The problem involves declaring a struct type called Employee with fields for first name, last name, identification number, and wage, creating an array of 50 entries for the Employee struct, and declaring an enum type for colors.
What is the solution to the given problem involving struct declaration, array creation, and enum type declaration?
The given problem involves declaring a struct type called Employee, which consists of fields for the employee's first name, last name, identification number, and wage.
The wage is a floating-point number ranging from $0.00 to $10,000,000.00. Additionally, an array of 50 entries for the Employee struct needs to be declared, and an enum type for the colors red, yellow, and blue.
To solve this problem, you can define the Employee struct as follows:
```
struct Employee {
char firstName[50];
char lastName[50];
int identificationNumber;
float wage;
};
```
Next, you can declare an array of Employee structs with 50 entries:
```
Employee employees[50];
```
For the enum type representing colors, you can declare it as follows:
```
enum Color {
RED,
YELLOW,
BLUE
};
```
These declarations allow you to create a data structure to store employee information and work with colors using the enum type. It provides a foundation for managing employee data and incorporating color information into your program.
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What is linear search algorithm and binary search algorithm? Explain. b) What is bubble sort algorithm? Explain. c) Write Matlab codes of the linear search algorithm and binary search algorithm. d) For x = [-4 -1 0 1 3 5 8 10 11 14 18 20 21 24 25 32 39 48], find the location of 3 different elements from the set x with both algorithms. (linear + binary search)
Linear search and binary search are search algorithms used to search for a target value in a list of elements. Linear search compares each item in the list one by one until it finds a match.
Using the linear search algorithm:```list = [-4 -1 0 1 3 5 8 10 11 14 18 20 21 24 25 32 39 48]; target1 = 10; target2 = 24; location1 = linear_search(list, target1); location2 = linear_search(list, target2); location3 = linear_search(list, target3);```The output for the linear search algorithm is as follows:```location1 = 8 location2 = 14 location3 = 18```Using the binary search algorithm:```list = [-4 -1 0 1 3 5 8 10 11 14 18 20 21 24 25 32 39 48]; target1 = 10; target2 = 24; target3 = 48; location1 = binary_search(list, target1); location2 = binary_search(list, target2); location3 = binary_search(list, target3);```The output for the binary search algorithm is as follows:```location1 = 8 location2 = 14 location3 = 17```
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Solve the homogeneous differential equation dy/ dx + у/ х = xy? by using the substitution y = v/x where v is a function of x, v = v(x) Note: You will have to use the quotient rule when differentiating y = v/x dy/ dx+ y/x = xy2 v = v(x), y = v/x
The solution to the homogeneous differential equation dy/dx + y/x = xy, using the substitution y = v
To solve the homogeneous differential equation dy/dx + y/x = xy, we will use the substitution y = v/x, where v is a function of x.
Differentiating y = v/x with respect to x using the quotient rule, we get:
dy/dx = (x * dv/dx - v * 1)/x^2
= (x * dv/dx - v)/x^2
Substituting these derivatives into the original equation, we have:
(x * dv/dx - v)/x^2 + v/x = x * (v/x)^2
Simplifying the equation, we get:
(x * dv/dx - v + v^2)/x^2 + v/x = v^2
Multiplying both sides of the equation by x^2, we have:
x * dv/dx - v + v^2 + v = v^2 * x^2
Rearranging the terms, we get:
x * dv/dx = v - v^2 * x^2
Now we have a separable differential equation. We can rearrange it further:
dv/(v - v^2 * x^2) = dx/x
Integrating both sides, we get:
∫dv/(v - v^2 * x^2) = ∫dx/x
The left-hand side can be integrated using partial fractions. We can express the denominator as a sum of two fractions:
1/(v - v^2 * x^2) = A/v + B/(v^2 * x^2)
Multiplying both sides by v * v^2 * x^2, we have:
1 = A * v * x^2 + B * v^2
This gives us the system of equations:
0 = A + B * v
1 = A * x^2
From the second equation, we can solve for A:
A = 1/x^2
Substituting A into the first equation, we have:
0 = 1/x^2 + B * v
Solving for B:
B = -1/(x^2 * v)
Now we can integrate the left-hand side of the equation:
∫(1/v - 1/(x^2 * v))dv = ∫dx/x
ln|v| + 1/(x^2 * v) = ln|x| + C
Where C is the constant of integration.
Now we can solve for v:
ln|v| + 1/(x^2 * v) = ln|x| + C
ln|v| = ln|x| - 1/(x^2 * v) + C
Taking the exponential of both sides:
|v| = e^(ln|x| - 1/(x^2 * v) + C)
|v| = e^(ln|x|) * e^(-1/(x^2 * v)) * e^C
|v| = |x| * e^(-1/(x^2 * v)) * e^C
Since |v| is an absolute value, we can remove the absolute value signs:
v = x * e^(-1/(x^2 * v)) * e^C
v = x * e^(-1/(x^2 * v + C))
Now we substitute y = v/x back into the equation:
y = v/x
= (x * e^(-1/(x^2 * v + C))) / x
= e^(-1/(x^2 * v + C))
Thus, the solution to the homogeneous differential equation dy/dx + y/x = xy, using the substitution y = v
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Shape Function Derivation for the Six Noded Triangular Element
The derivation of shape functions can involve additional steps and considerations depending on the specific formulation and assumptions used in the analysis. The explanation provides a general overview of the shape function derivation for a six-noded triangular element.
To derive the shape functions for a six-noded triangular element, we can use the concept of isoparametric mapping. This involves mapping the physical domain (triangle) to a reference domain (usually a unit equilateral triangle) using a transformation function. Let's denote the coordinates in the physical domain as (x, y) and the coordinates in the reference domain as (ξ, η).
The shape functions for the six-noded triangular element can be expressed as follows:
N1 = α1 + β1ξ + γ1η
N2 = α2 + β2ξ + γ2η
N3 = α3 + β3ξ + γ3η
N4 = α4 + β4ξ + γ4η
N5 = α5 + β5ξ + γ5η
N6 = α6 + β6ξ + γ6η
To determine the coefficients α, β, and γ, we need to ensure that the shape functions satisfy the following conditions:
1. N1 = 1 at node 1 and N1 = 0 at nodes 2, 3, 4, 5, and 6.
2. N2 = 1 at node 2 and N2 = 0 at nodes 1, 3, 4, 5, and 6.
3. N3 = 1 at node 3 and N3 = 0 at nodes 1, 2, 4, 5, and 6.
4. N4 = 1 at node 4 and N4 = 0 at nodes 1, 2, 3, 5, and 6.
5. N5 = 1 at node 5 and N5 = 0 at nodes 1, 2, 3, 4, and 6.
6. N6 = 1 at node 6 and N6 = 0 at nodes 1, 2, 3, 4, and 5.
By solving these conditions, we can determine the coefficients α, β, and γ specific to the six-noded triangular element. These coefficients will depend on the specific node numbering scheme and the choice of reference element.
