1. Analysing the AoA network Project Management

2. Total Project Time The minimum time in which the project can be completed. Calculation: forward pass Forward pass: calculating the earliest event times (EETs) and the earliest start times (ESTs) of all activities. Earliest Finishing Time = EST + Duration

3. Critical path Path: continuous series of project activities connected by logical relationships as designated in the project schedule network diagram. Critical path: sequence of activities that has no float time, and that determines the duration of the project. It is the longest path. Activities on the critical path are the critical activities. The critical path can be identified by a backward pass, calculating the Latest Event Times (LETs) and the Latest Finishing Times (LFTs). Latest Starting Time = Latest Finishing Time - Duration

4. Activity times & event times EET = EST of all emerging activities LET = LFT of all entering activities Activity identifier Duration 12 EET LET EET LET Deadline

5. TPT a 14 12 0 0 TPT = 14 EST 0 LST 0 LFT 14 EFT 14

6. Float Float on activity ‘a’: a 14 12 0 6 20 Float: 6 EST 0 EFT 14 LFT 20 LST 6

7. 4 22 ? 5 24 ? 6 ? 34 d e 8 10 Calculate the… EET of event 6 LETs of event 4 and 5 ESTs and EFTs of activity ‘d’ and ‘e’ LSTs and LFTs of activity ‘d’ and ‘e’

8. EST and EFT of ‘d’: 22 and 30 EST and EFT of ‘e’: 24 and 34 LST and LFT of ‘d’: 26 and 34 LST and LFT of ‘e’: 24 and 34 4 22 26 5 24 6 34 d e 8 10 24

9. 10 ? 25 ? 20 ? 35 15 ? 30 a b c 10 12 8

10. 10 17 25 18 25 20 22 35 15 20 30 a b c 10 12 8

11. Calculate all event and activity times and the float-times, find the critical path

12. 0 16 382035 51 16 31 2038 30 35 51 30 51 50 2135 39 24 829 36 3520 00

13. Activity times and float for the previous diagram ActivityESTLSTEFTLFTFloat A B C D E G H J K

14. Activity times and float for the previous diagram ActivityESTLSTEFTLFTFloat A0816248 B0020 0 C021305121 D20 35 0 E202930399 G353638391 H35 51 0 J162431398 K383950511

15. Four characteristics of the critical path 1.It starts at the first node 2.It is continuous 3.It ends at the last node 4.It has no float

16. Floats in the AoA network

17. Activity and event times Activity times: –EST –LST –EFT –LFT Event times: –EET –LET

18. Float Time available for an activity or path in addition to its duration. It can be positive or negative It is a property of activities (and available only at given activities) In case of more succeeding activities, the minimum have to be taken. Types of float: –Total float –Free float

19. Total float The total float possessed by an activity. Calculation for activity j: Total float j = LFT j – EFT j Identify the activity possessing float below, and calculate the total float. 1 0 ? 2 ? ? a 12 4 ? 35 c 10 3 ? ? b 15 d 8

20. Total float The total float possessed by an activity. Calculation for activity j: Total float j = LFT j – EFT j = LFT j – (EST j + D j ) Identify the activities possessing float below, and calculate the total float for each. 1 0 0 2 5 11 a 5 4 23 c 12 3 15 b d 8 Total float = 23 – (5+12) = 6 Total float = = 11 – (0+5) = 6

21. Calculating free float The float possessed by an activity which, if used, will not change the float in later activities. Free float j = EST j+1 – EFT j = EET head – EET tail - D 1 0 0 2 5 11 a 5 4 23 c 12 3 15 b d 8 Free float = 23 – 5 – 12 = 6 Free float = = 5 – 0 – 5 = 0

22. Negative float If the target time (deadline) for the project (or for a part of the project) is grater than TPT, than the float will appear on the critical path or even on some other paths. Negative float = the time by which activities on the path or paths concerned must be reduced if the TPT is to be met. Negative float is a type of the Total Float. 1 0 ? 2 ? ? a 10 3 ? ? c 15

23. Negative float If the target time (deadline) for the project (or for a part of the project) is grater than TPT, than the float will appear on the critical path or even on some other paths. Negative float = the time by which activities on the path or paths concerned must be reduced if the TPT is to be met. Negative float is a type of the Total Float. 1 0 -5 2 10 5 a 3 20 15 c 10 15 Total float: -5

