BGP Routing Protocol Practice Lab 01

at October 18, 2025

BGP Routing Protocol Practice Lab 01

Lab 1: MED and AS-Path Prepend

Basic configuration

R1:

interface Loopback0

ip address 1.1.1.1 255.255.255.255

interface FastEthernet0/0

ip address 150.1.1.1 255.255.255.0

no shut

interface Serial0/0

ip address 10.0.0.1 255.255.255.252

no shut

R2:

interface Loopback0

ip address 2.2.2.2 255.255.255.255

interface Loopback192

ip address 192.1.1.1 255.255.255.0

interface Loopback193

ip address 193.1.1.1 255.255.255.0

interface Loopback194

ip address 194.1.1.1 255.255.255.0

interface Loopback195

ip address 195.1.1.1 255.255.255.0

interface Serial0/0

ip address 10.0.0.2 255.255.255.252

no shut !

interface Serial0/1

ip address 10.0.0.9 255.255.255.252

no shut

R3:

interface Loopback0

ip address 3.3.3.3 255.255.255.255

interface FastEthernet0/0 ip address 150.3.3.3 255.255.255.0 no shut

interface Serial0/1

ip address 10.0.0.10 255.255.255.252

no shut !

interface Serial0/2

ip address 10.0.0.13 255.255.255.252

no shut !

interface Serial0/3

ip address 10.0.0.17 255.255.255.252

no shut

R4:

interface Loopback0

ip address 4.4.4.4 255.255.255.255

interface FastEthernet0/0 ip address 150.1.1.4 255.255.255.0 no shut

interface Serial0/0

ip address 10.0.0.14 255.255.255.252

no shut !

interface Serial0/1

ip address 10.0.0.18 255.255.255.252

no shut

Configure BGP as illustrated in the topology. Use the Loopback 0 addresses for peering. Do NOT configure any IGPs. Instead, use static routes only. R1 should peer with R2 and R4. R2 should peer with R1 and R3. R3 should peer with R2 and R4. R4 should peer with R1 and R3.

R1(config)#ip route 2.2.2.2 255.255.255.255 serial 0/0

R1(config)#ip route 4.4.4.4 255.255.255.255 fastethernet 0/0 150.1.1.4

R1(config)#router bgp 1
R1(config-router)#neighbor 2.2.2.2 remote-as 2

R1(config-router)#neighbor 2.2.2.2 update-source loopback 0

R1(config-router)#neighbor 2.2.2.2 ebgp-multihop 3

R1(config-router)#neighbor 4.4.4.4 remote-as 4

R1(config-router)#neighbor 4.4.4.4 update-source loopback 0

R1(config-router)#neighbor 4.4.4.4 ebgp-multihop 3

R2(config)#ip route 1.1.1.1 255.255.255.255 serial 0/0

R2(config)#ip route 3.3.3.3 255.255.255.255 serial 0/1

R2(config)#router bgp 2

R2(config-router)#neighbor 1.1.1.1 remote-as 1

R2(config-router)#neighbor 1.1.1.1 update-source loopback 0

R2(config-router)#neighbor 1.1.1.1 ebgp-multihop 3

R2(config-router)#neighbor 3.3.3.3 remote-as 3

R2(config-router)#neighbor 3.3.3.3 update-source loopback 0

R2(config-router)#neighbor 3.3.3.3 ebgp-multihop 3

R3(config)#ip route 2.2.2.2 255.255.255.255 serial 1/1

R3(config)#ip route 4.4.4.4 255.255.255.255 serial 1/2

R3(config)#ip route 4.4.4.4 255.255.255.255 serial 1/3

R3(config)#router bgp 3

R3(config-router)#neighbor 2.2.2.2 remote-as 2

R3(config-router)#neighbor 2.2.2.2 update-source loopback 0

R3(config-router)#neighbor 2.2.2.2 ebgp-multihop 3

R3(config-router)#neighbor 4.4.4.4 remote-as 4

R3(config-router)#neighbor 4.4.4.4 update-source loopback 0

R3(config-router)#neighbor 4.4.4.4 ebgp-multihop 3

R4(config)#ip route 1.1.1.1 255.255.255.255 fastethernet 0/0 150.1.1.1

R4(config)#ip route 3.3.3.3 255.255.255.255 serial 0/0

R4(config)#ip route 3.3.3.3 255.255.255.255 serial 0/1

R4(config)#router bgp 4

R4(config-router)#neighbor 1.1.1.1 remote-as 1

R4(config-router)#neighbor 1.1.1.1 update-source loopback 0

R4(config-router)#neighbor 1.1.1.1 ebgp-multihop 3

R4(config-router)#neighbor 3.3.3.3 remote-as 3

R4(config-router)#neighbor 3.3.3.3 update-source loopback 0

R4(config-router)#neighbor 3.3.3.3 ebgp-multihop 3

In order to ensure that the ORIGIN code is INCOMPLETE, you need to redistribute the LAN subnets into BGP. However, you can also use the network statement in conjunction with a route map and set the ORIGIN code within the route map.

