Skip to content
Related Articles

Related Articles

Completion time of a given process in round robin
  • Difficulty Level : Medium
  • Last Updated : 30 Apr, 2021

We are given n-processes with their completion times in form of an array. We need to find the time instant when a given process p ends if the scheduling process is round robin and time slice is 1-sec. 
note : Array index start with 0.
Examples : 
 

Input : arr[] = {3, 2, 4, 2}, p = 1
Output : Completion time = 6
Explanation : Snap of process for every second is as:
Time    |   Process Array
0       |   {3, 2, 4, 2}
1       |   {2, 2, 4, 2}
2       |   {2, 1, 4, 2}
3       |   {2, 1, 3, 2}
4       |   {2, 1, 3, 1}
5       |   {1, 1, 3, 1}
6       |   {1, 0, 3, 1}

Input : arr[] = {2, 4, 1, 3}, p = 2
Output :Completion time = 3
Explanation : Snap of process for every second is as:
Time    |   Process Array
0       |   {2, 4, 1, 3}
1       |   {1, 4, 1, 3}
2       |   {1, 3, 1, 3}
3       |   {1, 3, 0, 3}

Brute Force :The basic approach for solving this problem is to apply round robin algorithm with time slice 1. But the time complexity of that approach will be O(ΣAi) i.e. summation of all process’s time, which is quite high.
Efficient Approach: The idea is based on below observations. 
1) All processes with CPU time less than arr[p] would complete before arr[p]. We simply need to add time of these processes. 
2) We also need to add time of arr[p]. 
3) For every process x with CPU time more than arr[p], two cases arise : 
…..(i) If x is on left of arr[p] (scheduled before arr[p]), then this process takes arr[p] time of CPU before p finishes. 
…..(ii) If x is on right of arr[p] (scheduled after arr[p]), then this process takes arr[p]-1 time of CPU before p finishes.
Algorithm : 
 

time_req = 0;

// Add time for process on left of p 
// (Scheduled before p in a round of 
// 1 unit time slice)
for (int i=0; i<p; i++)
{
    if (arr[i] < arr[p])
        time_req += arr[i];
    else
        time_req += arr[p];
}

// step 2 : Add time of process p
time_req += arr[p];

// Add time for process on right
// of p (Scheduled after p in
// a round of 1 unit time slice)
for (int i=p+1; i<n; i++)
{
    if (arr[i] < arr[p])
        time_req += arr[i];
    else
        time_req += arr[p]-1;
}

 

 

C++




// Program to find end time of a process
// p in round robin scheduling with unit
// time slice.
#include <bits/stdc++.h>
using namespace std;
 
// Returns completion time of p.
int completionTime(int arr[], int n, int p) {
 
  // Initialize result
  int time_req = 0;
 
  // Step 1 : Add time of processes on left
  //  of p (Scheduled before p)
  for (int i = 0; i < p; i++) {
    if (arr[i] < arr[p])
      time_req += arr[i];
    else
      time_req += arr[p];
  }
 
  // Step 2 : Add time of p
  time_req += arr[p];
 
  // Step 3 : Add time of processes on right
  //  of p (Scheduled after p)
  for (int i = p + 1; i < n; i++) {
    if (arr[i] < arr[p])
      time_req += arr[i];
    else
      time_req += arr[p] - 1;
  }
 
  return time_req;
}
 
// driver program
int main() {
  int arr[] = {3, 5, 2, 7, 6, 1};
  int n = sizeof(arr) / sizeof(arr[0]);
  int p = 2;
  cout << "Completion time = "
       << completionTime(arr, n, p);
  return 0;
}

Java




// Program to find end time of a process
// p in round robin scheduling with unit
// time slice.
class GFG
{
    // Returns completion time of p.
    static int completionTime(int arr[], int n, int p) {
         
        // Initialize result
        int time_req = 0;
         
        // Step 1 : Add time of processes on left
        // of p (Scheduled before p)
        for (int i = 0; i < p; i++) {
            if (arr[i] < arr[p])
                time_req += arr[i];
            else
                time_req += arr[p];
        }
         
        // Step 2 : Add time of p
        time_req += arr[p];
         
        // Step 3 : Add time of processes on right
        // of p (Scheduled after p)
        for (int i = p + 1; i < n; i++) {
            if (arr[i] < arr[p])
                time_req += arr[i];
            else
                time_req += arr[p] - 1;
        }
         
        return time_req;
    }
         
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = {3, 5, 2, 7, 6, 1};
        int n =arr.length;;
        int p = 2;
         
        System.out.print("Completion time = "+
            completionTime(arr, n, p));
    }
}
 
// This code is contributed by Anant Agarwal.

