Scheduling priority tasks in limited time and minimizing loss

Given n tasks with arrival time, priority and number of time units they need. We need to schedule these tasks on a single resource. The objective is to arrange tasks such that maximum priority tasks are taken. Objective is to minimize sum of product of priority and left time of tasks that are not scheduled due to limited time. This criteria simply means that the scheduling should cause minimum possible loss.


Input : Total time = 3
        Task1: arrival = 1, units = 2, priority = 300
        Task2: arrival = 2, units = 2, priority = 100
Output : 100
Explanation : Two tasks are given and time to finish 
them is 3 units. First task arrives at time 1 and it
needs 2 units. Since no other task is running at that
time, we take it. 1 unit of task 1 is over. Now task 2 
arrives. The priority of task 1 is more than task 2 
(300 > 100) thus 1 more unit  of task 1 is employed. 
Now 1 unit of time is left and we have 2 units of task
2 to be done. So simply 1 unit of task 2 is done and 
the answer is ( units of task 2 left X priority of 
task 2 ) = 1 X 100 = 100

Input : Total Time = 3
        Task1: arrival = 1, units = 1, priority = 100
        Task2: arrival = 2, units = 2, priority = 300
Output : 0

We use a priority queue and schedule one unit of task at a time.

  1. Initialize total loss as sum of each priority and units. The idea is to initialize result as maximum, then one by one subtract priorities of tasks that are scheduled.
  2. Sort all tasks according to arrival time.
  3. Process through each unit of time from 1 to total time. For current time, pick the highest priority task that is available. Schedule 1 unit of this task and subtract its priority from total loss.

Below is C++ implementation of above steps.

// C++ code for task scheduling on basis
// of priority and arrival time
#include "bits/stdc++.h"
using namespace std;

// t is total time. n is number of tasks.
// arriv[] stores arrival times. units[]
// stores units of time required. prior[]
// stores priorities.
int minLoss(int n, int t, int arriv[],
            int units[],  int prior[])
    // Calculating maximum possible answer
    // that could be calculated. Later we
    // will subtract the tasks from it
    // accordingly.
    int ans = 0;
    for (int i = 0; i < n; i++)
        ans += prior[i] * units[i];

    // Pushing tasks in a vector so that they
    // could be sorted according with their
    // arrival time
    vector<pair<int, int> > v;
    for (int i = 0; i < n; i++)
        v.push_back({ arriv[i], i });

    // Sorting tasks in vector identified by
    // index and arrival time with respect
    // to their arrival time
    sort(v.begin(), v.end());

    // Priority queue to hold tasks to be
    // scheduled.
    priority_queue<pair<int, int> > pq;

    // Consider one unit of time at a time.
    int ptr = 0; // index in sorted vector
    for (int i = 1; i <= t; i++) {

        // Push all tasks that have arrived at
        // this time. Note that the tasks that
        // arrived earlier and could not be scheduled
        // are already in pq.
        while (ptr < n and v[ptr].x == i) {
            pq.push({ prior[v[ptr].y], units[v[ptr].y] });

        // Remove top priority task to be scheduled
        // at time i.
        if (!pq.empty()) {
            auto tmp =;

            // Removing 1 unit of task
            // from answer as we could
            // schedule it.
            ans -= tmp.x;
            if (tmp.y)

    return ans;

// driver code
int main()
    int n = 2, t = 3;
    int arriv[] = { 1, 2 };
    int units[] = { 2, 2 };
    int prior[] = { 100, 300 };

    printf("%d\n",  minLoss(n, t, arriv, units, prior));
    return 0;



This article is contributed by Parth Trehan. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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