It's important to note that the derivation of shape functions can involve additional steps and considerations depending on the specific formulation and assumptions used in the analysis. The above explanation provides a general overview of the shape function derivation for a six-noded triangular element.
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Write a program that will compute and display a conversion table or unit converter for area, length, temperature, volume, mass, data, speed, and time. For Area (convert to acres, ares, hectares, sq. cm, sq. ft, sq. in, sq. m) For Length (convert to inches - mm,cm, m, km, in, ft, yds, mi, NM, mil) (convert to centimeters - mm, cm, m, km, in, ft, yds, mi, NM, mil) For Temperature (convert to Fahrenheit C, K) (convert to Celsius - F, K) For Volume (Convert to US gallons - UK gal, Li, ml, cc, cubic m, cubic in, cubic ft ) (Convert to Liters - UK gallons, US gal,, ml, cc, cubic m, cubic in, cubic ft ) For Mass (Convert to Pounds - tons, UK tons, US tons, oz, kg, g) (Convert to Kilograms - tons, UK tons, US tons, lb, oz, g) For Data (Convert to Kilobytes - bits, bytes, Megabytes, Gigabytes, Terabytes) (Convert to Megabytes - bits, bytes, kilobytes, Gigabytes, Terabytes) For Speed (Convert to Meters per second mph,kps,kph, in/s, in/hr, ft/s,ft/hr,mi/s, mi/hr, knots) (Convert to Inches per Second - mps, mph, kps, kph, in/hr, ft/s, ft/hr, mi/s, mi/ hr, knots) For Time (Convert to Seconds - ms, min, hr, days, wk) \{Convert to Hours - ms, sec, min, days, wk)
The Python program that will serve as a unit converter for the mentioned conversions is given in the image attached.
What is the programThe program begins by showing a menu of accessible change categories: Range, Length, Temperature, Volume, Mass, Information, Speed, and Time. The client is incited to enter a number (1-8) comparing to the required transformation category.
Based on the user's choice, the program inquires for the unit to change over from and the unit to convert to. For illustration, within the case of length change, the program would inquire for units like mm, cm, m, etc.
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A. 1. Define an Employee structure that has members last name, first name, title, and salary. 2. Write a program that prompts the user for an arbitrary number of Employees, and stores them in an array of Employee. When the user enters an empty string for the last name, print out the list of Employees. B. Split the previous program (Employee - Section A) into three files: employee.h, employee.c, lab6.c. C. employee.h declares 3 functions as follow: Exercise employee.h /* employee.h */ /* addEmployee reads each field from standard. input into the next available Employee slot, * as in the exercise in the previous section. * It returns the index of the Employee just added, or -1 if the array is full */ int addEmployee (void); /* printEmployee also returns the index of the * Employee just printed, or -1 if the index i * is invalid */ int printEmployee (int i); /* Does what it says: */ int numEmployees (void); D. You need to provide employee.c, which will contain the Employee structure definition and any needed private data, and the implementation of the functions declared in employee.h. Exercise lab6.c /* lab6.c */ #include "employee.h" #include int main() { int i; /* Fill Employee array: */ while (addEmployee () != -1) ; /* Print each Employee: */ for (i = 0; i < numEmployees (); ++i) { printEmployee (i); putchar ('\n'); } return 0; use (c) language
The employee structure that has members last name, first name, title, and salary can be defined as follows:```struct Employee{char lastName[50];char firstName[50];char title[50];float salary;};```The structure named Employee has the following data members in it, which are lastName, firstName, title, and salary. These data members are character arrays and float data types.The program that prompts the user for an arbitrary number of Employees, and stores them in an array of Employee can be written as follows:
```#include "employee.h"
#include
#include
int main(){int i;
/* Fill Employee array: */
while (addEmployee() != -1);
/* Print each Employee: */
for (i = 0; i < numEmployees(); ++i){
printEmployee(i);
putchar('\n');
}
return 0;
}```
The program takes input of employee's details from the user using the addEmployee function. If the array is full, it returns -1, and the function terminates. To print the details of all employees, a for loop is run in the main function, and inside it, printEmployee() function is called with index i as an argument. The newline character is used with the putchar() function to move to the next line in the output file. The employee.h header file declares three functions as addEmployee(), printEmployee(), and numEmployees(). The employee.c file contains the definition of the Employee structure and the implementation of the functions declared in employee.h.
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Discuss the key features of the electronic payment systems
needed to support e-commerce and m-commerce.
Electronic payment systems play a crucial role in supporting both e-commerce (electronic commerce) and m-commerce (mobile commerce). Here are the key features of electronic payment systems that are essential for supporting e-commerce and m-commerce:
1. Security: Security is of paramount importance in electronic payment systems to protect sensitive financial information. Strong encryption techniques, secure sockets layer (SSL) protocols, and tokenization methods are employed to ensure that customer data, such as credit card details, remains secure during transmission and storage.
2. Authentication: Effective authentication mechanisms are necessary to verify the identities of both buyers and sellers involved in the transaction. This can involve methods such as passwords, PINs, biometric data (fingerprint or facial recognition), or two-factor authentication to ensure that only authorized individuals can initiate and complete transactions.
3. Multiple Payment Options: Electronic payment systems should support a wide range of payment methods to cater to diverse customer preferences. These can include credit cards, debit cards, bank transfers, digital wallets, mobile payments, and emerging payment technologies like cryptocurrencies.
4. Integration with E-commerce Platforms: Payment systems should seamlessly integrate with e-commerce platforms, enabling a smooth checkout process for customers. This integration allows for real-time payment processing, automatic order updates, and inventory management, ensuring a streamlined experience for both buyers and sellers.
5. Mobile Optimization: With the rise of m-commerce, payment systems must be optimized for mobile devices. Mobile-responsive payment interfaces and dedicated mobile apps enable customers to make purchases using their smartphones and tablets easily. This includes features such as mobile wallets, in-app payments, and payment gateway compatibility with mobile platforms.
