1. Resource analysis Project management (lecture + seminar)

2. Introduction Sometimes one or more resources (especially skilled workers) are equally or more important than time. According to priority, there are: – time-limited and – resource-limited projects. Priority can change over time but it is not wise to have double priorities. Basic tools came from production engineering: loading (resource allocation)

3. Resources Definition: anything that is scarce and required for any activity in the project. Resources are constraints for the project. Resources can be: – Non-storable: has to be renewed for each period e.g. work – Storable: depleted only by usage (remains available if not used) e.g. money The most common resource typology, the 4Ms: – Men – Machines – Money (cost) – Material – Other

4. Loading (resource allocation) The assignment of work to an worker, machine or unit (generally: to a workstation) in time. A workstation can be: – underloaded (load < capacity) – fully loaded (load = capacity) – overloaded (load > capacity) Fully loading is nearly impossible to reach except in flow production. Underloading is the most common, because it respects time. Overloading leads to be late.

5. Defining resources for projects SOW WBS Task list – Resource needs given in resource-hours (e.g. man-hour, machine-hour) – Two forms of resource specification: Rate-constant (can be changed to a constant function/pattern): constant usage rate defines the duration, too Total constant: to finish the activity

6. What to do with non-linear duration-resource functions? Use a computer Focus on quasi-linear parts of the functions

7. Capacity Be realistic: – Usual efficiency – Estimated absenteeism, sickness, holidays – Existing commitments – Ancillary tasks and their resource needs – Any additional constraints (like methodology) and limitations (like work contracts) Also calculate with the possibilities (cost, time, trade-offs) to increase capacity

8. Optimum seeking procedures Constraints: – Resource-limited projects – Time-limited projects – Resource-limited AND time-limited projects (question of priority) Methods: – Linear programming – Levelling – Allocation – Smoothing

9. Linear programming Successful only for small networks (up to about 200 activities) Need for precise data

10. Levelling (simplest technique) Need for a previously produced starting schedule. Attempts to level out peaks and valleys in resource requirements by rescheduling some activities. Difficulty of interactions between activities.

11. Allocation (for resource limited cases) Allocation resources ‘Splitting’ an activity: stopping an activity, which is currently in progress, by the removal of its resources for use on an activity of higher priority. Two procedures for allocation: – Serial: if only a few activities are splitable – Parallel: if many activities are splitable; more complicated method that needs more time and data

12. Allocation when resources are limited Serial procedure: 1.All activities in the project are ranked using constant priority rule. The most frequently used rule: ascending order of the LSTs with a tiebreaker of the ascending order of total floats (a kind of ‘ urgency ’ ). 2.Activities are scheduled in the above order at the earliest possible point in time consistent with the availability of resources and the precedence requirements. Step-by-step process: – Draw the network diagram for the logical connections – Compute the activity times and total floats – Plot a Gantt or time-scaled network in tandem with a resource- histogram

13. Allocation when resources are limited Parallel procedure: Activities are considered sequentially (sub-lists). Unscheduled activities are retained and ranked in the next period with new activities.

14. START a b c d g f e h k j i l FINISH

15. The time-limited case Balancing the S-curve with milestones: Forcing early activities to start sooner

16. Smoothing (time limited) Aim is to produce a feasible schedule within the time constraints & provide as smooth a resource requirement profile as possible. Informational needs of smoothing: –Start time and timeframe (TPT or deadline) –Priority order of the resources required Prioritising activities and selecting them in order for scheduling (critical activities have supreme priority). Finding the best place for the activity (placing). Repeat the process with the next activity.

17. Prioritising activities Considerables: Resource type(s) and importance Total work content (resource units per time multiplied with the duration for all resources used on the activity) Available float Formula if no initial scheduling exists: (Total work content) / (Float remaining) In descending order (greatest first).

