Mini Assignment Project Study Notes

MINI ASSIGNMENT PROJECT (50% Mark)

1. Components of a Data Communication System (5 marks)

  • Data Source: The origin of the data that is to be transmitted.
  • Transmitter: Converts the data into a signal suitable for transmission.
  • Transmission Medium: The physical path through which the signal travels (e.g. copper wires, fiber optics).
  • Receiver: Receives the signal and converts it back into the original data format.
  • Data Sink: The destination where the data is to be received and used.
  • Diagram: A block diagram showing the interaction between these components, highlighting the flow from Data Source → Transmitter → Transmission Medium → Receiver → Data Sink.

2. Static IP Addressing Issues (10 marks)

a. Problems Encountered with Static IP Addressing
  • IP Address Exhaustion: In larger organizations, the allocation of static IPs can lead to a shortage of available addresses, making it difficult to add new devices without reconfiguration.
  • Difficulty in Management: Static IPs require manual configuration for each device, which can lead to errors in assignments, difficulties in tracking IP addresses, and increased administrative overhead.
b. Bit Rate Calculation for a Noisy Channel (5 marks)
  • Given: SNR (Signal-to-Noise Ratio) = 400, Bandwidth = 5000 Hz.
  • Formula: The bit rate (C) can be calculated using the Shannon-Hartley theorem:
    C=Bimesextlog2(1+extSNR)C = B imes ext{log}_2(1 + ext{SNR})
      Where:
      - C is the channel capacity in bits per second (bps).
      - B is the bandwidth in Hz.
      - SNR is the signal-to-noise ratio.
  • Calculation:
      1. Calculate ext{log}_2(1 + SNR):
    extlog2(401)ext(Usingcalculatororlogarithmtable)extworksouttoapproximately8.64.ext{log}_2(401) ext{ (Using calculator or logarithm table)} ext{ works out to approximately } 8.64.
      2. Bit Rate:
    C=5000imes8.64extbpsext(Approximately43200bps).C = 5000 imes 8.64 ext{ bps} ext{ (Approximately 43200 bps)}.

3. Setup for New Wide/Local Area Network (27 marks)

  • Given IP Address: 192.168.10.20/27
      - Class: C
  • Components:
      - Block Diagram of IP Address (2 marks): Show the network address, mask, and hosts in a simplified network.
      - Usable Hosts Calculation:
        - Formula: 2n−22^n - 2 (where n = number of bits reserved for host addresses).
        - For /27, there are 5 bits for hosts:
          - 25−2=32−2=30extusablehosts.2^5 - 2 = 32 - 2 = 30 ext{ usable hosts}.
      - Number of Sub-networks Available:
        - Calculation:
          - For /27, we have reserved 3 bits for subnetting in a class C range. Total networks = 23=8extsub−networks.2^3 = 8 ext{ sub-networks}.
      - Subnet Mask for Last Octet:
        - The subnet mask corresponding to /27 is:
          - 255.255.255.224255.255.255.224
      - Block Size Calculation:
        - The block size is 2(32−n)2^{(32-n)} where n is the number of bits in the subnet.
        - Block size = 2(32−27)=25=32.2^{(32-27)} = 2^5 = 32.
      - First and Last Valid IP Addresses and Broadcast IDs of the Third Subnet:
        - Subnet Addresses:
          - Third Subnet starts at: 192.168.10.64 (Calculating from the first subnet: 192.168.10.0, 192.168.10.32, then 192.168.10.64)
          - First Valid IP: 192.168.10.65
          - Last Valid IP: 192.168.10.94
          - Broadcast Address: 192.168.10.95

4. Switching and Routing Concepts in Data and Networking Communication Systems (10 marks)

  • Switching: The process of directing data packets between devices on the same network.
      - Types of Switching:
        - Circuit Switching: Establishes a dedicated communication path between two nodes, e.g., traditional telephony.
        - Packet Switching: Data is divided into packets and each packet can take a different route to the destination, e.g., Ethernet and IP networks.
  • Routing: Refers to the process of selecting paths in a network along which to send network traffic.
      - Routing Methods:
        - Static Routing: Manual intervention is required; routes do not change unless manually changed.
        - Dynamic Routing: Routes are automatically adjusted based on network topology changes using protocols like RIP (Routing Information Protocol) or OSPF (Open Shortest Path First).
  • Corresponding Protocols:
      - RIP: A distance-vector routing protocol that employs the hop count as a routing metric.
      - OSPF: A link-state routing protocol that uses the shortest path first (SPF) algorithm.

5. Illustration of Data Packet Movement Between Networks (10 marks)

  • Diagram: Create a flowchart or network diagram showing nodes A and B in separate networks connected via a router. Indicate the path of a data packet from one network to another, showcasing interfaces and routing decisions.
  • Annotated Steps:
       1. Data packet is generated at Node A.
       2. The packet is sent to the local router where the switching process occurs.
       3. Router uses routing protocols to determine the next hop based on destination IP address.
       4. The packet traverses through potential intermediate routers until it reaches Node B.
       5. Node B receives the packet and processes it accordingly.

Submission Details

  • Date: 9th May 2026
  • Format: Hard copy on campus and Softcopy on LMS.