Answer:
Windows includes a tool called Microsoft System Information (Msinfo32.exe). This tool gathers information about your computer and displays a comprehensive view of your hardware, system components, and software environment, which you can use to diagnose computer issues.
Explanation:
The idea is to simulate random (x, y) points in a 2-D plane with domain as a square of side 1 unit. Imagine a circle inside the same domain with same diameter and inscribed into the square. We then calculate the ratio of number points that lied inside the circle and total number of generated points.
Answer:
For calculating Pi (π), the first and most obvious way is to measure the circumference and diameter.
π = Circumference / diameter
For the random points in the 2-D plane, we calculate the ratio of number points inside the circle and generate random pairs and then check
x² + y² < 1
If the given condition is true, then increment the points and used them. In randomized and simulation algorithms like Monte Carlo. In this case, if the number of the iteration is more then more accurate, the result is.
Explanation:
For simulating the random points in 2-D using Monte Carlo algorithms:
Monte Carlo
In this algorithm, we calculate the ratio of the number of points that lied in the circle. We do not need any graphics or simulations to generated points. In this, we generate random pairs and check x² + y² < 1.
π ≅ 4 × N inner / N total
The other way for calculating Pi (π) is to use Gregory-Leibniz Series.
Series which converges more quickly is Nilakangtha Series, which is developed in the 15th century. It is the other way to calculate Pi (π).
2. Consider the two-dimensional array A: int A[][] = new int[100][100]; where A[0][0] is at location 200 in a paged memory system with pages of size 200. Asmall process that manipulates the matrix resides in page 0 (locations 0 to 199). Thus, every instruction fetch will be from page 0. For three page frames, how many page faults are generated by the following array-initialization loops? Use LRU replacement, and assume that page frame 1 contains the process and the other two are initially empty. a. for (int j = 0; j < 100; j++) for (int i = 0; i < 100; i++) A[i][j] = 0; b. for (int i = 0; i < 100; i++) for (int j = 0; j < 100; j++) A[i][j] = 0;
Final answer:
The problem is about calculating the number of page faults for two different loop orderings in a paged memory system under LRU replacement. Loop (a) has a column-major ordering, likely causing a new page fault for each new column accessed, while loop (b) has a row-major ordering, causing fewer page faults due to contiguous memory access.
Explanation:
A student has asked about the number of page faults generated by array-initialization loops in a paged memory system, specific to a given scenario with LRU replacement policy. We have two different loop orderings to consider with three page frames available.
For the first loop (part a), which is column-major order:
Page fault will occur for each new column since each element of a column is on a different page (due to row-major storage in C-like languages).
There are 100 columns, so initially, we will have 100 page faults for the first 100 pages.
As pages are brought into memory and replaced using LRU, subsequent accesses to those pages within the loop may not cause additional faults if they remain in the frame.
For the second loop (part b), which is row-major order:
Since each row is contiguous in memory, fewer page faults will occur.
The loop accesses 200 int elements before moving to a new page, thus causing a page fault.
We can fill two page frames with array data since one frame is occupied by the process.
Write syntactically correct Javascript code to sort an array of sub-arrays containing integers using the sort and reduce array functions as described below. You will sort the sub-arrays in ascending order by the maximum value found in each sub-array. Your solution would be executed in the following manner:
The question is incomplete. The complete question is:
Write syntactically correct Javascript code to sort an array of sub-arrays containing integers using the sort and reduce array functions as described below. You will sort the sub-arrays in ascending order by the maxium value found in each sub-array. Your solution would be executed in the following manner:
var x = [ [ 2 , 5 , 1 ], [ 1 , 23 ] , [ 3 ] , [ 22, 16, 8 ] ] ;
x.sort(compare);
will result in variable x containing [ 3 ] , [ 2, 5, 1], [ 22 , 16 , 8], [ 1 , 23 ] ]. The maximum value of each sub-array is shown here in bold. Your solution must be written so that when run with the code above, x will contain the sorted array.
Answer:
function compare(a , b) {
var max1 = Math.max(...a); //get the max in array a
var max2 = Math.max(...b); //get the max in array b
return max1 - max2;
}
var x = [ [ 2 , 5 , 1 ], [ 1 , 23 ] , [ 3 ] , [ 22, 16, 8 ] ] ;
x.sort(compare);
Explanation:
Given an array A positive integers, sort the array in ascending order such that element in given subarray which start and end indexes are input in unsorted array stay unmoved and all other elements are sorted.
An array is a special kind of object. With the square brackets, access to a property arr[0] actually come from the object syntax. This is essentially the same as obj[key], where arr is the object, while numbers are used as keys.
Therefore, sorting sub-array in ascending order by the maxium value found in each sub-array is done:
function compare(a , b) {
var max1 = Math.max(...a); //get the max in array a
var max2 = Math.max(...b); //get the max in array b
return max1 - max2;
}
var x = [ [ 2 , 5 , 1 ], [ 1 , 23 ] , [ 3 ] , [ 22, 16, 8 ] ] ;
x.sort(compare);
JAVA
In this exercise, you are given a word or phrase and you need to return that word or phrase with the word ‘like’ in front of it.
If the phrase already starts with like, you should not add another one, just return the phrase as is.
Example:
like("totally awesome") --> "like totally awesome"
like("like cool dude") --> "like cool dude"
like("I like you") -->"like I like you"
Answer:
Answer in the screenshot provided.
Explanation:
Check if the String starts with like. If it does not then prepend it.
2- There are many different design parameters that are important to a cache’s overall performance. Below are listed parameters for different direct-mapped cache designs. Cache Data Size: 32 KiB Cache Block Size: 2 words Cache Access Time: 1 cycle Calculate the total number of bits required for the cache listed above, assuming a 32-bit address. Given that total size, find the total size of the closest direct-mapped cache with 16-word blocks of equal size or greater. Explain why the second cache, despite its larger data size, might provide slower performance than the first cache. You must show your work
Answer:
1. 2588672 bits
2. 4308992 bits
3. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
Explanation:
1. Number of bits in the first cache
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^15 (1+14+(32*2^1)) = 2588672 bits
2. Number of bits in the Cache with 16 word blocks
Using the formula: (2^index bits) * (valid bits + tag bits + (data bits * 2^offset bits))
total bits = 2^13(1 +13+(32*2^4)) = 4308992 bits
3. Caches are used to help achieve good performance with slow main memories. However, due to architectural limitations of cache, larger data size of cache are not as effective than the smaller data size. A larger cache will have a lower miss rate and a higher delay. The larger the data size of the cache, the larger the area of memory you will need to "search" making the access time and performance slower than the a cache with a smaller data size.
