CN Lab Manual

Practical 01

Aim: Introduction to data Transmission in network.

Exercise:

1. Write a program to convert character into binary and binary digits into character.

# Convert string to binary
string_input = input("Enter a string: ")
binary_output = ' '.join(format(ord(char), '08b') for char in string_input)
print("Binary value:", binary_output)
 
# Convert binary to string
binary_input = input("Enter binary (separate each 8-bit by space): ")
string_output = ''.join(chr(int(b, 2)) for b in binary_input.split())
print("String:", string_output)
 

Conclusion: This experiment shows how characters are converted to binary and back, which is the basic principle of data transmission in networks.

Practical 02

Aim: Study of networking devices

Exercise:

1. Define topology & list names of different topologies.

   • A network topology is the arrangement of devices and connections in a network. It defines how devices communicate and impacts performance and reliability. Common topologies include: Bus, Star, Ring, Mesh, Tree, and Hybrid.

2. List out different kinds of addresses at layers with example.

3. What does RJ45 stand for?

   • RJ45 stands for Registered Jack 45. It is a standard connector used for Ethernet cables to connect computers and networking devices.

4. Define mapping of:

   • ARP: Address Resolution Protocol; maps IP address → MAC address
   • RARP: Reverse Address Resolution Protocol; maps MAC address → IP address

5. Difference between HUB and SWITCH?

Conclusion: This experiment helps understand basic networking devices, their functions, and addressing, which are essential for network communication.

Practical 03

Aim: To study various networking commands

Exercise:

1. Why we are using ping command?

   • The ping command is used to test the connectivity between two network devices. It checks whether a device (like a computer or server) is reachable over the network and measures the round-trip time for messages sent.

2. Explain arp command.

   • The arp (Address Resolution Protocol) command is used to view and manage the ARP table on a device. It shows the mapping between IP addresses and MAC addresses, helping in resolving network layer addresses to link layer addresses.

Conclusion: This experiment helps understand basic networking commands, their purpose, and how they assist in network troubleshooting and device communication.

Practical 04

Aim: Introduction/installation of CISCO packet tracer, understanding of point to point network with net map, create network topology with HUB/Switch and simulate it.

Transfer PDU and view scenario graphically.

1. Which protocol is used to test scenario while simulation?

• The protocol used is ICMP (Internet Control Message Protocol). It is mainly used by the ping command to check connectivity between devices in a network. When one PC sends an ICMP Echo Request, the other PC replies with an ICMP Echo Reply. If the reply is received successfully, it confirms that the network connection is working properly.

Exercise: Create and Simulate a Network Topology Using a Hub

Objective: Create a network topology with more than two devices connected using a hub. Simulate the topology using any network simulation software (e.g., Cisco Packet Tracer). Take a screenshot of your network setup, print it, and paste the printed screenshot as required.

Steps:

1. Open Packet Tracer → Add 1 Hub & 3 PCs.
2. Connect each PC to Hub using Copper Straight cable.
3. Assign IP addresses:
   • Right-click PC → Desktop → IP Configuration
   • Set IPs: 10.0.0.2, 10.0.0.3, 10.0.0.4 (Subnet: 255.0.0.0).
4. Test connectivity using ping command in Command Prompt.
5. Transfer PDU → View results in Simulation.

Exercise: Create and Simulate a Network Topology Using a Switch

Objective: Create a network topology with more than two devices connected using a switch. Simulate the topology using any network simulation software (e.g., Cisco Packet Tracer). Take a screenshot of your network setup, print it, and paste the printed screenshot as required.

Steps:

1. Open Packet Tracer → Add 1 Switch & 3 PCs.
2. Connect each PC to Switch using Copper Straight cable.
3. Assign IP addresses:
   • Right-click PC → Desktop → IP Configuration
   • Set IPs: 192.168.1.2, 192.168.1.3, 192.168.1.4 (Subnet: 255.255.255.0).
4. Test connectivity using ping command in Command Prompt.
5. Transfer PDU → View results in Simulation.

Conclusion: In this practical, we created and simulated point-to-point, hub, and switch-based networks using Cisco Packet Tracer. IP addresses were assigned, and connectivity was successfully tested using the ping (ICMP) protocol.