R1(config)#route-map CONNECTED permit 10

R1(config-route-map)#match interface fastethernet 0/0

R1(config-route-map)#exit

R1(config)#route-map CONNECTED deny 20

R1(config-route-map)#exit

R1(config)#router bgp 1

R1(config-router)#redistribute connected route-map CONNECTED R1(config-router)#exit

You can verify the ORIGIN code by looking at the prefix entry in the BGP Tables. The ORIGIN code of INCOMPLETE is denoted by a question mark (?) in the output of the show ip bgp command. You can view additional detail on a per-prefix basis also when using this command

Configure BGP, so that R4 prefers the path via R3 to reach any subnet

In the output of the show ip bgp command on R4 we can see that the preferred route to reach 150.3.3.0 is via R3, however the preferred route to reach 150.2.2.0 is via R1 (the lowest routerid), also, to ensure that the subnet 150.1.1.0 will be reached via R3, configure BGP on R1 to advertise all prefixes with a longer AS-PATH to influence the path selection as follow:

R1(config)#route-map PREP permit 10

R1(config-route-map)#set as-path prepend 1 1 1 1 R1(config-route-map)#exit

R1(config)#router bgp 1

R1(config-router)#neighbor 4.4.4.4 route-map PREP out R1(config-router)#exit

Notice now the preferred path to reach both prefixes 150.3.3.0 and 150.2.2.0 is via R3 with the next-hop 3.3.3.3 because the shortest AS-PATH length:

Configure R4 so that it sends all updates to R3 with a MED of 4. Configure R2 so that it sends all updates to R3 with a MED of 2. Ensure that R3 prefers all routes with the better (lower) MED value.

Before configuring the MED let's verify the BGP RIBs on R3:

The preferred path to reach the prefix 150.1.1.0 is via R4, we should see all routes with the next-hop R2:

Let's configure MED on R3:

R4(config)#route-map MED permit 10

R4(config-route-map)#set metric 4

R4(config-route-map)#exit

R4(config)#router bgp 4

R4(config-router)#neighbor 3.3.3.3 route-map MED out

R4(config-router)#exit

R2(config)#route-map MED permit 10

R2(config-route-map)#set metric 2

R2(config-route-map)#exit

R2(config)#router bgp 2

R2(config-router)#neighbor 3.3.3.3 route-map MED out

R2(config-router)#exit

Let's verify the BGP RIBs of R3:

We have still the best path to reach 150.1.1.0 via R4 as shown by the show ip bgp command on R3 below, so the problem is not resolved even if R2 advertises the lowest MED comparing with R4.

The reason is: we met two issues in this case:

-the first issue is: by default, the MED is only compared for path received from the same AS ,in this case R3 receives two values of MED from two routers (R2 and R4) configured in different AS.

-The second issue: the MED is compared after the AS-PATH in the BGP decision process. In this case R3 will select the path via R4 as the best path to the 150.1.1.0/24 prefix because of the shorter AS-PATH length.

To override the two issues, configure the bgp always-compare-med command to avoid the first issue so always compare the MED even if MED is received from Different AS. And bgp bestpath as-path ignore command to avoid the second issue so that R3 override the BGP decision process by ignoring the step of the AS-PATH in the BGP Decision Process:

Let's configure these two commands:

R3(config)#router bgp 3

R3(config-router)#bgp bestpath as-path ignore R3(config-router)#bgp always-compare-med

We can see for the prefix 150.1.1.0 that the path with the longer AS-PATH length is preferred because the lowest MED even if the AS-PATH takes precedence over the MED in the order of the path selection in BGP:

Another way to verify all BGP RIBs with do show ip bgp, R3 prefers all routes from R2 because the lowest MED:

#BGP #LAB #CCNA #CCNP #CCIE #cisco #gns3 #solution

Campus Event Management System in JAVA

at August 29, 2024

Campus Event Management System

#java #event #fyp #project #source #code #learn #free

1. Introduction

Problem Statement: Managing campus events manually can be chaotic and time-consuming. The need for a streamlined digital platform is essential for enhancing user experience and improving event management efficiency.
Objective: To create a Java-based Campus Event Management System that allows users to create, manage, and register for campus events seamlessly.
Scope: The system will have modules for user registration, event creation, event registration, notifications, and administrative management.

2. Features

User Management: Registration, login, and profile management for students and organizers.
Event Management: Create, update, delete, and view events.
Registration System: Students can register for events, view the status of their registration, and receive notifications.
Notification System: Send email or SMS notifications for event updates, cancellations, and reminders.
Feedback System: Post-event feedback collection and analysis.

3. System Requirements

Software Requirements:
- JDK 8 or later
- IDE: Eclipse/IntelliJ IDEA
- Database: MySQL or PostgreSQL
- Apache Tomcat for web-based deployment
Hardware Requirements:
- RAM: 4GB minimum
- Processor: Dual Core Processor or higher
- Disk Space: 200MB for software and dependencies

4. System Design

Architecture: MVC (Model-View-Controller) architecture will be used for the web-based version of the application.
UML Diagrams:
- Use Case Diagram: Illustrates the interaction between users (students, organizers, and admin) and the system.
- Class Diagram: Shows the classes involved and their relationships.
- Sequence Diagram: Represents the flow of actions for key functionalities like event registration and notification.
Database Design:
- Tables for Users, Events, Registrations, Notifications, and Feedback.

