Completion time of a given process in round robin

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 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);   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



Output :

Completion time = 9

Time Complexity : O(n)

My Personal Notes arrow_drop_up Discovering ways to develop a plane for soaring career goals

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : vt_m

Article Tags :
Practice Tags :

Be the First to upvote.

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.