Job sequencing in Data Strcture
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The document discusses job sequencing with deadlines. It presents an algorithm to find the optimal sequence of jobs that maximizes profit while meeting all deadlines. The algorithm has a time complexity of O(n2) as it uses two nested loops. An example applies the algorithm to sequence 6 jobs with deadlines between 1-5 to find the maximum profit of 990 units.
![Algorithm
4
Algorithm Job_seq (d, 1, n){
d [0] = 0;
JS [1] = 1;
i = 1;
for j= 2 to n do;{
k = i;
while ((d[JS[k] > d[j]]) and (d[JS[k]] != k)) do
k = k -1;
if ((d[JS[k] < d[j]]) and (d[j] > k)) then {
For m= i to (k+1) step-1 do
JS [m+1] = JS [m];
JS [K+1] = j;
i = i+1;
}
}
Return i;
}
1
1
1
n
1
2*n
2
2*n
n*n
1*n*n
1*n*n
2*n*n
Time complexity = O(n2)](https://image.slidesharecdn.com/seconddaapresentation-190220000800/85/Job-sequencing-in-Data-Strcture-4-320.jpg)
Job sequencing in Data Strcture
- 1. JOB SEQUENCING 1 Presented by: Rabin BK BSc.CSIT 5th Semester
- 2. Introduction Algorithm Example References 2
- 3. Introduction 3 The objective is to find a sequence of jobs, which is completed within their deadlines with maximum profit Because we are using two loops one within another, the complexity of this algorithm is O(n2)
- 4. Algorithm 4 Algorithm Job_seq (d, 1, n){ d [0] = 0; JS [1] = 1; i = 1; for j= 2 to n do;{ k = i; while ((d[JS[k] > d[j]]) and (d[JS[k]] != k)) do k = k -1; if ((d[JS[k] < d[j]]) and (d[j] > k)) then { For m= i to (k+1) step-1 do JS [m+1] = JS [m]; JS [K+1] = j; i = i+1; } } Return i; } 1 1 1 n 1 2*n 2 2*n n*n 1*n*n 1*n*n 2*n*n Time complexity = O(n2)
- 5. Example 5 Value of maximum deadline = 5 • First, we take job J4. • Since, its deadline is 2, so we place it in the first empty cell before deadline 2 as- • Now, we take job J1. • Since, its deadline is 5, so we place it in the first empty cell before deadline 5 as- • Now, we take job J3. • Since, its deadline is 3, so we place it in the first empty cell before deadline 3 as-
- 6. Example 6 Value of maximum deadline = 5 • Now, we take job J2. • Since, its deadline is 3, so we place it in the first empty cell before deadline 3. • Since, the second and third cells are already filled, so we place job J2 in the first cell as- • Now, we take job J5. • Since, its deadline is 4, so we place it in the first empty cell before deadline 4 as- • The only job left is job J6 whose deadline is 2. • All the slots before deadline 2 are already occupied. • Thus, job J6 can not be completed. • Maximum earned profit = Sum of profits of all jobs in optimal schedule = Profit of job J2 + Profit of job J4 + Profit of job J3 + Profit of job J5 + Profit of job J1 = 180 + 300 + 190 + 120 + 200 = 990 units
- 7. References • https://www.gatevidyalay.com/job-sequencing-with-deadlines/ • https://www.includehelp.com/operating-systems/job-sequencing.aspx • https://www.geeksforgeeks.org/job-sequencing-problem-set-1-greedy-algorithm/ • http://www.cs.sfu.ca/~kabanets/307/sched307.pdf • https://www.techiedelight.com/job-sequencing-problem-deadlines/ • http://ggn.dronacharya.info/CSEDept/Downloads/QuestionBank/Even/VI%20sem/ ADA/Section-B/job-scheduling1.pdf • http://www.ques10.com/p/9083/write-short-notes-on-job-sequencing-with- deadlines/ 7
- 8. Queries 8