Shortest Job First (or SJF) CPU Scheduling Non-preemptive algorithm using Segment Tree

17 Jan, 2022
Shortest job first (SJF) or shortest job next, is a scheduling policy that selects the waiting process with the smallest execution time to execute next. SJN is a non-preemptive algorithm.  
Shortest Job first has the advantage of having a minimum average waiting time among all scheduling algorithms.
It is a Greedy Algorithm.
It may cause starvation if shorter processes keep coming. This problem can be solved using the concept of ageing.
It is practically infeasible as Operating System may not know burst time and therefore may not sort them. While it is not possible to predict execution time, several methods can be used to estimate the execution time for a job, such as a weighted average of previous execution times. SJF can be used in specialized environments where accurate estimates of running time are available.
For example:  
In the above example, since the arrival time of all the processes is 0, the execution order of the process is the ascending order of the burst time of the processes. The burst time is given by the column duration. Therefore, the execution order of the processes is given by:  
P4 -> P1 -> P3 -> P2
One implementation for this algorithm has already been discussed in the article with the help of Naive Approach. In this article, the algorithm is implemented by using the concept of a segment tree.  
Approach: The following is the approach used for the implementation of the shortest job first:  
As the name suggests, the shortest job first algorithm is an algorithm which executes the process whose burst time is least and has arrived before the current time. Therefore, in order to find the process which needs to be executed, sort all the processes from the given set of processes according to their arrival time. This ensures that the process with the shortest burst time which has arrived first is executed first.Instead of finding the minimum burst time process among all the arrived processes by iterating the whole struct array, the range minimum of the burst time of all the arrived processes upto the current time is calculated using segment tree.After selecting a process which needs to be executed, the completion time, turn around time and waiting time is calculated by using arrival time and burst time of the process. The formulae to calculate the respective times are: Completion Time: Time at which process completes its execution.  
As the name suggests, the shortest job first algorithm is an algorithm which executes the process whose burst time is least and has arrived before the current time. Therefore, in order to find the process which needs to be executed, sort all the processes from the given set of processes according to their arrival time. This ensures that the process with the shortest burst time which has arrived first is executed first.
Instead of finding the minimum burst time process among all the arrived processes by iterating the whole struct array, the range minimum of the burst time of all the arrived processes upto the current time is calculated using segment tree.
After selecting a process which needs to be executed, the completion time, turn around time and waiting time is calculated by using arrival time and burst time of the process. The formulae to calculate the respective times are: Completion Time: Time at which process completes its execution.  
Completion Time: Time at which process completes its execution.  
Completion Time = Start Time + Burst Time
Turn Around Time: Time Difference between completion time and arrival time.  
Turn Around Time = Completion Time – Arrival Time
Waiting Time(W.T): Time Difference between turn around time and burst time.  
Waiting Time = Turn Around Time – Burst Time
After calculating, the respective times are updated in the array and the burst time of the executed process is set to infinity in the segment tree base array so that it is not considered as the minimum burst time in the further queries.
After calculating, the respective times are updated in the array and the burst time of the executed process is set to infinity in the segment tree base array so that it is not considered as the minimum burst time in the further queries.
