Job Sequencing with Deadline


Problem Statement

In job sequencing problem, the objective is to find a sequence of jobs, which is completed within their deadlines and gives maximum profit.


Let us consider, a set of n given jobs which are associated with deadlines and profit is earned, if a job is completed by its deadline. These jobs need to be ordered in such a way that there is maximum profit.

It may happen that all of the given jobs may not be completed within their deadlines.

Assume, deadline of ith job Ji is di and the profit received from this job is pi. Hence, the optimal solution of this algorithm is a feasible solution with maximum profit.

Thus, $D(i) > 0$ for $1 \leqslant i \leqslant n$.

Initially, these jobs are ordered according to profit, i.e. $p_{1} \geqslant p_{2} \geqslant p_{3} \geqslant \:... \: \geqslant p_{n}$.

Algorithm: Job-Sequencing-With-Deadline (D, J, n, k) 
D(0) := J(0) := 0 
k := 1 
J(1) := 1   // means first job is selected 
for i = 2 … n do 
   r := k 
   while D(J(r)) > D(i) and D(J(r)) ≠ r do 
      r := r – 1 
   if D(J(r)) ≤ D(i) and D(i) > r then 
      for l = k … r + 1 by -1 do 
         J(l + 1) := J(l) 
         J(r + 1) := i 
         k := k + 1 


In this algorithm, we are using two loops, one is within another. Hence, the complexity of this algorithm is $O(n^2)$.


Let us consider a set of given jobs as shown in the following table. We have to find a sequence of jobs, which will be completed within their deadlines and will give maximum profit. Each job is associated with a deadline and profit.

Job J1 J2 J3 J4 J5
Deadline 2 1 3 2 1
Profit 60 100 20 40 20


To solve this problem, the given jobs are sorted according to their profit in a descending order. Hence, after sorting, the jobs are ordered as shown in the following table.

Job J2 J1 J4 J3 J5
Deadline 1 2 2 3 1
Profit 100 60 40 20 20

From this set of jobs, first we select J2, as it can be completed within its deadline and contributes maximum profit.

  • Next, J1 is selected as it gives more profit compared to J4.

  • In the next clock, J4 cannot be selected as its deadline is over, hence J3 is selected as it executes within its deadline.

  • The job J5 is discarded as it cannot be executed within its deadline.

Thus, the solution is the sequence of jobs (J2, J1, J3), which are being executed within their deadline and gives maximum profit.

Total profit of this sequence is 100 + 60 + 20 = 180.