The total number of bits required for the given cache is 335,872 bits. Despite the larger data size, the second cache with 16-word blocks might provide slower performance due to higher block transfer times and possible increased cache miss rates.
Given the cache parameters:
Cache Data Size: 32 KiBCache Block Size: 2 wordsCache Access Time: 1 cycleLet's calculate the total number of bits required for this cache. Assume a 32-bit address.
Calculate the number of blocks in the cache:To find the total size of the closest direct-mapped cache with 16-word blocks: A 16-word block = 16 × 4 bytes = 64 bytes. Number of blocks = 32768 bytes (32 KiB) / 64 bytes = 512 blocks.
Repeat the bit calculation with the new configuration:
Offset bits = log2(64) = 6 bitsIndex bits = log2(512) = 9 bitsTag bits = 32 - (Index bits + Offset bits) = 32 - 15 = 17 bitsTotal bits per block = 1 Valid bit + 17 Tag bits + 64 bytes × 8 bits/byte = 1 + 17 + 512 = 530 bits.Total bits for the new cache: 512 blocks × 530 bits/block = 271,360 bits.
Despite the larger data size, the second cache with 16-word blocks might provide slower performance due to higher block transfer times and possible increased cache miss rates. Larger blocks increase the penalty for cache misses and might not be fully utilized, leading to inefficiency.
What is required for real-time surveillance of a suspects computer activity?/p Group of answer choices a. Blocking data transmissions between a suspect’s computer and a network server. b. Poisoning data transmissions between a suspect’s computer and a network server. c. Preventing data transmissions between a suspect’s computer and a network server. d. Sniffing data transmissions between a suspect’s computer and a network server.
Answer:
c Preventing data transmissions between a suspect’s computer and a network server
Explanation:
A developer has the following class and trigger code:public class InsuranceRates {public static final Decimal smokerCharge = 0.01;}trigger ContactTrigger on Contact (before insert) {InsuranceRates rates = new InsuranceRates();Decimal baseCost = XXX;}Which code segment should a developer insert at the XXX to set the baseCost variable to the value of the class variable smokerCharge?
A. Rates.smokerCharge
B. Rates.getSmokerCharge()
C. ContactTrigger.InsuranceRates.smokerCharge
D. InsuranceRates.smokerCharge
Answer:
InsuranceRates.smokerCharge
Explanation:
To set the baseCost variable to the value of the class variable smokerCharge, the developer needs to replace
Decimal baseCost = XXX;
with
Decimal baseCost = InsuranceRates.smokerCharge;
This is done using the class reference type.
The smokerCharge should be declared in the InsuranceRates class, at first.
For example Ball b;
And then, a new instance of the object from the class is created, using the new keyword along with the class name: b = new Ball();
In comparing to the example I gave in the previous paragraph, the object reference in the program is:
InsuranceRates rates = new InsuranceRates();
Because the smokerCharge is coming from a different class, it can only be assigned to the variable baseCost via the InsuranceRates class to give:
Decimal baseCost = InsuranceRates.smokerCharge;
The correct code segment to set the baseCost variable to the value of the class variable smokerCharge is D. InsuranceRates.smokerCharge.
Explanation:The correct code segment to set the baseCost variable to the value of the class variable smokerCharge is D. InsuranceRates.smokerCharge.
In the given code, the smokerCharge variable is declared as a static final variable in the InsuranceRates class. This means that it can be accessed directly using the class name followed by the variable name.
Therefore, to set the baseCost variable to the value of smokerCharge, we use InsuranceRates.smokerCharge.
Write a function with local functions that calculates the volume, side areas, and total surface area of a box based on the lengths of the sides. It will have 3 inputs: the length, width, and height of the box. There will be no outputs since it will simply print the results to the screen.
Answer:
def calcBox(l, w, h): vol = l*w*h print("Volume is :" + str(vol)) sideArea1 = h*w sideArea2 = h*w sideArea3 = h*l sideArea4 = h*l sideArea5 = w*l sideArea6 = w*l print("Side Area 1: " + str(sideArea1)) print("Side Area 2: " + str(sideArea2)) print("Side Area 3: " + str(sideArea3)) print("Side Area 4: " + str(sideArea4)) print("Side Area 5: " + str(sideArea5)) print("Side Area 6: " + str(sideArea6)) surfAreas = 2*h*w + 2*h*l + 2*w*l print("Total surface areas is: " + str(surfAreas)) calcBox(5, 8, 12)Explanation:
Firstly, create a function calcBox that takes three inputs, l, w, and h which denote length, width and height, respectively.
Apply formula to calculate volume and print it out (Line 2-3).
Since there are six side areas of a box, and we define six variables to hold the calculated value of area for one side respectively (Line 5 - 10) and print them out (Line 11-16).
Apply formula to calculate surface areas and print it out (Line 18 - 19)
At last test the function (Line 21).
JAVA
Write a method that takes an int[] array and an int value and returns the first index at which value appears in the array.
If the value does not exist in the array, return -1
For example:
search(new int[]{10, 20, 30, 20}, 20)
Should return 1
Because 20 first appears at index 1 in the array.
search(new int[]{10, 20, 30, 20}, 40)
Should return -1
public int search(int[] array, int value)
{
}
Answer:
public static void main()
{
Scanner console = new Scanner(System.in);
System.out.println("Enter in a value");
int a = console.nextInt();
int[] array = {1,2,3,4,5};
int a1 = search(array, a);
System.out.println(a1);
}
public static int search(int[] array, int value)
{
boolean okay = true;
int b = 0;
for(b = 0; b<=array.length-1; b++)
{
if(value==array[0])
{
okay=true;
}
else
{
okay = false;
}
}
if(okay = true)
{
return b;
}
else
{
return -1;
}
}
Explanation:
Answer:
Answer is in the attached screenshot!