18. Placing The best position is the one that gives the lowest usage increase in the time span of the activity. Two ways of finding this position: –Visual –Sum of squares: Calculate the sum of squared resource needs of each period within the questionable time frame (between the ES and LF time of the given activity) for every possible positioning Choose the position with the lowest value

19. seminar

20. Defining resources for projects 2 SOW WBS Task list Logical connections (PNT) Gantt chart and histogram Levelling

21. Using the bar chart Set up and analyse the network – Assign the resource data to the activities Draw the Gantt chart – Aggregate each resource time period by time period throughout the total project Cumulating (Summation or S Curve): Use levelling the load for optimization

22. Network with single resource data START (0) 0 00 0 0 0 0 e (1) 10 130 10 13 3 0 d (2) 2 73 5 10 5 3 c (3) 2 100 2 8 0 b (4) 0 55 5 10 5 5 a (1) 0 20 0 2 2 0 FINISH (0) 13 0 0 0

23. Aggregation with a bar chart (single resource, earliest start) activity12345678910111213 a11 b44444 c33333333 d22222 e111 Res. aggr. 5599955333111 Cum. res. 5101928374247505356575859 Resource units 11 10 9 8 7 6 5 4 3 2 1 123456789 111213 Time

24. Aggregation with a bar chart (single resource, latest start) activity12345678910111213 a11 b44444 c33333333 d22222 e111 Res. aggr. 1133399999111 Cum. res. 1258112029384756575859 Resource units 11 10 9 8 7 6 5 4 3 2 1 123456789 111213 Time

25. The S Curve analysis The minimum slope level is the less ‘critical’ from the viewpoint of availability

26. S Curve of the example

27. Scheduling with constraints Three approaches: Allocation: Activities are scheduled so that an initially defined limit of resources or time is not exceeded. Levelling: A previously generated schedule must be given. Then it levels out the peaks and troughs without changing the TPT. Smoothing: A start time, a TPT and resource priorities.

28. Other possibilities Alternative resources Alternative methods Alternative sequences (if there is no technical dependency)

29. Levelling the load We must have a starting allocation of activities over time and a resource constraint (previous example). Resource units 11 10 9 8 7 6 5 4 3 2 1 123456789 111213 Time Trying to keep the original TPT unchanged means that critical activities should not be moved. Thus try to move activities with free float. Resource units 11 10 9 8 7 6 5 4 3 2 1 123456789 111213 Time

30. Solution There are only 2 activities with free float: b & d Which one to move and to where? Moving activity d 3 days in advance is eliminating the peak. Resource units 11 10 9 8 7 6 5 4 3 2 1 123456789 111213 Time

31. activity12345678910111213 a11 b44444 c33333333 d --- 22222 e111 Res. aggr. 55 777 55 555 111 Cum. res. 5101724313641465156575859

32. S Curve

33. Effect of levelling New „activity”: waiting for the resource (it is a lag, not a true activity) START (0) 0 00 0 0 0 0 e (1) 10 130 10 13 3 0 d (2) 5 100 5 5 0 c (3) 2 100 2 8 0 b (4) 0 50 0 5 5 0 a (1) 0 20 0 2 2 0 FINISH (0) 13 0 0 0 Changes: new precedence relationship, floats, late start and finish times

34. Network with single resource data START (0) 0 00 0 0 0 0 e (1) 10 130 10 13 3 0 d (2) 2 73 5 10 5 3 c (3) 2 100 2 8 0 b (4) 0 55 5 10 5 5 a (2) 0 20 0 2 2 0 FINISH (0) 13 0 0 0 Resource limit: 5

35. Network with multiple resource data START (0) 0 00 0 0 0 0 e (3B) 10 130 10 13 3 0 d (2B) 2 73 5 10 5 3 c (3A) 2 100 2 8 0 b (4B) 0 55 5 10 5 5 a (2A) 0 20 0 2 2 0 FINISH (0) 13 0 0 0 Resource limits: 5A, 5B f (3A) 5 85 10 13 3 5

36. Reading Lockyer – Gordon (2005): Chapter 17 & 18

37. Thanks for your attention