Project Structure:

Backend (Spring Boot Application)
- Models
- Repositories
- Services
- Controllers
- Configuration
Frontend (JavaFX Application)

UI Layouts
Controllers
Utils

1. Backend: Spring Boot Application

Step 1: Create a Spring Boot Project

Use Spring Initializr to generate a new Spring Boot project with the following dependencies:

Spring Web
Spring Data JPA
MySQL Driver
Lombok

Step 2: Create Models

user.java

package com.campus.eventmanagement.model;

import lombok.AllArgsConstructor;

import lombok.Data;

import lombok.NoArgsConstructor;

import javax.persistence.*;

@Entity

@Data

@NoArgsConstructor

@AllArgsConstructor

public class User {

@Id

@GeneratedValue(strategy = GenerationType.IDENTITY)

private Long id;

private String username;

private String password;

private String email;

private String role; // "STUDENT", "ORGANIZER", "ADMIN"

}

event.java

package com.campus.eventmanagement.model;

import lombok.AllArgsConstructor;

import lombok.Data;

import lombok.NoArgsConstructor;

import javax.persistence.*;

import java.util.Date;

@Entity

@Data

@NoArgsConstructor

@AllArgsConstructor

public class Event {

@Id

@GeneratedValue(strategy = GenerationType.IDENTITY)

private Long eventId;

private String eventName;

private String location;

@Temporal(TemporalType.DATE)

private Date eventDate;

private String organizer;

private String description;

}

UserRepository.java

package com.campus.eventmanagement.repository;

import com.campus.eventmanagement.model.User;

import org.springframework.data.jpa.repository.JpaRepository;

import org.springframework.stereotype.Repository;

@Repository

public interface UserRepository extends JpaRepository<User, Long> {

User findByUsername(String username);

}

EventRepository.java

package com.campus.eventmanagement.repository;

import com.campus.eventmanagement.model.Event;

import org.springframework.data.jpa.repository.JpaRepository;

import org.springframework.stereotype.Repository;

@Repository

public interface EventRepository extends JpaRepository<Event, Long> {

// Custom query methods (if needed)

}

UserService.java

package com.campus.eventmanagement.service;

import com.campus.eventmanagement.model.User;

import com.campus.eventmanagement.repository.UserRepository;

import org.springframework.beans.factory.annotation.Autowired;

import org.springframework.stereotype.Service;

import java.util.List;

@Service

public class UserService {

@Autowired

private UserRepository userRepository;

public User registerUser(User user) {

return userRepository.save(user);

}

public User getUserByUsername(String username) {

return userRepository.findByUsername(username);

}

public List<User> getAllUsers() {

return userRepository.findAll();

}

EventService.java

package com.campus.eventmanagement.service;

import com.campus.eventmanagement.model.Event;

import com.campus.eventmanagement.repository.EventRepository;

import org.springframework.beans.factory.annotation.Autowired;

import org.springframework.stereotype.Service;

import java.util.List;

@Service

public class EventService {

@Autowired

private EventRepository eventRepository;

public Event createEvent(Event event) {

return eventRepository.save(event);

}

public List<Event> getAllEvents() {

return eventRepository.findAll();

}

public void deleteEvent(Long eventId) {

eventRepository.deleteById(eventId);

}

UserController.java

package com.campus.eventmanagement.controller;

import com.campus.eventmanagement.model.User;

import com.campus.eventmanagement.service.UserService;

import org.springframework.beans.factory.annotation.Autowired;

import org.springframework.web.bind.annotation.*;

import java.util.List;

@RestController

@RequestMapping("/api/users")

public class UserController {

@Autowired

private UserService userService;

@PostMapping("/register")

public User registerUser(@RequestBody User user) {

return userService.registerUser(user);

}

@GetMapping("/{username}")

public User getUserByUsername(@PathVariable String username) {

return userService.getUserByUsername(username);

}

@GetMapping("/all")

public List<User> getAllUsers() {

return userService.getAllUsers();

}

EventController.java

package com.campus.eventmanagement.controller;

import com.campus.eventmanagement.model.Event;

import com.campus.eventmanagement.service.EventService;

import org.springframework.beans.factory.annotation.Autowired;

import org.springframework.web.bind.annotation.*;

import java.util.List;

@RestController

@RequestMapping("/api/events")

public class EventController {

@Autowired

private EventService eventService;

@PostMapping("/create")

public Event createEvent(@RequestBody Event event) {

return eventService.createEvent(event);

}

@GetMapping("/all")

public List<Event> getAllEvents() {

return eventService.getAllEvents();

}

@DeleteMapping("/{eventId}")

public void deleteEvent(@PathVariable Long eventId) {

eventService.deleteEvent(eventId);

}

Database Configuration

spring.datasource.url=jdbc:mysql://localhost:3306/event_management_db

spring.datasource.username=root

spring.datasource.password=yourpassword

spring.jpa.hibernate.ddl-auto=update

spring.jpa.show-sql=true

spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.MySQL5Dialect

2. Frontend: JavaFX Application

Step 1: Create the JavaFX Project Structure

Use an IDE like IntelliJ IDEA or Eclipse with JavaFX support.

Step 2: Design UI using FXML

Create a basic UI for the JavaFX application using FXML files. For simplicity, we'll focus on the main screens.