24. Slack Refers to events and not to activities. Slack = LET - EET Slack: 0 1 0 0 2 5 11 a 5 4 23 c 12 3 15 b d 8 Total float = 23 – (5+12) = 6 Total float = = 11 – (0+5) = 6 Slack: 0 Slack: 6

25. Readings Lockyer – Gordon (2005) Chapter 13

26. Analysing the AoN network Project Management (seminar)

27. Activity label & description Data on the activity node ESTEFT Duration LSTLFT Total Float

28. Total Project Time The shortest time in which the project can be completed. Determined by the critical path. Calculation: forward pass Forward pass: The earliest start times (EST) of all activities are calculated. Trom these the earliest finishing times (EFT) are also calculated

29. Critical path sequence of activities that has no float time, and that determines the duration of the project. It is the longest path. Activities on the critical path are the critical activities. The critical path can be identified by a backward pass, calculating the Latest Finishing Times (LFT), and from these the Latest Starting Times (LST).

30. Floats in AoN Total float: the time by which an activity can be delayed or extended without affecting the TPT. It can be used to delay the start of an activity or to increase its duration. TF = LST - EST Free float: the time by which an activity can be delayed or extended without affecting the start of any succeding activity. FF = EST j+1 - EFT j

31. Example: organising a conference Objectives: to organise a 3 days long open scientific conference with 100-200 participants, 30-50 lectures, buffet reception, a conference book of the best studies and TV and radio interviews with some of the most known lecturers. Create the WBS chart and create the task list with estimated durations and precedence relations (in a table form) Plot both the AoA and AoN diagram Calculate the TPT, identify the critical path the total and the free float times.

32. Example: WBS Project BookMarketing Event management Organising interviews Editing Collecting articles Publishing Peer reviewing Arranging event Infra- structure Organising participants StaffFacilitiesMaterials Invitation and marketing

33. Activity label Task descriptionDurationImmediate predecessors a Invitation b Organising participants c Facilities d Staffing e Materials h Collecting articles j Peer reviewing k Publishing l Organising interviews m Arranging event Activity label Task descriptionDuration (weeks) Immediate predecessors a Invitation2 b Organising participants4 c Facilities3 d Staffing4 e Materials3 h Collecting articles6 j Peer reviewing3 k Publishing5 l Organising interviews1 m Arranging event1 Activity label Task descriptionDuration (weeks) Immediate predecessors a Invitation2– b Organising participants4a c Facilities3b d Staffing4b e Materials3b f Collecting articles6b g Peer reviewing3f h Organising interviews1c, d, e i Publishing5g j Arranging event1h, i Task list with precedence relations

34. AoA 1 0 0 3 6 6 2 2 2 4 9 19 5 10 19 6 10 19 7 12 8 15 9 20 10 21 2a2a 3e3e 4d4d 3 4b4b c f 6 3g3g i 1h1h 5 1j1j TPT = 21 CP: a-b-f-g-i-j

35. AoN a 0 20 0 2 2 b 2 60 2 6 4 f 6 120 6 6 e 6 910 16 19 3 d 6 106 15 19 4 c 6 910 16 19 3 g 12 150 12 15 3 i 200 15 20 5 j 210 20 21 1 h 10 119 19 20 1 1 0 0 0 0 0 0 0 1 9 TPT: 21 CP: a-b-f-g-i-j

36. Activity times for the previous diagram (finalize individually) ActivityDurationESTLSTEFTLFTTotal float Free float a2020200 b4262600 c3691619101 d e f g h i j

37. Example 2 (for individual practice) a) Draw the AoA and AoN diagram with the data below: Activity label Duration (weeks) Immediate predecessors a 1– b 2a c 5a d 3a e 2b f 2e g 4f, c, d b) Determine the TPT and the critical path and activity floats. c) Compute the EETs, LETs and slack for every node in the AoA & ESTs, LSTs, EFTs, LFTs in the AoN diagram.