Explanation:
Just iterate through the values in the list - if the value at the index is equal to the one you are searching for, just return that index. If no value is returned then return with -1!
For each of the two questions below, decide whether the answer is (i) "Yes," (ii) "No," or (iii) "Unknown, because it would resolve the question of whether P = NP." Give a brief explanation of your answer. (a) Let’s define the decision version of the Interval Scheduling Prob- lem from Chapter 4 as follows: Given a collection of intervals on a time-line, and a bound k, does the collection contain a subset of nonoverlapping intervals of size at least k? Question: Is it the case that Interval Scheduling ≤P Vertex Cover? (b) Question: Is it the case that Independent Set ≤P Interval Scheduling?
Final answer:
The original student question pertains to computational complexity and the relationship between different problems within the NP complexity class, seeking to understand polynomial-time reductions between Interval Scheduling, Vertex Cover, and Independent Set.
Explanation:
The questions provided revolve around computational complexity, particularly the concepts surrounding the complexity classes P, NP, and the relations between different computational problems. Unfortunately, the rest of the information provided is not related to the P vs NP question raised and therefore not relevant for forming an answer to the question at hand. The original question seems to ask whether 'Interval Scheduling' can be polynomial-time reduced to 'Vertex Cover' (Interval Scheduling ≤P Vertex Cover), and whether 'Independent Set' can be polynomial-time reduced to 'Interval Scheduling' (Independent Set ≤P Interval Scheduling). These questions are seeking to understand the computational complexity relationship between different NP problems.
1. Compare the similarities and differences between traditional computing and the computing clouds launched in recent years. Also discuss possible convergence of the two computing paradigms in the future.
2. An increasing number of organizations in industry and business adopt cloud systems. Answer the following questions regarding cloud computing:
a. List and describe the main characteristics of cloud computing systems.
b. Discuss key enabling technologies in cloud computing systems.
c. Discuss different ways for cloud service providers to maximize their revenues.
d. Characterize the three cloud computing models using suitable examples.
Answer and explanation:
(1)
Resilience and Elasticity
The information and applications hosted in the cloud are evenly distributed across all the servers, which are connected to work as one. Therefore, if one server fails, no data is lost and downtime is avoided. The cloud also offers more storage space and server resources, including better computing power. This means your software and applications will perform faster.
Flexibility and Scalability
Cloud hosting offers an enhanced level of flexibility and scalability in comparison to traditional data centres. The on-demand virtual space of cloud computing has unlimited storage space and more server resources. Cloud servers can scale up or down depending on the level of traffic your website receives, and you will have full control to install any software as and when you need to. This provides more flexibility for your business to grow.
Automation
A key difference between cloud computing and traditional IT infrastructure is how they are managed. Cloud hosting is managed by the storage provider who takes care of all the necessary hardware, ensures security measures are in place, and keeps it running smoothly. Traditional data centres require heavy administration in-house, which can be costly and time consuming for your business. Fully trained IT personnel may be needed to ensure regular monitoring and maintenance of your servers – such as upgrades, configuration problems, threat protection and installations.
Running Costs
Cloud computing is more cost effective than traditional IT infrastructure due to methods of payment for the data storage services. With cloud based services, you only pay for what is used – similarly to how you pay for utilities such as electricity. Furthermore, the decreased likelihood of downtime means improved workplace performance and increased profits in the long run.
Security
Cloud computing is an external form of data storage and software delivery, which can make it seem less secure than local data hosting. Anyone with access to the server can view and use the stored data and applications in the cloud, wherever internet connection is available. Choosing a cloud service provider that is completely transparent in its hosting of cloud platforms and ensures optimum security measures are in place is crucial when transitioning to the cloud.
2(a)The five main characteristics of cloud computing systems are given below:
On-demand self-service: That is used when needed
Broad network access: That is ubiquitous, that is, it is everywhere
Multi-tenancy and resource pooling: That is sharing of resources
Rapid elasticity and scalability : Deploy only the amount of resources needed at a time
Measured service: Pay for what you use
(b)The five key enabling technologies in cloud computing systems are given below:
Client server using Application Control Interfaces Server Virtualization : Allows platform independence sort of scenario. Aggregation of computers into server based data centers: I.e high availability of computing resources Storage technologies supporting logical shards: Replication and safety of data Network virtualization(c). The three different ways for cloud service providers to maximize their revenues:
Development : Here the revenue comes from subscriptions paid by the customers.These also testing a new market for development.
Reselling : Using a pre-developed service or a cloud-based application- if it works really well. Then it can be ported to an on-premises solution thus bringing is a large possibility of prospective consumers.These also testing a new market for deployment.
Hosting: Server hardware is very expensive thus cloud computing is marketed as a good solution for disaster recovery solutions for customer data and profits are made as per subscription or dynamic virtual resource renting approach
(d)The characteristics of three cloud computing models using suitable examples is given below:
IAAS: This provides virtualized computing resources over cloud i.e over the internet. Here the cloud provider will host the infrastructure components which are traditionally present in an on-premises data center, hardware,servers etc. For example: Google Compute Engine, Linode
PAAS: This usually provides a platform as a service which allows the customers to develop, run, and manage applications. The users do not have to undergo the the complexity of building and maintaining the infrastructure which is typically associated with developing and launching an app if they are using a PAAS. Examples are AWS Elastik beanstalk, Apache Stratos
SAAS:In this model model the customer is provided with access to application software which is often referred to as "on-demand software". For example Google Drive, PayPal etc
Similarities and differences in computing.
Computing is the goal, oriented activity that needs benefits from creating the computing machinery that helps is study and measuring the algorithms process of development. Traditional and cloud-based computing refers to the delivery of services through data and the program on the internet.
Thus the answer is technology, security, infrastructure.
The difference is related to the local servers and the cloud or internet system. The convergence of the two technologies show the dynamic possibilities for the future of the technology leads to enhanced security and information maintenance. The provision of further technology of switches and routers.The increase in the number of organizations and businesses over the recent years has led to the growth of cloud computing and many services related to the international domain.This has led to the growth of technology and infrastructure.