Main.fxml

<?xml version="1.0" encoding="UTF-8"?>

<?import javafx.scene.control.*?>

<?import javafx.scene.layout.*?>

<AnchorPane xmlns:fx="http://javafx.com/fxml" fx:controller="com.campus.eventmanagement.controller.MainController">

<children>

<VBox spacing="10">

<Label text="Campus Event Management System" style="-fx-font-size: 20px;"/>

<Button text="Register for Event" onAction="#handleRegisterForEvent"/>

<Button text="View Events" onAction="#handleViewEvents"/>

<Button text="Create Event" onAction="#handleCreateEvent"/>

</VBox>

</children>

</AnchorPane>

MainController.java

package com.campus.eventmanagement.controller;

import javafx.event.ActionEvent;

import javafx.fxml.FXML;

import javafx.scene.control.Alert;

public class MainController {

@FXML

private void handleRegisterForEvent(ActionEvent event) {

// Implementation of registering for an event

showInfoAlert("Feature Coming Soon!");

}

@FXML

private void handleViewEvents(ActionEvent event) {

// Implementation of viewing events

showInfoAlert("Feature Coming Soon!");

}

@FXML

private void handleCreateEvent(ActionEvent event) {

// Implementation of creating an event

showInfoAlert("Feature Coming Soon!");

}

private void showInfoAlert(String message) {

Alert alert = new Alert(Alert.AlertType.INFORMATION);

alert.setTitle("Information");

alert.setHeaderText(null);

alert.setContentText(message);

alert.showAndWait();

}

MainApp.java

package com.campus.eventmanagement;

import javafx.application.Application;

import javafx.fxml.FXMLLoader;

import javafx.scene.Parent;

import javafx.scene.Scene;

import javafx.stage.Stage;

public class MainApp extends Application {

@Override

public void start(Stage primaryStage) throws Exception {

Parent root = FXMLLoader.load(getClass().getResource("/Main.fxml"));

primaryStage.setTitle("Campus Event Management System");

primaryStage.setScene(new Scene(root));

primaryStage.show();

}

public static void main(String[] args) {

launch(args);

}

Step 5: Build and Run

Make sure your JavaFX libraries are properly set up in your IDE.
Run the Spring Boot backend by executing MainApp.java to start the JavaFX UI.

Summary

This code provides a comprehensive backend using Spring Boot and a basic frontend using JavaFX. The backend includes CRUD operations for users and events, while the frontend includes basic JavaFX layout and event handlers.

ipv4 subnetting

at August 29, 2024

In this blog post, we dive deep into the art of subnetting IPv4 addresses, a crucial skill for network administrators and engineers. We start with the fundamentals of subnetting, explaining how IP addresses are divided into network and host portions. The post includes a variety of practice questions, each accompanied by detailed explanations to help you master the concepts. Whether you're preparing for certification exams or just brushing up on your skills, this guide will provide you with the knowledge and confidence you need to tackle subnetting challenges.

#Subnetting #IPv4 #Networking #IP Addressing #CCNA #Network Administration #IT_Certification #Practice Questions #NetworkEngineering #Subnetting #Explained

(Solutions Provided at End)

Question #1

What is the range of assignable IP addresses for a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.0.1 – 172.16.31.254

b. 172.16.0.1 – 172.16.63.254

c. 172.16.0.0 – 172.16.31.255

d. 172.16.0.1 – 172.16.31.255

e. 172.16.0.0 – 172.16.63.254

Question #2

You are assigning IP addresses to hosts in the 192.168.4.0 /26 subnet. Which two of the following IP addresses are assignable IP addresses that reside in that subnet?

a. 192.168.4.0

b. 192.168.4.63

c. 192.168.4.62

d. 192.168.4.32

e. 192.168.4.64

Question #3

A host in your network has been assigned an IP address of 192.168.181.182 /25. What is the subnet to which the host belongs?

a. 192.168.181.128 /25

b. 192.168.181.0 /25

c. 192.168.181.176 /25

d. 192.168.181.192 /25

e. 192.168.181.160 /25

Question #4

You are working with a Class B network with the private IP address of 172.16.0.0 /16. You need to maximize the number of broadcast domains, where each broadcast domain can accommodate 1000 hosts. What subnet mask should you use?

a. /22

b. /23

c. /24

d. /25

e. /26

Question #5

What is the directed broadcast address of a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.15.255

b. 172.16.31.255

c. 172.16.255.255

d. 172.16.95.255

e. 172.16.0.255

Question #6

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many subnets can be created by using this subnet mask?

a. 32

b. 16

c. 30

d. 8

e. 14

Question #7

Given a subnet of 172.16.56.0 /21, identify which of the following IP addresses belong to this subnet. (Select 2.)

a. 172.16.54.129

b. 172.16.62.255

c. 172.16.61.0

d. 172.16.65.255

e. 172.16.64.1

Question #8

What is the subnet address of the IP address 192.168.5.55 with a subnet mask of 255.255.255.224?

a. 192.168.5.0 /27

b. 192.168.5.16 /27

c. 192.168.5.32 /27

d. 192.168.5.48 /27

e. 192.168.5.64 /27

Question #9

You are working for a company that will be using the 192.168.1.0 /24 private IP address space for IP addressing inside their organization.

They have multiple geographical locations and want to carve up the 192.168.1.0 /24 address space into subnets. Their largest subnet will need 13 hosts.

