The document discusses scheduling projects with limited resources. It provides an example project schedule graph and manpower loading charts to illustrate how resource leveling techniques can be used to smooth peak resource requirements. Integer programming formulations are also presented as an optimal approach to solve the constrained project scheduling problem.

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 :