The document describes an algorithm for job sequencing with deadlines. It takes as input the deadline array D and job array J of size n. It assigns jobs to time slots T while respecting the deadlines, with the goal of maximizing profit. The algorithm initializes the time slots to empty, then iterates through jobs to find the earliest slot k meeting the deadline and assigns the job if the slot is empty, else moves to the next slot. This produces a job sequence with maximum profit respecting all deadlines. An example is provided to illustrate the algorithm.

1. Algorithm analysis
and design
Job sequencing with deadlines.

2. Algorithm stated
1. Algorithm JS(d,j,n)
2. {
3. D[0] : =T[0] : =0; / * This fictious job allows easy insertion into position 1 of time slot*/
4. T[1] : =1; / *assigning job 1 to time slot*/
5. k:=1;
6. For i:=2 to n do
7. {
8. r:=k;
9. while((D[T[r]]>D[i]) and (D[T[r]]!=r))do r:=r-1;
10. if ((D[T[r]]<=D[i]) and (D[i]>r))then
11. {
12. for q:=k to (r+1) step-1 do T[q+1]:=T[q]; /*”step-1” implies making the
consecutive slots empty*/
13. T[r+1]:= i; k:=k+1;
14. }
15. }
16. Return k;
17. }

3. Example:
Q. Let n=5;p=(20,15,10,5,1);d=(2,2,1,3,3)
Ans:-
Given:
Jobs J1 J2 J3 J4 J5
Profits 20 15 10 5 1
Deadline 2 2 1 3 3

4. Now , we give index to jobs: n=5
i
Index 1 2 3 4 5
Jobs J1 J2 J3 J4 J5
Profits 20 15 10 5 1
Deadline 2 2 1 3 3

5. The maximum deadline given for a job in the egs is:
3.
Hence, Dmax=3.
So we can get a maximum of 3 time slots(initially-1 job-
1 unit of time).
Henceforth , we create a table for time slot ,T:
Timeslot 1 2 3
Status

6. From the algo we are solving the problem:
 Look at index 1 of job array , its deadline is 2 ,tf it has to
come in the first slot.
Now, assigning i=1 =>J1 to timeslot array, T[].
Then , assigning other slots to be Empty[-1].
T[1] : =1; / *assigning job 1 to time slot*/
for q:=k to (r+1) step-1 do T[q+1]:=T[q]; /*”step-1” implies making
the consecutive slots empty*/
T[r+1]:= i; k:=k+1;
Table will be now,

7. Now , i becomes 2
Formula of k is min(Dmax, deadline(i))
k checks which is the min among the values
 k =min(3,2)=2;
Checking if k>=1.
The solution is a yes .
Hence, checking timeslot(k)==-1or empty
If yes fill it with the corresponding job.
Now , i becomes 3
Formula of k is min(Dmax, deadline(i))
k checks which is the min among the values
 k =min(3,1)=1;
Checking if k>=1.
The solution is a yes .
Hence, checking timeslot(k)==-1or empty
It’s a no,hence decrementing k by 1 and j3 rejected.

8. Now , i becomes 4
Formula of k is min(Dmax, deadline(i))
k checks which is the min among the values
 k =min(3,3)=3;
Checking if k>=1.
The solution is a yes .
Hence, checking timeslot(k)==-1or empty
If yes fill it with the corresponding job.(j4)
Now , i becomes 5
Formula of k is min(Dmax, deadline(i))
k checks which is the min among the values
 k =min(3,3)=3;
Checking if k>=1.
The solution is a yes .
Hence, checking timeslot(k)==-1or empty
It’s a no, hence decrement k by 1 and reject j5 and exit for
loop.

9. Final timeslot table results give a sequece of jobs with max
profit which meet the deadlines.
Therefore the sequence of jobs is :
J1->J2->J4

10. •Algorithm that can
be prepared from
the example(for
easy
implementation)

11. 1. Algorithm JS(D,J,n)
2. {
3. D[0] :=J[0] :=0;
4. for i:=1 to n
5. {
6. T[i]=-1;
7. }
8. J[1]:=1;
9. for i:=2 to n
10. {
11. k:= min(Dmax,D(i));
12. if(k>=1) then
13. {
14. if(T[k]==-1) then
15. {
16. T[k]:=J[k];
17. }
18. else
19. k--;
20. }
21. }