Time left 1:28If a time quantum used in round-robin preemptive CPU scheduling is allowedto grow too large, the algorithm degenerates to:O A. Longest time firstO B. PriorityO C. FCFSO D. SJFNext pagO.S. Structures ►2Jump to...

Given If a time quantum used in round-robin preemptive CPU scheduling is allowed to grow too large,…

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Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Complete the following analysis of an algorithm such that the time complexity, T(n), of the given algorithm is determined. The analysis uses instruction count (i.e. number of times the statement is executed) as the measure for the analysis. The instruction count of each instruction in the algorithm is indicated at the right of the statement after the //. You should replace each of the question marks(?) by the corresponding instruction count in order to complete the analysis. The total instruction count is obtained by adding all the instruction counts. Note that in this exercise, the linear search algorithm is analyzed in a case where the item to be searched is assumed not found in the array (i.e.Worst Case Analysis) int linearSearch(item, array) { // worst case instruction countint index = 0; // ?while (index < array.length) { // ?if (array[index] == item) { // ?return index; // ?}index++; // ?}return -1; // ?} // ------// T(n) = ? where n is the array size
A CPU performs 103 operations per second. Kim proposes an algorithm that has a time complexity of nlog10n. What would be the run time of the algorithm in minutes on the CPU for a problem size of n = 106?
Dynamic Programming Exercise Problem
An algorithm that takes 2(n) time ( always / sometimes / never ) takes more time to run than an algorithm that takes w(n) time on the same input and hardware, where n is the length of the input. sometimes never always
3. Define the time complexity of this ABC algorithm in Big-O notation.
The instruction count of each instruction in the algorithm is indicated at the right of the statement after the //. You should replace each of the question marks(?) by the corresponding instruction count in order to complete theanalysis. The total instruction count is obtained by adding all the instruction counts. Note that in this exercise, the linear search algorithm is analyzed in a case where the item to be searched is assumed to be found in the first cell of the array (i.e. Best Case Analysis). int linearSearch(item, array) { // best case instruction countint index = 0; // ?while (index < array.length) { // ?if (array[index] == item) { // ?return index; // ?}index++; //?}return -1; // ?} // ------// T(n) = ? where n is the array size
Complete the following analysis of an algorithm such that the time complexity, T(n), of the given algorithm is determined. The analysis uses instruction count (i.e. number of times the statement is executed) as the measure for the analysis. The instruction count of each instruction in the algorithm is indicated at the right of the statement after the //. You should replace each of the question marks(?) by the corresponding instruction count in order to complete the analysis. The total instruction count is obtained by adding all the instruction counts. Note that in this exercise, the binary search algorithm is analyzed in a case where the item to be searched is assumed to be NOT found in the sorted array (i.e. Worst Case Analysis). // Worst Case Analysis (e.g., item not in array)int binarySearch(item, sortedArray) { // Instruction Countint low = 0; // 1int high = sortedArray.length - 1; // 1int mid = 0; // 1while (low <= high) { // ≈ log(n) + 1mid = (low + high) / 2; // ≈ log(n)if…
For the following function, select the most appropriate time complexity in terms of n: def rate(n): i = 0 total = 0 while in: j= n while j > 1: total + j j // = 2 i += 1 return total IMPORTANT: The penalty scheme for this question is 50, 100%. Select one: O a. Time complexity = O(n) O b. Time complexity = O(n^2) O c. Time complexity = O(2n) O d. Time complexity = O(log n) e. Time complexity = O(n log n) Check
Three major steps of a divide-and-conquer algorithm are: • Divide the problem in to two or more parts. • Solve each part separately. • Combine the partial solution. In the Quicksort algorithm, which of these steps represent most of the work. Please explain
General Computer science questions
OS Let's Try This One ... Process P1 P2 P3 P4 P5 CPU-burst 7 4 5 4 2 Arrival Time 0 2 2 4 6 (a) Solve using FCFS (b) Solve using non-preemptive SJF (c) Solve using preemptive SJF 59
Compute time complexity of the given program. The complexity computation should be for the complete program. Map it on its relevant standard function. Computation should be clear. void swap(int *xp, int *yp) { int temp = *xp; *xp = *yp; *yp = temp; } // A function to implement bubble sort void bubbleSort(int arr[], int n) { int i, j; for (i = 0; i < n-1; i++) // Last i elements are already in place for (j = 0; j < n-i-1; j++) if (arr[j] > arr[j+1]) swap(&arr[j], &arr[j+1]); } /* Function to print an array */ void printArray(int arr[], int size) { int i; for (i = 0; i < size; i++) cout << arr[i] << " "; cout << endl; }

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If a time quantum used in round-robin preemptive CPU scheduling is allowed to grow too large, the algorithm degenerates to: O A. Longest time first O B. Priority O C. FCFS O D. SJF