Conversation with Merlin [email protected] · Sat Nov 04 2023

Question 1 [30 marks] a) A university has 80 LANs with 150 hosts in each LAN. Suppose the university has one class B IP-address: i. Using subnetting, give the university a suitable class B IP-address and a subnet mask which can accommodate the 80 LANs [4] ii. List the network addresses and the broadcast addresses of the first 4 subnets (Show your working) [5] b) Regarding network topologies i. Explain the difference between a ring topology, a star topology and a tree topology. Additionally, give one disadvantage for each of the three topologies [6] ii. Explain two advantages and two disadvantages of a bus topology [4] c) Regarding the OSI Model: i. Explain three functions of Layer 2 [3] ii. Describe check-pointing and adjournment regarding Layer 5 [4] d) Explain any four factors that affect cable attenuation [4] SECTION B (ANSWER ANY TWO QUESTIONS) 25 MARKS EACH Question 2 [25 marks] a) Ethernet technology uses a CDMA-CD or CDMA-CA algorithm to manage collisions. Explain the difference between the two modes of operation. [4] b) Regarding MAN technologies: i. Explain one advantage and one disadvantage of SONET ii. Explain one advantage and one disadvantage of ATM [2] iii. Explain two advantages that Gigabit Ethernet has over SONET and ATM [2] [2]

a)

i. To accommodate the 80 LANs with 150 hosts in each LAN, a suitable class B IP address and a subnet mask can be used. In a class B IP address, the first two octets are reserved for the network portion, while the last two octets are used for host addresses. Since each LAN has 150 hosts, which can be represented by 8 bits (2^8 = 256, minus the network and broadcast), we need to allocate enough bits for the host portion.

By using subnetting, we can divide the class B network into multiple subnets, each with its own network and broadcast addresses. To 80 LANs, we need to allocate enough bits for the subnet portion.

Let's calculate the number of bits needed for the subnet portion. To accommodate 80 LANs, we need at least 7 bits (27 = 128, which is greater than 80)., we need to account for host addresses as well. Since each LAN has 150 hosts (which requires 8 bits), we can use the remaining 1 bit for the subnet portion.

Therefore, a suitable class B IP address for the university could be 172.16.0.0, and the subnet mask would be 255.255.128.0 (/17 subnet mask).

ii. To calculate the network and broadcast addresses of the first 4 subnets, we can use the following formula:

Network address = IP address AND Subnet mask Broadcast address = Network address (with all host portion bits to 1)

Using the class B IP address 172.16.0.0 and the subnet mask 255.255.128.0 (/17 subnet mask), we can calculate the network and broadcast addresses for the first 4 subnets:

Subnet 1: Network address: 172.16.0.0 Broadcast address: 172.16.127.255

Subnet 2: Network address: 172.16.128.0 Broadcast address: 172.16.255.255

Subnet 3: Network address: 172.17.0.0 Broadcast address: 172.17.127.255

Subnet 4: Network address: 172.17.128.0 Broadcast address: 172.17.255.255

b)

i. Ring Topology: In a ring topology, the devices are connected in a circular loop, where each device is connected to exactly two other devices. Data is transmitted in one direction around the ring. One disadvantage of a ring topology is that if one device or connection fails, the entire network can be disrupted.

Star Topology: In a star topology, all devices are connected to a central device, such as a switch or hub. Each device has a dedicated connection to the central device. One disadvantage of a star topology is that if the central device fails, the entire network can be affected.

Tree Topology: A tree topology, also known as a hierarchical topology, is a combination of multiple star topologies. Devices are arranged in a hierarchical structure, with a central root device and multiple levels of cascading branches. One disadvantage of a tree topology is that if the root device fails, it can disrupt the entire network.

ii. Bus Topology:

Advantages:

  1. Simple and inexpensive to implement: A bus topology requires less cabling compared to other topologies, making it cost-effective to implement.
  2. Easy to add or remove devices: Adding or devices in a bus topology is relatively easy since each device is connected to the main bus.

Disadvantages:

  1. Limited scalability: A bus topology has physical limitations in terms of the number of devices that can be connected to the bus. As more devices are added, the overall performance of the network can degrade.
  2. Single point of failure: If the main bus cable fails, the entire network can be disrupted. It is essential to have backup measures in place to prevent downtime.

c)

i. Layer 2, also known as the Data Link Layer, performs the following functions:

  1. Framing: The Data Link Layer encapsulates data from the Network Layer into frames by adding header and trailer information. Frames are then transmitted across the physical network.
  2. Media Access ControlMAC): The Data Link Layer controls access to the network medium and ensures that only one device transmits at a time to prevent collisions.
  3. Error Detection and Correction: The Data Link Layer detects and corrects transmission errors by using error detection codes, such as cyclic redundancy check (CRC).

ii. Check-pointing and adjournment are not specific functions of Layer 5. Check-pointing is a technique used in fault-tolerant systems, where the system periodically saves its state to a stable storage location to enable recovery from failures. Adjournment refers to the process of ending a session or connection gracefully.

d)

Four factors that affect cable attenuation are:

  1. Cable length: As