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Hoochie [10]
3 years ago
13

Consider a single CPU system with an active process A. Explain what happens in the following circumstances including any interru

pts, system calls, etc. and how they are handled until a process is back to running again?
a. Process A forks a new process B
b. Process A is running, and needs to read from a file
c. A timer interrupt occurs while A is running.
Computers and Technology
1 answer:
Anika [276]3 years ago
5 0

Answer:

The answer to this question as follows:

Explanation:

In option a, When we use fork, a new mechanism, that uses fork() method, which is in the parental process, to replicate all sites. It has been installed in a space-differentiating operating system.

In option b, It is the present system cycle that scan and wait for any of the system processor to be installed.

In option c,  The time delay happens when A operates. It is a global that make issues, which are cleared and slowed down when an interrupt happens. In this process, there are not any distractions. It splits into slowly as it heads into the ISR. It helps to understand the code easily.

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Create an array of numbers filled by the random number generator. (value = (int)(Math.random() * 100 + 1);) Print the array and
Y_Kistochka [10]

Answer:

Explanation:

the following is the code to run this (JAVA)

MeanStandardDev.java

import java.util.Random;

import java.util.Scanner;

public class MeanStandardDev {

public static void main(String[] args) {

// Declaring variables

int N;

double lower, upper, min, max, mean, stdDev;

/*

* Creating an Scanner class object which is used to get the inputs

* entered by the user

*/

Scanner sc = new Scanner(System.in);

// Getting the input entered by the user

System.out.print(" How many Random Numbers you want to generate :");

N = sc.nextInt();

System.out.print("Enter the Lower Limit in the Range :");

lower = sc.nextDouble();

System.out.print("Enter the Upper Limit in the Range :");

upper = sc.nextDouble();

// Creating Random class object

Random rand = new Random();

double nos[] = new double[N];

// this loop generates and populates 10 random numbers into an array

for (int i = 0; i < nos.length; i++) {

nos[i] = lower + (upper - lower) * rand.nextDouble();

}

//calling the methods

min = findMinimum(nos);

max = findMaximum(nos);

mean = calMean(nos);

stdDev = calStandardDev(nos, mean);

//Displaying the output

System.out.printf("The Minimum Number is :%.1f\n",min);

System.out.printf("The Maximum Number is :%.1f\n",max);

System.out.printf("The Mean is :%.2f\n",mean);

System.out.printf("The Standard Deviation is :%.2f\n",stdDev);

}

//This method will calculate the standard deviation

private static double calStandardDev(double[] nos, double mean) {

//Declaring local variables

double standard_deviation=0.0,variance=0.0,sum_of_squares=0.0;

/* This loop Calculating the sum of

* square of eeach element in the array

*/

for(int i=0;i<nos.length;i++)

{

/* Calculating the sum of square of

* each element in the array    

*/

sum_of_squares+=Math.pow((nos[i]-mean),2);

}

//calculating the variance of an array

variance=((double)sum_of_squares/(nos.length-1));

//calculating the standard deviation of an array

standard_deviation=Math.sqrt(variance);

return standard_deviation;

}

//This method will calculate the mean

private static double calMean(double[] nos) {

double mean = 0.0, tot = 0.0;

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Calculating the sum of all the elements in the array

tot += nos[i];

}

mean = tot / nos.length;

return mean;

}

//This method will find the Minimum element in the array

private static double findMinimum(double[] nos) {

double min = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] < min)

min = nos[i];

}

return min;

}

//This method will find the Maximum element in the array

private static double findMaximum(double[] nos) {

double max = nos[0];

// This for loop will find the minimum and maximum of an array

for (int i = 0; i < nos.length; i++) {

// Finding minimum element

if (nos[i] > max)

max = nos[i];

}

return max;

}

}

the OUTPUT should give;

How many Random Numbers you want to generate :10

Enter the Lower Limit in the Range :1.0

Enter the Upper Limit in the Range :10.0

The Minimum Number is :1.1

The Maximum Number is :9.9

The Mean is :6.30

The Standard Deviation is :2.98

cheers i hope this helps!!!

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Images help tell a story where describing with words is either too lengthy, or practically impossible. For instance, you could have a map of a location and various arrows and other markings to describe movements of troops during a battle of the civil war. This is one example of many that you could have as an image on a website. Describing the troop movements with words only may be really difficult to do. Plus many people are visually oriented learners, so they benefit with images every now and then. Of course, it's best not to overdo things and overload the site with too many images. A nice balance is needed.

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__________ is a term used to indicate any unwanted event that takes place outside normal daily security operations. This type of
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The answer is “a security event”
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Which Application program saves data automatically as it is entered?
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The application program that saves data automatically as it is entered is the MS Access.
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2 years ago
Draw a timeline for each of the following scheduling algorithms. (It may be helpful to first compute a start and finish time for
12345 [234]

Answer:

See explanation below

Explanation:

Previos concepts

First Come First Serve (FCFS) "is an operating system scheduling algorithm that automatically executes queued requests and processes in order of their arrival".

Shortest job next (SJN), or the shortest job first (SJF) or shortest "is a scheduling policy that selects for execution the waiting process with the smallest execution time".

Shortest remaining time (SRF) "is a scheduling method that is a preemptive version of shortest job next scheduling'".

Round robin (RR) is an algorithm where the time parts "are assigned to each process in equal portions and in circular order, handling all processes without priority"

Solution for the problem

Assuming the dataset given on the plot attached.

Part a

For this algorithm the result would be:

Job A   0-6

Job B   6-(6+3) = 6-9

Job C   9-(9+1) = 9-10

Job D   10-(10+4) = 10-14

Part b

For this algorithm the result would be:

Job A   0-6

Job C   6-(6+1) = 6-7

Job B   7-(7+3) = 7-10

Job D   10-(10+4) = 10-14

Part c

For this algorithm the result would be:

Job A   0-1  until 14

Job B   2-(2+3) = 2-5

Job C   3-(3+2) = 3-5

Job D   9-(9+5) = 9-14

Part d

For this algorithm the result would be:

Job A   0-2 , 7-9, 12-14

Job B   2-4, 9-10

Job C   4-(4+1) = 4-5

Job D   5-7, 10-12

8 0
3 years ago
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