38. Solution: AoA 1 0 0 4 4 7 5 5 5 3 3 3 2 1 1 6 7 7 1a1a 2e2e 2b2b 5c5c 3d3d 4g4g 2f2f 7 11 TPT: 11 CP: a-b-e-f-g Float: 0 T. float: 3 F. float: 3 Float: 0 Slack: 0 Slack: 3 Slack: 0 T. float: 1 F. float: 1 0

39. Solution: AoN a 0 10 0 1 1 0 b 1 30 1 3 2 c 1 61 2 7 5 0 d 1 43 4 7 3 3 f 5 70 5 7 2 e 3 50 3 2 2 1 g 7 110 7 4 0 0 0 TPT: 11 CP: a-b-e-f-g

40. Example 3 1 0 0 4 2 5 3 5 5 2 2 2 5 8 8 2a2a 3e3e 2b2b 5c5c 3d3d 4g4g 7 14 6 12 1f1f 2h2h a)Calculate the EETs and LETs. b)Create a precedence table (with task, duration, immediate predecessor, total and free floats). 0

41. Solution ActivityDurationTotal float Free float a200 b200 c511 d333 e300 f155 g400 h200

42. ‘Crashing’ – reducing task durations by increased costs

43. Definition of crashing Obtaining reduction in time at an increased cost (increasing the employed resources). Cost-slope: the cost of reducing duration time by unit time. Let’s see the following example: a 0 0 4 b 2 c 2 e 2 d 5 f 3 a 0 40 0 4 4 0 b 4 60 4 6 2 0 c 6 81 7 9 2 0 e 8 101 9 11 2 1 d 6 0 6 5 0 f 140 11 14 3 0

44. Procedure for crashing 1.Crash one time unit at a time 2.Only the crashing of critical activities has any effect on TPT 3.Crash that activity first that is the cheapest to reduce in time 4.Be aware of multiple critical paths 5.Stop crashing when: the crash-time is reached at every ‘crashable’ activity, benefits of possible crashing are lower than crashing costs.

45. Crashing table If the costs to reduce times are known, then a table can be set up showing the relative costs for the reduction in time of each activity by a constant amount. Crash-time is the minimum duration of an activity. It is given by technical factors. Activity (label) Duration (day) Float (day) Crash time Cost-slope (€/day) a402100 b202150 c211110 d503200 e211160 f302500 Benefit of reducing TPT by one day: 400 €/day

46. Solution method 1.step: identify the critical activities 2.step: find the critical activity with cheapest crash cost, and if its cost slope is lower than the daily benefit from crashing, reduce its duration with one day. If there is no activity to crash, or it is too costly, stop crashing and go to step 4. 3.step: reidentify the critical path, and go back to step two. 4.step: identify the final critical path(s), TPT and the total net benefit of crashing.

47. Solution Path durations Path / activity crached normalstep 1step 2step 3step 4step 5 Cost: Cumulated net benefit: Path durations Path / activity crashed normalstep 1step 2step 3step 4step 5 –aadd, cnone a-b-c-e-f131211 10– a-b-d-f1413121110– Cost:–100 200310– Cumulated net benefit: –300600800890– After crashing: – there are two critical paths – TPT is 10 days – total benefit of crashing is €890

48. Example 2 (for individual work) Identify the critical path and the TPT. b 2 a 0 0 3 c 3 d 2 e 5 f 3 g 3 7

49. Example 2 (for individual work) Critcal: a-b-d-e-g TPT: 15 Using tbe table on the next slide, calculate the optimal TPT with crashing. b 3 50 3 5 2 0 a 0 30 0 3 3 0 c 3 61 4 7 3 0 d 5 70 5 7 2 0 e 7 120 7 5 0 f 6 93 9 3 3 g 150 12 15 3 0 7

50. Activity (label) Normal duration (day) Float (day) Crash time Cost- slope (€/day) a31500 b21550 c21150 d53900 e54400 f32100 g33200 Benefit of reducing TPT by one day: 1200 €/day What is the new TPT? What is the total profit on crashing? 10 days €3000 Activity (label) Normal duration (day) Float (day) Crash time Cost- slope (€/day) a301500 b201550 c211150 d503900 e504400 f332100 g303200

51. Reading Lockyer – Gordon (2005) Chapter 8 pp. 61-63. & Chapter 14

52. Thanks for the attention!