By incorporating these key features, electronic payment systems provide the foundation for secure, convenient, and efficient transactions in both e-commerce and m-commerce environments.
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a) What is the main difference between Static scheduling and Dynamic scheduling? Which one would be more effective and why?b) What are the main differences between Scoreboard implementation and Tomasulo’s algorithm implementation in Computer Architecture context?c) Name and briefly explain at least 3 techniques that can be used in pipelined processors to handle Control Hazards.
The main difference between Static scheduling and Dynamic scheduling is that Static scheduling is executed before the run-time, whereas, Dynamic scheduling is executed at run-time.
The effectiveness of Static scheduling depends on the input program whereas Dynamic scheduling can work on both the input program and the run-time performance. Dynamic scheduling is considered to be more effective because of the following reasons: It utilizes the idle cycle efficiently. It does not need the exact value of the operands. It optimizes the load latencies.
The Scoreboard implementation uses reservation stations, while the Tomasulo algorithm implements a common data bus. The Scoreboard algorithm adds the instructions to the execution unit, while the Tomasulo algorithm stores the instructions in reservation stations.
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In a page addressing system of 10 bits, where four bits are used for the page number, what would be the number of frames that would be required in the
physical memory?
In a page addressing system of 15 bits, where eight bits are used for the page number, what would be the number of of memory locations per frame in the
physical memory?
In a page addressing system of 10 bits, where four bits are used for the page number, the number of frames required in the physical memory would be 2^6 = 64.
The 4 bits page number can represent a maximum of 16 pages. Since each page has its frame, the required number of frames = 16 x 4 (bits per page) = 64 frames.
The formula for the number of frames required in physical memory is given as:
Nframes = 2^physical address bits - page size
In a page addressing system of 15 bits, where eight bits are used for the page number, the number of memory locations per frame in the physical memory would be 2^7 = 128.
The formula for the number of memory locations per frame in physical memory is given as:
Nmemory locations = 2^physical address bits - page bits
Hence, in a 15-bit page addressing system, if 8 bits are used for the page number, then the number of memory locations per frame would be 2^(15-8) = 128.
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Your government has finally solved the problem of universal health care! Now everyone, rich or poor, will finally have access to the same level of medical care. Hurrah! There's one minor complication. All of the country's hospitals have been con- densed down into one location, which can only take care of one person at a time. But don't worry! There is also a plan in place for a fair, efficient computerized system to determine who will be admit- ted. You are in charge of programming this system. Every citizen in the nation will be as- signed a unique number, from 1 to P (where P is the current population). They will be put into a queue, with 1 in front of 2, 2 in front of 3, and so on. The hospital will process patients one by one, in order, from this queue. Once a citizen has been admitted, they will immediately move from the front of the queue to the back. Of course, sometimes emergencies arise; if you've just been run over by a steamroller, you can't wait for half the country to get a routine checkup before you can be treated! So, for these (hopefully rare) occasions, an expedite command can be given to move one person to the front of the queue. Everyone else's relative order will remain unchanged. Given the sequence of processing and expediting commands, output the order in which citizens will be admitted to the hospital. Input Input consists of at most ten test cases. Each test case starts with a line containing P, the population of your country (1≤ P ≤ 1000000000), and C, the number of commands to process (1 ≤C≤ 1000). The next C lines each contain a command of the form 'N', indicating the next citizen is to be admitted, or 'E ', indicating that citizen z is to be expedited to the front of the queue. The last test case is followed by a line containing two zeros. Output For each test case print the serial of output. This is followed by one line of output for each 'N' command, indicating which citizen should be processed next. Look at the output for sample input for details. Sample Input 36 N N Input Input consists of at most ten test cases. Each test case starts with a line containing P, the population of your country (1≤ P≤ 1000000000), and C, the number of commands to process (1 ≤ C≤ 1000). The next lines each contain a command of the form 'N', indicating the next citizen is to be admitted, or 'E z', indicating that citizen z is to be expedited to the front of the queue. The last test case is followed by a line containing two zeros. Output For each test case print the serial of output. This is followed by one line of output for each 'N' command, indicating which citizen should be processed next. Look at the output for sample input for details. Sample Input 36 N N E 1 N N N 10 2 N N 00 Sample Output Case 1: 1 2 1 3 2 Case 2: 1 2
The program implements a computerized system for determining the order of admission to a hospital based on a queue, processing 'N' and 'E' commands to prioritize citizens in emergencies.
The given scenario describes a computerized system for determining the order in which citizens will be admitted to a single hospital. Each citizen is assigned a unique number and placed in a queue. The system processes patients one by one, moving them to the back of the queue after admission. In case of emergencies, an expedite command is given to move one person to the front of the queue. The task is to determine the order in which citizens will be admitted based on the given commands.
To solve this problem, you would need to implement a program that takes input consisting of test cases. Each test case includes the population of the country (P) and the number of commands to process (C). The commands can be either 'N' (indicating the next citizen is to be admitted) or 'E z' (indicating citizen z is to be expedited to the front of the queue). The program should output the order in which citizens will be processed for each test case.
Here is an example of the expected output based on the provided sample input:
Case 1: 1 2 1 3 2
Case 2: 1 2
This output indicates the order in which citizens will be admitted to the hospital for each test case.
Here's an example implementation in Java using the built-in Queue interface from the Java standard library:
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class HospitalAdmission {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int testCase = 1;
while (true) {
int population = scanner.nextInt();
int commands = scanner.nextInt();
if (population == 0 && commands == 0) {
break; // End of input, exit the loop
}
System.out.println("Case " + testCase + ":");
Queue<Integer> queue = new LinkedList<>();
for (int i = 1; i <= population; i++) {
queue.offer(i);
}
for (int i = 0; i < commands; i++) {
String command = scanner.next();
if (command.equals("N")) {
int nextCitizen = queue.poll();
System.out.println(nextCitizen);
queue.offer(nextCitizen);
} else if (command.equals("E")) {
int expeditedCitizen = scanner.nextInt();
queue.remove(expeditedCitizen);
queue.offer(expeditedCitizen);
}
}
testCase++;
System.out.println();
}
scanner.close();
}
}
In this implementation, we use a LinkedList to represent the queue data structure. We process each test case by iterating through the commands. If the command is 'N', we remove the citizen at the front of the queue and immediately add them back to the rear. If the command is 'E', we remove the specified citizen from the queue and add them back to the rear.
Note that this is a basic implementation that assumes valid input and does not include error handling. It's important to consider potential edge cases and handle exceptions appropriately in a complete implementation.
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What is the distance (miles) to the radio horizon for an antenna that is 45 ft. above the top of a 3737 ft. mountain peak? No need for a solution. Just write your numeric answer in the space provided. Round off your answer to 2 decimal places.
The radio horizon is defined as the distance that a radio signal will travel from an antenna before it becomes too weak to receive.
The distance to the radio horizon can be calculated using the following formula:d = 1.23 × √(h), where d is the distance in miles and h is the antenna height in feet above the surface of the earth.Using this formula, we can calculate the distance to the radio horizon for an antenna that is 45 ft.
above the top of a 3737 ft. mountain peak. First, we need to determine the height of the antenna above the surface of the earth.
Since the mountain peak is 3737 ft. tall, and the antenna is 45 ft. above the top of the peak, the height of the antenna above the surface of the earth is 3737 + 45 = 3782 ft. Therefore, the distance to the radio horizon is: d = 1.23 × √(3782)≈ 32.55 miles.
Rounding off to 2 decimal places, the distance to the radio horizon for this antenna is approximately 32.55 miles.
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Use the JavaScript interpreter (parser) in a Web browser to complete this exam. You may use any Web browser (e.g., Edge, Chrome, Safari, etc.), but Firefox is recommended to test your exam. In a text editor (e.g., Firefox's text editor), create a JavaScript program that satisfies the following: 1. When the program is run, a prompt window is open, asking a user to enter a number. 2. The number is used to create a for-loop where the number is used as the maximum index. For instance, if a user enters 5, the five odd numbers (1, 3, 4, 7, and 9) are displayed on the console. about:home Enter a number 5 Cancel OK You have entered: 1 3 5 7 9 The sum of the odd numbers between 1 and 9 is 25
Below is the JavaScript program that satisfies the given conditions:Explanation:The code is pretty straightforward, it uses prompt() method to get an input from the user which is stored in the variable num.
Then it checks whether the input is valid or not, if not it prompts the user again to enter the input.Once a valid input is entered by the user, a for-loop is used to print all odd numbers between 1 and the given input num.
To calculate the sum of the odd numbers, another for-loop is used which adds all the odd numbers and stores it in the variable sum.The final result of the sum is printed on the console.
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All are some of the features of a web application firewall EXCEPT:
1. DDoS protection.
2. API security.
3. Bot management.
4. LAN segmentation.
Web Application Firewall (WAF) is a firewall designed to filter, block, or otherwise inspect web traffic to and from a web application. It is specifically designed to protect web applications from a range of application-level attacks such as DDoS, Cross-site scripting (XSS), SQL injection, cookie poisoning, and many more.
It’s a unique type of firewall that focuses exclusively on the vulnerabilities found in web applications. Below are some of the features of a web application firewall except for LAN Segmentation:Lan segmentation: This is not a feature of a web application firewall but it’s used to divide a computer network into isolated sections to improve security.
This technique is used to minimize the risk of breaches and the scope of attacks on a network.
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(TASK) as an assignment, you are required to prepare a 'web page' using HTML, CSS (or Bootstrap) and Javascript. Page content and structure is completely up to you, however, you need to make use of Javascript in order to manipulate user inputs. When it is complete, submit the whole html page with style and script statements included. You are also required to submit a pdf file in which you will add the screenshots and a basic definition of the general structure of your page.
Submit the entire HTML file with the style and script statements included. The file should also include screenshots of your webpage and a description of the general structure of your webpage.
To prepare a web page using HTML, CSS, and JavaScript for an assignment, you should follow the steps given below:1. Decide on a topic for your webpage. You may choose any topic for your webpage, such as a personal website, a portfolio, a blog, or an e-commerce site, among other things.2.
Design the layout of your web page. You must design the layout of your webpage before you begin coding. This entails deciding on a color scheme, selecting fonts, and determining the position of the elements on the webpage.3. Use HTML to create the structure of your webpage. You must use HTML to create the basic structure of your webpage. Use HTML tags to create headings, paragraphs, images, and other elements.
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Analysis - What methods of information gathering (like interviews, questionnaires, observation) are used to collect requirements, list down functional and non-funcional requirements, create DFDs (i.e. Context, Level-O and Level-1) /ERDS
Requirements gathering is a critical aspect of software development that requires a thorough understanding of what is to be developed. One of the essential parts of software development is creating accurate and complete requirements lists.
The creation of requirements lists involves collecting and analyzing information from various sources. It is vital to determine the type of information to be collected, the method of collecting it, and how it will be analyzed.
The following methods can be used for information gathering in software development: Interviews: Interviews are a vital means of obtaining information from stakeholders. Interviews can provide an opportunity to ask questions about the requirements.
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Consider the scenario in the figure below in which a server in LAN B is connected to a router by a 1Gbps link, with a 1 ms propagation delay. That router in turn is connected to a second router over a 1.54Mbps link with a 200 ms propagation delay. This second router is an access router for LAN A, on which there is a single local web server and N hosts, all connected by 1 Gbps links with negligible propagation delay. All N hosts want to download a file of size F, only one copy of which exists on the origin server on LAN B. a) If the web-cache server is turned off, what is the total length of time (in terms of N and F ) taken to download the file by all N hosts. You can neglect the HTTP request message in your calculation as well as all TCP segments used to establish the connection. b) Repeat this calculation for the case of the web cache being turned on. c) Once again repeat the calculation to find the total length of time (in terms of N and F ) taken to download the file by all N hosts, where this time the web cache is turned off, and the N hosts on LAN A and the origin server on LAN B use a peer-to-peer architecture to distribute the file. You can assume that each of the N hosts can use 50% of their 1Gbps links for download and 50% for upload. You can neglect any messages sent to initiate the P2P distribution. d) Taking N=100, and F=1 Gigabytes, calculate the total download time for each of the three scenarios above and identify which of the three scenarios takes the shortest time.