What subnet mask should you use to accommodate at least 13 hosts per subnet, while maximizing the number of subnets that can be created?

a. 255.255.255.248

b. 255.255.255.224

c. 255.255.255.252

d. 255.255.255.192

e. 255.255.255.240

Question #10

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many assignable addresses are available in each of the subnets?

a. 32

b. 16

c. 30

d. 8

e. 14

Question #11

An IP address of 192.168.0.100 /27 belongs to which of the following subnets?

a. 192.168.0.92

b. 192.168.0.128

c. 192.168.0.64

d. 192.168.0.96

e. 192.168.0.32

Question #12

What subnet mask should be used to subnet the 192.168.10.0 network to support the number of subnets and IP addresses per subnet shown in the following topology?

a. 255.255.255.0

b. 255.255.255.128

c. 255.255.255.192

d. 255.255.255.224

e. 255.255.255.240

Solutions

Question #1

What is the range of assignable IP addresses for a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.0.1 – 172.16.31.254

b. 172.16.0.1 – 172.16.63.254

c. 172.16.0.0 – 172.16.31.255

d. 172.16.0.1 – 172.16.31.255

e. 172.16.0.0 – 172.16.63.254

Answer: a

To determine the subnets, assignable IP address ranges, and directed broadcast addresses created by the 19-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 19-bit subnet mask, which is written in binary as:

11111111 11111111 11100000 00000000

The interesting octet is the third octet, because the third octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 19-bit subnet mask can be written in dotted decimal notation as: 255.255.224.0

Since the third octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

172.16.0.0 /19

We then count by the block size (of 32) in the interesting octet (the third octet in this question) to determine the remaining subnets:

172.16.32.0 /19

172.16.64.0 /19

172.16.96.0 /19

172.16.128.0 /19

172.16.160.0 /19

172.16.192.0 /19

172.16.224.0 /19

Step #5: Identify the subnet address, the directed broadcast address, and the usable range of addresses.

Looking through the subnets created by the 19-bit subnet mask reveals that the IP address of 172.16.1.10 resides in the 172.16.0.0 /19 subnet.

The directed broadcast address, where all host bits are set to a 1, is 1 less than the next subnet address.

The next subnet address is 172.16.32.0. So, the directed broadcast address for the 172.16.0.0 /19 subnet is 1 less than 172.16.32.0, which is:

172.16.31.255

The usable IP addresses are all the IP addresses between the subnet address and the directed broadcast address. Therefore, in this example, the assignable IP address range for the 172.16.0.0 /19 network is:

172.16.0.1 – 172.16.31.254

Question #2

You are assigning IP addresses to hosts in the 192.168.4.0 /26 subnet. Which two of the following IP addresses are assignable IP addresses that reside in that subnet?

a. 192.168.4.0

b. 192.168.4.63

c. 192.168.4.62

d. 192.168.4.32

e. 192.168.4.64

Answer: c and d

To determine subnets and usable address ranges created by the 26-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 26-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 11000000

The interesting octet is the forth octet, because the forth octet (i.e. 11000000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 26-bit subnet mask can be written in dotted decimal notation as: 255.255.255.192

Since the forth octet is the interesting octet, the decimal value in the interesting octet is 192.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 192 = 64

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.4.0 /26

We then count by the block size (of 64) in the interesting octet (the forth octet in this question) to determine the remaining subnets:

192.168.4.64 /26

192.168.4.128 /26

192.168.4.192 /26

Step #5:

This question is asking about the 192.168.4.0 /26 subnet. From the above list of subnets, we can determine that the assignable range of IP addresses for this subnet is 192.168.4.1 – 192.168.4.62. We can also determine that 192.168.4.0 is the network address, and 192.168.4.63 is the directed broadcast address.

From the assignable range of IP addresses we have calculated, we can determine that the two assignable IP addresses given as options in this question are: 192.168.4.62 and 192.168.4.32.

Question #3

A host in your network has been assigned an IP address of 192.168.181.182 /25. What is the subnet to which the host belongs?

a. 192.168.181.128 /25

b. 192.168.181.0 /25

c. 192.168.181.176 /25

d. 192.168.181.192 /25

e. 192.168.181.160 /25

Answer: a

To determine subnets and usable address ranges created by the 25-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 25-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 10000000

The interesting octet is the forth octet, because the forth octet (i.e. 10000000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 25-bit subnet mask can be written in dotted decimal notation as: 255.255.255.128

Since the forth octet is the interesting octet, the decimal value in the interesting octet is 128.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 128 = 128

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.181.0 /25

We then count by the block size (of 128) in the interesting octet (the forth octet in this question) to determine the remaining subnets, or in this case just a single additional subnet.

192.168.181.128 /25

Now that we have our two subnets identified, we can determine the subnet in which the IP address of 192.168.181.182 resides.

Since the usable range of IP addresses for the 192.168.181.128 /25 network is 192.168.181.129 – 192.168.181.254 (because 192.168.181.128 is the network address, and 192.168.181.255 is the directed broadcast address), and since 192.168.181.182 is in that range, the subnet to which 192.168.181.182 /25 belongs is:

192.168.181.128 /25

Question #4

a. /22

b. /23

c. /24

d. /25

e. /26

Answer: a

In addition to testing your knowledge of subnetting, this question is also making sure you understand that a subnet is a broadcast domain. This should not be confused with a collision domain (i.e. each port on a switch is in its own collision domain).