Learn more about the compare the similarities.
brainly.com/question/24507709.
What are the disadvantages of a Cessna 172?
Answer:
The 172 accounted for 17-percent of the active fleet and flew 16-percent of the hours flown while accounting for six-percent of the fatal accidents.
Explanation:
In a two-year period there was but one fatal 172 accident that was due to a mechanical failure. That was an engine failure related to a valve. There were no fatal accidents related to fuel exhaustion or starvation.
Despite the good record in that area, the 172 is probably involved in just as many forced landings as any like airplane. It just appears more adaptable to impromptu arrivals than some other airplanes. The low landing speed contributes to this. There is no available statistic on this, but I would bet that most 172 forced landings don’t result in what the NTSB classifies as an accident.
I looked at fatal 172 accidents that occurred during two more recent years (2012 and 2013) when virtually all the NTSB reports were final as opposed to preliminary. There were 25 such accidents in the 48 contiguous states. If the methodology I used years ago is applied to that number, the 172 safety record appears to have improved, maybe substantially.
Write a C++ program that reads students' names followed by their test scores. The program should output each students' name followed by the test scores and relevant grade. It should also find and print the highest test score and the name of the students having the highest test score. Student data should be stores in a struct variable of type studentType, which has four components: studentFName and studentLname of type string, testScore of type int (testScore is between 0 and 100), and and grade of type char. Suppose that the class has 20 students. Use an array of 20 components of type studentType.
Your Program must contain at least the following functions:
a. A function to read the students' data into the array.
b. A function to assign the relevant grade to each student.
c. A function to find the highest test score.
d. A function to print the names of the students having the highest test score.
Answer:
#include<iostream>
#include<conio.h>
using namespace std;
struct studentdata{
char Fname[50];
char Lname[50];
int marks;
char grade;
};
main()
{
studentdata s[20];
for (int i=0; i<20;i++)
{
cout<<"\nenter the First name of student";
cin>>s[i].Fname;
cout<<"\nenter the Last name of student";
cin>>s[i].Lname;
cout<<"\nenter the marks of student";
cin>>s[i].marks;
}
for (int j=0;j<20;j++)
{
if (s[j].marks>90)
{
s[j].grade ='A';
}
else if (s[j].marks>75 && s[j].marks<90)
{
s[j].grade ='B';
}
else if (s[j].marks>60 && s[j].marks<75)
{
s[j].grade ='C';
}
else
{
s[j].grade ='F';
}
}
int highest=0;
int z=0;
for (int k=0;k<20; k++)
{
if (highest<s[k].marks)
{
highest = s[k].marks;
z=k;
}
}
cout<<"\nStudents having highest marks"<<endl;
cout<<"Student Name"<< s[z].Fname<<s[z].Lname<<endl;
cout<<"Marks"<<s[z].marks<<endl;
cout<<"Grade"<<s[z].grade;
getch();
}
Explanation:
This program is used to enter the information of 20 students that includes, their first name, last name and marks obtained out of 100.
The program will compute the grades of the students that may be A,B, C, and F. If marks are greater than 90, grade is A, If marks are greater than 75 and less than 90, grade is B. For Mark from 60 to 75 the grade is C and below 60 grade is F.
The program will further found the highest marks and than display the First name, last name, marks and grade of student who have highest marks.
A C++ program prompts for student names and test scores, storing data in a structure. Functions assign grades, find the highest score, and print students with the highest score. It utilizes arrays and functions to manage student data effectively within the specified requirements.
Here's a C++ program that fulfills the requirements you specified:
```cpp
#include <iostream>
#include <string>
using namespace std;
struct studentType {
string studentFName;
string studentLName;
int testScore;
char grade;
};
const int MAX_STUDENTS = 20;
// Function prototypes
void readStudentData(studentType students[], int numStudents);
void assignGrades(studentType students[], int numStudents);
int findHighestScore(const studentType students[], int numStudents);
void printHighestScorers(const studentType students[], int numStudents, int highestScore);
int main() {
studentType students[MAX_STUDENTS];
int numStudents;
cout << "Enter the number of students (up to 20): ";
cin >> numStudents;
if (numStudents > MAX_STUDENTS || numStudents < 1) {
cout << "Invalid number of students. Exiting program.";
return 1;
}
readStudentData(students, numStudents);
assignGrades(students, numStudents);
int highestScore = findHighestScore(students, numStudents);
printHighestScorers(students, numStudents, highestScore);
return 0;
}
void readStudentData(studentType students[], int numStudents) {
for (int i = 0; i < numStudents; ++i) {
cout << "Enter student " << i + 1 << "'s first name: ";
cin >> students[i].studentFName;
cout << "Enter student " << i + 1 << "'s last name: ";
cin >> students[i].studentLName;
cout << "Enter student " << i + 1 << "'s test score: ";
cin >> students[i].testScore;
}
}
void assignGrades(studentType students[], int numStudents) {
for (int i = 0; i < numStudents; ++i) {
if (students[i].testScore >= 90)
students[i].grade = 'A';
else if (students[i].testScore >= 80)
students[i].grade = 'B';
else if (students[i].testScore >= 70)
students[i].grade = 'C';
else if (students[i].testScore >= 60)
students[i].grade = 'D';
else
students[i].grade = 'F';
}
}
int findHighestScore(const studentType students[], int numStudents) {
int highestScore = students[0].testScore;
for (int i = 1; i < numStudents; ++i) {
if (students[i].testScore > highestScore)
highestScore = students[i].testScore;
}
return highestScore;
}
void printHighestScorers(const studentType students[], int numStudents, int highestScore) {
cout << "Students with the highest score of " << highestScore << ":\n";
for (int i = 0; i < numStudents; ++i) {
if (students[i].testScore == highestScore)
cout << students[i].studentFName << " " << students[i].studentLName << endl;
}
}
```
This program reads student data, assigns grades, finds the highest test score, and prints the names of students with the highest score. It utilizes structures and functions as required.
