P.S I did not create this ppt.

1. Network Scheduling with Limited Resources

3. Schedule Graph , Unlimited Resources Labor available Project completion time 18 10 12 6 6 1

4. Let assume following constraint 10 labor available on any one day Schedule Graph , Resource Limit Of 10 Men Cont.. 10 10 10 10 6 6 1

8. Early start schedule graph Manpower loading chart Maximum slack 7 days so postpone activity 9 Peak requirement 24 Number over activity indicates the activity number as well as resource.

9. Schedule Graph Manpower loading chart . Peak requirement reduce to 15 Cont..

10. Schedule Graph Manpower Loading Chart Activity 8 is delay by full slack of 2 days Peak requirement occur at days 1, 2 Cont..

11. Schedule Graph Manpower Loading Chart. Best result obtain by delaying activity 3 by 2 days Smoothen histogram Cont..

12. Smoothing Peak Resource Requirement

14. Day 1 - only one job 7 is available to start in this period and there are sufficient men to schedule it. Schedule job 7 , 3 men remain available Only 10 men are available for the project on any day. Resource Day

15. Day 2- Job 1,3,4,5 can be started on day 2 , but there are not enough men to schedule them Then second heuristic is applied for scheduling the jobs with least slack first Schedule job 3( slack=0); 7 men remain Schedule job 4( slack=0); 3 men remain Schedule job 1( slack=3); 2 men remain Job 5 must be delayed , as there are just [2=10-(4+3+1)] men unassigned Postponed the job 5 ( slack =4) Cont..

16. Day 3- continue job 3(slack =0); 7 men remain continue job 4(slack =0); 3men remain continue job 1(slack =3); 2men remain Postpone job 5 (slack =3) Cont..

17. Day 4 Continue job 4 (slack =0); 6 men remain Continue job 1 (slack =3); 5 men remain Job 6 has no slack and hence is critical , but only 5 men are still unassigned Then 3 rd heuristic is applied – job 1 could be postponed without delaying the project Reschedule job 1 ; 6 men remain Schedule job 6 (slack =0); 0 men remain Postpone Job 5 (slack =2) Cont..

18. Cont.. Day Manpower/ resource requirement

19. Day 5 Continue job 6 (slack=0); 4 men remain Schedule Job 2 (slack =0); 2 men remain Schedule Job 1 ( slack = 0); 1 men remain Postpone job 5 (slack =1) Day 6 Continue Job 6 (slack =0); 4 Men remain . Continue Job 2 (slack =0); 2 Men remain . Continue Job 1 (slack =0); 1 Men remain . Postpone job 5 (slack =0) Now job 5 is critical but no active job can be postponed without delaying project Cont..

20. Cont.. Day Manpower

21. Day 7 Continue Job 1 (slack =1); 9 Men remain Schedule Job 5 ( slack =0); 4 Men remain Postpone job 9 ( slack =1) Day 8 Continue Job 5 (slack =0); 5 men remain Postpone job 9 (slack =0); No noncritical job can be rescheduled Postpone Job 8 (Slack =2) Day Manpower

22. Day 9– Schedule job 9 ( slack =0); 1 man remain . Postpone job 8 (slack =2) Day 10 – Schedule job 10 ( slack =0); no man remain. Postpone Job 8 (slack =1) Day 11 – Continue job 10 (slack =0); no men remain . Postpone Job 8 (slack =0)

23. Day 12— Continue job 10 (slack =1); no men remain Postpone Job 8( slack =0) Day 13— Schedule job 8 ( slack =0); 2 men remain Day 14 – Continue job 8 ( slack = 0); 2 men remain Cont.. Manpower Day Final schedule limited resource 1 2 3 4 5 6 7 8 9 10 11 12 13 14 7 7 7 10 9 9 6 6 9 10 10 10 8 8

27. Conceptual Problem of Critical Path Analysis When Resource Are Limited

28. Early start schedule Late Start schedule Cont.. Day Resource Day Resource 1 2 3 4 5 18 10 10 8 7 1 2 3 4 5 4 4 5 18 22

29. If resources are limited up to 10 men then optimal early start schedule and Late start schedule is as follows Early start schedule Day Resource 1 2 3 4 5 6 10 10 10 9 7 7

30. Late Start schedule Day Resource 1 2 3 4 5 6 10 10 8 10 7 8

38. 2 . Job will be performed : 3. Capacity of shop will not be exceeded 4.No job will be started before its predecessors are completed d = 1,……,z AND j= 1,….,n except beginning job

39. 5 . No jobs will be split :