To determine how many host bits are required to support 1000 hosts, we can create a table from the following formula:

Number of Hosts = 2h – 2, where h is the number of host bits

From this formula, we can create the following table:

1 Host Bit => 0 Hosts

2 Host Bits => 2 Hosts

3 Host Bits => 6 Hosts

4 Host Bits => 14 Hosts

5 Host Bits => 30 Hosts

6 Host Bits => 62 Hosts

7 Host Bits => 126 Hosts

8 Host Bits => 254 Hosts

9 Host Bits => 510 Hosts

10 Host Bits => 1022 Hosts

This table tells us that a subnet with 10 host bits will accommodate the requirement of 1000 hosts. If we have 10 host bits, then we have a 22-bit subnet mask (i.e. 32 – 10 = 22). Also, by not using more host bits than we need, we are maximizing the number of subnets that can be created.

Question #5

What is the directed broadcast address of a subnet containing an IP address of 172.16.1.10 /19?

a. 172.16.15.255

b. 172.16.31.255

c. 172.16.255.255

d. 172.16.95.255

e. 172.16.0.255

Answer: b

To determine the subnets, assignable IP address ranges, and directed broadcast addresses created by the 19-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 19-bit subnet mask, which is written in binary as:

11111111 11111111 11100000 00000000

The interesting octet is the third octet, because the third octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 19-bit subnet mask can be written in dotted decimal notation as: 255.255.224.0

Since the third octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

172.16.0.0 /19

We then count by the block size (of 32) in the interesting octet (the third octet in this question) to determine the remaining subnets:

172.16.32.0 /19

172.16.64.0 /19

172.16.96.0 /19

172.16.128.0 /19

172.16.160.0 /19

172.16.192.0 /19

172.16.224.0 /19

Step #5: Identify the subnet address, the directed broadcast address, and the usable range of addresses.

Looking through the subnets created by the 19-bit subnet mask reveals that the IP address of 172.16.1.10 resides in the 172.16.0.0 /19 subnet.

The directed broadcast address, where all host bits are set to a 1, is 1 less than the next subnet address.

The next subnet address is 172.16.32.0. So, the directed broadcast address for the 172.16.0.0 /19 subnet is 1 less than 172.16.32.0, which is:

172.16.31.255

172.16.0.1 – 172.16.31.254

Question #6

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many subnets can be created by using this subnet mask?

a. 32

b. 16

c. 30

d. 8

e. 14

Answer: b

The subnet in this question is a Class C network, because there is a 192 in the first octet. A class C network has a natural mask of 24 bits. However, this network has a 28-bit subnet mask. Therefore, we have 4 borrowed bits, which are network bits added to a network’s natural mask (i.e. 28 – 24 = 4). The number of subnets can be calculated as follows:

Number of Subnets = 2s, where s is the number of borrowed bits.

Therefore, in this question, the number of created subnets is 16:

Number of Subnets = 24 = 16

Question #7

Given a subnet of 172.16.56.0 /21, identify which of the following IP addresses belong to this subnet. (Select 2.)

a. 172.16.54.129

b. 172.16.62.255

c. 172.16.61.0

d. 172.16.65.255

e. 172.16.64.1

Answer: b, c

To determine subnets and usable address ranges created by the 21-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 21-bit subnet mask, which is written in binary as:

11111111 11111111 11111000 00000000

The interesting octet is the third octet, because the third octet (i.e. 11111000) is the first octet to contain a 0 in the binary subnet mask.

Step #2: Identify the decimal value in the interesting octet of the subnet mask. A 21-bit subnet mask can be written in dotted decimal notation as: 255.255.248.0

Since the third octet is the interesting octet, the decimal value in the interesting octet is 248.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 248 = 8

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

172.16.0.0 /21

We then count by the block size (of 8) in the interesting octet (the third octet in this question) to determine the remaining subnets:

172.16.8.0 /21 172.16.16.0 /21 172.16.24.0 /21 172.16.32.0 /21 172.16.40.0 /21 172.16.48.0 /21 172.16.56.0 /21 172.16.64.0 /21 ... SUBNETS OMITTED ...

We can stop counting after we pass the subnet we are being asked about. Specifically, in this question, we’re being asked about 172.16.56.0 /21.

Step #5: Identify the subnet address, the directed broadcast address, and the usable range of addresses.

The subnet address, where all host bits are set to a 0, is given:

172.16.56.0 /24

The directed broadcast address, where all host bits are set to a 1, is 1 less than the next subnet address.

The next subnet address is 172.16.64.0. So, the directed broadcast address for the 172.16.54.0 /21 subnet is 1 less than 172.16.64.0, which is: 172.16.63.255

The usable IP addresses are all the IP addresses between the subnet address and the directed broadcast address. Therefore, in this example, the usable IP address range for the 172.16.56.0 /21 network is:

172.16.56.1 – 172.16.63.254

The only IP addresses in this question that reside in this range are:

172.16.62.255 172.16.61.0

WARNING: Many CCNA R&S candidates look at IP addresses like these and immediately assume they are not usable IP addresses, because they have a 0 or a 255 in the forth octet. They argue that 172.16.61.0 is a subnet address and that 172.16.62.255 is a directed broadcast address.

