transport Layer: Responsible for end-to-end communication and error recovery. Key protocols are TCP (reliable) and UDP (unreliable).
- Internet Layer: Responsible for logical addressing and routing packets across different networks. The key protocol is IP (Internet Protocol).
- Network Access/Link Layer: Responsible for the physical transmission of data over the network hardware (e.g., Ethernet, Wi-Fi). It deals with MAC addresses and physical media.**
QUESTION FIVE
(30 Marks)
a) List four advantages of layered approach in computer communication.
(8 Marks)
Answer (Click to show)
1. Modularity and Ease of Maintenance: Each layer can be developed and updated independently.
2. Interoperability: Standards defined for each layer allow products from different vendors to work together.
3. Easier Troubleshooting: Problems can be isolated to a specific layer, simplifying diagnosis and repair.
4. Technology Independence: Layers hide the implementation details of the layer below, allowing underlying technologies to change without affecting upper layers.
b) Find classes of the following IP addresses:
(6 Marks)
i. 00000101 00010101 00001100 00001110
Answer (Click to show)
The first octet in binary is 00000101. The first bit is 0, so this is a Class A address.
ii. 11110011 10011011 11001100 00000001
Answer (Click to show)
The first octet in binary is 11110011. The first four bits are 1111, which falls in the range for Class E (1111).
c) Write the subnet-network address, broadcast address, and the valid host range for the IP address 192.16.10.33/27.
(9 Marks)
Answer (Click to show)
**- Subnet Mask: 255.255.255.224
- Block Size: 256 - 224 = 32
- Subnet Network Address: 192.16.10.32
- Broadcast Address: 192.16.10.63
- Valid Host Range: 192.16.10.33 to 192.16.10.62**
d) What is the subnetwork address if the destination address is 200.45.34.56 and the subnet mask is 255.255.240.0?
(3 Marks)
Answer (Click to show)
**- The mask 255.255.240.0 in the third octet has a block size of 16 (256-240).
- The interesting octet is the third octet (34).
- 34 / 16 = 2 (subnet number), so the subnet starts at 32.
- Subnetwork Address: 200.45.32.0**
e) 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?
(4 Marks)
Answer (Click to show)
**- For 12 subnets, we need at least 4 bits (2⁴ = 16 subnets).
- For 10 hosts per subnet, we need at least 4 host bits (2⁴ - 2 = 14 hosts).
- A Class C default mask is /24. Borrowing 4 bits gives us /28 (255.255.255.240).
- This provides 16 subnets and 14 hosts per subnet, meeting the requirements.
- Subnet Mask: 255.255.255.240**
QUESTION SIX
(30 Marks)
a) With the aid of diagram clearly explain the advantages and disadvantages of at least four types of network physical topology.
(12 Marks)
Answer (Click to show)
**1. Bus Topology (Diagram: Single backbone cable with connected nodes):
- Advantages: Simple, low cost, easy to install.
- Disadvantages: Single point of failure (backbone), difficult to troubleshoot, performance degrades with heavy traffic.
- Star Topology (Diagram: Central hub/switch with nodes connected radially):
- Advantages: Easy to install and manage, failure of one node doesn’t affect others, easy to add new nodes.
- Disadvantages: Central device is a single point of failure, requires more cable than bus.
- Ring Topology (Diagram: Nodes connected in a closed loop):
- Advantages: Ordered network access, no data collisions.
- Disadvantages: A single node failure can break the entire ring, difficult to troubleshoot, adding/removing nodes disrupts the network.
- Mesh Topology (Diagram: Every node connected to every other node):
- Advantages: Highly reliable and redundant, multiple paths for data.
- Disadvantages: Very high cost, complex installation and management, requires many cables.**
b) What Is Ipconfig Command? Why It Is Used?
(4 Marks)
Answer (Click to show)
**
ipconfig
is a command-line utility in Microsoft Windows operating systems.
It is used to:
- Display the current TCP/IP network configuration values (IP address, subnet mask, default gateway).
- Refresh Dynamic Host Configuration Protocol (DHCP) and Domain Name System (DNS) settings.
- It is a primary tool for troubleshooting network connectivity problems.**
c) What is the purpose of cables being shielded and having twisted pairs?
(4 Marks)
Answer (Click to show)
**- Shielding: The purpose of a shield (usually a metallic braid or foil) is to protect the internal conductors from External Electromagnetic Interference (EMI) and to prevent the signals within the cable from radiating EMI to other devices.
- Twisted Pairs: The purpose of twisting the pairs of wires together is to cancel out crosstalk (interference from adjacent pairs) and external EMI. The twisting ensures that both wires in the pair are equally affected by noise, making the receiver able to filter it out effectively.**
d) Subnet the Class C IP Address 195.1.1.0 So that you have 10 subnets each with a maximum 12 hosts on each subnet. For each of the subnetwork list the first and last host address.
(10 Marks)
Answer (Click to show)
**- Requirements: 10 subnets, 12 hosts/subnet.
- For 10 subnets, we need 4 bits (2⁴ = 16). For 12 hosts, we need 4 host bits (2⁴ - 2 = 14).
- Subnet Mask: 255.255.255.240 (/28)
- Block Size: 16
Subnet Network Address First Host Last Host Broadcast Address 1 195.1.1.0 195.1.1.1 195.1.1.14 195.1.1.15 2 195.1.1.16 195.1.1.17 195.1.1.30 195.1.1.31 3 195.1.1.32 195.1.1.33 195.1.1.46 195.1.1.47 4 195.1.1.48 195.1.1.49 195.1.1.62 195.1.1.63 5 195.1.1.64 195.1.1.65 195.1.1.78 195.1.1.79 6 195.1.1.80 195.1.1.81 195.1.1.94 195.1.1.95 7 195.1.1.96 195.1.1.97 195.1.1.110 195.1.1.111 8 195.1.1.112 195.1.1.113 195.1.1.126 195.1.1.127 9 195.1.1.128 195.1.1.129 195.1.1.142 195.1.1.143 10 195.1.1.144 195.1.1.145 195.1.1.158 195.1.1.159
END OF EXAMINATION PAPER