Below is a sequence of 32-bit memory address references observed during the execution of an application, and given as word addresses.
3, 180, 43, 2, 191, 88, 190, 14, 181, 44, 186, 253
For each of these references, identify the binary address, the tag, and the index given a direct mapped cache with 16 one-word blocks. Also list if each reference is a hit or a miss, assuming the cache is initially empty
Answer:
a. 1 word
b. 2 words
Explanation:
We say that a computer has a 32-bit memory address when it allows 32-bit memory addresses. This entails that a byte-addressable 32-bit computer can address 232 = 4,294,967,296 bytes of memory, or 4 gibibytes (GiB). This allows one memory address to be efficiently stored in one word.
See attachment for the analysis on a table.
Each memory address results in a miss because the cache is initially empty. The address is broken down into tag, index, and the hit/miss status is determined. This process helps in understanding the functioning of a direct mapped cache.
Given a sequence of 32-bit memory address references and a direct mapped cache with 16 one-word blocks, let's analyze each memory address:
**Memory Address 3**:Since the cache is initially empty, all references result in a miss. For a 32-bit address, the index is derived from the next 4 bits (considering 16 blocks = 2^4), and the tag is derived from the remaining higher order bits.
Write programs for two MSP430 boards to do the following:
Board A: If the P1.1 button is pressed, send the message 0xAA via the UART at 9600 baud.
Board B: The red LED shall initially be off. However, if the Board B micro-controller receives a UART message of 0xAA at 9600 baud, turn on the Board B red LED.
Answer:
Explanation:
#include "msp430g2553.h"
#define TXLED BIT0
#define RXLED BIT6
#define TXD BIT2
#define RXD BIT1
const char string[] = { "Hello World\r\n" };
unsigned int i; //Counter
int main(void)
{
WDTCTL = WDTPW + WDTHOLD; // Stop WDT
DCOCTL = 0; // Select lowest DCOx and MODx settings
BCSCTL1 = CALBC1_1MHZ; // Set DCO
DCOCTL = CALDCO_1MHZ;
P2DIR |= 0xFF; // All P2.x outputs
P2OUT &= 0x00; // All P2.x reset
P1SEL |= RXD + TXD ; // P1.1 = RXD, P1.2=TXD
P1SEL2 |= RXD + TXD ; // P1.1 = RXD, P1.2=TXD
P1DIR |= RXLED + TXLED;
P1OUT &= 0x00;
UCA0CTL1 |= UCSSEL_2; // SMCLK
UCA0BR0 = 0x08; // 1MHz 115200
UCA0BR1 = 0x00; // 1MHz 115200
UCA0MCTL = UCBRS2 + UCBRS0; // Modulation UCBRSx = 5
UCA0CTL1 &= ~UCSWRST; // **Initialize USCI state machine**
UC0IE |= UCA0RXIE; // Enable USCI_A0 RX interrupt
__bis_SR_register(CPUOFF + GIE); // Enter LPM0 w/ int until Byte RXed
while (1)
{ }
}
#pragma vector=USCIAB0TX_VECTOR
__interrupt void USCI0TX_ISR(void)
{
P1OUT |= TXLED;
UCA0TXBUF = string[i++]; // TX next character
if (i == sizeof string - 1) // TX over?
UC0IE &= ~UCA0TXIE; // Disable USCI_A0 TX interrupt
P1OUT &= ~TXLED; }
#pragma vector=USCIAB0RX_VECTOR
__interrupt void USCI0RX_ISR(void)
{
P1OUT |= RXLED;
if (UCA0RXBUF == 'a') // 'a' received?
{
i = 0;
UC0IE |= UCA0TXIE; // Enable USCI_A0 TX interrupt
UCA0TXBUF = string[i++];
}
P1OUT &= ~RXLED;
}
The Online Shopping system facilitates the customer to view the products, inquire about the product details, and product availability. It allows the customer to get register in order to purchase products. The customer can search products by browsing different product categories or by entering search keywords. Customer can place order and pay online. There are two acceptable payment methods. These are (1) pay by credit card and (2) pay by PayPal. The system provide service to seller to place the products for selling. The seller creates account to become the member and places his products under suitable product category. The systems allows the administrator to manage the products. It facilitates the administrator to modify the existing products categories or to add new products categories. The system facilitate site manager to view different reports including (1) order placed by customers, (2) products added by sellers, and (3) accounts created by users. Question 3.4.1 List software system actors. (10 points) Question 3.4.2 Write use cases associated with each stakeholder. (Note: each use case starts with an action word) (10 points)
Answer:
see explaination
Explanation:
We will consider the five Actors in the prearranged Online shopping software system:
Customer:
The scheme allows the customer to do below mention actions:
View goods and inquire about the niceties of products and their ease of use.
To create version to be able to purchase invention from the structure.
Browse crop through search category or shortest search alternative.
Place order for the necessary crop.
Make sum for the order(s) positioned.
Payment System:
Payment system allows client to pay using subsequent two methods:
Credit card.
PayPal.
Seller:
System allow seller to perform underneath actions:
Place the foodstuffs for selling under apposite product category.
Create account to happen to a member.
Administrator:
Following actions are perform by Admin:
Manage the goods posted in the system.
Modify existing manufactured goods categories or add novel categories for foodstuffs.
Site Manager:
System privileges Site director with the following role in the system:
View information on:
Orders Placed by customer
Products added by seller
Accounts created by users
check attachment
Write a program that will open a file named thisFile.txt and write every other line into the file thatFile.txt. Assume the input file (thisFile.txt) resides in the same directory as your code file, and that your output file (thatFile.txt) will reside in the same location. Do not attempt to read from or write to other locations.
Answer:
Check the explanation
Explanation:
The programmable code to solve the question above will be written in a python programming language (which is a high-level, interpreted, general-purpose programming language.)
f = open('thisFile.txt', 'r')
w = open('thatFile.txt', 'w')
count = 0
for line in f:
if count % 2 == 0:
w.write(line)
count += 1
w.close()
f.close()
Implement a superclass Appointment and subclasses Onetime, Daily, and Monthly. An appointment has the date of an appointment, the month of an appointment, the year of an appointment, and the description of an appointment (e.g., "see the dentist"). In other words, four major instant variables in these classes. Then, write a method occcursOn(year, month, day) that checks whether the appointment occurs on that date. Ensure to have a test program to test your classes. In your test program, you should create a list of appointments using the concept of polymorphism as your appointment pool. In other words, creating multiple objects using your subclasses of Onetime, Daily, and Monthly. Once you have those appointments available, allow the user to input day, month, and year to check whether the day of the month matches and print out the information of that matched appointment. For example, for a monthly appointment, you must check whether the day of the month matches. Have the user enter a date and print out all appointments that occur on that date.
Final answer:
The question involves creating an appointment system in object-oriented programming, where a superclass 'Appointment' and its subclasses 'Onetime', 'Daily', and 'Monthly' are defined, each with a method to check if an appointment occurs on a specified date.
Explanation:
Implementing Appointment Superclass and Subclasses
To model an appointment system, we implement a superclass named Appointment with subclasses called Onetime, Daily, and Monthly. Each class contains instance variables for the date, month, year, and a description of the appointment. The key method, occursOn(year, month, day), determines if an appointment is scheduled for a specific date by checking the year, month, and day against the appointment's details.
Appointment Superclass
The superclass includes the four main instance variables and provides the foundational behavior for its subclasses.
Onetime, Daily, and Monthly Subclasses
The subclasses override the occursOn() method to provide specific checking logic for one-time, daily, and monthly appointments.
A test program can be created to instantiate different types of appointments and check if they occur on a user-input date. By utilizing polymorphism, we can maintain a list of appointments of various subclasses. The test program will iterate through this list, applying the occursOn() method to find and print out matching appointments for the given date.
5. Modify the file by allowing the owner to remove data from the file: a. Ask the owner to enter a description to remove b. If the description exists, remove the coffee name and the quantity c. If the description is not found, display the message: That item was not found in the file. 6. Modify the file by allowing the owner to delete data from the file: a. Ask the owner to enter a description to delete b. If the description exists, delete the coffee name and the quantity c. Replace the name and quantity of the coffee removed in step b by asking the user to enter a new coffee name and quantity d. If the description i
Answer:
Check the explanation
Explanation:
Code:
# 3------reading Coffe shop Inventory file
f = open('coffeeInventory.txt','r+')#opening file in reading mode
f1 = f.readlines()
numPounds = 0
for i in range(len(f1)):
print(f1[i].split('\n')[0])
if i!=0:
numPounds+=(int((f1[i].split('\n')[0]).split(';')[1]))
#Total pounds of coffee
print("Total pounds of Coffee:",numPounds)
f.close()#closing file
#4---------
f = open('coffeeInventory.txt','a+')#opening file in appending mode
#appending records
f.write('\nGuatemala Antigua;22')
f.write('\nHouse Blend;25')
f.write('\nDecaf House Blend;16')
f.close()#closing file
#5--------
#removing the record as per owner's description
Description = input("enter Description to remove:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
#storing the previous data
f2= open('coffeeInventory_pre.txt','w+')
f2.writelines(f1)
f2.close()
#removing the data if found
if found==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
#modifying file after removing
modify_file = open('coffeeInventory.txt','w+')
for x in data_list:
modify_file.write(x)
modify_file.close()
f.close()
#6--------------
#deleting the record as per owner's description
Description = input("enter Description to delete:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
# deleting in the pre_Coffee_inventory file
#finding record
f2 = open('coffeeInventory_pre.txt','r')
f3 =f2.readlines()
found = -1
for i in range(len(f3)):
line = f3[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f3.remove(f3[found])#deleting record
data_list1 = f3
f2.close()
#modifying pre file
f2 = open('coffeeInventory_pre.txt','w+')
f2.writelines(data_list1)
f2.close()
#opening file in writing mode
modify_file = open('coffeeInventory.txt','w+')
modify_file.writelines(data_list)
modify_file.close()
Answer:
See explaination
Explanation:
#5
#removing the record as per owner's description
Description = input("enter Description to remove:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
#storing the previous data
f2= open('coffeeInventory_pre.txt','w+')
f2.writelines(f1)
f2.close()
#removing the data if found
if found==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
#modifying file after removing
modify_file = open('coffeeInventory.txt','w+')
for x in data_list:
modify_file.write(x)
modify_file.close()
f.close()
#6
#deleting the record as per owner's description
Description = input("enter Description to delete:")
f = open('coffeeInventory.txt','r')#opening files in reading mode
f1 = f.readlines()
found = -1
for i in range(len(f1)):
line = f1[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f1.remove(f1[found])
data_list = f1
f.close()
# deleting in the pre_Coffee_inventory file
#finding record
f2 = open('coffeeInventory_pre.txt','r')
f3 =f2.readlines()
found = -1
for i in range(len(f3)):
line = f3[i].split('\n')[0]
des = line.split(';')[0]
if des==Description:
found = i
if i==-1:
print("The item was not found in the file")
else:
f3.remove(f3[found])#deleting record
data_list1 = f3
f2.close()
#modifying pre file
f2 = open('coffeeInventory_pre.txt','w+')
f2.writelines(data_list1)
f2.close()
#opening file in writing mode
modify_file = open('coffeeInventory.txt','w+')
modify_file.writelines(data_list)
modify_file.close()
a user's given name followed by the user's age from standard input. Then use an ofstream object named outdata (which you must declare) to write this information separated by a space into a file called outdata. Assume that this is the extent of the output that this program will do. Declare any variables that you need.
Answer:
See explaination for program code
Explanation:
#include <iostream>
#include <string>
#include <fstream>
using namespace std;
int main() {
string name;
int age;
cout << "Enter name of user: ";
getline(cin, name);
cout << "Enter age of user: ";
cin >> age;
ofstream outdata("outdata");
outdata << name << " " << age << endl;
outdata.close();
return 0;
}
Design state machines to control the minutes and hours of a standard 24 hour clock. Your clock should include an AM/PM indicator light, 2-digit hours (1 – 12), 2-digit minutes (00 – 59). (There is no need to include seconds; if we wanted to display seconds, it would be the same state control unit as minutes.)