While that would only be true of the subnet mask were 24-bits, remember that, by definition, a subnet address has all of its host bits set to a 0, and a directed broadcast address has all of its host bits set to a 1. In this question, we have 11 host bits (i.e. 32 – 21 = 11), not 8 host bits. So, 172.16.62.255 and 172.16.61.0 are actually usable IP addresses.

Question #8

What is the subnet address of the IP address 192.168.5.55 with a subnet mask of 255.255.255.224?

a. 192.168.5.0 /27

b. 192.168.5.16 /27

c. 192.168.5.32 /27

d. 192.168.5.48 /27

e. 192.168.5.64 /27

Answer: c

To determine subnets and usable address ranges created by the 27-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 27-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 11100000

The interesting octet is the forth octet, because the forth octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 27-bit subnet mask can be written in dotted decimal notation as: 255.255.255.224

Since the forth octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.5.0 /27

We then count by the block size (of 32) in the interesting octet (the forth octet in this question) to determine the remaining subnets:

192.168.5.32 /27

192.168.5.64 /27

192.168.5.96 /27

192.168.5.128 /27

192.168.5.160 /27

192.168.5.192 /27

192.168.5.224 /27

Now that we have all of our subnets identified, we can determine the subnet in which the IP address of 192.168.5.55 resides.

Since the usable range of IP addresses for the 192.168.5.32 /27 network is 192.168.5.33 – 192.168.5.62 (because 192.168.5.32 is the network address, and 192.168.5.63 is the directed broadcast address), and since 192.168.5.55 is in that range, the subnet to which 192.168.5.55 /27 belongs is:

192.168.5.32 /27

Question #9

You are working for a company that will be using the 192.168.1.0 /24 private IP address space for IP addressing inside their organization.

They have multiple geographical locations and want to carve up the 192.168.1.0 /24 address space into subnets. Their largest subnet will need 13 hosts.

What subnet mask should you use to accommodate at least 13 hosts per subnet, while maximizing the number of subnets that can be created?

a. 255.255.255.248

b. 255.255.255.224

c. 255.255.255.252

d. 255.255.255.192

e. 255.255.255.240

Answer: e

We can determine the maximum number of hosts allowed in a subnet by raising the number 2 to the power of the number of host bits and then subtracting 2. So, the formula looks like this:

Maximum Number of Hosts per Subnet = 2h – 2, where h is the number of host bits.

Why are we subtracting two? Well, there are two IP addresses in the subnet that cannot be assigned. These addresses are: (1) the network address, where all of the host bits are set to a 0 and (2) the directed broadcast address, where all of the host bits are set to a 1.

In the actual exam, if you are given scratch paper or access to a note taking application, you might want to write out a table such as the following for your reference:

1 Host Bit: 21 – 2 = 0

2 Host Bits: 22 – 2 = 2

3 Host Bits: 23 – 2 = 6

4 Host Bits: 24 – 2 = 14

5 Host Bits: 25 – 2 = 30

6 Host Bits: 26 – 2 = 62

7 Host Bits: 27 – 2 = 126

8 Host Bits: 28 – 2 = 254

In this question, we’re asked to determine a subnet mask that accommodates at least 13 hosts per subnet. By looking at the reference table we created, we can see that 4 host bits (which support 14 hosts) would work, while 3 host bits (which supports only 6 hosts) would not be enough.

So, we need a subnet with 4 host bits, which are enough host bits to meet the design goal, but not more than we need. Using more host bits than we need would violate the requirement to maximize the number of subnets.

A subnet mask with 4 host bits has 28 network bits (i.e. 32 – 4 = 28), and therefore a 28-bit subnet mask. A 28-bit subnet mask can be written as:

255.255.255.240

Question #10

A customer is using a Class C network of 192.168.10.0 subnetted with a 28-bit subnet mask. How many assignable addresses are available in each of the subnets?

a. 32

b. 16

c. 30

d. 8

e. 14

Answer: e

An IPv4 address contains a total of 32 bits. Since, in this question, we have 28 subnet bits, the number of host bits is 4 (i.e. 32 – 28 = 4). The number of assignable IP addresses in a subnet can be calculated as follows:

Number of Assignable IP Addresses = 2h – 2, where h is the number of host bits.

Therefore, in this question, each subnet has 14 assignable IP addresses:

Number of Assignable IP Addresses = 24 – 2 = 16 – 2 = 14

Question #11

An IP address of 192.168.0.100 /27 belongs to which of the following subnets?

a. 192.168.0.92

b. 192.168.0.128

c. 192.168.0.64

d. 192.168.0.96

e. 192.168.0.32

Answer: d

To determine the subnets created by the 27-bit subnet mask we perform the following steps:

Step #1: Identify the interesting octet (i.e. the octet that contains the first zero in the binary subnet mask).

In this question, we have a 19-bit subnet mask, which is written in binary as:

11111111 11111111 11111111 11100000

The interesting octet is the forth octet, because the forth octet (i.e. 11100000) is the first octet to contain a 0 in the binary.

Step #2: Identify the decimal value in the interesting octet of the subnet mask.

A 27-bit subnet mask can be written in dotted decimal notation as: 255.255.255.224

Since the forth octet is the interesting octet, the decimal value in the interesting octet is 224.

Step #3: Determine the block size by subtracting the decimal value of the interesting octet from 256.

Block Size = 256 – 224 = 32

Step #4: Determine the subnets by counting by the block size in the interesting octet, starting at 0.