Below is the implementation of the shortest job first using the concept of segment tree:  
C++
Java
// C++ implementation of shortest job first// using the concept of segment tree  #include <bits/stdc++.h>using namespace std;#define ll long long#define z 1000000007#define sh 100000#define pb push_back#define pr(x) printf("%d ", x)  struct util {      // Process ID    int id;    // Arrival time    int at;    // Burst time    int bt;    // Completion time    int ct;    // Turnaround time    int tat;    // Waiting time    int wt;}  // Array to store all the process information// by implementing the above struct utilar[sh + 1];  struct util1 {      // Process id    int p_id;    // burst time    int bt1;};  util1 range;  // Segment tree array to// process the queries in nlognutil1 tr[4 * sh + 5];  // To keep an account of where// a particular process_id is// in the segment tree base arrayint mp[sh + 1];  // Comparator function to sort the// struct array according to arrival timebool cmp(util a, util b){    if (a.at == b.at)        return a.id < b.id;    return a.at < b.at;}  // Function to update the burst time and process id// in the segment treevoid update(int node, int st, int end,            int ind, int id1, int b_t){    if (st == end) {        tr[node].p_id = id1;        tr[node].bt1 = b_t;        return;    }    int mid = (st + end) / 2;    if (ind <= mid)        update(2 * node, st, mid, ind, id1, b_t);    else        update(2 * node + 1, mid + 1, end, ind, id1, b_t);    if (tr[2 * node].bt1 < tr[2 * node + 1].bt1) {        tr[node].bt1 = tr[2 * node].bt1;        tr[node].p_id = tr[2 * node].p_id;    }    else {        tr[node].bt1 = tr[2 * node + 1].bt1;        tr[node].p_id = tr[2 * node + 1].p_id;    }}  // Function to return the range minimum of the burst time// of all the arrived processes using segment treeutil1 query(int node, int st, int end, int lt, int rt){    if (end < lt || st > rt)        return range;    if (st >= lt && end <= rt)        return tr[node];    int mid = (st + end) / 2;    util1 lm = query(2 * node, st, mid, lt, rt);    util1 rm = query(2 * node + 1, mid + 1, end, lt, rt);    if (lm.bt1 < rm.bt1)        return lm;    return rm;}  // Function to perform non_preemptive// shortest job first and return the// completion time, turn around time and// waiting time for the given processesvoid non_preemptive_sjf(int n){      // To store the number of processes    // that have been completed    int counter = n;      // To keep an account of the number    // of processes that have been arrived    int upper_range = 0;      // Current running time    int tm = min(INT_MAX, ar[upper_range + 1].at);      // To find the list of processes whose arrival time    // is less than or equal to the current time    while (counter) {        for (; upper_range <= n;) {            upper_range++;            if (ar[upper_range].at > tm || upper_range > n) {                upper_range--;                break;            }              update(1, 1, n, upper_range,                   ar[upper_range].id, ar[upper_range].bt);        }          // To find the minimum of all the running times        // from the set of processes whose arrival time is        // less than or equal to the current time        util1 res = query(1, 1, n, 1, upper_range);          // Checking if the process has already been executed        if (res.bt1 != INT_MAX) {            counter--;            int index = mp[res.p_id];            tm += (res.bt1);              // Calculating and updating the array with            // the current time, turn around time and waiting time            ar[index].ct = tm;            ar[index].tat = ar[index].ct - ar[index].at;            ar[index].wt = ar[index].tat - ar[index].bt;              // Update the process burst time with            // infinity when the process is executed            update(1, 1, n, index, INT_MAX, INT_MAX);        }        else {            tm = ar[upper_range + 1].at;        }    }}  // Function to call the functions and perform// shortest job first operationvoid execute(int n){      // Sort the array based on the arrival times    sort(ar + 1, ar + n + 1, cmp);    for (int i = 1; i <= n; i++)        mp[ar[i].id] = i;      // Calling the function to perform    // non-preemptive-sjf    non_preemptive_sjf(n);}  // Function to print the required values after// performing shortest job firstvoid print(int n){      cout << "ProcessId  "         << "Arrival Time  "         << "Burst Time  "         << "Completion Time  "         << "Turn Around Time  "         << "Waiting Time\n";    for (int i = 1; i <= n; i++) {        