Practical 05

Aim: Introduction to different types of wired transmission media

Exercise:

1. Compare different types of light sources used in optical fiber.

2. List out different types of wired media used in computer lab with their specifications.

   • Twisted Pair Cable (UTP/STP): Consists of pairs of insulated copper wires twisted together to reduce interference. Used in LANs. Categories include Cat5e and Cat6. Maximum distance ~100m, speed up to 1 Gbps (Cat5e) and 10 Gbps (Cat6).

   • Coaxial Cable: Single copper conductor with insulating layer, metal shield, and outer insulation. Used in older networks and cable TV. Moderate bandwidth, distance up to 500m, resistant to interference.

   • Fiber Optic Cable: Uses glass or plastic fibers to transmit data as light pulses. Supports very high speeds (up to 100 Gbps), long distances (several kilometers), and is immune to electromagnetic interference. Used in backbone networks.

   • Shielded Twisted Pair (STP): Twisted pair cable with extra shielding to protect against electromagnetic interference. Slightly more expensive than UTP, used in industrial environments.

   • Ethernet Cable (Cat5e / Cat6 / Cat6a): Standard cable for LAN connections, supports high-speed data transmission. Cat5e: up to 1 Gbps, Cat6: up to 10 Gbps for short distances. Easy to install and widely used in labs.

3. Give difference between following:

(A) Twisted Pair Cable and Coaxial Cable

(B) Twisted Pair Cable and Fiber Optic Cable

(C) Fiber Optic Cable and Coaxial Cable

(D) Wired Communication and Wireless Communication

Conclusion: This experiment helps understand different types of wired transmission media, their characteristics, applications, and the key differences between various cables and communication methods, which are essential for designing and maintaining efficient networks.

Practical 06

Aim: Implementation of Dynamic Routing (Using RIP)

Exercise: Connect Two Networks Using Three Routers with RIP

Router1:

1. Open CLI → enable → configure terminal
2. Set hostname → hostname Router1
3. Configure interfaces:
   • interface fastethernet 0/0 → ip address 10.1.1.100 255.0.0.0 → no shutdown
   • interface fastethernet 1/0 → ip address 20.1.1.100 255.0.0.0 → no shutdown
4. Enable RIP →
   • router rip
   • network 10.0.0.0
   • network 20.0.0.0

Router2:

1. Open CLI → enable → configure terminal
2. Set hostname → hostname Router2
3. Configure interfaces:
   • interface fastethernet 0/0 → ip address 20.1.1.200 255.0.0.0 → no shutdown
   • interface fastethernet 1/0 → ip address 30.1.1.100 255.0.0.0 → no shutdown
4. Enable RIP →
   • router rip
   • network 20.0.0.0
   • network 30.0.0.0

Router3:

1. Open CLI → enable → configure terminal
2. Set hostname → hostname Router3
3. Configure interfaces:
   • interface fastethernet 0/0 → ip address 30.1.1.200 255.0.0.0 → no shutdown
   • interface fastethernet 1/0 → ip address 40.1.1.100 255.0.0.0 → no shutdown
4. Enable RIP →
   • router rip
   • network 30.0.0.0
   • network 40.0.0.0

Verification:

• Use ping <destination IP> from PCs connected to the network to ensure connectivity.
• Check routing table with show ip route to confirm RIP routes are learned.

Conclusion: In this practical, three routers were successfully configured to connect two different networks using RIP dynamic routing. IP addresses and RIP networks were correctly set, allowing communication between all networks. The routing tables updated automatically, demonstrating the functionality of the distance vector protocol.

Practical 07

Aim: Study of subnetting with an examples.

1. If your client needs three groups, how will you create the groups? Find the IP range, host range, network ID, and broadcast ID of each group. (IP: 200.1.2.0)

Ans.

• IP 200.1.2.0 is a Class C network.
• Required groups = 3 → Use formula 2ⁿ ≥ groups → 2² = 4 ≥ 3 → 2 bits needed for subnetting.
• New subnet mask = /26 → 255.255.255.192
• Each subnet has 2^(8–2) = 64 addresses → 62 usable hosts

2. In a Class B network on the internet has a subnet mask 255.255.240.0. What is the max. number of hosts per subnet?

Ans.

• Given subnet mask = 255.255.240.0 → /20 (because 255.255.240.0 = 11111111.11111111.11110000.00000000 → 20 network bits)
• Number of host bits = 32 – 20 = 12
• Maximum addresses per subnet = 2^12 = 4096
• Usable hosts per subnet = 2^12 – 2 = 4094
• So, 4094 usable hosts per subnet

3. Given IP Address 172.16.0.0/25, find the number of subnets and the number of hosts per subnet. Also, for the first subnet block, find the subnet address, first host ID, last host ID and broadcast address.

Ans.

• IP 172.16.0.0 is a Class B network.
• Given subnet mask = /25 → 255.255.255.128
• Number of subnet bits = 25 – 16 = 9 bits (because Class B default mask is /16)
• Number of subnets = 2^9 = 512 subnets
• Number of host bits = 32 – 25 = 7 bits
• Number of hosts per subnet = 2^7 – 2 = 126 usable hosts

First Subnet Block:

4. In the network 200.10.11.144/27, the fourth octet (in decimal) of the last IP address of the network which can be assigned to a host is:

• Ans: (A) 158

5. In the IPv4 addressing format, the number of networks allowed under Class C addresses is:

• Ans: (C) 2²¹

6. Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same net mask N. Which of the values of N given below should not be used if A and B should belong to the same network?

• Ans: (D) 255.255.255.224

7. If a Class B network on the Internet has a subnet mask of 255.255.248.0, what is the maximum number of hosts per subnet?

• Ans: (C) 2046

8. What is the broadcast address of the network 172.31.180.131/25 OR 172.31.180.128/25?

Ans.

• Case 1: 172.31.180.131/25

  • Subnet mask = 255.255.255.128 → /25 → 128 addresses per subnet
  • Subnet range: 172.31.180.128 – 172.31.180.255
  • Broadcast address = 172.31.180.255
  • Usable host range = 172.31.180.129 – 172.31.180.254

• Case 2: 172.31.180.128/25

  • Subnet mask = 255.255.255.128 → /25 → 128 addresses per subnet
  • Subnet range: 172.31.180.128 – 172.31.180.255
  • Broadcast address = 172.31.180.255
  • Usable host range = 172.31.180.129 – 172.31.180.254

9. What valid host range is the IP address 10.254.201.56/20 a part of?

Ans.

• /20 → 255.255.240.0 → 4096 addresses per subnet
• Network address: 10.254.192.0 (because 201 falls in 192–207 range for fourth octet)
• Broadcast address: 10.254.207.255
• Usable host range: 10.254.192.1 – 10.254.207.254

10. How many subnets and hosts per subnet can you get from the network 172.28.0.0/23?

Ans.

• IP 172.28.0.0 is a Class B network.
• Given subnet mask = /23 → 255.255.254.0
• Default Class B mask = /16 → Subnet bits = 23 – 16 = 7 bits
• Number of subnets = 2^7 = 128 subnets
• Host bits = 32 – 23 = 9 bits
• Number of hosts per subnet = 2^9 – 2 = 510 usable hosts
• So, Number of subnets: 128 and Hosts per subnet: 510

11. What is the last valid host on the subnetwork 172.19.156.0/23?

Ans.

• IP 172.19.156.0 is a Class B network.
• Subnet mask = /23 → 255.255.254.0
• Number of host bits = 32 – 23 = 9 bits → 2^9 = 512 addresses per subnet
• Network range: 172.19.156.0 – 172.19.157.255
• Broadcast address = 172.19.157.255
• Last valid host = broadcast – 1 = 172.19.157.254

12. What valid host range is the IP address 192.168.206.28 and subnet mask 255.255.255.248 a part of?

Ans.

• IP 192.168.206.28 is a Class C network.
• Subnet mask = 255.255.255.248 → /29 → 8 addresses per subnet (2^3 = 8)
• Network address = 192.168.206.24 (because 28 falls in 24–31 range)
• Broadcast address = 192.168.206.31
• Usable host range = 192.168.206.25 – 192.168.206.30

Conclusion: In this practical, we studied subnetting by dividing larger networks into smaller subnets. We learned how to calculate network ID, broadcast ID, usable host range, number of subnets, and hosts per subnet for Class B and Class C networks. Subnetting helps efficiently utilize IP addresses, reduces broadcast traffic, and improves network management.

Practical 08

Aim: Understanding NAT (Network Address Translation) with example.

Exercise :

1. What do you mean by NAT?

   • Network Address Translation (NAT) is a technique used in computer networks to translate private IP addresses into public IP addresses so that devices within a private network can communicate with external networks like the Internet. NAT helps conserve public IP addresses and provides a layer of security by hiding internal IP addresses.

   • Types of NAT:

     1. Static NAT: One-to-one mapping between a private IP and a public IP.
     2. Dynamic NAT: Private IPs are mapped to a pool of public IPs dynamically.
     3. PAT (Port Address Translation) / NAT Overload: Multiple private IPs share a single public IP using different ports.

   • Example:

     • Private IP: 192.168.1.10 → Public IP: 203.1.1.5 (Static NAT)
     • This allows a device in the private network to access the Internet.

Conclusion:

• NAT allows multiple devices in a private network to share a single public IP.
• It provides security by hiding internal IP addresses.
• NAT is widely used in home, office, and enterprise networks to connect private networks to the Internet.

Practical 09

Aim: Perform routing using OSPF in packet tracer

Exercise:

1. Which kind of protocol can be used by OSPF?

   • OSPF (Open Shortest Path First) is a link-state routing protocol that operates within an Autonomous System (AS). It uses the Dijkstra algorithm to calculate the shortest path to each network and exchanges routing information with other OSPF routers.

2. Differentiate OSPF with other protocols (RIP, IGRP, EIGRP, OSPF).

Conclusion: In this practical, we learned how OSPF works as a link-state routing protocol. OSPF provides faster convergence, better scalability, and efficient routing compared to distance vector protocols like RIP and IGRP, making it suitable for large networks.

Practical 10

Aim: Configuration of DHCP server in packet tracer

Exercise:

• IP Range: 192.168.0.100 to 192.168.0.124
• Network ID: 192.168.0.0
• Gateway and its subnet: 192.168.0.1 and 255.255.255.0
• Excluded IP range from DHCP: 192.168.0.1
• IP range for DHCP: 192.168.0.100 to 192.168.0.124

1. What is scope?

   • Scope in computer networks means the range where an address, message, or protocol is valid or can be used. For example:
   • Link-local scope → only within the same network segment.
   • Global scope → across the internet.

2. List types of IP address allocation methods with its brief

   • Static Allocation:
     • IP address is manually assigned to a device.
     • It doesn’t change unless configured again.
     • Used for servers, printers, or devices needing a fixed address.
   • Dynamic Allocation (via DHCP):
     • IP addresses are automatically assigned from a pool.
     • The address can change over time.
     • Used in most networks for client devices like laptops and smartphones.
   • Automatic Allocation:
     • Device generates its own IP address (e.g., using APIPA in Windows).
     • Used when a DHCP server is unavailable.
   • Manual Allocation:
     • Similar to static allocation, but done by network administrators with specific planning.
     • Ensures organized address management.
   • Public vs Private Allocation:
     • Public IP: Assigned by ISPs, reachable over the internet.
     • Private IP: Used within local networks, not directly accessible from the internet.
   • Reserved Allocation:
     • Certain IP ranges are reserved for special purposes (e.g., loopback, multicast).
     • Not assigned to typical devices.

3. What is IP reservation?

   • IP Reservation is a method where a specific IP address is permanently assigned to a particular device within a network, but the assignment is done through the Dynamic Host Configuration Protocol (DHCP) rather than manually configuring the device.
   • The device requests an IP from the DHCP server.
   • The DHCP server always assigns the same IP to that device based on its MAC address.
   • It combines the flexibility of DHCP with the stability of a static IP.

Conclusion: The practical successfully demonstrated the configuration of DHCP and its function in a network.