Answer:
1 Hz clock generator to produce 1 PPS (beat every second) sign to the seconds square.
SECONDS square - contains a partition by 10 circuit followed by a gap by 6 circuit. Will produce a 1 PPM (beat every moment) sign to the minutes square. The BCD yields associate with the BCD to Seven Segment circuit to show the seconds esteems.
MINUTES square - indistinguishable from the seconds square it contains 2 dividers; a partition by 10 followed by a separation by 6. Will produce a 1 PPH (beat every hour) sign to the HOURS square. The BCD yields interfaces with the BCD to Seven Segment circuit to show the minutes esteems.
HOURS square - relying upon whether it is a 12 or 24H clock, will have a partition 24 or separation by 12. For 24H, it will tally from 00 to 23. For 12H, it will tally from 00 to 11. The BCD yields associates with the BCD to Seven Segment circuit to show the hours esteems.
Note that: To rearrange the table, K is Q0 of IC1 (ones), G is Q0 of IC2 (tens, etc.
For the 12H clock, when the include in BCD comes to
0A, IC1 must be cleared (Y=1)
12, IC1 must be cleared (Y=1) and IC2 must be cleared (X=1)
Utilizing SOP (entirety of items), we get
Y = HJ + GJ where Y is the IC1 MR1, MR2 inputs associated together.
X = GJ where X is the IC2 MR1, MR2 inputs associated together.
Explanation:
Server farms such as Google and Yahoo! provide enough compute capacity for the highest request rate of the day. Imagine that most of the time these servers operate at only 60% capacity. Assume further that the power does not scale linearly with the load; that is, when the servers are operating at 60% capacity, they consume 90% of maximum power. The servers could be turned off, but they would take too long to restart in response to more load. A new system has been proposed that allows for a quick restart but requires 20% of the maximum power while in this "barely alive" state. a. How much power savings would be achieved by turning off 60% of the servers? b. How much power savings would be achieved by placing 60% of the servers in the "barely alive" state? c. How much power savings would be achieved by reducing the voltage by 20% and frequency by 40%? d. How much power savings would be achieved by placing 30% of the servers in the "barely alive" state and 30% off?
Answer:
a) Power saving = 26.7%
b) power saving = 48%
c) Power saving = 61%
d) Power saving = 25.3%
Explanation:
During routine maintenance on your server, you discover that one of the hard drives on your RAID array is starting to go bad. After you replace the drive and rebuild the data, you run diagnostics on the RAID controller card with no problems, so you believe you fixed the problem. The next morning you arrive at work and discover that the RAID has completely failed. None of the disks are working. Your server is still running because you installed the OS on a non-RAID disk, but all the data is missing, and the server shows that none of your RAID disks are found.What might be the problem with your server? (Select all that apply.)asked Aug 13, 2019 in Computer Science & Information Technology by fili1001A. MotherboardB. CPUC. RAMD. RAID controllerE. Power supply
Answer:
RAID controller or Motherboard ie. A and D
Explanation:
From the explaination on the question the two most likely system parts that may be affecting the server are the Raid controller and the Motherboard.
The raid controller, this is because after the other drive has been replaced, the Raid still failed, it needs to be checked properly and replaced.
The motherboard, this is because it is the core centre of all boards.
7.8.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new = (old + 1) * 2. Ex: If xVal = 3 and yVal = 4, then xValNew is 8 and yValNew is 10. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 #include using namespace std; /* Your solution goes here */ int main() { int xValNew; int yValNew; int xValUser; int yValUser; cin >> xValUser; cin >> yValUser; CoordTransform(xValUser, yValUser, xValNew, yValNew); cout << "(" << xValUser << ", " << yValUser << ") becomes (" << xValNew << ", " << yValNew << ")" << endl; return 0; } 1 test passed All tests passed Run Feedback? How was this section?
Answer:
Here is the CoordTransform() function:
void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){
xValNew = (xVal + 1) * 2;
yValNew = (yVal + 1) * 2; }
The above method has four parameters xVal and yVal that are used as input parameters and xValNew yValNew as output parameters. This function returns void and transforms its first two input parameters xVal and yVal into two output parameters xValNew and yValNew according to the formula: new = (old + 1) *
Here new variables are xValNew and yValNew and old is represented by xVal and yVal.
Here the variables xValNew and yValNew are passed by reference which means any change made to these variables will be reflected in main(). Whereas variables xVal and yVal are passed by value.
Explanation:
Here is the complete program:
#include <iostream>
using namespace std;
void CoordTransform(int xVal, int yVal, int &xValNew, int &yValNew){
xValNew = (xVal + 1) * 2;
yValNew = (yVal + 1) * 2;}
int main()
{ int xValNew;
int yValNew;
int xValUser;
int yValUser;
cin >> xValUser;
cin >> yValUser;
CoordTransform(xValUser, yValUser, xValNew, yValNew);
cout << "(" << xValUser << ", " << yValUser << ") becomes (" << xValNew << ", " << yValNew << ")" << endl;
return 0; }
The output is given in the attached screenshot
JAVA
Write a method that takes 2 parameters: a String[] array, and an int numRepeats representing the number of times to repeat each element in the array.
Return a new array with each element repeated numRepeats times.
For example:
repeatElements(new String[]{"hello", "world"}, 3)
Should return a new array with the elements:
["hello", "hello", "hello", "world", "world", "world"]
public String[] repeatElements(String[] array, int numRepeats){
}
Answer:
Answer is in the provided screenshot! This was a lot of fun to make!
Explanation:
We need to create a new Array which has to be the size of amount * original - as we now that we are going to have that many elements. Then we just iterate through all the values of the new array and set them equal to each of the elements in order.
Ignore the code in my main function, this was to print to the terminal the working code - as you can see from the output at the bottom!
P5. (5pts) Recall that we saw the Internet checksum being used in both transport-layer segment (in UDP and TCP headers) and in network-layer datagrams (IP header). Now consider a transport layer segment encapsulated in an IP datagram. Are the checksums in the segment header and datagram header computed over any common bytes in the IP datagram
Answer:
see explaination
Explanation:
No common bytes are used to compute the checksum in the IP datagram. Only the IP header is used to compute the checksum at the network Layer.