cout << ar[i].id << " \t\t "             << ar[i].at << " \t\t "             << ar[i].bt << " \t\t "             << ar[i].ct << " \t\t "             << ar[i].tat << " \t\t "             << ar[i].wt << " \n";    }}  // Driver codeint main(){    // Number of processes    int n = 5;      // Initializing the process id    // and burst time    range.p_id = INT_MAX;    range.bt1 = INT_MAX;      for (int i = 1; i <= 4 * sh + 1; i++) {        tr[i].p_id = INT_MAX;        tr[i].bt1 = INT_MAX;    }      // Arrival time, Burst time and ID    // of the processes on which SJF needs    // to be performed    ar[1].at = 1;    ar[1].bt = 7;    ar[1].id = 1;      ar[2].at = 2;    ar[2].bt = 5;    ar[2].id = 2;      ar[3].at = 3;    ar[3].bt = 1;    ar[3].id = 3;      ar[4].at = 4;    ar[4].bt = 2;    ar[4].id = 4;      ar[5].at = 5;    ar[5].bt = 8;    ar[5].id = 5;      execute(n);      // Print the calculated time    print(n);}
// Java implementation of shortest job first // using the concept of segment tree import java.util.*;  class GFG {      static int z = 1000000007;    static int sh = 100000;      static class util {          // Process ID        int id;        // Arrival time        int at;        // Burst time        int bt;        // Completion time        int ct;        // Turnaround time        int tat;        // Waiting time        int wt;    }      // Array to store all the process information    // by implementing the above struct util    static util[] ar = new util[sh + 1];    static {        for (int i = 0; i < sh + 1; i++) {            ar[i] = new util();        }    }      static class util1 {          // Process id        int p_id;        // burst time        int bt1;    };      static util1 range = new util1();      // Segment tree array to    // process the queries in nlogn    static util1[] tr = new util1[4 * sh + 5];    static {        for (int i = 0; i < 4 * sh + 5; i++) {            tr[i] = new util1();        }    }      // To keep an account of where    // a particular process_id is    // in the segment tree base array    static int[] mp = new int[sh + 1];      // Comparator function to sort the    // struct array according to arrival time      // Function to update the burst time and process id    // in the segment tree    static void update(int node, int st, int end,                         int ind, int id1, int b_t)    {        if (st == end) {            tr[node].p_id = id1;            tr[node].bt1 = b_t;            return;        }        int mid = (st + end) / 2;        if (ind <= mid)            update(2 * node, st, mid, ind, id1, b_t);        else            update(2 * node + 1, mid + 1, end, ind, id1, b_t);        if (tr[2 * node].bt1 < tr[2 * node + 1].bt1) {            tr[node].bt1 = tr[2 * node].bt1;            tr[node].p_id = tr[2 * node].p_id;        } else {            tr[node].bt1 = tr[2 * node + 1].bt1;            tr[node].p_id = tr[2 * node + 1].p_id;        }    }      // Function to return the range minimum of the burst time    // of all the arrived processes using segment tree    static util1 query(int node, int st, int end,                        int lt, int rt)     {        if (end < lt || st > rt)            return range;        if (st >= lt && end <= rt)            return tr[node];        int mid = (st + end) / 2;        util1 lm = query(2 * node, st, mid, lt, rt);        util1 rm = query(2 * node + 1, mid + 1, end, lt, rt);        if (lm.bt1 < rm.bt1)            return lm;        return rm;    }      // Function to perform non_preemptive    // shortest job first and return the    // completion time, turn around time and    // waiting time for the given processes    static void non_preemptive_sjf(int n) {          // To store the number of processes        // that have been completed        int counter = n;          // To keep an account of the number        // of processes that have been arrived        int upper_range = 0;          // Current running time        int tm = Math.min(Integer.MAX_VALUE, ar[upper_range + 1].at);          // To find the list of processes whose arrival time        // is less than or equal to the current time        while (counter != 0) {            for (; upper_range <= n;) {                upper_range++;                if (ar[upper_range].at > tm || upper_range > n) {                    upper_range--;                    break;                }                  update(1, 1, n, upper_range, ar[upper_range].id,                         ar[upper_range].bt);            }              // To find the minimum of all the running times            // from the set of processes