Placing a zero in the first interesting octet identifies the first subnet as:

192.168.0.0 /27

We then count by the block size (of 32) in the interesting octet (the forth octet in this question) to determine the remaining subnets:

192.168.0.32 /27

192.168.0.64 /27

192.168.0.96 /27

192.168.0.128 /27

192.168.0.160 /27

192.168.0.192 /27

192.168.0.224 /27

Step #5: Identify the subnet address of the IP address 192.168.0.100 /27.

Looking through the subnets created by the 27-bit subnet mask reveals that the IP address of 192.168.0.100 resides in the 192.168.0.96 subnet.

Question #12

What subnet mask should be used to subnet the 192.168.10.0 network to support the number of subnets and IP addresses per subnet shown in the following topology?

a. 255.255.255.0

b. 255.255.255.128

c. 255.255.255.192

d. 255.255.255.224

e. 255.255.255.240

Answer: c

To meet the design requirements, four subnets must be created, and each subnet must accommodate a maximum of 50 IP addresses.

We can begin by creating a listing of how many subnets are created from different numbers of borrowed bits, using the formula:

Number of Subnets Created = 2n, where n is the number of borrowed bits

1 borrowed bits => 2 subnets

2 borrowed bits => 4 subnets

3 borrowed bits => 8 subnets

4 borrowed bits => 16 subnets

5 borrowed bits => 32 subnets

6 borrowed bits => 64 subnets

7 borrowed bits => 128 subnets

From this, we can see we need at least 2 borrowed bits to accommodate 4 subnets. However, we need to make sure the subnet will accommodate 50 IP addresses. To determine this, we can use the formula:

Number of IP Addresses = 2h – 2, where h is the number of host bits

If we have 2 borrowed bits (i.e. the minimum number of borrowed bits required for 4 subnets), we have 6 host bits (i.e. 8 – 2 = 6). From the above formula, we can determine the number of IP addresses supported by 6 host bits.

Number of IP Addresses = 26 – 2 = 62

Since 6 host bits meet our requirement of at least 50 IP addresses per subnet, we can use a 26-bit subnet mask (i.e. 2 bits added to the Class C default mask (also known as the natural mask) of 24 bits). A 26-bit subnet mask can be written as:

255.255.255.192

Python Project: Audio Transcription and Text-to-Speech Conversion Using Wav2Vec2 and Pyttsx3

at August 29, 2024

Explore an advanced Python project that combines audio transcription and text-to-speech synthesis using state-of-the-art tools like Librosa, PyTorch, and Hugging Face's Transformers library. This script demonstrates how to load and resample audio files, transcribe speech to text using Facebook's Wav2Vec2 model, and convert text back to speech with customizable voice options using pyttsx3. Perfect for anyone interested in speech processing, AI-driven voice technology, or natural language processing projects. Ideal for enhancing your Python skills and diving into real-world applications of AI in audio analysis.

import librosa

from scipy.signal import resample

import torch

from transformers import Wav2Vec2ForCTC, Wav2Vec2Tokenizer

import pyttsx3

from scipy.signal import resample

# Load audio file

audio_file = "directory of audio file"

audio, sr = librosa.load(audio_file, sr=None)

def resample_audio(audio, orig_sr, target_sr):

duration = audio.shape[0] / orig_sr

target_length = int(duration * target_sr)

resampled_audio = resample(audio, target_length)

return resampled_audio

# Example usage:

# resampled_audio = resample_audio(audio, 48000, 16000)

# Resample if necessary

if sr != 16000:

audio = resample_audio(audio, sr, 16000)

sr = 16000

print("Audio loaded and resampled successfully.")

# Load Wav2Vec2 model and tokenizer

tokenizer = Wav2Vec2Tokenizer.from_pretrained("facebook/wav2vec2-base-960h")

model = Wav2Vec2ForCTC.from_pretrained("facebook/wav2vec2-base-960h")

print("Model loaded successfully.")

# Tokenize input

input_values = tokenizer(audio, return_tensors="pt").input_values

# Perform inference

with torch.no_grad():

logits = model(input_values).logits

# Get predicted ids

predicted_ids = torch.argmax(logits, dim=-1)

# Decode the ids to text

transcription = tokenizer.batch_decode(predicted_ids)[0]

print("Transcription: ", transcription)

# Text-to-Speech

def text_to_speech(text, voice_gender='female', rate=150):

engine = pyttsx3.init()

voices = engine.getProperty('voices')

if voice_gender == 'male':

engine.setProperty('voice', voices[0].id)

else:

engine.setProperty('voice', voices[1].id)

engine.setProperty('rate', rate)

engine.say(text)

engine.runAndWait()

# Example usage

long_text = "hi what happened"

text_to_speech(long_text, voice_gender='male', rate=150) # For male voice

text_to_speech(long_text, voice_gender='female', rate=180) # For female voice

print("Text-to-Speech conversion completed.")

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#AudioTranscription
#TextToSpeech
#Wav2Vec2
#PyTorch
#Librosa
#NLP
#SpeechRecognition
#VoiceSynthesis
#AIinPython
#NaturalLanguageProcessing
#SpeechToText
#Pyttsx3
#MachineLearning
#DeepLearning
#AudioProcessing
#PythonAI
#TransformersLibrary
#PythonCoding
#PythonTutorial

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