At TCP/UDP segment, the IP datagram may have different protocol stacks used.
Therefore, when a TCP datagram is entered in the IP datagram, it is not necessary to use common byte to compute checksum.
The checksums in the transport-layer (TCP/UDP) and network-layer (IP) headers are computed over different sets of bytes, ensuring the integrity of their respective areas independently.
The IP header checksum covers only the IP header, while the TCP/UDP checksum includes their headers and data, along with some IP header information.
The checksums in the transport-layer segment (like TCP and UDP) and the network-layer datagram (IP) headers are designed to ensure data integrity by detecting errors that may have occurred during transmission.
However, the checksums are not computed over any common bytes of the IP datagram. The IP header checksum is computed exclusively over the IP header, which includes the source and destination IP addresses, protocol, and other fields. Conversely, the checksum for the TCP or UDP segment is computed over the entire segment, which includes the payload (data) and a pseudo-header. The pseudo-header contains certain fields from the IP header, like source and destination addresses, protocol, and segment length, but the checksum itself does not overlap with that of the IP header. Example:
If an IP datagram carries a TCP segment, the IP checksum will only ensure the integrity of the IP header.The TCP checksum ensures the integrity of the TCP header and its data.To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method: 1. Create a list of consecutive integers from two to n: (2, 3, 4, ..., n), 2. Initially, let p equal 2, the first prime number, 3. While enumerating all multiples of p starting from p2, strike them off from the original list, 4. Find the first number remaining on the list after p (it's the next prime); let p equal this number, 5. Repeat steps 3 and 4 until p2 is greater than n. 6. All the remaining numbers in the list are prime.
Answer:
I am writing a Python program:
def Eratosthenes(n):
primeNo = [True for i in range(n+1)] # this is a boolean array
p = 2 # the first prime number is initialized as 2
while (p * p <= n): # enumerates all multiples of p
if (primeNo[p] == True):
for i in range(p * p, n+1, p): #update multiples
primeNo[i] = False
p = p + 1
for p in range(2, n): #display all the prime numbers
if primeNo[p]:
print(p),
def main(): #to take value of n from user and display prime numbers #less than or equal to n by calling Eratosthenes method
n= int(input("Enter an integer n: "))
print("The prime numbers less than or equal to",n, "are: ")
Eratosthenes(n)
main()
Explanation:
The program contains a Boolean type array primeNo that is initialized by True which means that any value i in prime array will be true if i is a prime otherwise it will be false. The while loop keeps enumerating all multiples of p starting from 2, and striking them off from the original array list and for loops keep updating the multiples. This process will continue till the p is greater than n. The last for loop displays all the prime numbers less than or equal to n which is input by user. main() function prompts user to enter the value of integer n and then calls Eratosthenes() function to print all the prime numbers less than or equal to a given integer n.
Encryption Using Rotate Operations Write a procedure that performs simple encryption by rotating each plaintext byte a varying number of positions in different directions. For example, in the following array that represents the encryption key, a negative value indicates a rotation to the left and a positive value indicates a rotation to the right. The integer in each position indicates the magnitude of the rotation: key BYTE -2, 4, 1, 0, -3, 5, 2, -4, -4, 6
o implement simple encryption using rotate operations in assembly language, you can create a procedure that takes a plaintext byte and a key array as parameters. The key array represents the rotation values for each position. Here's an example of how you can write such a procedure in x86 assembly:
section .data key db -2, 4, 1, 0, -3, 5, 2, -4, -4, 6 section .text
global rotate_encrypt
rotate_encrypt:
; Input parameters:
; rdi: pointer to plaintext byte
; rsi: index in the key array
; Load the plaintext byte mov al, [rdi]
; Load the rotation value from the key array mov cl, [key + rsi]; Check if the rotation is positive or negative cmp cl, 0jge rotate_right jl rotate_left
rotate_right: ; Rotate right by the magnitude specified in clshr al, cljmp done
rotate_left: ; Rotate left by the magnitude specified in cl (convert to positive)neg clshl al, cl
done: ; Return the encrypted byte mov [rdi], alret
The sum of the numbers 1 to n can be calculated recursively as follows: The sum from 1 to 1 is 1. The sum from 1 to n is n more than the sum from 1 to n-1 Write a function named sum that accepts a variable containing an integer value as its parameter and returns the sum of the numbers from 1 to to the parameter (calculated recursively).
Answer:
See explaination
Explanation:
The code
#function named sum that accepts a variable
#containing an integer value as its parameter
#returns the sum of the numbers from 1 to to the parameter
def sum(n):
if n == 0:
return 0
else:
#call the function recursively
return n + sum(n - 1)
#Test to find the sum()
#function with parameter 5.
print(sum(5))
A recursive function is simply a function that continues to execute itself until a condition is satisfied
The recursive function sum in Python, where comments are used to explain each line is as follows:
#This defines the sum function
def sum(n):
#The function returns n, when n is 1 or less
if n <= 1:
return n
#Otherwise, the sum is calculated recursively
else:
return n + sum(n - 1)
Read more about recursive functions at:
https://brainly.com/question/13657607
please identify three examples of how the United States is heteronormative
Final answer:
Heteronormativity in U.S. society is evident in legal recognition of heterosexual marriages, social expectations based on traditional gender roles, and media representations that prioritize heterosexual relationships.
Explanation:
Heteronormativity refers to the assumption that heterosexual orientation is the norm and other sexual orientations are considered abnormal. Here are three examples of how U.S. society is heteronormative:
Legal recognition: The U.S. government and many states have historically only recognized heterosexual marriages, excluding same-sex couples from legal recognition and benefits.Social expectations: Traditional gender roles and expectations often assume a heterosexual orientation, such as the expectation that a man and a woman will marry and have children.Media representations: Mainstream media tends to overwhelmingly portray heterosexual relationships as the default, downplaying or erasing the existence of LGBTQ+ relationships.These examples demonstrate how heteronormativity privileges and normalizes heterosexual orientations while marginalizing and stigmatizing non-heterosexual orientations.