Python




# Program to find end time of a process
# p in round robin scheduling with unit
# time slice.
 
# Returns completion time of p.
def completionTime(arr, n, p) :
 
    # Initialize result
    time_req = 0
     
    # Step 1 : Add time of processes on
    # left of p (Scheduled before p)
    for i in range(0, p):
        if (arr[i] < arr[p]):
            time_req += arr[i]
        else:
            time_req += arr[p]
     
     
    # Step 2 : Add time of p
    time_req += arr[p]
     
    # Step 3 : Add time of processes on
    # right of p (Scheduled after p)
    for i in range(p + 1, n):
        if (arr[i] < arr[p]):
            time_req += arr[i]
        else:
            time_req += arr[p] - 1
     
    return time_req
     
 
# driver program
arr = [3, 5, 2, 7, 6, 1]
n = len(arr)
p = 2
print("Completion time =",
        completionTime(arr, n, p))
 
 
# This code is contributed by
# Smitha Dinesh Semwal

C#




// C# program to find end time of a process
// p in round robin scheduling with unit
// time slice.
using System;
 
class GFG {
     
    // Returns completion time of p.
    static int completionTime(int []arr,
                            int n, int p)
    {
         
        // Initialize result
        int time_req = 0;
         
        // Step 1 : Add time of processes
        // on left of p (Scheduled before p)
        for (int i = 0; i < p; i++) {
            if (arr[i] < arr[p])
                time_req += arr[i];
            else
                time_req += arr[p];
        }
         
        // Step 2 : Add time of p
        time_req += arr[p];
         
        // Step 3 : Add time of processes on
        // right of p (Scheduled after p)
        for (int i = p + 1; i < n; i++) {
            if (arr[i] < arr[p])
                time_req += arr[i];
            else
                time_req += arr[p] - 1;
        }
         
        return time_req;
    }
         
    // Driver code
    public static void Main ()
    {
        int []arr = {3, 5, 2, 7, 6, 1};
        int n =arr.Length;;
        int p = 2;
         
        Console.WriteLine("Completion time = "+
                    completionTime(arr, n, p));
    }
}
 
// This code is contributed by vt_m.

PHP




<?php
// Program to find end time
// of a process p in round
// robin scheduling with
// unit time slice.
 
 
// Returns completion time of p.
function completionTime($arr, $n, $p)
{
 
// Initialize result
$time_req = 0;
 
// Step 1 : Add time of processes on
// left of p (Scheduled before p)
for ($i = 0; $i < $p; $i++)
{
    if ($arr[$i] < $arr[$p])
    $time_req += $arr[$i];
    else
    $time_req += $arr[$p];
}
 
// Step 2 : Add time of p
$time_req += $arr[$p];
 
// Step 3 : Add time of processes on
// right of p (Scheduled after p)
for ($i = $p + 1; $i < $n; $i++)
{
    if ($arr[$i] < $arr[$p])
    $time_req += $arr[$i];
    else
    $time_req += $arr[$p] - 1;
}
 
return $time_req;
}
 
// Driver Code
$arr = array(3, 5, 2, 7, 6, 1);
$n = count($arr);
$p = 2;
echo"Completion time = " ,
     completionTime($arr, $n, $p);
 
// This code is contributed by anuj_67.
?>

Javascript




<script>
 
// Program to find end time of a process
// p in round robin scheduling with unit
// time slice.
 
// Returns completion time of p.
function completionTime(arr, n, p) {
 
  // Initialize result
  var time_req = 0;
 
  // Step 1 : Add time of processes on left
  //  of p (Scheduled before p)
  for (var i = 0; i < p; i++) {
    if (arr[i] < arr[p])
      time_req += arr[i];
    else
      time_req += arr[p];
  }
 
  // Step 2 : Add time of p
  time_req += arr[p];
 
  // Step 3 : Add time of processes on right
  //  of p (Scheduled after p)
  for (var i = p + 1; i < n; i++) {
    if (arr[i] < arr[p])
      time_req += arr[i];
    else
      time_req += arr[p] - 1;
  }
 
  return time_req;
}
 
// driver program
var arr = [3, 5, 2, 7, 6, 1];
var n = arr.length;
var p = 2;
document.write( "Completion time = "
     + completionTime(arr, n, p));
 
// This code is contributed by noob2000.
</script>

Output : 
 

Completion time = 9

Time Complexity : O(n)
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :