Programming Assignment 6 Instructions NOTE: THIS MUST BE WRITTEN IN VISUAL STUDIO AS A WINDOWS FORM APP (.NET Framework) - VISUAL BASICAssignment instructions You have been tasked to write an application which will help your organization calculate the cost of your expenses. Write a Visual Basic program to implement the budget application in the diagram below. The program should have a class called BudgetReceipts that keeps track of the balance and provides the ability to add to the income and subtract to the expenses. Requirements:1. Form Setupa. You must save your project using your initials in the name** This is required and the project will not be accepted otherwise.b. Design your screen to look like the one below. c. The background image of the form will be set to an appropriate image of your choice.d. Use appropriate naming conventions for controls and variables. e. Include access keys for all buttons. f. The user will not be directed to the net balance field.g. The balance text box will be read only. h. Lock all controls.i. All values must be formatted in currencyj. On the form load, the user is prompted with an input box to enter the starting balance for the balance sheet. The user is prompted until a value is entered. k. The initial balance is saved by the class.2. Classa. A class is created which uses a get property for the balance, a sub that subtracts the expenses, and sub that controls the income.b. The class is created separate from the class form.c. The property balance will be called to display the balance each timed. The income sub will be used to add to the incomee. The expenses sub will be used to subtract the expenses3. Income Buttona. When the Income button is selected the application verifies a value is entered in the amount text box. b. The income button prompts the user for the description of the income the amount is for.c. If a value is not entered the application presents and error and the program does not continue. d. The program will call the class to add the value to the balance and then display the total balance to the user using the appropriate property. e. The income button presents the user with a description of the value and the value amountf. The textbox amount is cleared.4. Expense Buttona. When the expense button is selected the application verifies a value is entered in the amount text box. b. If a value is not entered the application presents and error and the program does not continue. c. The program will call the class to subtract the value to the balance and then display the balance to the user using the appropriate property. d. If the balance will go negative the user will be prompted with a message box but the program will continue.e. The income button presents the user with a description of the value and the value amountf. The textbox amount is cleared

Sources This assignment covers material presented in the chapter 'Memory Management & Operating Systems'. You can refer particularly to the following lessons in order to help you working through your solution: Memory Allocation Schemes: Definition & Uses Paged Memory Allocation: Definition, Purpose & Structure Memory Deallocation: Definition & Purpose Page Replacement: Definition & Algorithms Memory Management (MEM) Coding: Analysis & Example Prompt Consider the below settings to begin this project and keep them in mind: Total Memory size = 20 Kbyte Page size = 1 Kbyte Run the following list of jobs given the above considerations. Job IDStart TimeJob required sizeExecution intervalJob state at the end of the interval1127End2238Sleep3346End4436Sleep5529Sleep6636Sleep7726Sleep Using the functions you studied in the previously mentioned code that implement the different allocation, deallocation and replacement schemes, try three different scenarios in order to manage memory allocation for the following jobs Job IDStart TimeJob required sizeExecution intervalJob state at the end of the interval8834Sleep9955Sleep101028Sleep111146End121265Sleep21336End41334Sleep131353End71323End91744Sleep1019211End61936End520210Sleep421312Sleep1222613End82239End928511End53323Sleep434310End538210End Show the memory state for the different scenarios and analyze which choice is the best. Discuss the impact that Linux verses Windows might have on the ending memory state. Here's the sample code: #include <stdio.h> //Process struct to create an object with the attributes needed struct Process { int processID; int processingTime; int waitTime; int turnAroundTime; //add arrivingTime; int arrivingTime; //add is Processing int isProcessing; }; //This will calculate the Wait Time for each Process according to the algorithm struct Process * calculateWaitTime (struct Process queue[], int size) { for (int i = 1; i < size; i++) { queue[i].waitTime = queue[i - 1].processingTime + queue[i - 1].waitTime; } return queue; } //This will calculate how much time it takes for a process to wait and be executed struct Process * calculateTurnAround (struct Process queue[], int size) { for (int i = 0; i < size; i++) { queue[i].turnAroundTime = queue[i].processingTime + queue[i].waitTime; } return queue; } //This will give us an average of the waiting time for each process void averageWaitingTime (struct Process queue[], int size) { float average = 0; int count = 0; for (int i = 0; i < size; i++) { average += queue[i].waitTime; count++; } average = average / count; printf("Average Waiting Time: %f\n", average); } //This function helps to swap items in an array void swap(struct Process *a, struct Process *b) { struct Process t = *a; a = b; *b = t; } //This method aids the pivot for Quick Sort int partition (struct Process queue[], int low, int high) { int pivot = queue[high].processingTime; int i = low - 1; for (int j = low; j <= high - 1; j++) { if (queue[j].processingTime <= pivot) { i++; swap(&queue[i],&queue[j]); } } swap(&queue[i + 1], &queue[high]); return (i + 1); } //This is a recursive implementation of the Quick Sort void quickSort (struct Process queue[], int low, int high) { if (low < high) { int pi = partition(queue, low, high); quickSort(queue, low, pi - 1); quickSort(queue, pi + 1, high); } } //First Server First Come scheduling algorithm // void firstServeFirstCome (struct Process queue[], int size) { // queue = calculateWaitTime(queue, size); // queue = calculateTurnAround(queue, size); // averageWaitingTime(queue, size); // } //Shortest Job Next scheduling algorithm void shortestJobNext (struct Process queue[], int size) { quickSort(queue, 0, size - 1); queue = calculateWaitTime(queue, size); queue = calculateTurnAround(queue, size); averageWaitingTime(queue, size); } // Round Robin struct Process * roundRobin (struct Process queue[], int size, int quantum) { int iterator = 0; int flag = 0; int time = 0; //The time variable to manage waiting, turnaround and incoming processes quickSort(queue, 0, size - 1); //This while loop is to control the whole scheduling while (flag == 0) { //This loop is to make sure every process is only executed within the quantum for (int k = 0; k < quantum; k++) { queue[iterator].isProcessing = 1; int flagProcessingTime = 0; //Since we are not using Threading, this loop is to make sure every process is considered within each time unit for (int i = 0; i < size; i++) { //This will make sure that we only affect the process that has actually arrive at the scheduler if (queue[i].arrivingTime <= time && queue[i].processingTime > 0) { flagProcessingTime++; /** * If the process isProcessing, remove processing time. * It is not processing, just add to the wait time. */ switch (queue[i].isProcessing) { case 1: queue[i].processingTime--; queue[i].turnAroundTime++; break; case 0: queue[i].waitTime++; queue[i].turnAroundTime++; break; } } } time++; queue[iterator].isProcessing = 0; //Once all processes get to 0, this will finish execution of the scheduler if (flagProcessingTime == 0) { flag = 1; break; } } iterator++; iterator = iterator % size; } averageWaitingTime(queue, size); return 0; } //main int main() { //Array with processes with their id and processing time struct Process queue[] ={ {1, 6, 0, 0}, {2, 8, 0, 0}, {3, 7, 0, 0}, {4, 3, 0, 0}, {5, 1, 0, 0}}; //This will help determine the number of items in the array //int size = sizeof(queue)/sizeof(struct Process); int size = 5; //firstServeFirstCome(queue, size); shortestJobNext(queue, size); //roundRobin(queue, size, 1); printf("ProcessID Processing Time Wait Time Turnaround Time \n"); for (int i = 0; i < size; i++) { printf("\t %d \t\t\t %d \t\t\t\t %d \t\t\t %d \n", queue[i].processID, queue[i].processingTime, queue[i].waitTime, queue[i].turnAroundTime); } return 0; }  

1.Consider the below java code: import java.util.Scanner; import java.io.File; public class ScannerFromFile {   public static void main(String[] args) {      Scanner in = new Scanner(new File("test.in"));      // do something ...   } }               Write an exception to handle if the file “test.in” does not exist 2.Consider the below java code public class Sample {   public static void main(String[] args) {      int num = Integer.parseInt(args[0]);         ………………………………………………………..      // The execution terminates if the value of num      // is greater than 100       System.out.println("The assertion passed...");   } } Write an assertion that would terminate the program if the entered number is greater than 100 3.The Chinese animal zodiac is a repeating cycle of 12 years, with each year being represented by an animal and its related attributes. Traditionally these zodiac animals were used to date the years. In order, the 12 animals are: Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig. Each year is associated with a zodiac animal. 2019 is the year of the Pig, 2020 is the year of Rat, 2021 will be the year of Ox Use direct access method to implement the method YearToAnimalZodiac() 4.refactor the below pieces of code: public double areaOfCircle (double radius)  {     return Math.pow(radius * 3.14159, 2);  } There is 2  more that i will attch on word file