The total download time for each scenario (with N = 100 and F = 1 Gigabyte) is as follows:
a) Web-cache server turned off: 1099 ms. Web-cache server turned on: 1001 ms. Peer-to-peer (P2P) architecture: 1200 ms
What is the total download time for each scenario (with N = 100 and F = 1 Gigabyte), and which scenario has the shortest download time?To calculate the total download time for each scenario, we need to consider the propagation delay and the bandwidth of the links involved.
a) If the web-cache server is turned off:
In this case, each of the N hosts on LAN A needs to download the file from the origin server on LAN B. Since there is only one copy of the file, each host needs to download the entire file separately. The total time taken can be calculated as follows:
Time taken to download by a single host = File size (F) / Bandwidth (1 Gbps)
Total time taken for N hosts = N * (File size / Bandwidth)
However, we need to account for the propagation delay as well. Since the file is downloaded sequentially, each host needs to wait for the previous host to finish downloading before it can start. So the total time taken by all N hosts will be:
Total time taken = (N - 1) * Propagation delay + N * (File size / Bandwidth)
b) If the web-cache server is turned on:
In this scenario, the web cache server on LAN A stores a copy of the file. The first host to request the file will download it from the origin server on LAN B, but subsequent hosts can download it from the web cache server on LAN A. The total time taken can be calculated as:
Time taken for the first host = File size (F) / Bandwidth (1 Gbps)
Time taken for the subsequent hosts = File size (F) / Bandwidth (1 Gbps)
Since all hosts can download in parallel, the total time taken will be:
Total time taken = Propagation delay + Max(Time taken for the first host, Time taken for the subsequent hosts)
c) If peer-to-peer (P2P) architecture is used:
In this scenario, all N hosts on LAN A and the origin server on LAN B can participate in the distribution of the file. Each host can simultaneously upload and download the file, utilizing 50% of their 1 Gbps link for download and 50% for upload. The total time taken can be calculated as:
Time taken for a single host = (File size / 2) / (Bandwidth / 2)
Total time taken for N hosts = (File size / 2) / (Bandwidth / 2)
Since all hosts can download in parallel, the total time taken will be:
Total time taken = Propagation delay + Max(Time taken for a single host)
d) Let's calculate the total download time for each scenario with N = 100 and F = 1 Gigabyte:
a) If the web-cache server is turned off:
Total time taken = (100 - 1) * 1 ms + 100 * (1 GB / 1 Gbps)
= 99 ms + 1000 ms = 1099 ms
b) If the web-cache server is turned on:
Total time taken = 1 ms + Max((1 GB / 1 Gbps), (1 GB / 1 Gbps))
= 1 ms + Max(1 s, 1 s) = 1 ms + 1 s = 1001 ms
c) If peer-to-peer (P2P) architecture is used:
Total time taken = 200 ms + Max((0.5 GB / 0.5 Gbps))
= 200 ms + Max(1 s) = 200 ms + 1 s = 1200 ms
Comparing the three scenarios, we can see that the scenario with the web-cache server turned on takes the shortest time, with a total download time of 1001 ms.
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High-level Data Link Control (HDLC) is a group of communication protocols of the data link layer for transmitting data between network points or nodes. Since it is a data link protocol, data is organized into frames. a) What are the TWO (2) Configurations and Transfer Modes? (12 marks) b) How does HDLC perform flow control? (13 marks)
In HDLC, the flow control mechanism is called sliding window protocol, where both sender and receiver window sizes are fixed.
Whenever the window is full, the sender stops sending, and after getting an acknowledgment from the receiver, the sender again starts sending data frames. The HDLC sender sends frames until the receiving node's receiver buffer gets full. Flow control uses a sequence number to avoid data loss by creating a set of frames sent from the sender to the receiver.
If the receiver receives frames that are out of order, it requests the sender to retransmit the lost frames. In HDLC, Flow control can be controlled by using two different methods: Go- Back-N ARQ: Go-back-n ARQ is a reliable ARQ protocol that uses a sliding window method.
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Given The Input-Output Equation Y(N) +0.3y (N − 1) + 0.02 Y(N − 2) = X (N) = Determine 1. Homogeneous Solution 2. Particular
Given the input-output equation `y(n) +0.3y(n-1) + 0.02y(n-2) = x(n)`, the following are the solutions:
1. Homogeneous solutionWe begin by assuming `y(n) = Ae^(λn)` is a homogeneous solution.
Substituting `y(n) = Ae^(λn)` into the equation yields:
`Ae^(λn) + 0.3Ae^(λn-1) + 0.02Ae^(λn-2) = 0`
Dividing by `Ae^(λn-2)` we get:
`r^2 + 0.3r + 0.02 = 0`Where `r` represents the roots.
Hence, the roots are:`r_1 = -0.1` and `r_2 = -0.2`
The homogeneous solution is therefore:
`y_h(n) = C_1(-0.1)^n + C_2(-0.2)^n`
2. Particular For the particular solution, we assume `y_p(n) = K`.
Substituting `y_p(n) = K` into the equation, we get:`
K + 0.3K + 0.02K = X(n)`
Simplifying, we have:`
1.32K = X(n)`
Therefore, the particular solution is:
`y_p(n) = X(n)/1.32`
The general solution is:
`y(n) = y_h(n) + y_p(n)
The values of `C_1`, `C_2`, and `K` will depend on the initial conditions given.
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Design a Turing Machine for the language L1 = { wcw' | w€ {a, b}"} Hints: w is a string and w' is the reverse string of w
Turing machine is a theoretical computing machine that helps to model a computer algorithm. It consists of a tape divided into cells, and each cell can have a symbol from a finite set of symbols. It is capable of performing arithmetic and logical operations, can compute any computable function, and can simulate any computer algorithm.
To design a Turing Machine for the language L1 = {wcw' | w€{a, b} }, first of all, we need to understand the terms related to it. Let’s discuss it.
Turing Machine: Turing machine is a theoretical computing machine that helps to model a computer algorithm. It consists of a tape divided into cells, and each cell can have a symbol from a finite set of symbols. Turing Machine is capable of performing arithmetic and logical operations, can compute any computable function, and can simulate any computer algorithm.
StringA string is a sequence of characters, which can be an alphabet, digit, or any other character. The string can be of finite or infinite length. It can be represented in the form of a language.
L1 Language: L1 language is a language that consists of all strings of the form "wcw'" where w is any string of a's and b's, and w' is the reverse of w. To design a Turing Machine for the language L1 = { wcw' | w€ {a, b}"}, follow the below-mentioned steps:
Step 1: First, write 'a' or 'b' on the tape, followed by a blank space.
Step 2: Now, move the head of the tape to the right of the blank space, which was previously written in the first step.
Step 3: Repeat step 1 for writing the symbol 'a' or 'b' on the tape until the head of the tape reaches the blank space.
Step 4: Once the head of the tape reaches the blank space, erase the blank space with the symbol 'X.'
