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1 year ago

List and describe with details at least three approaches to selecting processes from priority-based multi-level queues.

1 answer:

6 0

Based on the information given, it is important to give higher priority to the queue that contains the high priority thread.

<h3></h3><h3>What is the Important Fact?</h3>

A lot of interrupts can take place at any time, and due to that, they cannot always be ignored as the part of code affected by interrupts need to be guarded from constant use.
So, the load-balancing requirements for keeping about the same number of threads would need to be taken or retained and the important case of top priority thread would also be kept.
An priority-based scheduling algorithm can handle this situation if one run queue had all high-priority threads and a second queue had all low-priority threads because if giving greater priority to the two queue that has the national priority comment section as well as, so, first method is the thread in all of the queue.
Multi-level queue scheduling algorithm is used in scenarios where the processes can be classified into groups based on property like process type, CPU time, IO access, memory size, etc. One general classification of the processes is foreground processes and background processes.

To learn more about multi-level refer to:

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A computer retail store has 15 personal computers in stock. A buyer wants to purchase 3 of them. Unknown to either the retail st

Dima020 [189]

Answer:

a. 1365 ways

b. Probability = 0.4096

c. Probability = 0.5904

Explanation:

Given

PCs = 15

Purchase = 3

Solving (a): Ways to select 4 computers out of 15, we make use of Combination formula as follows;

$^nC_r = \frac{n!}{(n-r)!r!}$

Where $n = 15\ and\ r = 4$

$^{15}C_4 = \frac{15!}{(15-4)!4!}$

$^{15}C_4 = \frac{15!}{11!4!}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12 * 11!}{11! * 4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{15 * 14 * 13 * 12}{4 * 3 * 2 * 1}$

$^{15}C_4 = \frac{32760}{24}$

$^{15}C_4 = 1365$

<em>Hence, there are 1365 ways </em>

Solving (b): The probability that exactly 1 will be defective (from the selected 4)

First, we calculate the probability of a PC being defective (p) and probability of a PC not being defective (q)

<em>From the given parameters; 3 out of 15 is detective;</em>

So;

$p = 3/15$

$p = 0.2$

$q = 1 - p$

$q = 1 - 0.2$

$q = 0.8$

Solving further using binomial;

$(p + q)^n = p^n + ^nC_1p^{n-1}q + ^nC_2p^{n-2}q^2 + .....+q^n$

Where n = 4

For the probability that exactly 1 out of 4 will be defective, we make use of

$Probability = ^nC_3pq^3$

Substitute 4 for n, 0.2 for p and 0.8 for q

$Probability = ^4C_3 * 0.2 * 0.8^3$

$Probability = \frac{4!}{3!1!} * 0.2 * 0.8^3$

$Probability = 4 * 0.2 * 0.8^3$

$Probability = 0.4096$

Solving (c): Probability that at least one is defective;

In probability, opposite probability sums to 1;

Hence;

<em>Probability that at least one is defective + Probability that at none is defective = 1</em>

Probability that none is defective is calculated as thus;

$Probability = q^n$

Substitute 4 for n and 0.8 for q

$Probability = 0.8^4$

$Probability = 0.4096$

Substitute 0.4096 for Probability that at none is defective

Probability that at least one is defective + 0.4096= 1

Collect Like Terms

Probability = 1 - 0.4096

Probability = 0.5904

8 0

3 years ago

The director of HR realizes that the KSAs currently used for hiring entry-level engineers are outdated. In order to establish wh

Vaselesa [24]

Answer:

Job analysis

Explanation:

The HR should preferably use job analysis.

6 0

3 years ago

Write a c++ program to print even numbers from 1 to 20

Novay_Z [31]

Answer:

#include <bits/stdc++.h>

using namespace std;

// Function to print even numbers

void printEvenNumbers(int N)

{

cout << "Even: ";

for (int i = 1; i <= 2 * N; i++) {

// Numbers that are divisible by 2

if (i % 2 == 0)

cout << i << " ";

}

// Function to print odd numbers

void printOddNumbers(int N)

{

cout << "\nOdd: ";

for (int i = 1; i <= 2 * N; i++) {

// Numbers that are not divisible by 2

if (i % 2 != 0)

cout << i << " ";

}

// Driver code

int main()

{

int N = 20;

printEvenNumbers(N);

printOddNumbers(N);

return 0;

}

Explanation:

Note: This will find both odd and even numbers, you have to change the number above to the number of your choice

For even numbers

Even number are numbers that are divisible by 2.

To print even numbers from 1 to N, traverse each number from 1.

Check if these numbers are divisible by 2.

If true, print that number.

For odd numbers

Odd number are numbers that are not divisible by 2.

To print Odd numbers from 1 to N, traverse each number from 1.

Check if these numbers are not divisible by 2.

If true, print that number

8 0

3 years ago

What should I do when the computer doesn't display Korean？（They are all □□）

bogdanovich [222]

Answer:

I Think You Could Go To Settings And Change It

Explanation:

I Think This Helps

5 0

3 years ago

List 3 characteristics of the ideal encryption scheme

JulsSmile [24]

Three characteristics of an ideal encrytion scheme are:
1. The encryption sheme should be strong: the algorithm is imprevious to direct attack and attempts are derived.
2. The encryption scheme should create unique ciphertext from the same plaintext for each key permutation, among other traits.
3. It should take at least millions of years to break ideal encryption scheme, based on mathematical predictions.

6 0

3 years ago

Read 2 more answers