whose arrival time is            // less than or equal to the current time            util1 res = query(1, 1, n, 1, upper_range);              // Checking if the process has already been executed            if (res.bt1 != Integer.MAX_VALUE) {                counter--;                int index = mp[res.p_id];                tm += (res.bt1);                  // Calculating and updating the array with                // the current time, turn around time and waiting time                ar[index].ct = tm;                ar[index].tat = ar[index].ct - ar[index].at;                ar[index].wt = ar[index].tat - ar[index].bt;                  // Update the process burst time with                // infinity when the process is executed                update(1, 1, n, index, Integer.MAX_VALUE, Integer.MAX_VALUE);            } else {                tm = ar[upper_range + 1].at;            }        }    }      // Function to call the functions and perform    // shortest job first operation    static void execute(int n) {          // Sort the array based on the arrival times        Arrays.sort(ar, 1, n, new Comparator<util>() {            public int compare(util a, util b) {                if (a.at == b.at)                    return a.id - b.id;                return a.at - b.at;            }        });        for (int i = 1; i <= n; i++)            mp[ar[i].id] = i;          // Calling the function to perform        // non-preemptive-sjf        non_preemptive_sjf(n);    }      // Function to print the required values after    // performing shortest job first    static void print(int n) {          System.out.println("ProcessId Arrival Time Burst Time" +                " Completion Time Turn Around Time Waiting Time");        for (int i = 1; i <= n; i++) {            System.out.printf("%d\t\t%d\t\t%d\t\t%d\t\t%d\t\t%d\n",                  ar[i].id, ar[i].at, ar[i].bt, ar[i].ct, ar[i].tat,                    ar[i].wt);        }    }      // Driver Code    public static void main(String[] args)     {        // Number of processes        int n = 5;          // Initializing the process id        // and burst time        range.p_id = Integer.MAX_VALUE;        range.bt1 = Integer.MAX_VALUE;          for (int i = 1; i <= 4 * sh + 1; i++)        {            tr[i].p_id = Integer.MAX_VALUE;            tr[i].bt1 = Integer.MAX_VALUE;        }          // Arrival time, Burst time and ID        // of the processes on which SJF needs        // to be performed        ar[1].at = 1;        ar[1].bt = 7;        ar[1].id = 1;          ar[2].at = 2;        ar[2].bt = 5;        ar[2].id = 2;          ar[3].at = 3;        ar[3].bt = 1;        ar[3].id = 3;          ar[4].at = 4;        ar[4].bt = 2;        ar[4].id = 4;          ar[5].at = 5;        ar[5].bt = 8;        ar[5].id = 5;          execute(n);          // Print the calculated time        print(n);    }}  // This code is contributed by// sanjeev2552
ProcessId  Arrival Time  Burst Time  Completion Time  Turn Around Time  Waiting Time
1          1          7          8          7          0 
2          2          5          16          14          9 
3          3          1          9          6          5 
4          4          2          11          7          5 
5          5          8          24          19          11
 
Time Complexity: In order to analyze the running time of the above algorithm, the following running times needs to be understood first:  
The time complexity to construct a segment tree for N processes is O(N).
The time complexity to update a node in a segment tree is given by O(log(N)).
The time complexity to perform a range minimum query in a segment tree is given by O(log(N)).
Since the update operation and queries are performed for given N processes, the total time complexity of the algorithm is O(N*log(N)) where N is the number of processes.
This algorithm performs better than the approach mentioned in this article because it takes O(N2) for execution.
 
sanjeev2552
varshagumber28
simmytarika5
cpu-scheduling
Operating Systems-CPU Scheduling
Greedy
Mathematical
Operating Systems
Tree
Operating Systems
Greedy
Mathematical
Tree
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Dijkstra's shortest path algorithm | Greedy Algo-7
Program for array rotation
Prim’s Minimum Spanning Tree (MST) | Greedy Algo-5
Write a program to print all permutations of a given string
Kruskal’s Minimum Spanning Tree Algorithm | Greedy Algo-2
Program for Fibonacci numbers
Set in C++ Standard Template Library (STL)
Write a program to print all permutations of a given string
C++ Data Types
Merge two sorted arrays