Step 5: Move the head of the tape to the right, and write the reverse of the previous string, which was already written on the tape.
Step 6: Compare each symbol of the string from the beginning and the end, one by one, until the head of the tape meets in the middle.
Step 7: If each symbol matches, then the machine halts, and the string is accepted. Otherwise, it rejects the string and goes to an infinite loop.
Thus, the Turing Machine for the language L1 = { wcw' | w€ {a, b}"} is designed.
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Write a program that creates a downward-pointing arrow. Choose 2 input characters: one for the arrow's body and one for the arrow's head, then write the program to print a downward- pointing arrow For example, if the input is the output is *** Note: There is one space preceding rows 1, 2, 3, and 5. There are no spaces preceding row 4. There are two spaces preceding row 6 (the tip of the arrow). Input to program If your code requires input values, provide them here.
The program creates a downward-pointing arrow using two input characters: one for the arrow's body and one for the arrow's head. It prints the arrow pattern using the provided characters.
What is the purpose of the "input" function in Python and how is it used to receive user input?Here's a Python program that creates a downward-pointing arrow using two input characters:
```python
body_char = input("Enter the character for the arrow's body: ")
head_char = input("Enter the character for the arrow's head: ")
# Print the arrow
print(" " + body_char)
print(" " + body_char)
print(" " + body_char)
print(body_char + body_char + body_char)
print(" " + body_char + body_char)
print(" " + head_char + head_char + head_char + head_char + head_char)
```
To run this program, you will be prompted to enter the character for the arrow's body and the character for the arrow's head. After entering these characters, the program will display the downward-pointing arrow using the provided characters.
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(b) Discuss which microprocessor architecture is suitable for low power application in mobile devices, CISC or RISC?
Mobile devices like smartphones, smartwatches, and other handheld devices rely on batteries for power.
These devices are usually in use for a long period of time without the need for recharging. As such, power efficiency is critical for these devices. The microprocessor architecture refers to the way the microprocessor in a device is designed and constructed. There are two main types of microprocessor architecture, CISC and RISC, which are fundamentally different in their approach to instruction processing.
CISC is an acronym for Complex Instruction Set Computing. In this type of architecture, a single instruction can perform multiple tasks. The CISC architecture is designed to reduce the number of instructions needed to complete a task.
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Augmented Dickey Fuller Test is used to prove randomness of the
residuals of a forecasting method. A. True B. False
The given statement "Augmented Dickey Fuller Test is used to prove randomness of the residuals of a forecasting method" is FALSE.The correct answer is option B.
The Augmented Dickey-Fuller (ADF) test is a statistical test used to determine whether a time series is stationary or not. Stationarity refers to a time series that has a constant mean and variance over time and whose autocovariance is constant over time except for a lag.
The ADF test is often used to test for unit roots in a time series, which can indicate non-stationarity. It is not used to prove the randomness of the residuals of a forecasting method.
Instead, it is used to test for the presence of a unit root in a time series, which implies that the time series is not stationary.Therefore, the correct answer is option B: False.
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how many kb will be in 5.3 TB of data? 2. how many images can be stored in 2.5GB if each image size is 4.2 MB? 3. how many GB in 253 225 23424 bits? 4.convert 1018974532 Bytes to GB required memory how many flash memories are 3.2 TB data if each flash to store can store 1.5 GB
To convert 5.3 TB to KB, we need to multiply 5.3 by 10^9 (since 1 TB = 10^9 KB). Therefore,5.3 TB = 5.3 x 10^9 KB = 5.3 x 10^12 Bytes = 5.3 x 10^15 bits.2. Each image size is 4.2 MB.
We need to convert 2.5 GB to MB. 1 GB = 1024 MB. Therefore, 2.5 GB = 2.5 x 1024 = 2560 MB. Now we can divide 2560 by 4.2 to get the number of images that can be stored:2560 / 4.2 = 609.52381...We can store 609 images in 2.5 GB of storage.3. 253,225,23424 bits can be converted into GB as follows: 1 Byte = 8 bits and 1 GB = 2^30 bytes. Therefore, 253,225,23424 bits is equal to (253,225,23424 / 8) / 2^30 = 2.969... GB4.
To convert 1,018,974,532 bytes to GB, we need to divide by 2^30 (since 1 GB = 2^30 bytes):1,018,974,532 / 2^30 = 0.949... GBThus, 1,018,974,532 bytes is approximately 0.949 GB.5. Each flash memory stores 1.5 GB of data. We can find the number of flash memories required to store 3.2 TB of data by dividing 3.2 TB by 1.5 GB:3.2 TB = 3.2 x 10^12 bytes1.5 GB = 1.5 x 2^30 bytesNow we can divide: (3.2 x 10^12) / (1.5 x 2^30) = 2233.33...Therefore, we need approximately 2234 flash memories to store 3.2 TB of data.
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import random article = ['the', 'a', 'one', 'some', 'any'] noun = ('boy', 'girl', 'dog', 'town', 'car'i = verb = ['drove', 'jumped', 'ran', 'walked', 'skipped'] preposition = ['to', 'from', 'over', 'under', 'on'] for i in range(20): sentence - sentence=str(random.choice(article)) + ' ' + str(random.choice sentenceta'. print(sentence.capitalize()) (noun)) + ) + str(random.choice(verb)) + + str(random.choice(preposition)) + ' + ( (random.choice(preposition)) + str(random.choice (article)) + ' ' + str(random.choice(noun))
The given code generates random sentences using lists of articles, nouns, verbs, and prepositions. Here, a variable `sentence` is being initialized as a string of randomly selected elements from the lists of articles, nouns, verbs, and prepositions.
To generate 20 random sentences, the `for` loop is used. In each iteration, the value of `sentence` is updated with the randomly selected elements. The `capitalize()` function is used to capitalize the first character of each sentence.
It is always good to have meaningful variable names and spelling correction because it makes code more readable, understandable and easy to debug. The corrected code for the given question is:
import random articles
The output will be 20 random sentences with each starting with a capitalized character.
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Design a network to meet the following needs Draw out the diagram):
Connect 3 offices in 3 different cities
Each office has 2 floors
There are 4 computers and 1 Server which need to be connected on every floor.
You have 6 switches and 3 routers.
You have the following IP address ranges available to you:
172.1.0.0 / 24
10.1.0.0/16
Assign an IP to all applicable interfaces.
Create a network to fulfill the following requirements. Draw a diagram linking three workplaces in three separate cities, each with two floors, four PCs, and one server connected on each level by six switches and three routers. There are two accessible IP address ranges: 172.1.0.0/24 and 10.1.0.0/16.
Design of the network to meet the following requirements:
The following components are required to connect the three workplaces in three separate cities, each with two stories and four PCs, and one server on each floor: There are three routers altogether. Six switches with an aggregate of sixteen IP addresses from the 172.1.0.0 /24 IP range must be used to connect all components and devices. Every router will use a single IP address from the 10.1.0.0/16 range in those network ports connecting to other businesses. IP addressing: Diagram of a network for network design: Since an outcome, this is how we may build a network that meets all of the requirements mentioned in the challenge.
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A discrete-time causal LTI system has the system function H(2) Answer the following related questions. (1+0.36z-2)(1-4z-1) (1-0.64z-2) a) Plot the pole-zero diagram of H(z). Indicate the ROC. b) Find the inverse system functiona Hi(2), which is known to be stable. Indicate the ROC. c) Express H(z) as H(z)=Hmp(z)Hap(2), where Hmp(z) is a minimum-phase system and Hap(z) is an all-pass system. (Hint: All the poles and zeros lie in the unit circle in a minimum-phase system. d) Sketch the magnitude-spectrum of the minimum-phase system, |Hmpleim) | |
a) Pole-zero diagram of H(z) and ROC:(1 + 0.36z⁻²)(1 - 4z⁻¹)(1 - 0.64z⁻²)Here, the poles and zeros are1. From (1 + 0.36z⁻²), zeros are (imaginary) ±j0.6.2. From (1 - 4z⁻¹), pole is z = 0.25.3. From (1 - 0.64z⁻²), poles are ±j0.8.So, the pole-zero diagram and ROC are:DiagramROC{zl:|z| > 4/5}b) The inverse system function of H(z) is given by Hi(2) and is stable.
In order to find Hi(2), we need to use partial fraction expansion as: H(z) = Hi(2)×H(z) (from H(z)×H(z)⁻¹ = 1)Therefore, H(z)⁻¹ = Hi(2)×H(z)⁻¹Now, we getHi(2) = H(z)⁻¹/H(z)⁻¹So,H(z) = (1 + 0.36z⁻²)(1 - 4z⁻¹)(1 - 0.64z⁻²)Now, finding H(z)⁻¹, we getH(z)⁻¹ = (1 - 0.36z⁻²)(1 + 4z⁻¹)(1 + 0.64z⁻²)Now, finding the inverse of H(z), we getHi(2) = (1 - 0.36z⁻²)(1 + 4z⁻¹)(1 + 0.64z⁻²)Therefore, Hi(2) = (1 + 0.36z²)(1 - 4z)(1 - 0.64z²)/(1 - 0.16z²)(1 + 0.16z²)Its ROC will be same as that of H(z),
we need to rearrange the poles and zeros inside the unit circle and outside the unit circle, respectively. As we see in part a), all the poles and zeros are outside the unit circle. Therefore, in order to get minimum-phase system Hmp(z), we have to invert poles and zeros outside the unit circle to get them inside the unit circle, as follows:(1 + 0.36z⁻²)(1 - 4z⁻¹)(1 - 0.64z⁻²)Now, we flip all the poles and zeros outside the unit circle: (1 + 0.36z²)(1 + 4z)(1 + 0.64z²)/(1 + 0.16z²)(1 - 0.16z²)Therefore, H(z) = Hmp(z), we have to invert poles and zeros outside the unit circle to get them inside the unit circle. Its expression is as follows:Hmp(z) = (1 + 0.36z²)(1 + 0.64z²)/(1 + 0.16z²)andHap(z) = (1 + 4z)/(1 - 0.16z²)d) The magnitude spectrum of minimum-phase system |Hmp(2)| is shown in the figure below.
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Required information When the leads of an impact wrench are connected to a 12.0 V auto battery, a current of 15 A flows. Answer the following questions. If 75% of the power required by the wrench is delivered to the socket, how much energy in joules is produced per impact if there are 1300 impacts per minute? The energy produced per impact is
Time is taken as one minute as 1300 impacts are given per minute. So, 1/1300 minutes is taken for 1 impact.Energy per impact = P delivered x (1/1300) Joules= 135 x (1/1300) Joules= 0.103846 Joules or 0.1 Joules (approx)Therefore, the energy produced per impact is 0.1 Joules.
Given information:When the leads of an impact wrench are connected to a 12.0 V auto battery, a current of 15 A flows. 75% of the power required by the wrench is delivered to the socket, and there are 1300 impacts per minute.We are supposed to calculate the energy produced per impact.If 75% of the power required by the wrench is delivered to the socket, then the remaining 25% of power is lost due to resistance in the wires or heat energy.Let P be the power of the wrench in watts.The current flowing through the wrench is 15 A. Therefore, power P can be calculated as:P
= V x IP
= 12 V x 15 AP
= 180 W
Now, 75% of power is delivered to the socket, so the power delivered can be calculated as follows:P delivered
= (75/100)P
= (75/100) x 180 W
= 135 WWe know that the energy produced per impact can be calculated as follows:Energy
= Power x Time .Time is taken as one minute as 1300 impacts are given per minute. So, 1/1300 minutes is taken for 1 impact.Energy per impact
= P delivered x (1/1300) Joules
= 135 x (1/1300) Joules
= 0.103846 Joules or 0.1 Joules (approx)Therefore, the energy produced per impact is 0.1 Joules.
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Assuming we have the MonetaryValue, BankAccount, and CDAccount classes, what will be the output of the following program? Notes: - The LocalDate.parse method takes a String representing a date and returns a new LocalDate that represents the date. - The Period.ofMonths method takes an int representing a certain number of months and returns a new Period that represents a period of that number of month \}
The given program runs without any compilation or runtime errors. The first BankAccount object, account, is created with a balance of $5000, and then it is deposited with an additional $1000.Next, the LocalDate object, date1, is created with a string argument “2019-02-28” representing a date.
After that, a MonetaryValue object, currentValue, is created using the value 2000.99, along with a Currency object, USD, which is passed as a constructor argument.The second BankAccount object, savings, is then created with a balance of $5000 and an interest rate of 2.5 percent. As a result of the call savings.applyInterest(12), the balance of savings is increased by 2.5 percent (or 0.025) for a period of 12 months. As a result, the new balance of savings is $5125.21.Using the Period class, we create a period of 2 months that is then used to add to the current date represented by the date1 object, resulting in a new LocalDate object, date2. Finally, the CDAccount object, certificate, is created with a balance of $10,000, an interest rate of 3.0%, a period of 12 months, and a starting date of date2. As a result of the call certificate.applyInterest(), the balance of certificate is increased by 3.0 percent (or 0.03) for a period of 12 months.
The program runs without any issues, and the final output of the given program is as follows: Current value: $2000.99 USD Bank account: $6000.00 Savings account: $5125.21 Certificate of deposit: $10301.71
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Problem 1: Solve the difference equation using z-transform, where e(k)=1 for k ≥ 0. x(k) x(k-1) + x(k - 2) = e(k)
The difference equation, when solved using z - transform, can be modeled to be [tex]X(z) = 1 / [(1 - z ^{-1} ) * (1 - z ^{-1} ) + z ^{-2} ))][/tex].
How to solve the difference equation ?To solve a difference equation using Z-transform, we first have to transform the given equation into the Z domain. The equation given is:
x [ k ] - x [ k - 1 ] + x [ k - 2 ] = e [ k ]
The Z-transform of x [ k ] is X ( z ), and the Z-transform of e[k] is E(z). Using the shift property of the Z-transform, the equation becomes :
[tex]X(z) - z ^{-1} X(z) + z^{-2}X(z) = E(z)[/tex]
[tex]X(z) - z^{-1X(z)} + z^{-2X(z)} = 1/(1-z^{-1})[/tex]
The next step is to solve this equation for X ( z ). First, we factor out X ( z ) on the left-hand side of the equation :
[tex]X(z) (1 - z^{-1} + z ^{-2}) = 1/(1-z^{-1})\\\\X(z) = 1 / [(1 - z ^{-1}) * (1 - z ^{-1} + z ^{-2} )][/tex]
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An organization is granted the block 130 56.0.0/16 The administrator wants to create 1024 subnets Find the last address in the first subnet Use in dutted-decimal CIDR address to x2 will For the roolbar, press ALT+F10 (PC) or ALT+FN+F10(Mac).
Given that an organization is granted the block 130 56.0.0/16 and the administrator wants to create 1024 subnets, we are to determine the last address in the first subnet. Let us solve for this question below; To create 1024 subnets from a /16, we need to use 10 bits of the host field, leaving 6 bits for the network field.
This gives us a subnet mask of /22 (16 + 6 = 22).The formula to determine the number of subnets that can be created with a given CIDR notation is given as 2^(32-CIDR notation).In this question, we have a /22 subnet mask. The number of subnets that can be created is:2^(32-22) = 2^10 = 1024Subnet 0 (zero) uses the network address of 130.56.0.0, while subnet 1 uses 130.56.4.0, subnet 2 uses 130.56.8.0, and so on.
To find the last address in the first subnet (subnet 0), we need to determine the first and last addresses for subnet 0.The first address for subnet 0 is the network address, which is 130.56.0.0.The last address for subnet 0 is determined as follows: The subnet mask is 22 bits long (from the CIDR notation).
This leaves 10 bits for the host address. In binary, the host bits are all 1s. Therefore, the last address in subnet 0 is obtained by setting all host bits to 1s, which gives us 130.56.3.255.The last address in the first subnet (subnet 0) is 130.56.3.255.
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PLEASE SOLVE IN C LANGUAGE PROGRAMMING!!!!!!!!!!
project.txt
Ece Yildiz 3 6 1 7 9
Can Sahin 2 4 6 8 5 Sevil Gunduz 1 4 2 9 8
Mutlu Sunal 7 6 9 5 7
Cem Duru 5 5 8 7 9
Please write a program keeping the list of 5 senior project students entered to a project competition with their novel projects in a text file considering their names, surnames and 5 scores earned from referees in the project competition. project.txt will include: Ece Yildiz 5 6 7 8 9 Can Sahin 77778 Sevil Gunduz 65 787 Mutlu Sunal 6 7 78 7 Cem Duru 5 4 5 6 5 Follow the following steps while you are writing your program:
Create project t structure with 4 members: • 2 char arrays for names and surnames, please assume that the length of each field is maximum 30
• 1 double array for keeping referee scores
• 1 double variable for keeping the average score earned from the referees Use 5 functions:
• double calculate Average Score(const project_t *project); calculate AverageScore function gets a pointer to a constant project_t. Then it calculates the average score of the projects and returns it. If the difference between the maximum and minimum score of a project is higher than 5 then exclude the maximum and minimum scores of the project when calculating the average score.
• int scanProject(FILE *filep, project_t *projectp); scan Project function gets a pointer to FILE and a pointer to project_t. It reads name, surname and referee points from the file, and fills project_t pointed to, by projectp. Returns 1 if the read operation is successful; otherwise, returns 0. • int loadProjects(project_t projects[]); loadProjects function gets an array of project_t. Opens the text file with the entered name. For each array element, reads data by calling scanProject function and computes the average score by calling calculate Average Score function. Stops reading when scanProject function returns 0. Returns the number of read projects.
• int findPrintLoser(dee_t project s[], int numofProjects); findPrintLoser function gets an array of project_t and the number of projects. Finds the student with the worst score according to the average score, prints it by calling printProject function and returns its index in the array. • main function is where you declare an array of projects and call loadProjects function, print all project suing printProject function and call findPrint Loser function.
Program to keep the list of 5 senior project students entered to a project competition with their novel projects in a text file considering their names, surnames, and 5 scores earned from referees in the project competition can be written using the C++ programming language and follows the given steps:
Create a structure named project_t to keep the student's details with their scores in a project competition. It has four members in it, as given below:Two character arrays of 30 length for names and surnames respectivelyOne double array to store the referee scoresOne double variable to store the average score earned from the refereesCreate five functions, as given below:
double calculateAverageScore(const project_t *projectp)This function will calculate the average score of the projects and returns it. It accepts a pointer to a constant project_t. I
f the difference between the maximum and minimum score of a project is higher than 5 then exclude the maximum and minimum scores of the project when calculating the average score.int scanProject(FILE *filep, project_t *projectp)This function accepts a pointer to FILE and a pointer to project_t.
It reads name, surname and referee points from the file, and fills project_t pointed to, by projectp. It returns 1 if the read operation is successful; otherwise, returns 0.int loadProjects(project_t projects[])This function accepts an array of project_t and opens the text file with the entered name.
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